Chapter 3 Matrices

Solution:

We are asked to construct a $2 \times 2$ matrix $A = [a_{ij}]$ where the elements $a_{ij}$ are given by the formula $a_{ij} = e^{2ix} \sin jx$.

A $2 \times 2$ matrix has 2 rows and 2 columns. The indices $i$ and $j$ represent the row number and column number, respectively. For a $2 \times 2$ matrix, $i$ can take values 1 or 2, and $j$ can take values 1 or 2.

The matrix A will be of the form:

$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$

We need to find the value of each element $a_{ij}$ by substituting the corresponding values of $i$ and $j$ into the given formula $a_{ij} = e^{2ix} \sin jx$.

For the element in the first row and first column ($a_{11}$), we have $i=1$ and $j=1$:

$a_{11} = e^{2(1)x} \sin (1x) = e^{2x} \sin x$

For the element in the first row and second column ($a_{12}$), we have $i=1$ and $j=2$:

$a_{12} = e^{2(1)x} \sin (2x) = e^{2x} \sin 2x$

For the element in the second row and first column ($a_{21}$), we have $i=2$ and $j=1$:

$a_{21} = e^{2(2)x} \sin (1x) = e^{4x} \sin x$

For the element in the second row and second column ($a_{22}$), we have $i=2$ and $j=2$:

$a_{22} = e^{2(2)x} \sin (2x) = e^{4x} \sin 2x$

Now, substituting these calculated values back into the matrix form, we get the matrix A:

$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} e^{2x} \sin x & e^{2x} \sin 2x \\ e^{4x} \sin x & e^{4x} \sin 2x \end{pmatrix}$

Thus, the constructed matrix A is:

$\begin{pmatrix} e^{2x} \sin x & e^{2x} \sin 2x \\ e^{4x} \sin x & e^{4x} \sin 2x \end{pmatrix}$

Solution:

We are given four matrices:

$A = \begin{bmatrix} 2&3 \\ 1&2 \end{bmatrix}$

$B = \begin{bmatrix} 1&3&2 \\ 4&3&1 \end{bmatrix}$

$C = \begin{bmatrix}1 \\ 2 \end{bmatrix}$

$D = \begin{bmatrix} 4&6&8 \\ 5&7&9 \end{bmatrix}$

First, let's determine the dimensions (order) of each matrix.

The dimension of matrix A is $2 \times 2$ (2 rows and 2 columns).

The dimension of matrix B is $2 \times 3$ (2 rows and 3 columns).

The dimension of matrix C is $2 \times 1$ (2 rows and 1 column).

The dimension of matrix D is $2 \times 3$ (2 rows and 3 columns).

Two matrices can be added if and only if they have the same dimensions. We need to check the dimensions for each given sum.

Sum A + B:

Dimension of A is $2 \times 2$.

Dimension of B is $2 \times 3$.

Since the dimensions of A and B are not the same ($2 \times 2 \neq 2 \times 3$), the sum A + B is not defined.

Sum B + C:

Dimension of B is $2 \times 3$.

Dimension of C is $2 \times 1$.

Since the dimensions of B and C are not the same ($2 \times 3 \neq 2 \times 1$), the sum B + C is not defined.

Sum C + D:

Dimension of C is $2 \times 1$.

Dimension of D is $2 \times 3$.

Since the dimensions of C and D are not the same ($2 \times 1 \neq 2 \times 3$), the sum C + D is not defined.

Sum B + D:

Dimension of B is $2 \times 3$.

Dimension of D is $2 \times 3$.

Since the dimensions of B and D are the same ($2 \times 3 = 2 \times 3$), the sum B + D is defined.

Out of the given sums, only B + D is defined because matrices B and D have the same dimensions.

Solution:

Let $A$ be a matrix.

By definition, a matrix $A$ is called symmetric if its transpose $A^T$ is equal to the matrix itself.

So, if $A$ is symmetric, we have:

Also by definition, a matrix $A$ is called skew-symmetric if its transpose $A^T$ is equal to the negative of the matrix itself.

So, if $A$ is skew-symmetric, we have:

We are given that the matrix $A$ is both symmetric and skew-symmetric.

Therefore, from equation (1) and equation (2), we must have:

$A = A^T$

and

$A^T = -A$

Combining these two equations, we get:

Now, let's rearrange this equation to solve for $A$. We can add $A$ to both sides of the equation:

This simplifies to:

Here, $0$ on the right-hand side represents the zero matrix of the same dimensions as $A$.

To isolate $A$, we can multiply both sides by $\frac{1}{2}$:

This gives us:

This result shows that the matrix $A$ must be the zero matrix, i.e., a matrix where every element is zero.

Thus, a matrix which is both symmetric and skew-symmetric must be a zero matrix.

Solution:

We are given the matrix equation:

We need to find the value of $x$ that satisfies this equation.

The equation involves the product of three matrices. We can perform the matrix multiplication step by step. Let's first multiply the first two matrices.

The first matrix is a $1 \times 2$ matrix $\begin{bmatrix} 2x & 3 \end{bmatrix}$.

The second matrix is a $2 \times 2$ matrix $\begin{bmatrix} 1&2 \\ −3&0 \end{bmatrix}$.

The product of a $1 \times 2$ matrix and a $2 \times 2$ matrix will be a $1 \times 2$ matrix.

Let's calculate the elements of the resulting matrix:

First element (Row 1, Column 1): $(2x)(1) + (3)(-3) = 2x - 9$

Second element (Row 1, Column 2): $(2x)(2) + (3)(0) = 4x + 0 = 4x$

So, the product of the first two matrices is $\begin{bmatrix} 2x - 9 & 4x \end{bmatrix}$.

Now, we multiply this resulting $1 \times 2$ matrix by the third matrix.

The resulting matrix is $\begin{bmatrix} 2x - 9 & 4x \end{bmatrix}$.

The third matrix is a $2 \times 1$ matrix $\begin{bmatrix} x\\8 \end{bmatrix}$.

The product of a $1 \times 2$ matrix and a $2 \times 1$ matrix will be a $1 \times 1$ matrix.

Let's calculate the single element of the final resulting matrix:

Element (Row 1, Column 1): $(2x - 9)(x) + (4x)(8)$

Expanding this expression:

$(2x)(x) - 9(x) + (4x)(8) = 2x^2 - 9x + 32x$

Combining like terms:

$2x^2 + (-9 + 32)x = 2x^2 + 23x$

So, the final resulting matrix is $\begin{bmatrix} 2x^2 + 23x \end{bmatrix}$.

The given equation (1) states that this final matrix is equal to the zero matrix. For a $1 \times 1$ matrix, the zero matrix is simply the scalar 0.

Therefore, we have the equation:

Equating the elements of the matrices, we get a scalar equation:

This is a quadratic equation in $x$. We can solve it by factoring.

Factor out the common term $x$ from the expression $2x^2 + 23x$:

For the product of two terms to be zero, at least one of the terms must be zero. So, we have two possible cases:

Case 1: $x = 0$

Case 2: $2x + 23 = 0$

Subtract 23 from both sides:

$2x = -23$

Divide by 2:

Thus, the values of $x$ that satisfy the given matrix equation are $x = 0$ and $x = -\frac{23}{2}$.

Solution:

Given:

$A$ is a $3 \times 3$ invertible matrix.

$k$ is a non-zero scalar ($k \neq 0$).

To Prove:

$kA$ is invertible and $(kA)^{-1} = \frac{1}{k} A^{-1}$.

Proof:

By the definition of an invertible matrix, since $A$ is an invertible matrix, there exists a matrix $A^{-1}$ of the same dimension ($3 \times 3$) such that:

and

where $I$ is the $3 \times 3$ identity matrix.

To show that $kA$ is invertible, we need to find a matrix $X$ such that $(kA)X = I$ and $X(kA) = I$.

Let's consider the matrix $X = \frac{1}{k} A^{-1}$. Since $k$ is a non-zero scalar, $\frac{1}{k}$ is well-defined. Also, $A^{-1}$ is a $3 \times 3$ matrix, so $\frac{1}{k} A^{-1}$ is also a $3 \times 3$ matrix.

Let's evaluate the product $(kA) \left(\frac{1}{k} A^{-1}\right)$. Using the property $(c B) C = c (BC)$ and $B (cC) = c (BC)$ for a scalar $c$ and matrices $B, C$ (where multiplication is defined), we can group the scalars:

Since $k \neq 0$, we have $k \cdot \frac{1}{k} = 1$. Substituting this and equation (1):

The product of the scalar 1 and the identity matrix $I$ is just $I$:

Now, let's evaluate the product in the reverse order: $\left(\frac{1}{k} A^{-1}\right) (kA)$. Again, grouping the scalars:

Since $k \neq 0$, we have $\frac{1}{k} \cdot k = 1$. Substituting this and equation (2):

This simplifies to:

From equations (3) and (4), we have found a matrix $\frac{1}{k} A^{-1}$ such that when multiplied by $kA$ from either the left or the right, the result is the identity matrix $I$.

This confirms that $kA$ is invertible, and its inverse is indeed $\frac{1}{k} A^{-1}$.

Therefore, it is shown that if A is a $3 \times 3$ invertible matrix, then for any scalar $k$ (non-zero), $kA$ is invertible and $(kA)^{-1} = \frac{1}{k} A^{-1}$.

Given:

Matrix A is

$A = \begin{bmatrix}2&4&−6 \\ 7&3&5 \\ 1&−2&4 \end{bmatrix}$

To Express:

Express matrix A as the sum of a symmetric and a skew-symmetric matrix.

Solution:

We know that any square matrix can be uniquely expressed as the sum of a symmetric and a skew-symmetric matrix. If A is a square matrix, then $A = P + Q$, where $P = \frac{1}{2}(A+A')$ is the symmetric part and $Q = \frac{1}{2}(A-A')$ is the skew-symmetric part.

Given matrix A is:

$A = \begin{bmatrix}2&4&−6 \\ 7&3&5 \\ 1&−2&4 \end{bmatrix}$

The transpose of A is A':

$A' = \begin{bmatrix}2&7&1 \\ 4&3&−2 \\ −6&5&4 \end{bmatrix}$

First, let's calculate the symmetric part $P = \frac{1}{2}(A+A')$:

Find $A+A'$:

$A+A' = \begin{bmatrix}2&4&−6 \\ 7&3&5 \\ 1&−2&4 \end{bmatrix} + \begin{bmatrix}2&7&1 \\ 4&3&−2 \\ −6&5&4 \end{bmatrix} = \begin{bmatrix}2+2&4+7&−6+1 \\ 7+4&3+3&5+(-2) \\ 1+(-6)&−2+5&4+4 \end{bmatrix} = \begin{bmatrix}4&11&−5 \\ 11&6&3 \\ −5&3&8 \end{bmatrix}$

Now, calculate P:

$P = \frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix}4&11&−5 \\ 11&6&3 \\ −5&3&8 \end{bmatrix} = \begin{bmatrix}\frac{4}{2}&\frac{11}{2}&\frac{−5}{2} \\ \frac{11}{2}&\frac{6}{2}&\frac{3}{2} \\ \frac{−5}{2}&\frac{3}{2}&\frac{8}{2} \end{bmatrix} = \begin{bmatrix}2&\frac{11}{2}&−\frac{5}{2} \\ \frac{11}{2}&3&\frac{3}{2} \\ −\frac{5}{2}&\frac{3}{2}&4 \end{bmatrix}$

Let's verify if P is symmetric ($P' = P$):

$P' = \begin{bmatrix}2&\frac{11}{2}&−\frac{5}{2} \\ \frac{11}{2}&3&\frac{3}{2} \\ −\frac{5}{2}&\frac{3}{2}&4 \end{bmatrix}' = \begin{bmatrix}2&\frac{11}{2}&−\frac{5}{2} \\ \frac{11}{2}&3&\frac{3}{2} \\ −\frac{5}{2}&\frac{3}{2}&4 \end{bmatrix} = P$.

Hence, P is a symmetric matrix.

Next, let's calculate the skew-symmetric part $Q = \frac{1}{2}(A-A')$:

Find $A-A'$:

$A-A' = \begin{bmatrix}2&4&−6 \\ 7&3&5 \\ 1&−2&4 \end{bmatrix} - \begin{bmatrix}2&7&1 \\ 4&3&−2 \\ −6&5&4 \end{bmatrix} = \begin{bmatrix}2-2&4-7&−6-1 \\ 7-4&3-3&5-(-2) \\ 1-(-6)&−2-5&4-4 \end{bmatrix} = \begin{bmatrix}0&−3&−7 \\ 3&0&7 \\ 7&−7&0 \end{bmatrix}$

Now, calculate Q:

$Q = \frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix}0&−3&−7 \\ 3&0&7 \\ 7&−7&0 \end{bmatrix} = \begin{bmatrix}\frac{0}{2}&\frac{−3}{2}&\frac{−7}{2} \\ \frac{3}{2}&\frac{0}{2}&\frac{7}{2} \\ \frac{7}{2}&\frac{−7}{2}&\frac{0}{2} \end{bmatrix} = \begin{bmatrix}0&−\frac{3}{2}&−\frac{7}{2} \\ \frac{3}{2}&0&\frac{7}{2} \\ \frac{7}{2}&−\frac{7}{2}&0 \end{bmatrix}$

Let's verify if Q is skew-symmetric ($Q' = -Q$):

$Q' = \begin{bmatrix}0&−\frac{3}{2}&−\frac{7}{2} \\ \frac{3}{2}&0&\frac{7}{2} \\ \frac{7}{2}&−\frac{7}{2}&0 \end{bmatrix}' = \begin{bmatrix}0&\frac{3}{2}&\frac{7}{2} \\ −\frac{3}{2}&0&−\frac{7}{2} \\ −\frac{7}{2}&\frac{7}{2}&0 \end{bmatrix}$

$-Q = -\begin{bmatrix}0&−\frac{3}{2}&−\frac{7}{2} \\ \frac{3}{2}&0&\frac{7}{2} \\ \frac{7}{2}&−\frac{7}{2}&0 \end{bmatrix} = \begin{bmatrix}0&\frac{3}{2}&\frac{7}{2} \\ −\frac{3}{2}&0&−\frac{7}{2} \\ −\frac{7}{2}&\frac{7}{2}&0 \end{bmatrix}$.

Since $Q' = -Q$, Q is a skew-symmetric matrix.

Finally, we express A as the sum of P and Q:

$A = P + Q$

$A = \begin{bmatrix}2&\frac{11}{2}&−\frac{5}{2} \\ \frac{11}{2}&3&\frac{3}{2} \\ −\frac{5}{2}&\frac{3}{2}&4 \end{bmatrix} + \begin{bmatrix}0&−\frac{3}{2}&−\frac{7}{2} \\ \frac{3}{2}&0&\frac{7}{2} \\ \frac{7}{2}&−\frac{7}{2}&0 \end{bmatrix}$

Let's add P and Q to verify:

$P+Q = \begin{bmatrix}2+0&\frac{11}{2}−\frac{3}{2}&−\frac{5}{2}−\frac{7}{2} \\ \frac{11}{2}+\frac{3}{2}&3+0&\frac{3}{2}+\frac{7}{2} \\ −\frac{5}{2}+\frac{7}{2}&\frac{3}{2}−\frac{7}{2}&4+0 \end{bmatrix} = \begin{bmatrix}2&\frac{8}{2}&−\frac{12}{2} \\ \frac{14}{2}&3&\frac{10}{2} \\ \frac{2}{2}&−\frac{4}{2}&4 \end{bmatrix} = \begin{bmatrix}2&4&−6 \\ 7&3&5 \\ 1&−2&4 \end{bmatrix} = A$

Thus, the matrix A is expressed as the sum of a symmetric matrix and a skew-symmetric matrix as:

$A = \underbrace{\begin{bmatrix}2&\frac{11}{2}&−\frac{5}{2} \\ \frac{11}{2}&3&\frac{3}{2} \\ −\frac{5}{2}&\frac{3}{2}&4 \end{bmatrix}}_{\text{Symmetric}} + \underbrace{\begin{bmatrix}0&−\frac{3}{2}&−\frac{7}{2} \\ \frac{3}{2}&0&\frac{7}{2} \\ \frac{7}{2}&−\frac{7}{2}&0 \end{bmatrix}}_{\text{Skew-Symmetric}}$

The final answer is:

$A = \begin{bmatrix}2&\frac{11}{2}&−\frac{5}{2} \\ \frac{11}{2}&3&\frac{3}{2} \\ −\frac{5}{2}&\frac{3}{2}&4 \end{bmatrix} + \begin{bmatrix}0&−\frac{3}{2}&−\frac{7}{2} \\ \frac{3}{2}&0&\frac{7}{2} \\ \frac{7}{2}&−\frac{7}{2}&0 \end{bmatrix}$

Given:

The matrix $A = \begin{bmatrix} 1&3&2 \\ 2&0&−1 \\ 1&2&3 \end{bmatrix}$.

To Show:

A satisfies the equation $A^3 – 4A^2 – 3A + 11I = O$, where I is the identity matrix of order 3 and O is the zero matrix of order 3.

Solution:

We need to calculate $A^2$, $A^3$, $4A^2$, $3A$, and $11I$ and then evaluate the expression $A^3 – 4A^2 – 3A + 11I$.

First, calculate $A^2 = A \times A$:

$A^2 = \begin{bmatrix} 1&3&2 \\ 2&0&−1 \\ 1&2&3 \end{bmatrix} \begin{bmatrix} 1&3&2 \\ 2&0&−1 \\ 1&2&3 \end{bmatrix}$

$A^2 = \begin{bmatrix} 1(1)+3(2)+2(1) & 1(3)+3(0)+2(2) & 1(2)+3(−1)+2(3) \\ 2(1)+0(2)+(−1)(1) & 2(3)+0(0)+(−1)(2) & 2(2)+0(−1)+(−1)(3) \\ 1(1)+2(2)+3(1) & 1(3)+2(0)+3(2) & 1(2)+2(−1)+3(3) \end{bmatrix}$

$A^2 = \begin{bmatrix} 1+6+2 & 3+0+4 & 2-3+6 \\ 2+0-1 & 6+0-2 & 4+0-3 \\ 1+4+3 & 3+0+6 & 2-2+9 \end{bmatrix}$

$A^2 = \begin{bmatrix} 9&7&5 \\ 1&4&1 \\ 8&9&9 \end{bmatrix}$

Next, calculate $A^3 = A^2 \times A$:

$A^3 = \begin{bmatrix} 9&7&5 \\ 1&4&1 \\ 8&9&9 \end{bmatrix} \begin{bmatrix} 1&3&2 \\ 2&0&−1 \\ 1&2&3 \end{bmatrix}$

$A^3 = \begin{bmatrix} 9(1)+7(2)+5(1) & 9(3)+7(0)+5(2) & 9(2)+7(−1)+5(3) \\ 1(1)+4(2)+1(1) & 1(3)+4(0)+1(2) & 1(2)+4(−1)+1(3) \\ 8(1)+9(2)+9(1) & 8(3)+9(0)+9(2) & 8(2)+9(−1)+9(3) \end{bmatrix}$

$A^3 = \begin{bmatrix} 9+14+5 & 27+0+10 & 18-7+15 \\ 1+8+1 & 3+0+2 & 2-4+3 \\ 8+18+9 & 24+0+18 & 16-9+27 \end{bmatrix}$

$A^3 = \begin{bmatrix} 28&37&26 \\ 10&5&1 \\ 35&42&34 \end{bmatrix}$

Calculate $4A^2$:

$4A^2 = 4 \begin{bmatrix} 9&7&5 \\ 1&4&1 \\ 8&9&9 \end{bmatrix} = \begin{bmatrix} 4 \times 9 & 4 \times 7 & 4 \times 5 \\ 4 \times 1 & 4 \times 4 & 4 \times 1 \\ 4 \times 8 & 4 \times 9 & 4 \times 9 \end{bmatrix} = \begin{bmatrix} 36&28&20 \\ 4&16&4 \\ 32&36&36 \end{bmatrix}$

Calculate $3A$:

$3A = 3 \begin{bmatrix} 1&3&2 \\ 2&0&−1 \\ 1&2&3 \end{bmatrix} = \begin{bmatrix} 3 \times 1 & 3 \times 3 & 3 \times 2 \\ 3 \times 2 & 3 \times 0 & 3 \times (−1) \\ 3 \times 1 & 3 \times 2 & 3 \times 3 \end{bmatrix} = \begin{bmatrix} 3&9&6 \\ 6&0&−3 \\ 3&6&9 \end{bmatrix}$

Calculate $11I$ (where $I = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}$):

$11I = 11 \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 11 \times 1 & 11 \times 0 & 11 \times 0 \\ 11 \times 0 & 11 \times 1 & 11 \times 0 \\ 11 \times 0 & 11 \times 0 & 11 \times 1 \end{bmatrix} = \begin{bmatrix} 11&0&0 \\ 0&11&0 \\ 0&0&11 \end{bmatrix}$

Now substitute these results into the expression $A^3 – 4A^2 – 3A + 11I$:

$A^3 – 4A^2 – 3A + 11I = \begin{bmatrix} 28&37&26 \\ 10&5&1 \\ 35&42&34 \end{bmatrix} - \begin{bmatrix} 36&28&20 \\ 4&16&4 \\ 32&36&36 \end{bmatrix} - \begin{bmatrix} 3&9&6 \\ 6&0&−3 \\ 3&6&9 \end{bmatrix} + \begin{bmatrix} 11&0&0 \\ 0&11&0 \\ 0&0&11 \end{bmatrix}$

$A^3 – 4A^2 – 3A + 11I = \begin{bmatrix} (28-36-3+11) & (37-28-9+0) & (26-20-6+0) \\ (10-4-6+0) & (5-16-0+11) & (1-4-(-3)+0) \\ (35-32-3+0) & (42-36-6+0) & (34-36-9+11) \end{bmatrix}$

$A^3 – 4A^2 – 3A + 11I = \begin{bmatrix} (28-36-3+11) & (37-28-9) & (26-20-6) \\ (10-4-6) & (5-16+11) & (1-4+3) \\ (35-32-3) & (42-36-6) & (34-36-9+11) \end{bmatrix}$

$A^3 – 4A^2 – 3A + 11I = \begin{bmatrix} (28+11-36-3) & (37-(28+9)) & (26-(20+6)) \\ (10-(4+6)) & (5+11-16) & (1+3-4) \\ (35-(32+3)) & (42-(36+6)) & (34+11-36-9) \end{bmatrix}$

$A^3 – 4A^2 – 3A + 11I = \begin{bmatrix} (39-39) & (37-37) & (26-26) \\ (10-10) & (16-16) & (4-4) \\ (35-35) & (42-42) & (45-45) \end{bmatrix}$

$A^3 – 4A^2 – 3A + 11I = \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$

This is the zero matrix O.

Thus, $A^3 – 4A^2 – 3A + 11I = O$ is satisfied.

Given:

The matrix $A = \begin{bmatrix}2&3 \\ −1&2 \end{bmatrix}$.

To Show:

A satisfies the equation $A^2 – 4A + 7I = O$, where I is the identity matrix of order 2 and O is the zero matrix of order 2.

To Find:

Calculate $A^5$ using the result $A^2 – 4A + 7I = O$.

Solution:

First, we calculate $A^2 = A \times A$:

$A^2 = \begin{bmatrix}2&3 \\ −1&2 \end{bmatrix} \begin{bmatrix}2&3 \\ −1&2 \end{bmatrix}$

$A^2 = \begin{bmatrix}2(2)+3(−1)&2(3)+3(2) \\ (−1)(2)+2(−1)&(−1)(3)+2(2) \end{bmatrix}$

$A^2 = \begin{bmatrix}4-3&6+6 \\ -2-2&-3+4 \end{bmatrix}$

$A^2 = \begin{bmatrix}1&12 \\ -4&1 \end{bmatrix}$

Next, we calculate $4A$:

$4A = 4 \begin{bmatrix}2&3 \\ −1&2 \end{bmatrix} = \begin{bmatrix}4 \times 2&4 \times 3 \\ 4 \times (−1)&4 \times 2 \end{bmatrix} = \begin{bmatrix}8&12 \\ -4&8 \end{bmatrix}$

The identity matrix of order 2 is $I = \begin{bmatrix}1&0 \\ 0&1 \end{bmatrix}$. Calculate $7I$:

$7I = 7 \begin{bmatrix}1&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix}7 \times 1&7 \times 0 \\ 7 \times 0&7 \times 1 \end{bmatrix} = \begin{bmatrix}7&0 \\ 0&7 \end{bmatrix}$

Now substitute these into the expression $A^2 – 4A + 7I$:

$A^2 – 4A + 7I = \begin{bmatrix}1&12 \\ -4&1 \end{bmatrix} - \begin{bmatrix}8&12 \\ -4&8 \end{bmatrix} + \begin{bmatrix}7&0 \\ 0&7 \end{bmatrix}$

$A^2 – 4A + 7I = \begin{bmatrix}1-8+7&12-12+0 \\ -4-(-4)+0&1-8+7 \end{bmatrix}$

$A^2 – 4A + 7I = \begin{bmatrix}1-8+7&12-12 \\ -4+4&1-8+7 \end{bmatrix}$

$A^2 – 4A + 7I = \begin{bmatrix}0&0 \\ 0&0 \end{bmatrix}$

This is the zero matrix O.

Thus, $A^2 – 4A + 7I = O$ is shown.

Now we use the result $A^2 – 4A + 7I = O$ to calculate $A^5$.

From the equation, we can write:

$A^2 = 4A - 7I$

Multiply by A to find $A^3$:

$A^3 = A \times A^2 = A(4A - 7I)$

$A^3 = 4A^2 - 7AI$

Since AI = A:

$A^3 = 4A^2 - 7A$

Substitute the expression for $A^2$ ($A^2 = 4A - 7I$):

$A^3 = 4(4A - 7I) - 7A$

$A^3 = 16A - 28I - 7A$

$A^3 = (16-7)A - 28I$

$A^3 = 9A - 28I$

Multiply $A^3$ by A to find $A^4$:

$A^4 = A \times A^3 = A(9A - 28I)$

$A^4 = 9A^2 - 28AI$

$A^4 = 9A^2 - 28A$

Substitute the expression for $A^2$ ($A^2 = 4A - 7I$):

$A^4 = 9(4A - 7I) - 28A$

$A^4 = 36A - 63I - 28A$

$A^4 = (36-28)A - 63I$

$A^4 = 8A - 63I$

Multiply $A^4$ by A to find $A^5$:

$A^5 = A \times A^4 = A(8A - 63I)$

$A^5 = 8A^2 - 63AI$

$A^5 = 8A^2 - 63A$

Substitute the expression for $A^2$ ($A^2 = 4A - 7I$):

$A^5 = 8(4A - 7I) - 63A$

$A^5 = 32A - 56I - 63A$

$A^5 = (32-63)A - 56I$

$A^5 = -31A - 56I$

Now substitute the matrices for A and I:

$A^5 = -31 \begin{bmatrix}2&3 \\ −1&2 \end{bmatrix} - 56 \begin{bmatrix}1&0 \\ 0&1 \end{bmatrix}$

$A^5 = \begin{bmatrix}-31 \times 2&-31 \times 3 \\ -31 \times (−1)&-31 \times 2 \end{bmatrix} - \begin{bmatrix}56 \times 1&56 \times 0 \\ 56 \times 0&56 \times 1 \end{bmatrix}$

$A^5 = \begin{bmatrix}-62&-93 \\ 31&-62 \end{bmatrix} - \begin{bmatrix}56&0 \\ 0&56 \end{bmatrix}$

$A^5 = \begin{bmatrix}-62-56&-93-0 \\ 31-0&-62-56 \end{bmatrix}$

$A^5 = \begin{bmatrix}-118&-93 \\ 31&-118 \end{bmatrix}$

The final answer is:

$A^2 – 4A + 7I = O$ has been shown.

$A^5 = \begin{bmatrix}-118&-93 \\ 31&-118 \end{bmatrix}$

Given:

A and B are square matrices of the same order.

To Find:

The expression equal to $(A + B) (A – B)$.

Solution:

We need to expand the product $(A + B) (A – B)$.

Using the distributive property of matrix multiplication, we multiply each term in the first matrix by each term in the second matrix:

$(A + B) (A – B) = A(A – B) + B(A – B)$

Now distribute A and B:

$A(A – B) = A \times A - A \times B = A^2 - AB$

$B(A – B) = B \times A - B \times B = BA - B^2$

Substitute these back into the expression:

$(A + B) (A – B) = (A^2 - AB) + (BA - B^2)$

$(A + B) (A – B) = A^2 - AB + BA - B^2$

Unlike scalar multiplication where $(a+b)(a-b) = a^2 - b^2$, for matrices, $AB$ is generally not equal to $BA$. Therefore, the $-AB$ and $+BA$ terms do not cancel each other out unless AB = BA (i.e., A and B commute).

The expanded expression is $A^2 - AB + BA - B^2$.

Comparing this result with the given options:

(A) $A^2 – B^2$ (Incorrect)

(B) $A^2 – BA – AB – B^2$ (Incorrect signs for BA and AB)

(C) $A^2 – B^2 + BA – AB$ (Correct terms, just rearranged order: $A^2 - AB + BA - B^2$)

(D) $A^2 – BA + B^2 + AB$ (Incorrect signs for $B^2$, incorrect signs/order for BA and AB)

The correct option is (C).

The final answer is:

$(A + B) (A – B) = A^2 - AB + BA - B^2$

The correct option is (C) $A^2 – B^2 + BA – AB$.

Given:

Matrix $A = \begin{bmatrix} 2&−1&3 \\ −4&5&1 \end{bmatrix}$ and Matrix $B = \begin{bmatrix} 2&3 \\ 4&−2 \\ 1&5 \end{bmatrix}$.

To Find:

Determine which matrix products (AB, BA) are defined.

Solution:

For the product of two matrices, say PQ, to be defined, the number of columns in the first matrix (P) must be equal to the number of rows in the second matrix (Q).

First, let's determine the dimensions (order) of matrices A and B.

Matrix A has 2 rows and 3 columns. So, the order of A is $2 \times 3$.

Matrix B has 3 rows and 2 columns. So, the order of B is $3 \times 2$.

Now, let's check if the product AB is defined:

Matrix A is of order $2 \times 3$.

Matrix B is of order $3 \times 2$.

Number of columns in A = 3.

Number of rows in B = 3.

Since the number of columns in A (3) is equal to the number of rows in B (3), the product AB is defined. The order of the resulting matrix AB will be $2 \times 2$.

Next, let's check if the product BA is defined:

Matrix B is of order $3 \times 2$.

Matrix A is of order $2 \times 3$.

Number of columns in B = 2.

Number of rows in A = 2.

Since the number of columns in B (2) is equal to the number of rows in A (2), the product BA is defined. The order of the resulting matrix BA will be $3 \times 3$.

Since both AB and BA are defined, the correct option is (C).

The final answer is:

AB is defined (order $2 \times 2$).

BA is defined (order $3 \times 3$).

The correct option is (C) AB and BA both are defined.

Given:

The matrix $A = \begin{bmatrix} 0&0&5 \\ 0&5&0 \\ 5&0&0 \end{bmatrix}$.

To Find:

Identify the type of the given matrix A.

Solution:

Let's examine the given matrix $A = \begin{bmatrix} 0&0&5 \\ 0&5&0 \\ 5&0&0 \end{bmatrix}$.

The matrix has 3 rows and 3 columns. Since the number of rows is equal to the number of columns, it is a square matrix.

Let's check the other options:

• A diagonal matrix is a square matrix where all non-diagonal elements are zero. In matrix A, the elements $a_{13}=5$ and $a_{31}=5$ are non-diagonal and non-zero. So, it is not a diagonal matrix.
• A scalar matrix is a diagonal matrix where all diagonal elements are equal. Since A is not a diagonal matrix, it cannot be a scalar matrix. Also, the diagonal elements (0, 5, 0) are not equal.
• A unit matrix (or identity matrix) is a scalar matrix where all diagonal elements are 1. Since A is not a scalar matrix and its diagonal elements are not all 1, it is not a unit matrix.

Based on the definitions, the matrix A fits the definition of a square matrix.

The final answer is:

The matrix $A = \begin{bmatrix} 0&0&5 \\ 0&5&0 \\ 5&0&0 \end{bmatrix}$ is a square matrix.

The correct option is (D) square matrix.

Given:

A and B are symmetric matrices of the same order.

This means $A' = A$ and $B' = B$.

To Find:

The type of matrix (AB′ – BA′).

Solution:

Let $C = AB' – BA'$.

To determine the type of matrix C, we need to find its transpose, $C'$, and compare it with C.

We use the properties of matrix transpose:

1. $(X - Y)' = X' - Y'$

2. $(XY)' = Y'X'$

3. $(X')' = X$

Using property 1:

$C' = (AB' – BA')' = (AB')' – (BA')'$

Using property 2 for $(AB')'$ and $(BA')'$:

$(AB')' = (B')'A'$

$(BA')' = (A')'B'$

So,

$C' = (B')'A' – (A')'B'$

Using property 3, $(B')' = B$ and $(A')' = A$. Substitute these into the expression for $C'$:

$C' = BA' – AB'$

We are given that A and B are symmetric matrices, which means $A' = A$ and $B' = B$. Substitute these into the expression for $C'$:

$C' = BA – AB$

Now, let's compare $C'$ with $C = AB' – BA'$. Since $A'=A$ and $B'=B$, the original matrix is $C = AB – BA$.

We have $C' = BA – AB$.

Let's consider $-C$:

$-C = -(AB – BA) = -AB + BA = BA – AB$

We observe that $C' = BA – AB$ and $-C = BA – AB$.

Therefore, $C' = -C$.

A matrix C for which $C' = -C$ is defined as a skew-symmetric matrix.

Thus, $(AB' – BA')$ is a skew-symmetric matrix.

The correct option is (A).

The final answer is:

The matrix $(AB' – BA')$ is a skew-symmetric matrix.

This is because $(AB' – BA')' = (B')'A' - (A')'B' = BA' - AB' = BA - AB = -(AB - BA) = -(AB' - BA')$.

Let $M = AB' - BA'$. Then $M' = -(AB' - BA') = -M$.

$\underbrace{(AB' – BA')}_{\text{Skew-Symmetric}}$

The correct option is (A) Skew symmetric matrix.

Given:

A and B are skew symmetric matrices of the same order.

By the definition of a skew symmetric matrix, we have:

$A' = -A$

$B' = -B$

To Find:

The condition for the product AB to be a symmetric matrix.

Solution:

For the matrix AB to be a symmetric matrix, its transpose must be equal to itself.

$(AB)' = AB$

We use the property of transpose of a product of matrices, which states that $(XY)' = Y'X'$.

Applying this property to $(AB)'$:

$(AB)' = B'A'$

Now, substitute the given conditions for A and B being skew symmetric ($A' = -A$, $B' = -B$) into this expression:

$(AB)' = (-B)(-A)$

Since $(-B)(-A) = BA$, we have:

$(AB)' = BA$

For AB to be symmetric, we must have $(AB)' = AB$.

Equating the two expressions for $(AB)'$:

$BA = AB$

Thus, the product AB of two skew symmetric matrices A and B of the same order is a symmetric matrix if and only if A and B commute (i.e., $AB = BA$).

The final answer is:

If A and B are two skew symmetric matrices of same order, then AB is symmetric matrix if $AB = BA$.

Given:

A and B are matrices of the same order.

To Find:

The expression equal to $(3A – 2B)′$.

Solution:

We use the properties of matrix transpose:

1. The transpose of the sum or difference of matrices is the sum or difference of their transposes: $(X \pm Y)' = X' \pm Y'$.

2. The transpose of a scalar multiple of a matrix is the scalar multiple of the transpose of the matrix: $(kX)' = kX'$, where $k$ is a scalar.

Applying these properties to the expression $(3A – 2B)′$:

$(3A – 2B)′ = (3A)′ – (2B)′$

Now apply the property for scalar multiples:

$(3A)′ = 3A′$

$(2B)′ = 2B′$

Substitute these back into the expression:

$(3A – 2B)′ = 3A′ – 2B′$

The final answer is:

If A and B are matrices of same order, then $(3A – 2B)′$ is equal to $3A′ – 2B′$.

Given:

A statement about the condition for matrix addition to be defined.

To Find:

The condition on the order of matrices for their addition to be defined.

Solution:

For the addition or subtraction of two matrices, the matrices must have the same number of rows and the same number of columns. In other words, their orders must be identical.

For example, if matrix A is of order $m \times n$, and matrix B is of order $p \times q$, then A + B (or A - B) is defined only if $m = p$ and $n = q$.

The final answer is:

Addition of matrices is defined if order of the matrices is same.

Given:

A and B are two matrices of the same order.

To Evaluate:

Determine if the statement $2A + B = B + 2A$ is true or false.

Solution:

We are given that A and B are matrices of the same order. Let their order be $m \times n$.

The expression $2A$ represents scalar multiplication of matrix A by the scalar 2. The resulting matrix, $2A$, will also be of order $m \times n$.

The expression $2A + B$ represents the addition of matrix $2A$ (of order $m \times n$) and matrix B (of order $m \times n$). Matrix addition is defined because the orders are the same. The resulting matrix $2A + B$ is also of order $m \times n$.

The expression $B + 2A$ represents the addition of matrix B (of order $m \times n$) and matrix $2A$ (of order $m \times n$). Matrix addition is defined, and the result $B + 2A$ is of order $m \times n$.

Matrix addition is commutative. For any two matrices X and Y of the same order, we have $X + Y = Y + X$.

Let $X = 2A$ and $Y = B$. Since $2A$ and B are matrices of the same order, the commutative property of matrix addition applies:

$2A + B = B + 2A$

This equation is always true for any two matrices A and B of the same order.

The final answer is:

The statement is True.

Given:

The statement "Matrix subtraction is associative".

To Evaluate:

Determine if the statement is true or false.

Solution:

For an operation to be associative, the grouping of the operands does not affect the result. For matrices X, Y, and Z of the same order, associativity of subtraction would mean $(X - Y) - Z = X - (Y - Z)$.

Let's evaluate both sides of the potential associative property:

Left side: $(X - Y) - Z = X - Y - Z$

Right side: $X - (Y - Z) = X - Y + Z$

Comparing the left side ($X - Y - Z$) and the right side ($X - Y + Z$), we can see that they are not generally equal unless Z is the zero matrix. For example, if Z is a non-zero matrix, then $-Z \neq +Z$.

Therefore, matrix subtraction is not associative.

As a concrete example, let X, Y, and Z be $1 \times 1$ matrices (which are just scalars):

$X = [1]$, $Y = [2]$, $Z = [3]$

$(X - Y) - Z = ([1] - [2]) - [3] = [-1] - [3] = [-4]$

$X - (Y - Z) = [1] - ([2] - [3]) = [1] - [-1] = [1] + [1] = [2]$

Since $[-4] \neq [2]$, the associative property does not hold for matrix subtraction.

The final answer is:

The statement is False.

Given:

A is a non-singular matrix.

This implies that the inverse of A, denoted by $A^{-1}$, exists.

To Evaluate:

Determine if the statement $(A′)^{–1} = (A^{–1})′$ is true or false.

Solution:

A matrix X is the inverse of a matrix Y if and only if $XY = YX = I$, where I is the identity matrix of the same order.

Here, we want to show that $(A^{-1})'$ is the inverse of $A'$. This means we need to verify if the following conditions hold:

$A'(A^{-1})' = I$

and

$(A^{-1})'A' = I$

We know that since $A^{-1}$ is the inverse of A, we have:

Take the transpose of both sides of equation (i):

$(AA^{-1})' = I'$

Using the property of transpose of a product $(XY)' = Y'X'$, we get:

$(A^{-1})'A' = I'$

Since the transpose of an identity matrix is the identity matrix itself ($I' = I$), we have:

$(A^{-1})'A' = I$

Now, take the transpose of both sides of equation (ii):

$(A^{-1}A)' = I'$

Using the property of transpose of a product $(XY)' = Y'X'$, we get:

$A'(A^{-1})' = I'$

Since $I' = I$, we have:

$A'(A^{-1})' = I$

We have shown that $A'(A^{-1})' = I$ and $(A^{-1})'A' = I$. By the definition of an inverse, this means that $(A^{-1})'$ is the inverse of $A'$. Since the inverse of a matrix is unique, $(A')^{-1}$ must be equal to $(A^{-1})'$.

The final answer is:

The statement is True.

Given:

A statement claiming that if $AB = AC$ for any three matrices A, B, and C of the same order, then $B = C$.

To Evaluate:

Determine if the statement is true or false.

Solution:

The statement suggests that a cancellation law holds for matrix multiplication, meaning we can "cancel" A from both sides of the equation $AB = AC$. However, the cancellation law in matrix multiplication only holds under certain conditions.

Specifically, if A is a non-singular matrix (i.e., its determinant is non-zero, $\text{det}(A) \neq 0$), then its inverse $A^{-1}$ exists. In this case, if $AB = AC$, we can pre-multiply both sides by $A^{-1}$:

$A^{-1}(AB) = A^{-1}(AC)$

By the associative property of matrix multiplication:

$(A^{-1}A)B = (A^{-1}A)C$

Since $A^{-1}A = I$ (the identity matrix):

$IB = IC$

Since $IB = B$ and $IC = C$:

$B = C$

So, if A is non-singular, the cancellation law holds, and $AB = AC$ implies $B = C$.

However, the statement says "for any three matrices of same order". This includes cases where A is a singular matrix (i.e., $\text{det}(A) = 0$, and $A^{-1}$ does not exist).

Let's consider an example where A is a singular matrix:

Let $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, and $C = \begin{bmatrix} 1 & 2 \\ 5 & 6 \end{bmatrix}$.

A, B, and C are all $2 \times 2$ matrices (same order).

Calculate AB:

$AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1 \times 1 + 0 \times 3 & 1 \times 2 + 0 \times 4 \\ 0 \times 1 + 0 \times 3 & 0 \times 2 + 0 \times 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$

Calculate AC:

$AC = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 1 \times 1 + 0 \times 5 & 1 \times 2 + 0 \times 6 \\ 0 \times 1 + 0 \times 5 & 0 \times 2 + 0 \times 6 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$

We see that $AB = AC = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$.

However, $B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 2 \\ 5 & 6 \end{bmatrix}$. Clearly, $B \neq C$.

This counterexample shows that the statement "AB = AC ⇒ B = C for any three matrices of same order" is false, as it does not hold when matrix A is singular.

The final answer is:

The statement is False.

Given:

Number of elements in a matrix is 28.

Number of elements in another matrix is 13.

To Find:

The possible orders of a matrix with 28 elements.

The possible orders of a matrix with 13 elements.

Solution:

The order of a matrix is represented as $m \times n$, where $m$ is the number of rows and $n$ is the number of columns.

The total number of elements in a matrix of order $m \times n$ is the product of the number of rows and the number of columns, i.e., $m \times n$.

Therefore, to find the possible orders of a matrix with a given number of elements, we need to find all possible pairs of positive integers $(m, n)$ such that the product $m \times n$ is equal to the given number of elements.

Case 1: Matrix has 28 elements.

We need to find all pairs of positive integers $(m, n)$ such that $m \times n = 28$. These pairs correspond to the factors of 28.

The positive integer factors of 28 are 1, 2, 4, 7, 14, and 28.

The possible pairs $(m, n)$ such that $m \times n = 28$ are:

If $m=1$, then $n=28$. Order: $1 \times 28$.

If $m=2$, then $n=14$. Order: $2 \times 14$.

If $m=4$, then $n=7$. Order: $4 \times 7$.

If $m=7$, then $n=4$. Order: $7 \times 4$.

If $m=14$, then $n=2$. Order: $14 \times 2$.

If $m=28$, then $n=1$. Order: $28 \times 1$.

So, the possible orders for a matrix with 28 elements are $1 \times 28$, $2 \times 14$, $4 \times 7$, $7 \times 4$, $14 \times 2$, and $28 \times 1$. There are 6 possible orders.

Case 2: Matrix has 13 elements.

We need to find all pairs of positive integers $(m, n)$ such that $m \times n = 13$.

The number 13 is a prime number. Its only positive integer factors are 1 and 13.

The possible pairs $(m, n)$ such that $m \times n = 13$ are:

If $m=1$, then $n=13$. Order: $1 \times 13$.

If $m=13$, then $n=1$. Order: $13 \times 1$.

So, the possible orders for a matrix with 13 elements are $1 \times 13$ and $13 \times 1$. There are 2 possible orders.

The final answer is:

If a matrix has 28 elements, the possible orders are $1 \times 28$, $2 \times 14$, $4 \times 7$, $7 \times 4$, $14 \times 2$, $28 \times 1$.

If a matrix has 13 elements, the possible orders are $1 \times 13$, $13 \times 1$.

Given:

The matrix $A = \begin{bmatrix} a&1&x \\ 2&\sqrt{3}&x^2−y \\ 0&5&\frac{−2}{5} \end{bmatrix}$.

To Find:

(i) The order of the matrix A.

(ii) The number of elements in A.

(iii) The values of elements $a_{23}$, $a_{31}$, and $a_{12}$.

Solution:

(i) The order of a matrix is given by the number of rows by the number of columns. In matrix A, there are 3 horizontal rows and 3 vertical columns.

So, the order of matrix A is $3 \times 3$.

(ii) The number of elements in a matrix is the product of the number of rows and the number of columns.

Number of elements = (Number of rows) $\times$ (Number of columns)

Number of elements = $3 \times 3 = 9$.

(iii) The element $a_{ij}$ is the element located in the $i$-th row and the $j$-th column of the matrix.

The given matrix is $A = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{bmatrix} = \begin{bmatrix} a&1&x \\ 2&\sqrt{3}&x^2−y \\ 0&5&−\frac{2}{5} \end{bmatrix}$.

The element $a_{23}$ is in the 2nd row and 3rd column. From the matrix A, $a_{23} = x^2 - y$.

The element $a_{31}$ is in the 3rd row and 1st column. From the matrix A, $a_{31} = 0$.

The element $a_{12}$ is in the 1st row and 2nd column. From the matrix A, $a_{12} = 1$.

The final answer is:

(i) The order of the matrix A is $3 \times 3$.

(ii) The number of elements in the matrix A is 9.

(iii) The elements are:

$a_{23} = x^2 - y$

$a_{31} = 0$

$a_{12} = 1$

Given:

We need to construct a $2 \times 2$ matrix A, where the element in the $i$-th row and $j$-th column is denoted by $a_{ij}$.

A $2 \times 2$ matrix has the form $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$.

To Construct:

A $2 \times 2$ matrix A for the given definitions of $a_{ij}$.

Solution:

(i) The formula for the elements is $a_{ij} = \frac{(i − 2j)^2}{2}$.

We calculate each element of the $2 \times 2$ matrix:

For $a_{11}$ (i=1, j=1):

$a_{11} = \frac{(1 - 2 \times 1)^2}{2} = \frac{(1 - 2)^2}{2} = \frac{(-1)^2}{2} = \frac{1}{2}$

For $a_{12}$ (i=1, j=2):

$a_{12} = \frac{(1 - 2 \times 2)^2}{2} = \frac{(1 - 4)^2}{2} = \frac{(-3)^2}{2} = \frac{9}{2}$

For $a_{21}$ (i=2, j=1):

$a_{21} = \frac{(2 - 2 \times 1)^2}{2} = \frac{(2 - 2)^2}{2} = \frac{0^2}{2} = \frac{0}{2} = 0$

For $a_{22}$ (i=2, j=2):

$a_{22} = \frac{(2 - 2 \times 2)^2}{2} = \frac{(2 - 4)^2}{2} = \frac{(-2)^2}{2} = \frac{4}{2} = 2$

So, the matrix for part (i) is:

$A = \begin{bmatrix} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \end{bmatrix}$

(ii) The formula for the elements is $a_{ij} = |-2i + 3j |$.

We calculate each element of the $2 \times 2$ matrix:

For $a_{11}$ (i=1, j=1):

$a_{11} = |-2(1) + 3(1)| = |-2 + 3| = |1| = 1$

For $a_{12}$ (i=1, j=2):

$a_{12} = |-2(1) + 3(2)| = |-2 + 6| = |4| = 4$

For $a_{21}$ (i=2, j=1):

$a_{21} = |-2(2) + 3(1)| = |-4 + 3| = |-1| = 1$

For $a_{22}$ (i=2, j=2):

$a_{22} = |-2(2) + 3(2)| = |-4 + 6| = |2| = 2$

So, the matrix for part (ii) is:

$A = \begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix}$

The final answer is:

(i) The matrix is $\begin{bmatrix} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \end{bmatrix}$.

(ii) The matrix is $\begin{bmatrix} 1 & 4 \\ 1 & 2 \end{bmatrix}$.

Given:

We need to construct a $3 \times 2$ matrix A, where the element in the $i$-th row and $j$-th column is denoted by $a_{ij}$.

The formula for the elements is $a_{ij} = e^{ix} \sin(jx)$.

To Construct:

A $3 \times 2$ matrix A with elements given by the formula $a_{ij} = e^{ix} \sin(jx)$.

Solution:

A $3 \times 2$ matrix has the form $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}$.

We calculate each element of the $3 \times 2$ matrix using the formula $a_{ij} = e^{ix} \sin(jx)$:

For $a_{11}$ (i=1, j=1):

$a_{11} = e^{1 \cdot x} \sin(1 \cdot x) = e^{x} \sin x$

For $a_{12}$ (i=1, j=2):

$a_{12} = e^{1 \cdot x} \sin(2 \cdot x) = e^{x} \sin 2x$

For $a_{21}$ (i=2, j=1):

$a_{21} = e^{2 \cdot x} \sin(1 \cdot x) = e^{2x} \sin x$

For $a_{22}$ (i=2, j=2):

$a_{22} = e^{2 \cdot x} \sin(2 \cdot x) = e^{2x} \sin 2x$

For $a_{31}$ (i=3, j=1):

$a_{31} = e^{3 \cdot x} \sin(1 \cdot x) = e^{3x} \sin x$

For $a_{32}$ (i=3, j=2):

$a_{32} = e^{3 \cdot x} \sin(2 \cdot x) = e^{3x} \sin 2x$

So, the $3 \times 2$ matrix is:

$A = \begin{bmatrix} e^{x} \sin x & e^{x} \sin 2x \\ e^{2x} \sin x & e^{2x} \sin 2x \\ e^{3x} \sin x & e^{3x} \sin 2x \end{bmatrix}$

The final answer is:

The $3 \times 2$ matrix is $\begin{bmatrix} e^{x} \sin x & e^{x} \sin 2x \\ e^{2x} \sin x & e^{2x} \sin 2x \\ e^{3x} \sin x & e^{3x} \sin 2x \end{bmatrix}$.

Given:

Matrices $A = \begin{bmatrix}a+4&3b \\ 8&−6\end{bmatrix}$ and $B = \begin{bmatrix} 2a+2&b^2+2 \\ 8&b^2−5b \end{bmatrix}$.

We are given that $A = B$.

To Find:

The values of a and b.

Solution:

Two matrices are equal if and only if they are of the same order and their corresponding elements are equal.

Both matrices A and B are of order $2 \times 2$. Since A = B, their corresponding elements must be equal.

Equating the corresponding elements, we get the following equations:

$a_{11} = b_{11} \Rightarrow a + 4 = 2a + 2$

$a_{12} = b_{12} \Rightarrow 3b = b^2 + 2$

$a_{21} = b_{21} \Rightarrow 8 = 8$ (This equation is trivially true and does not help in finding a or b)

$a_{22} = b_{22} \Rightarrow -6 = b^2 - 5b$

From the first equation ($a + 4 = 2a + 2$), we can solve for a:

$4 - 2 = 2a - a$

$2 = a$

Now consider the equation from the element $a_{12} = b_{12}$ ($3b = b^2 + 2$). Rearrange this into a quadratic equation:

$b^2 - 3b + 2 = 0$

Factor the quadratic equation:

$(b - 1)(b - 2) = 0$

This gives possible values for b:

$b - 1 = 0 \Rightarrow b = 1$

$b - 2 = 0 \Rightarrow b = 2$

So, $b$ could be 1 or 2 based on this equation.

Now consider the equation from the element $a_{22} = b_{22}$ ($-6 = b^2 - 5b$). Rearrange this into a quadratic equation:

$b^2 - 5b + 6 = 0$

Factor the quadratic equation:

$(b - 2)(b - 3) = 0$

This gives possible values for b:

$b - 2 = 0 \Rightarrow b = 2$

$b - 3 = 0 \Rightarrow b = 3$

So, $b$ could be 2 or 3 based on this equation.

For the matrices A and B to be equal, the value of b must satisfy both the equation from $a_{12} = b_{12}$ and the equation from $a_{22} = b_{22}$. The common value for b from these two sets of solutions is $b=2$.

Therefore, the required value of b is 2.

From equations (1) and (2), we have found the values of a and b.

The final answer is:

$a = 2$

$b = 2$

Given:

Matrices $A = \begin{bmatrix} \sqrt{3}&1 \\ 2&3 \end{bmatrix}$ and $B = \begin{bmatrix}x&y&z \\ a&b&6 \end{bmatrix}$.

To Find:

The sum of matrices A and B, if possible.

Solution:

For the addition of two matrices to be defined, the matrices must have the same order (i.e., the same number of rows and the same number of columns).

Let's determine the order of matrix A.

Matrix A has 2 rows and 2 columns. So, the order of A is $2 \times 2$.

Let's determine the order of matrix B.

Matrix B has 2 rows and 3 columns. So, the order of B is $2 \times 3$.

The order of matrix A ($2 \times 2$) is not the same as the order of matrix B ($2 \times 3$).

Since the orders of matrices A and B are different, their sum $A+B$ is not defined.

The final answer is:

The sum of matrices A and B is not defined because their orders are different.

Given:

Matrices $X = \begin{bmatrix} 3&1&−1 \\ 5&−2&−3 \end{bmatrix}$ and $Y = \begin{bmatrix} 2&1&−1 \\ 7&2&4 \end{bmatrix}$.

To Find:

(i) The sum of matrices X and Y.

(ii) The matrix $2X - 3Y$.

(iii) A matrix Z such that $X + Y + Z = O$, where O is the zero matrix.

Solution:

Both matrices X and Y are of order $2 \times 3$. Since their orders are the same, addition and subtraction are defined.

(i) Find X + Y:

To find the sum of two matrices of the same order, we add their corresponding elements.

$X + Y = \begin{bmatrix} 3&1&−1 \\ 5&−2&−3 \end{bmatrix} + \begin{bmatrix} 2&1&−1 \\ 7&2&4 \end{bmatrix}$

$X + Y = \begin{bmatrix} 3+2&1+1&−1+(-1) \\ 5+7&−2+2&−3+4 \end{bmatrix}$

$X + Y = \begin{bmatrix} 5&2&−2 \\ 12&0&1 \end{bmatrix}$

(ii) Find 2X – 3Y:

First, find 2X by multiplying each element of X by 2:

$2X = 2 \begin{bmatrix} 3&1&−1 \\ 5&−2&−3 \end{bmatrix} = \begin{bmatrix} 2 \times 3&2 \times 1&2 \times (−1) \\ 2 \times 5&2 \times (−2)&2 \times (−3) \end{bmatrix} = \begin{bmatrix} 6&2&−2 \\ 10&−4&−6 \end{bmatrix}$

Next, find 3Y by multiplying each element of Y by 3:

$3Y = 3 \begin{bmatrix} 2&1&−1 \\ 7&2&4 \end{bmatrix} = \begin{bmatrix} 3 \times 2&3 \times 1&3 \times (−1) \\ 3 \times 7&3 \times 2&3 \times 4 \end{bmatrix} = \begin{bmatrix} 6&3&−3 \\ 21&6&12 \end{bmatrix}$

Now, find 2X - 3Y by subtracting the corresponding elements of 3Y from 2X:

$2X – 3Y = \begin{bmatrix} 6&2&−2 \\ 10&−4&−6 \end{bmatrix} - \begin{bmatrix} 6&3&−3 \\ 21&6&12 \end{bmatrix}$

$2X – 3Y = \begin{bmatrix} 6-6&2-3&−2-(-3) \\ 10-21&−4-6&−6-12 \end{bmatrix}$

$2X – 3Y = \begin{bmatrix} 0&−1&−2+3 \\ −11&−10&−18 \end{bmatrix}$

$2X – 3Y = \begin{bmatrix} 0&−1&1 \\ −11&−10&−18 \end{bmatrix}$

(iii) Find a matrix Z such that X + Y + Z is a zero matrix.

The zero matrix O of order $2 \times 3$ is $\begin{bmatrix} 0&0&0 \\ 0&0&0 \end{bmatrix}$.

We are given $X + Y + Z = O$.

We can rearrange this equation to solve for Z:

$Z = O - (X + Y)$

$Z = -(X + Y)$

From part (i), we found $X + Y = \begin{bmatrix} 5&2&−2 \\ 12&0&1 \end{bmatrix}$.

Now, find - (X + Y) by negating each element of the sum matrix:

$Z = - \begin{bmatrix} 5&2&−2 \\ 12&0&1 \end{bmatrix} = \begin{bmatrix} -5&-2&−(−2) \\ -12&-0&-1 \end{bmatrix}$

$Z = \begin{bmatrix} -5&-2&2 \\ -12&0&-1 \end{bmatrix}$

The final answer is:

(i) $X + Y = \begin{bmatrix} 5&2&−2 \\ 12&0&1 \end{bmatrix}$

(ii) $2X – 3Y = \begin{bmatrix} 0&−1&1 \\ −11&−10&−18 \end{bmatrix}$

(iii) $Z = \begin{bmatrix} -5&-2&2 \\ -12&0&-1 \end{bmatrix}$

Given:

The matrix equation:

$x\begin{bmatrix} 2x&2 \\ 3&x \end{bmatrix} + 2 \begin{bmatrix} 8&5x \\ 4&4x \end{bmatrix} = 2 \begin{bmatrix} (x^2+8)&24 \\ (10)&6x \end{bmatrix}$.

To Find:

Non-zero values of x that satisfy the given matrix equation.

Solution:

Perform the scalar multiplications on both sides of the equation:

Left side:

$x\begin{bmatrix} 2x&2 \\ 3&x \end{bmatrix} = \begin{bmatrix} x(2x)&x(2) \\ x(3)&x(x) \end{bmatrix} = \begin{bmatrix} 2x^2&2x \\ 3x&x^2 \end{bmatrix}$

$2 \begin{bmatrix} 8&5x \\ 4&4x \end{bmatrix} = \begin{bmatrix} 2(8)&2(5x) \\ 2(4)&2(4x) \end{bmatrix} = \begin{bmatrix} 16&10x \\ 8&8x \end{bmatrix}$

Right side:

$2 \begin{bmatrix} (x^2+8)&24 \\ (10)&6x \end{bmatrix} = \begin{bmatrix} 2(x^2+8)&2(24) \\ 2(10)&2(6x) \end{bmatrix} = \begin{bmatrix} 2x^2+16&48 \\ 20&12x \end{bmatrix}$

Now, substitute the resulting matrices back into the equation:

$\begin{bmatrix} 2x^2&2x \\ 3x&x^2 \end{bmatrix} + \begin{bmatrix} 16&10x \\ 8&8x \end{bmatrix} = \begin{bmatrix} 2x^2+16&48 \\ 20&12x \end{bmatrix}$

Perform the matrix addition on the left side:

$\begin{bmatrix} 2x^2+16&2x+10x \\ 3x+8&x^2+8x \end{bmatrix} = \begin{bmatrix} 2x^2+16&48 \\ 20&12x \end{bmatrix}$

$\begin{bmatrix} 2x^2+16&12x \\ 3x+8&x^2+8x \end{bmatrix} = \begin{bmatrix} 2x^2+16&48 \\ 20&12x \end{bmatrix}$

For two matrices to be equal, their corresponding elements must be equal. This gives us a system of equations:

Element (1, 1): $2x^2 + 16 = 2x^2 + 16$ (This is always true)

Element (1, 2): $12x = 48$

Element (2, 1): $3x + 8 = 20$

Element (2, 2): $x^2 + 8x = 12x$

Let's solve each equation involving x:

From the (1, 2) element:

$12x = 48$

$x = \frac{48}{12}$

$x = 4$

From the (2, 1) element:

$3x + 8 = 20$

$3x = 20 - 8$

$3x = 12$

$x = \frac{12}{3}$

$x = 4$

From the (2, 2) element:

$x^2 + 8x = 12x$

$x^2 + 8x - 12x = 0$

$x^2 - 4x = 0$

Factor out x:

$x(x - 4) = 0$

This gives two possible solutions for x:

$x = 0$ or $x - 4 = 0 \Rightarrow x = 4$

For the matrix equation to be satisfied, the value of x must satisfy all the independent equations derived from the matrix equality. Comparing the solutions from equations (1), (2), and (3), the common non-zero value for x is 4.

The question asks for non-zero values of x, so x = 0 is excluded.

The final answer is:

The non-zero value of x satisfying the matrix equation is 4.

Given:

Matrices $A = \begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix}$ and $B = \begin{bmatrix} 0&−1 \\ 1&0 \end{bmatrix}$.

To Show:

$(A + B) (A – B) ≠ A^2 – B^2$.

Solution:

We need to calculate $(A + B)$, $(A - B)$, $(A + B)(A - B)$, $A^2$, $B^2$, and $A^2 - B^2$ and then compare $(A + B)(A - B)$ with $A^2 - B^2$.

First, calculate $(A + B)$:

$A + B = \begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix} + \begin{bmatrix} 0&−1 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 0+0&1+(-1) \\ 1+1&1+0 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 2&1 \end{bmatrix}$

Next, calculate $(A - B)$:

$A - B = \begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix} - \begin{bmatrix} 0&−1 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 0-0&1-(-1) \\ 1-1&1-0 \end{bmatrix} = \begin{bmatrix} 0&1+1 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 0&2 \\ 0&1 \end{bmatrix}$

Now, calculate $(A + B) (A – B)$:

$(A + B) (A – B) = \begin{bmatrix} 0&0 \\ 2&1 \end{bmatrix} \begin{bmatrix} 0&2 \\ 0&1 \end{bmatrix}$

$(A + B) (A – B) = \begin{bmatrix} 0(0)+0(0)&0(2)+0(1) \\ 2(0)+1(0)&2(2)+1(1) \end{bmatrix} = \begin{bmatrix} 0+0&0+0 \\ 0+0&4+1 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&5 \end{bmatrix}$

Now, calculate $A^2 = A \times A$:

$A^2 = \begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix} \begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix} = \begin{bmatrix} 0(0)+1(1)&0(1)+1(1) \\ 1(0)+1(1)&1(1)+1(1) \end{bmatrix} = \begin{bmatrix} 0+1&0+1 \\ 0+1&1+1 \end{bmatrix} = \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix}$

Calculate $B^2 = B \times B$:

$B^2 = \begin{bmatrix} 0&−1 \\ 1&0 \end{bmatrix} \begin{bmatrix} 0&−1 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 0(0)+(-1)(1)&0(-1)+(-1)(0) \\ 1(0)+0(1)&1(-1)+0(0) \end{bmatrix} = \begin{bmatrix} 0-1&0+0 \\ 0+0&-1+0 \end{bmatrix} = \begin{bmatrix} -1&0 \\ 0&-1 \end{bmatrix}$

Now, calculate $A^2 – B^2$:

$A^2 – B^2 = \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix} - \begin{bmatrix} -1&0 \\ 0&-1 \end{bmatrix} = \begin{bmatrix} 1-(-1)&1-0 \\ 1-0&2-(-1) \end{bmatrix} = \begin{bmatrix} 1+1&1 \\ 1&2+1 \end{bmatrix} = \begin{bmatrix} 2&1 \\ 1&3 \end{bmatrix}$

Compare the result from (1) and (2):

$(A + B) (A – B) = \begin{bmatrix} 0&0 \\ 0&5 \end{bmatrix}$

$A^2 – B^2 = \begin{bmatrix} 2&1 \\ 1&3 \end{bmatrix}$

Since $\begin{bmatrix} 0&0 \\ 0&5 \end{bmatrix} \neq \begin{bmatrix} 2&1 \\ 1&3 \end{bmatrix}$, we have shown that $(A + B) (A – B) \neq A^2 – B^2$.

This inequality holds because, for matrices, matrix multiplication is not commutative in general, meaning $AB \neq BA$. The expansion of $(A+B)(A-B)$ is $A^2 - AB + BA - B^2$. If $AB \neq BA$, then $-AB + BA \neq O$, and thus $A^2 - AB + BA - B^2 \neq A^2 - B^2$.

The final answer is:

Calculations show that $(A + B) (A – B) = \begin{bmatrix} 0&0 \\ 0&5 \end{bmatrix}$ and $A^2 – B^2 = \begin{bmatrix} 2&1 \\ 1&3 \end{bmatrix}$.

Since $\begin{bmatrix} 0&0 \\ 0&5 \end{bmatrix} \neq \begin{bmatrix} 2&1 \\ 1&3 \end{bmatrix}$, the statement $(A + B) (A – B) ≠ A^2 – B^2$ is shown to be true for the given matrices A and B.

Given:

The matrix equation $\begin{bmatrix} 1&x&1 \end{bmatrix} \begin{bmatrix} 1&3&2 \\ 2&5&1 \\ 15&3&2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = 0$.

To Find:

The value of x that satisfies the given matrix equation.

Solution:

We need to perform the matrix multiplications step by step. We will multiply the first two matrices first, and then multiply the result by the third matrix.

Let $M_1 = \begin{bmatrix} 1&x&1 \end{bmatrix}$, $M_2 = \begin{bmatrix} 1&3&2 \\ 2&5&1 \\ 15&3&2 \end{bmatrix}$, and $M_3 = \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}$.

The equation is $M_1 M_2 M_3 = 0$. Note that 0 on the right side represents the zero matrix of the appropriate order. The order of $M_1$ is $1 \times 3$, the order of $M_2$ is $3 \times 3$, and the order of $M_3$ is $3 \times 1$.

The order of the product $M_1 M_2$ will be $1 \times 3$.

The order of the product $(M_1 M_2) M_3$ will be $1 \times 1$. So the right side is a $1 \times 1$ zero matrix, i.e., $[0]$.

First, calculate $M_1 M_2$:

$M_1 M_2 = \begin{bmatrix} 1&x&1 \end{bmatrix} \begin{bmatrix} 1&3&2 \\ 2&5&1 \\ 15&3&2 \end{bmatrix}$

$M_1 M_2 = \begin{bmatrix} 1(1) + x(2) + 1(15) & 1(3) + x(5) + 1(3) & 1(2) + x(1) + 1(2) \end{bmatrix}$

$M_1 M_2 = \begin{bmatrix} 1 + 2x + 15 & 3 + 5x + 3 & 2 + x + 2 \end{bmatrix}$

$M_1 M_2 = \begin{bmatrix} 2x + 16 & 5x + 6 & x + 4 \end{bmatrix}$

Now, multiply this result by $M_3$:

$(M_1 M_2) M_3 = \begin{bmatrix} 2x + 16 & 5x + 6 & x + 4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}$

$(M_1 M_2) M_3 = \begin{bmatrix} (2x + 16)(1) + (5x + 6)(2) + (x + 4)(x) \end{bmatrix}$

$(M_1 M_2) M_3 = \begin{bmatrix} 2x + 16 + 10x + 12 + x^2 + 4x \end{bmatrix}$

$(M_1 M_2) M_3 = \begin{bmatrix} x^2 + 2x + 10x + 4x + 16 + 12 \end{bmatrix}$

$(M_1 M_2) M_3 = \begin{bmatrix} x^2 + 16x + 28 \end{bmatrix}$

The given equation is $(M_1 M_2) M_3 = 0$. So, we have:

$\begin{bmatrix} x^2 + 16x + 28 \end{bmatrix} = [0]$

Equating the elements of the two $1 \times 1$ matrices:

$x^2 + 16x + 28 = 0$

This is a quadratic equation in x. We can solve it by factoring or using the quadratic formula.

Find two numbers that multiply to 28 and add up to 16. These numbers are 2 and 14 ($2 \times 14 = 28$, $2 + 14 = 16$).

Factor the quadratic equation:

$(x + 2)(x + 14) = 0$

This gives two possible solutions for x:

$x + 2 = 0 \Rightarrow x = -2$

$x + 14 = 0 \Rightarrow x = -14$

Both values are non-zero and satisfy the original matrix equation.

The final answer is:

The values of x are -2 and -14.

Given:

The matrix $A = \begin{bmatrix} 5&3 \\ −1&−2 \end{bmatrix}$.

To Show:

A satisfies the equation $A^2 – 3A – 7I = O$, where I is the identity matrix of order 2 and O is the zero matrix of order 2.

To Find:

Find the inverse of matrix A ($A^{-1}$) using the given equation.

Solution (Part 1: Show the equation is satisfied):

We need to calculate $A^2$, $3A$, and $7I$ and then evaluate the expression $A^2 – 3A – 7I$.

First, calculate $A^2 = A \times A$:

$A^2 = \begin{bmatrix} 5&3 \\ −1&−2 \end{bmatrix} \begin{bmatrix} 5&3 \\ −1&−2 \end{bmatrix}$

$A^2 = \begin{bmatrix} 5(5)+3(−1)&5(3)+3(−2) \\ (−1)(5)+(−2)(−1)&(−1)(3)+(−2)(−2) \end{bmatrix}$

$A^2 = \begin{bmatrix} 25-3&15-6 \\ -5+2&-3+4 \end{bmatrix}$

$A^2 = \begin{bmatrix} 22&9 \\ -3&1 \end{bmatrix}$

Next, calculate $3A$:

$3A = 3 \begin{bmatrix} 5&3 \\ −1&−2 \end{bmatrix} = \begin{bmatrix} 3 \times 5&3 \times 3 \\ 3 \times (−1)&3 \times (−2) \end{bmatrix} = \begin{bmatrix} 15&9 \\ -3&-6 \end{bmatrix}$

The identity matrix of order 2 is $I = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$. Calculate $7I$:

$7I = 7 \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 7 \times 1&7 \times 0 \\ 7 \times 0&7 \times 1 \end{bmatrix} = \begin{bmatrix} 7&0 \\ 0&7 \end{bmatrix}$

Now substitute these into the expression $A^2 – 3A – 7I$:

$A^2 – 3A – 7I = \begin{bmatrix} 22&9 \\ -3&1 \end{bmatrix} - \begin{bmatrix} 15&9 \\ -3&-6 \end{bmatrix} - \begin{bmatrix} 7&0 \\ 0&7 \end{bmatrix}$

$A^2 – 3A – 7I = \begin{bmatrix} 22-15-7&9-9-0 \\ -3-(-3)-0&1-(-6)-7 \end{bmatrix}$

$A^2 – 3A – 7I = \begin{bmatrix} 22-(15+7)&(9-9)-0 \\ -3+3-0&1+6-7 \end{bmatrix}$

$A^2 – 3A – 7I = \begin{bmatrix} 22-22&0 \\ 0&7-7 \end{bmatrix}$

$A^2 – 3A – 7I = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$

This is the zero matrix O.

Thus, $A^2 – 3A – 7I = O$ is satisfied.

Solution (Part 2: Find A^–1):

We use the result $A^2 – 3A – 7I = O$.

Rearrange the equation to isolate the term with the identity matrix:

$A^2 – 3A = 7I$

We want to find $A^{-1}$. We can pre-multiply (or post-multiply) both sides of the equation by $A^{-1}$. Let's pre-multiply by $A^{-1}$:

$A^{-1}(A^2 – 3A) = A^{-1}(7I)$

Using the distributive property of matrix multiplication:

$A^{-1}A^2 – A^{-1}(3A) = A^{-1}(7I)$

Use the properties $A^{-1}A = I$ and $(A^{-1}A)A = IA = A$, and for scalar multiplication $k(XY) = (kX)Y = X(kY)$:

$(A^{-1}A)A – 3(A^{-1}A) = 7(A^{-1}I)$

Using $A^{-1}A = I$ and $A^{-1}I = A^{-1}$:

$IA – 3I = 7A^{-1}$

Using $IA = A$:

$A – 3I = 7A^{-1}$

Now, solve for $A^{-1}$ by multiplying both sides by $\frac{1}{7}$ (or dividing by 7):

$A^{-1} = \frac{1}{7}(A – 3I)$

Substitute the matrices for A and I:

$A^{-1} = \frac{1}{7}\left(\begin{bmatrix} 5&3 \\ −1&−2 \end{bmatrix} - 3\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}\right)$

Calculate 3I:

$3\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 3&0 \\ 0&3 \end{bmatrix}$

Calculate $A - 3I$:

$A – 3I = \begin{bmatrix} 5&3 \\ −1&−2 \end{bmatrix} - \begin{bmatrix} 3&0 \\ 0&3 \end{bmatrix} = \begin{bmatrix} 5-3&3-0 \\ -1-0&-2-3 \end{bmatrix} = \begin{bmatrix} 2&3 \\ -1&-5 \end{bmatrix}$

Finally, calculate $A^{-1}$:

$A^{-1} = \frac{1}{7}\begin{bmatrix} 2&3 \\ -1&-5 \end{bmatrix} = \begin{bmatrix} \frac{2}{7}&\frac{3}{7} \\ \frac{-1}{7}&\frac{-5}{7} \end{bmatrix}$

The final answer is:

The equation $A^2 – 3A – 7I = O$ has been shown to be satisfied.

Using this result, the inverse of A is $A^{-1} = \begin{bmatrix} \frac{2}{7}&\frac{3}{7} \\ \frac{-1}{7}&\frac{-5}{7} \end{bmatrix}$.

Given:

The matrix equation:

$\begin{bmatrix} 2&1 \\ 3&2 \end{bmatrix} A \begin{bmatrix} −3&2 \\ 5&−3 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$.

To Find:

The matrix A satisfying the given matrix equation.

Solution:

Let the given matrix equation be in the form $P A Q = I$, where:

$P = \begin{bmatrix} 2&1 \\ 3&2 \end{bmatrix}$

$Q = \begin{bmatrix} −3&2 \\ 5&−3 \end{bmatrix}$

and $I = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$ is the identity matrix of order 2.

To find matrix A, we need to isolate it in the equation $P A Q = I$. We can do this by pre-multiplying both sides by the inverse of P ($P^{-1}$) and post-multiplying both sides by the inverse of Q ($Q^{-1}$).

The equation becomes:

$P^{-1} (P A Q) Q^{-1} = P^{-1} I Q^{-1}$

Using the associative property of matrix multiplication $(XY)Z = X(YZ)$ and the property $M^{-1}M = MM^{-1} = I$:

$(P^{-1} P) A (Q Q^{-1}) = P^{-1} I Q^{-1}$

$I A I = P^{-1} Q^{-1}$

Since $IA = A$ and $AI = A$:

$A = P^{-1} Q^{-1}$

First, we need to find the inverse of matrix P. For a $2 \times 2$ matrix $\begin{bmatrix} a&b \\ c&d \end{bmatrix}$, the inverse is $\frac{1}{ad-bc} \begin{bmatrix} d&-b \\ -c&a \end{bmatrix}$, provided $ad-bc \neq 0$.

$P = \begin{bmatrix} 2&1 \\ 3&2 \end{bmatrix}$

The determinant of P is $\text{det}(P) = (2)(2) - (1)(3) = 4 - 3 = 1$.

Since $\text{det}(P) = 1 \neq 0$, $P^{-1}$ exists.

$P^{-1} = \frac{1}{1} \begin{bmatrix} 2&-1 \\ -3&2 \end{bmatrix} = \begin{bmatrix} 2&-1 \\ -3&2 \end{bmatrix}$

Next, we need to find the inverse of matrix Q.

$Q = \begin{bmatrix} −3&2 \\ 5&−3 \end{bmatrix}$

The determinant of Q is $\text{det}(Q) = (−3)(−3) - (2)(5) = 9 - 10 = -1$.

Since $\text{det}(Q) = -1 \neq 0$, $Q^{-1}$ exists.

$Q^{-1} = \frac{1}{-1} \begin{bmatrix} -3&-2 \\ -5&-3 \end{bmatrix} = -1 \begin{bmatrix} -3&-2 \\ -5&-3 \end{bmatrix} = \begin{bmatrix} 3&2 \\ 5&3 \end{bmatrix}$

Now, we can find matrix A by calculating the product $P^{-1} Q^{-1}$.

$A = P^{-1} Q^{-1} = \begin{bmatrix} 2&-1 \\ -3&2 \end{bmatrix} \begin{bmatrix} 3&2 \\ 5&3 \end{bmatrix}$

Perform the matrix multiplication:

$A = \begin{bmatrix} (2)(3) + (-1)(5) & (2)(2) + (-1)(3) \\ (-3)(3) + (2)(5) & (-3)(2) + (2)(3) \end{bmatrix}$

$A = \begin{bmatrix} 6 - 5 & 4 - 3 \\ -9 + 10 & -6 + 6 \end{bmatrix}$

$A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$

The final answer is:

$A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$

Given:

The matrix equation $\begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} A = \begin{bmatrix} −4&8&4 \\ −1&2&1 \\ −3&6&3 \end{bmatrix}$.

To Find:

The matrix A that satisfies the given equation.

Solution:

Let the given equation be $P A = R$, where $P = \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix}$ and $R = \begin{bmatrix} −4&8&4 \\ −1&2&1 \\ −3&6&3 \end{bmatrix}$.

The order of matrix P is $3 \times 1$. The order of matrix R is $3 \times 3$.

For the product PA to be defined, the number of columns in P must be equal to the number of rows in A. Since P has 1 column, A must have 1 row.

Let the order of matrix A be $1 \times n$.

The order of the product matrix PA is (number of rows in P) $\times$ (number of columns in A), which is $3 \times n$.

We are given that $PA = R$, so the order of PA must be equal to the order of R. Thus, $3 \times n = 3 \times 3$.

This implies that $n = 3$.

So, the order of matrix A must be $1 \times 3$. Let A be represented as:

$A = \begin{bmatrix} a&b&c \end{bmatrix}$

Substitute this into the given matrix equation:

$\begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} \begin{bmatrix} a&b&c \end{bmatrix} = \begin{bmatrix} −4&8&4 \\ −1&2&1 \\ −3&6&3 \end{bmatrix}$

Perform the matrix multiplication on the left side. The product of a $3 \times 1$ matrix and a $1 \times 3$ matrix is a $3 \times 3$ matrix:

$\begin{bmatrix} 4 \times a & 4 \times b & 4 \times c \\ 1 \times a & 1 \times b & 1 \times c \\ 3 \times a & 3 \times b & 3 \times c \end{bmatrix} = \begin{bmatrix} 4a & 4b & 4c \\ a & b & c \\ 3a & 3b & 3c \end{bmatrix}$

So the matrix equation becomes:

$\begin{bmatrix} 4a & 4b & 4c \\ a & b & c \\ 3a & 3b & 3c \end{bmatrix} = \begin{bmatrix} −4&8&4 \\ −1&2&1 \\ −3&6&3 \end{bmatrix}$

For two matrices to be equal, their corresponding elements must be equal. Equating the elements, we get:

From the element in the 2nd row and 1st column: $a = -1$

From the element in the 2nd row and 2nd column: $b = 2$

From the element in the 2nd row and 3rd column: $c = 1$

Let's check if these values are consistent with the other rows:

For the first row: $4a = 4(-1) = -4$ (Matches), $4b = 4(2) = 8$ (Matches), $4c = 4(1) = 4$ (Matches).

For the third row: $3a = 3(-1) = -3$ (Matches), $3b = 3(2) = 6$ (Matches), $3c = 3(1) = 3$ (Matches).

All elements are consistent. Thus, the values of a, b, and c are -1, 2, and 1 respectively.

Therefore, the matrix A is $\begin{bmatrix} -1&2&1 \end{bmatrix}$.

The final answer is:

$A = \begin{bmatrix} -1&2&1 \end{bmatrix}$

Given:

Matrices $A = \begin{bmatrix} 3&−4 \\ 1&1 \\ 2&0 \end{bmatrix}$ and $B = \begin{bmatrix} 2&1&2 \\ 1&2&4 \end{bmatrix}$.

To Verify:

$(BA)^2 \neq B^2A^2$.

Solution:

First, let's determine the orders of the matrices A and B.

Matrix A has 3 rows and 2 columns. So, the order of A is $3 \times 2$.

Matrix B has 2 rows and 3 columns. So, the order of B is $2 \times 3$.

Now, let's consider the left side of the inequality: $(BA)^2$.

First, calculate the product BA.

The order of B is $2 \times 3$ and the order of A is $3 \times 2$. Since the number of columns in B (3) is equal to the number of rows in A (3), the product BA is defined, and its order is $2 \times 2$.

$BA = \begin{bmatrix} 2&1&2 \\ 1&2&4 \end{bmatrix} \begin{bmatrix} 3&−4 \\ 1&1 \\ 2&0 \end{bmatrix}$

$BA = \begin{bmatrix} 2(3)+1(1)+2(2) & 2(−4)+1(1)+2(0) \\ 1(3)+2(1)+4(2) & 1(−4)+2(1)+4(0) \end{bmatrix}$

$BA = \begin{bmatrix} 6+1+4 & -8+1+0 \\ 3+2+8 & -4+2+0 \end{bmatrix}$

$BA = \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix}$

Now, calculate $(BA)^2 = (BA) \times (BA)$. Since BA is a $2 \times 2$ matrix, its square is defined and is also a $2 \times 2$ matrix.

$(BA)^2 = \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix} \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix}$

$(BA)^2 = \begin{bmatrix} 11(11)+(-7)(13) & 11(−7)+(-7)(−2) \\ 13(11)+(-2)(13) & 13(−7)+(-2)(−2) \end{bmatrix}$

$(BA)^2 = \begin{bmatrix} 121-91 & -77+14 \\ 143-26 & -91+4 \end{bmatrix}$

$(BA)^2 = \begin{bmatrix} 30 & -63 \\ 117 & -87 \end{bmatrix}$

Now, let's consider the right side of the inequality: $B^2A^2$.

First, consider $B^2 = B \times B$. The order of B is $2 \times 3$. For $B^2$ to be defined, the number of columns in B (3) must be equal to the number of rows in B (2). This is not the case.

Therefore, $B^2$ is not defined.

Next, consider $A^2 = A \times A$. The order of A is $3 \times 2$. For $A^2$ to be defined, the number of columns in A (2) must be equal to the number of rows in A (3). This is not the case.

Therefore, $A^2$ is not defined.

Since $B^2$ and $A^2$ are not defined, the product $B^2A^2$ is also not defined.

Comparing the results from (1) and (2):

From (1), $(BA)^2$ is a $2 \times 2$ matrix $\begin{bmatrix} 30 & -63 \\ 117 & -87 \end{bmatrix}$.

From (2), $B^2A^2$ is not defined.

A defined matrix cannot be equal to an undefined expression.

Therefore, $(BA)^2 \neq B^2A^2$ is verified.

The final answer is:

Calculations show that $(BA)^2 = \begin{bmatrix} 30 & -63 \\ 117 & -87 \end{bmatrix}$.

The terms $B^2$ and $A^2$ are not defined for the given matrices B (order $2 \times 3$) and A (order $3 \times 2$).

Since $B^2A^2$ is not defined and $(BA)^2$ is a defined matrix, we have verified that $(BA)^2 \neq B^2A^2$.

Given:

Matrices $A = \begin{bmatrix} 2&1&2 \\ 1&2&4 \end{bmatrix}$ and $B = \begin{bmatrix} 4&1 \\ 2&3 \\ 1&2 \end{bmatrix}$.

To Find:

If possible, find the matrix products BA and AB.

Solution:

For the product of two matrices, say XY, to be defined, the number of columns in the first matrix (X) must be equal to the number of rows in the second matrix (Y).

First, let's determine the dimensions (order) of matrices A and B.

Matrix A has 2 rows and 3 columns. So, the order of A is $2 \times 3$.

Matrix B has 3 rows and 2 columns. So, the order of B is $3 \times 2$.

Check if the product BA is defined:

Matrix B is of order $3 \times 2$.

Matrix A is of order $2 \times 3$.

Number of columns in B = 2.

Number of rows in A = 2.

Since the number of columns in B (2) is equal to the number of rows in A (2), the product BA is defined. The order of the resulting matrix BA will be $3 \times 3$.

Calculate BA:

$BA = \begin{bmatrix} 4&1 \\ 2&3 \\ 1&2 \end{bmatrix} \begin{bmatrix} 2&1&2 \\ 1&2&4 \end{bmatrix}$

$BA = \begin{bmatrix} 4(2)+1(1) & 4(1)+1(2) & 4(2)+1(4) \\ 2(2)+3(1) & 2(1)+3(2) & 2(2)+3(4) \\ 1(2)+2(1) & 1(1)+2(2) & 1(2)+2(4) \end{bmatrix}$

$BA = \begin{bmatrix} 8+1 & 4+2 & 8+4 \\ 4+3 & 2+6 & 4+12 \\ 2+2 & 1+4 & 2+8 \end{bmatrix}$

$BA = \begin{bmatrix} 9&6&12 \\ 7&8&16 \\ 4&5&10 \end{bmatrix}$

Check if the product AB is defined:

Matrix A is of order $2 \times 3$.

Matrix B is of order $3 \times 2$.

Number of columns in A = 3.

Number of rows in B = 3.

Since the number of columns in A (3) is equal to the number of rows in B (3), the product AB is defined. The order of the resulting matrix AB will be $2 \times 2$.

Calculate AB:

$AB = \begin{bmatrix} 2&1&2 \\ 1&2&4 \end{bmatrix} \begin{bmatrix} 4&1 \\ 2&3 \\ 1&2 \end{bmatrix}$

$AB = \begin{bmatrix} 2(4)+1(2)+2(1) & 2(1)+1(3)+2(2) \\ 1(4)+2(2)+4(1) & 1(1)+2(3)+4(2) \end{bmatrix}$

$AB = \begin{bmatrix} 8+2+2 & 2+3+4 \\ 4+4+4 & 1+6+8 \end{bmatrix}$

$AB = \begin{bmatrix} 12&9 \\ 12&15 \end{bmatrix}$

The final answer is:

$BA = \begin{bmatrix} 9&6&12 \\ 7&8&16 \\ 4&5&10 \end{bmatrix}$

$AB = \begin{bmatrix} 12&9 \\ 12&15 \end{bmatrix}$

Given:

We need to find two non-zero matrices A and B such that their product AB is the zero matrix O.

To Show (by Example):

Provide matrices A and B such that $A \neq O$, $B \neq O$, but $AB = O$.

Solution:

In general, for real numbers, if $a \neq 0$ and $b \neq 0$, then their product $ab \neq 0$. However, this property does not hold for matrix multiplication. It is possible for the product of two non-zero matrices to be the zero matrix.

Consider the following two $2 \times 2$ matrices:

Let $A = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix}$

Let $B = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$

Check if A and B are non-zero matrices:

Matrix A contains a non-zero element (the element in the first row and first column is 1). Therefore, $A \neq O$.

Matrix B contains a non-zero element (the element in the second row and second column is 1). Therefore, $B \neq O$.

Now, let's calculate the product AB:

$AB = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$

$AB = \begin{bmatrix} 1(0)+0(0)&1(0)+0(1) \\ 0(0)+0(0)&0(0)+0(1) \end{bmatrix}$

$AB = \begin{bmatrix} 0+0&0+0 \\ 0+0&0+0 \end{bmatrix}$

$AB = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$

The product AB is the zero matrix O.

This example shows two non-zero matrices A and B such that their product AB is the zero matrix.

Alternate Example:

Let $A = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$

Let $B = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$

Here, $A \neq O$ and $B \neq O$. Calculate AB:

$AB = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 0(0)+1(0)&0(1)+1(0) \\ 0(0)+0(0)&0(1)+0(0) \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$

This is another example where the product of two non-zero matrices is the zero matrix.

The final answer is:

The example $A = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix}$ and $B = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$ demonstrates that for $A \neq O$, $B \neq O$, it is possible for $AB = O$. Here, $AB = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$.

Given:

Matrices $A = \begin{bmatrix} 2&4&0 \\ 3&9&6 \end{bmatrix}$ and $B = \begin{bmatrix} 1&4 \\ 2&8 \\ 1&3 \end{bmatrix}$.

To Verify:

Verify if $(AB)′ = B′A′$. This is a fundamental property of matrix transpose for products.

Solution:

First, determine the orders of A and B.

Matrix A is of order $2 \times 3$.

Matrix B is of order $3 \times 2$.

Calculate the left side: $(AB)′$.

First, find the product AB. The number of columns in A (3) is equal to the number of rows in B (3), so AB is defined. The order of AB is $2 \times 2$.

$AB = \begin{bmatrix} 2&4&0 \\ 3&9&6 \end{bmatrix} \begin{bmatrix} 1&4 \\ 2&8 \\ 1&3 \end{bmatrix}$

$AB = \begin{bmatrix} 2(1)+4(2)+0(1) & 2(4)+4(8)+0(3) \\ 3(1)+9(2)+6(1) & 3(4)+9(8)+6(3) \end{bmatrix}$

$AB = \begin{bmatrix} 2+8+0 & 8+32+0 \\ 3+18+6 & 12+72+18 \end{bmatrix}$

$AB = \begin{bmatrix} 10&40 \\ 27&102 \end{bmatrix}$

Now, find the transpose of AB:

$(AB)′ = \begin{bmatrix} 10&40 \\ 27&102 \end{bmatrix}′ = \begin{bmatrix} 10&27 \\ 40&102 \end{bmatrix}$

Calculate the right side: $B′A′$.

First, find the transpose of A, $A′$. The order of A is $2 \times 3$, so the order of $A′$ is $3 \times 2$.

$A′ = \begin{bmatrix} 2&4&0 \\ 3&9&6 \end{bmatrix}′ = \begin{bmatrix} 2&3 \\ 4&9 \\ 0&6 \end{bmatrix}$

Next, find the transpose of B, $B′$. The order of B is $3 \times 2$, so the order of $B′$ is $2 \times 3$.

$B′ = \begin{bmatrix} 1&4 \\ 2&8 \\ 1&3 \end{bmatrix}′ = \begin{bmatrix} 1&2&1 \\ 4&8&3 \end{bmatrix}$

Now, find the product $B′A′$. The order of $B′$ is $2 \times 3$ and the order of $A′$ is $3 \times 2$. Since the number of columns in $B′$ (3) is equal to the number of rows in $A′$ (3), the product $B′A′$ is defined, and its order is $2 \times 2$.

$B′A′ = \begin{bmatrix} 1&2&1 \\ 4&8&3 \end{bmatrix} \begin{bmatrix} 2&3 \\ 4&9 \\ 0&6 \end{bmatrix}$

$B′A′ = \begin{bmatrix} 1(2)+2(4)+1(0) & 1(3)+2(9)+1(6) \\ 4(2)+8(4)+3(0) & 4(3)+8(9)+3(6) \end{bmatrix}$

$B′A′ = \begin{bmatrix} 2+8+0 & 3+18+6 \\ 8+32+0 & 12+72+18 \end{bmatrix}$

$B′A′ = \begin{bmatrix} 10&27 \\ 40&102 \end{bmatrix}$

Compare the results from (1) and (2):

$(AB)′ = \begin{bmatrix} 10&27 \\ 40&102 \end{bmatrix}$

$B′A′ = \begin{bmatrix} 10&27 \\ 40&102 \end{bmatrix}$

Since the resulting matrices are equal, we have verified that $(AB)′ = B′A′$ for the given matrices A and B.

The final answer is:

Yes, $(AB)′ = B′A′$.

Calculations show that $(AB)′ = \begin{bmatrix} 10&27 \\ 40&102 \end{bmatrix}$ and $B′A′ = \begin{bmatrix} 10&27 \\ 40&102 \end{bmatrix}$. Since these are equal, the property is verified.

Given:

The matrix equation $x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = O$, where O is the zero matrix.

To Find:

The values of x and y that satisfy the given matrix equation.

Solution:

The matrices in the equation are column matrices of order $2 \times 1$. Thus, the zero matrix O on the right side is $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

Perform the scalar multiplications on the terms involving x and y:

$x\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 2x \\ 1x \end{bmatrix} = \begin{bmatrix} 2x \\ x \end{bmatrix}$

$y\begin{bmatrix} 3 \\ 5 \end{bmatrix} = \begin{bmatrix} 3y \\ 5y \end{bmatrix}$

Substitute these into the matrix equation:

$\begin{bmatrix} 2x \\ x \end{bmatrix} + \begin{bmatrix} 3y \\ 5y \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Perform the matrix addition on the left side by adding the corresponding elements:

$\begin{bmatrix} 2x + 3y + (-8) \\ x + 5y + (-11) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$\begin{bmatrix} 2x + 3y - 8 \\ x + 5y - 11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

For two matrices to be equal, their corresponding elements must be equal. This gives us a system of two linear equations with two variables:

$2x + 3y - 8 = 0 \Rightarrow 2x + 3y = 8$

$x + 5y - 11 = 0 \Rightarrow x + 5y = 11$

We can solve this system using substitution or elimination. Let's use elimination.

Multiply equation (2) by 2 to make the coefficient of x the same as in equation (1):

$2(x + 5y) = 2(11)$

$2x + 10y = 22$

Subtract equation (1) from equation (3):

$(2x + 10y) - (2x + 3y) = 22 - 8$

$2x + 10y - 2x - 3y = 14$

$7y = 14$

$y = \frac{14}{7}$

$y = 2$

Now substitute the value of y = 2 into equation (2):

$x + 5(2) = 11$

$x + 10 = 11$

$x = 11 - 10$

$x = 1$

Thus, the values that satisfy the equation are x = 1 and y = 2.

The final answer is:

$x = 1$

$y = 2$

Given:

Two matrix equations involving $2 \times 2$ matrices X and Y:

$2X + 3Y = \begin{bmatrix} 2&3 \\ 4&0 \end{bmatrix}$

$3X + 2Y = \begin{bmatrix} −2&2 \\ 1&−5 \end{bmatrix}$

To Find:

The matrices X and Y that satisfy both equations.

Solution:

We can solve this system of matrix equations using methods similar to solving a system of linear algebraic equations, such as elimination or substitution.

Let's use elimination. We aim to eliminate either X or Y by multiplying the equations by appropriate scalars and subtracting them.

Multiply equation (1) by 3 and equation (2) by 2 to make the coefficients of X equal:

Multiply (1) by 3:

$3(2X + 3Y) = 3\begin{bmatrix} 2&3 \\ 4&0 \end{bmatrix}$

$6X + 9Y = \begin{bmatrix} 3 \times 2&3 \times 3 \\ 3 \times 4&3 \times 0 \end{bmatrix} = \begin{bmatrix} 6&9 \\ 12&0 \end{bmatrix}$

Multiply (2) by 2:

$2(3X + 2Y) = 2\begin{bmatrix} −2&2 \\ 1&−5 \end{bmatrix}$

$6X + 4Y = \begin{bmatrix} 2 \times (−2)&2 \times 2 \\ 2 \times 1&2 \times (−5) \end{bmatrix} = \begin{bmatrix} −4&4 \\ 2&−10 \end{bmatrix}$

Subtract equation (4) from equation (3):

$(6X + 9Y) - (6X + 4Y) = \begin{bmatrix} 6&9 \\ 12&0 \end{bmatrix} - \begin{bmatrix} −4&4 \\ 2&−10 \end{bmatrix}$

$6X + 9Y - 6X - 4Y = \begin{bmatrix} 6-(-4)&9-4 \\ 12-2&0-(-10) \end{bmatrix}$

$5Y = \begin{bmatrix} 6+4&5 \\ 10&0+10 \end{bmatrix} = \begin{bmatrix} 10&5 \\ 10&10 \end{bmatrix}$

Now, solve for Y by multiplying by $\frac{1}{5}$:

$Y = \frac{1}{5} \begin{bmatrix} 10&5 \\ 10&10 \end{bmatrix} = \begin{bmatrix} \frac{10}{5}&\frac{5}{5} \\ \frac{10}{5}&\frac{10}{5} \end{bmatrix} = \begin{bmatrix} 2&1 \\ 2&2 \end{bmatrix}$

Now that we have Y, we can substitute it back into either equation (1) or (2) to solve for X. Let's use equation (1):

$2X + 3Y = \begin{bmatrix} 2&3 \\ 4&0 \end{bmatrix}$

$2X + 3 \begin{bmatrix} 2&1 \\ 2&2 \end{bmatrix} = \begin{bmatrix} 2&3 \\ 4&0 \end{bmatrix}$

Calculate 3Y:

$3 \begin{bmatrix} 2&1 \\ 2&2 \end{bmatrix} = \begin{bmatrix} 3 \times 2&3 \times 1 \\ 3 \times 2&3 \times 2 \end{bmatrix} = \begin{bmatrix} 6&3 \\ 6&6 \end{bmatrix}$

Substitute 3Y into the equation:

$2X + \begin{bmatrix} 6&3 \\ 6&6 \end{bmatrix} = \begin{bmatrix} 2&3 \\ 4&0 \end{bmatrix}$

Subtract $\begin{bmatrix} 6&3 \\ 6&6 \end{bmatrix}$ from both sides:

$2X = \begin{bmatrix} 2&3 \\ 4&0 \end{bmatrix} - \begin{bmatrix} 6&3 \\ 6&6 \end{bmatrix}$

$2X = \begin{bmatrix} 2-6&3-3 \\ 4-6&0-6 \end{bmatrix}$

$2X = \begin{bmatrix} -4&0 \\ -2&-6 \end{bmatrix}$

Now, solve for X by multiplying by $\frac{1}{2}$:

$X = \frac{1}{2} \begin{bmatrix} -4&0 \\ -2&-6 \end{bmatrix} = \begin{bmatrix} \frac{-4}{2}&\frac{0}{2} \\ \frac{-2}{2}&\frac{-6}{2} \end{bmatrix} = \begin{bmatrix} -2&0 \\ -1&-3 \end{bmatrix}$

The final answer is:

$X = \begin{bmatrix} -2&0 \\ -1&-3 \end{bmatrix}$

$Y = \begin{bmatrix} 2&1 \\ 2&2 \end{bmatrix}$

Given:

Matrices $A = \begin{bmatrix} 3&5 \end{bmatrix}$ and $B = \begin{bmatrix} 7&3 \end{bmatrix}$.

The matrix equation $AC = BC$, where C is a non-zero matrix.

To Find:

A non-zero matrix C satisfying the equation $AC = BC$.

Solution:

Let the order of matrix A be $1 \times 2$ and the order of matrix B be $1 \times 2$.

For the product AC to be defined, the number of columns in A (which is 2) must be equal to the number of rows in C. Let the order of C be $2 \times n$.

The order of the product AC will be $(1 \times 2) \times (2 \times n) = 1 \times n$.

For the product BC to be defined, the number of columns in B (which is 2) must be equal to the number of rows in C. This is consistent with C being $2 \times n$.

The order of the product BC will be $(1 \times 2) \times (2 \times n) = 1 \times n$.

For the matrix equation $AC = BC$ to hold, both products must be defined and have the same order, which is $1 \times n$. We need to find a non-zero matrix C of order $2 \times n$. The simplest case is when $n=1$, meaning C is a $2 \times 1$ column matrix.

Let $C = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$. C must be non-zero, so at least one of $c_1$ or $c_2$ must be non-zero.

Calculate AC:

$AC = \begin{bmatrix} 3&5 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 3c_1 + 5c_2 \end{bmatrix}$

Calculate BC:

$BC = \begin{bmatrix} 7&3 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 7c_1 + 3c_2 \end{bmatrix}$

The equation $AC = BC$ means the corresponding elements are equal:

$\begin{bmatrix} 3c_1 + 5c_2 \end{bmatrix} = \begin{bmatrix} 7c_1 + 3c_2 \end{bmatrix}$

Equating the elements, we get a linear equation:

$3c_1 + 5c_2 = 7c_1 + 3c_2$

Rearrange the terms to solve for $c_1$ and $c_2$:

$5c_2 - 3c_2 = 7c_1 - 3c_1$

$2c_2 = 4c_1$

$c_2 = 2c_1$

We need to find a non-zero matrix C that satisfies this condition. We can choose any non-zero value for $c_1$ and find the corresponding value for $c_2$.

Let's choose $c_1 = 1$. Then $c_2 = 2(1) = 2$.

So, the matrix C is $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$. This matrix is non-zero.

Let's verify the solution:

$AC = \begin{bmatrix} 3&5 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3(1) + 5(2) \end{bmatrix} = \begin{bmatrix} 3 + 10 \end{bmatrix} = \begin{bmatrix} 13 \end{bmatrix}$

$BC = \begin{bmatrix} 7&3 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 7(1) + 3(2) \end{bmatrix} = \begin{bmatrix} 7 + 6 \end{bmatrix} = \begin{bmatrix} 13 \end{bmatrix}$

Since $AC = [13]$ and $BC = [13]$, the equation $AC = BC$ is satisfied by the non-zero matrix $C = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$.

Note that any matrix C of the form $\begin{bmatrix} k \\ 2k \end{bmatrix}$ where $k$ is a non-zero scalar will also satisfy the equation. For example, if $k=2$, $C = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$.

$AC = \begin{bmatrix} 3&5 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \end{bmatrix} = [3(2)+5(4)] = [6+20] = [26]$

$BC = \begin{bmatrix} 7&3 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \end{bmatrix} = [7(2)+3(4)] = [14+12] = [26]$

The final answer is:

A non-zero matrix C satisfying the equation $AC = BC$ is $C = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$.

Given:

We need to provide an example of three matrices A, B, and C such that A is a non-zero matrix, $AB = AC$, but $B \neq C$.

To Provide (by Example):

Matrices A, B, C satisfying the conditions $A \neq O$, $AB = AC$, and $B \neq C$.

Solution:

The property $AB = AC \Rightarrow B = C$ holds if matrix A is invertible (non-singular). To find an example where this cancellation property does not hold, we need to use a matrix A that is non-zero but not invertible (i.e., a singular matrix, with $\text{det}(A) = 0$).

Consider the following matrices:

Let $A = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix}$

Let $B = \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix}$

Let $C = \begin{bmatrix} 1&2 \\ 5&6 \end{bmatrix}$

Check the conditions:

1. Is A a non-zero matrix? Yes, A has non-zero elements, so $A \neq O$. The determinant of A is $(1)(0) - (0)(0) = 0$, so A is singular.

2. Are B and C different? Yes, $B = \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix}$ and $C = \begin{bmatrix} 1&2 \\ 5&6 \end{bmatrix}$. The element in the second row, first column is different (3 in B, 5 in C), so $B \neq C$.

3. Is $AB = AC$? Let's calculate both products.

Calculate AB:

$AB = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix} = \begin{bmatrix} 1(1)+0(3)&1(2)+0(4) \\ 0(1)+0(3)&0(2)+0(4) \end{bmatrix} = \begin{bmatrix} 1+0&2+0 \\ 0+0&0+0 \end{bmatrix} = \begin{bmatrix} 1&2 \\ 0&0 \end{bmatrix}$

Calculate AC:

$AC = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} 1&2 \\ 5&6 \end{bmatrix} = \begin{bmatrix} 1(1)+0(5)&1(2)+0(6) \\ 0(1)+0(5)&0(2)+0(6) \end{bmatrix} = \begin{bmatrix} 1+0&2+0 \\ 0+0&0+0 \end{bmatrix} = \begin{bmatrix} 1&2 \\ 0&0 \end{bmatrix}$

We see that $AB = \begin{bmatrix} 1&2 \\ 0&0 \end{bmatrix}$ and $AC = \begin{bmatrix} 1&2 \\ 0&0 \end{bmatrix}$. Thus, $AB = AC$.

The matrices A, B, and C chosen satisfy all the given conditions: A is non-zero, B is not equal to C, and yet $AB = AC$.

The final answer is:

An example of matrices A, B, and C satisfying the conditions is:

$A = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix}$

$B = \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix}$

$C = \begin{bmatrix} 1&2 \\ 5&6 \end{bmatrix}$

Here $A \neq O$, $B \neq C$, and $AB = AC = \begin{bmatrix} 1&2 \\ 0&0 \end{bmatrix}$.

Given:

Matrices $A = \begin{bmatrix} 1&2 \\ −2&1 \end{bmatrix}$, $B = \begin{bmatrix} 2&3 \\ 3&−4 \end{bmatrix}$, and $C = \begin{bmatrix} 1&0 \\ −1&0 \end{bmatrix}$.

To Verify:

(i) $(AB) C = A (BC)$ (Associativity of matrix multiplication)

(ii) $A (B + C) = AB + AC$ (Distributivity of matrix multiplication over addition)

Solution:

All given matrices A, B, and C are of order $2 \times 2$. The operations (multiplication and addition) are defined.

(i) Verify (AB) C = A (BC):

Calculate the left side: (AB) C.

First, calculate AB:

$AB = \begin{bmatrix} 1&2 \\ −2&1 \end{bmatrix} \begin{bmatrix} 2&3 \\ 3&−4 \end{bmatrix} = \begin{bmatrix} 1(2)+2(3)&1(3)+2(−4) \\ (−2)(2)+1(3)&(−2)(3)+1(−4) \end{bmatrix} = \begin{bmatrix} 2+6&3-8 \\ -4+3&-6-4 \end{bmatrix} = \begin{bmatrix} 8&−5 \\ −1&−10 \end{bmatrix}$

Now, calculate (AB) C:

$(AB) C = \begin{bmatrix} 8&−5 \\ −1&−10 \end{bmatrix} \begin{bmatrix} 1&0 \\ −1&0 \end{bmatrix} = \begin{bmatrix} 8(1)+(−5)(−1)&8(0)+(−5)(0) \\ (−1)(1)+(−10)(−1)&(−1)(0)+(−10)(0) \end{bmatrix} = \begin{bmatrix} 8+5&0+0 \\ -1+10&0+0 \end{bmatrix} = \begin{bmatrix} 13&0 \\ 9&0 \end{bmatrix}$

Calculate the right side: A (BC).

First, calculate BC:

$BC = \begin{bmatrix} 2&3 \\ 3&−4 \end{bmatrix} \begin{bmatrix} 1&0 \\ −1&0 \end{bmatrix} = \begin{bmatrix} 2(1)+3(−1)&2(0)+3(0) \\ 3(1)+(−4)(−1)&3(0)+(−4)(0) \end{bmatrix} = \begin{bmatrix} 2-3&0+0 \\ 3+4&0+0 \end{bmatrix} = \begin{bmatrix} −1&0 \\ 7&0 \end{bmatrix}$

Now, calculate A (BC):

$A (BC) = \begin{bmatrix} 1&2 \\ −2&1 \end{bmatrix} \begin{bmatrix} −1&0 \\ 7&0 \end{bmatrix} = \begin{bmatrix} 1(−1)+2(7)&1(0)+2(0) \\ (−2)(−1)+1(7)&(−2)(0)+1(0) \end{bmatrix} = \begin{bmatrix} -1+14&0+0 \\ 2+7&0+0 \end{bmatrix} = \begin{bmatrix} 13&0 \\ 9&0 \end{bmatrix}$

Compare the results from (1) and (2). Since $\begin{bmatrix} 13&0 \\ 9&0 \end{bmatrix} = \begin{bmatrix} 13&0 \\ 9&0 \end{bmatrix}$, we have verified that $(AB) C = A (BC)$.

(ii) Verify A (B + C) = AB + AC:

Calculate the left side: A (B + C).

First, calculate B + C:

$B + C = \begin{bmatrix} 2&3 \\ 3&−4 \end{bmatrix} + \begin{bmatrix} 1&0 \\ −1&0 \end{bmatrix} = \begin{bmatrix} 2+1&3+0 \\ 3+(-1)&-4+0 \end{bmatrix} = \begin{bmatrix} 3&3 \\ 2&−4 \end{bmatrix}$

Now, calculate A (B + C):

$A (B + C) = \begin{bmatrix} 1&2 \\ −2&1 \end{bmatrix} \begin{bmatrix} 3&3 \\ 2&−4 \end{bmatrix} = \begin{bmatrix} 1(3)+2(2)&1(3)+2(−4) \\ (−2)(3)+1(2)&(−2)(3)+1(−4) \end{bmatrix} = \begin{bmatrix} 3+4&3-8 \\ -6+2&-6-4 \end{bmatrix} = \begin{bmatrix} 7&−5 \\ −4&−10 \end{bmatrix}$

Calculate the right side: AB + AC.

From part (i), we already calculated AB:

$AB = \begin{bmatrix} 8&−5 \\ −1&−10 \end{bmatrix}$

From the calculation for part (i), we also need AC. Let's calculate AC:

$AC = \begin{bmatrix} 1&2 \\ −2&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ −1&0 \end{bmatrix} = \begin{bmatrix} 1(1)+2(−1)&1(0)+2(0) \\ (−2)(1)+1(−1)&(−2)(0)+1(0) \end{bmatrix} = \begin{bmatrix} 1-2&0+0 \\ -2-1&0+0 \end{bmatrix} = \begin{bmatrix} −1&0 \\ −3&0 \end{bmatrix}$

Now, calculate AB + AC:

$AB + AC = \begin{bmatrix} 8&−5 \\ −1&−10 \end{bmatrix} + \begin{bmatrix} −1&0 \\ −3&0 \end{bmatrix} = \begin{bmatrix} 8+(-1)&-5+0 \\ -1+(-3)&-10+0 \end{bmatrix} = \begin{bmatrix} 8-1&-5 \\ -1-3&-10 \end{bmatrix} = \begin{bmatrix} 7&−5 \\ −4&−10 \end{bmatrix}$

Compare the results from (3) and (4). Since $\begin{bmatrix} 7&−5 \\ −4&−10 \end{bmatrix} = \begin{bmatrix} 7&−5 \\ −4&−10 \end{bmatrix}$, we have verified that $A (B + C) = AB + AC$.

The final answer is:

(i) Calculations show that $(AB) C = \begin{bmatrix} 13&0 \\ 9&0 \end{bmatrix}$ and $A (BC) = \begin{bmatrix} 13&0 \\ 9&0 \end{bmatrix}$. Thus, $(AB) C = A (BC)$ is verified.

(ii) Calculations show that $A (B + C) = \begin{bmatrix} 7&−5 \\ −4&−10 \end{bmatrix}$ and $AB + AC = \begin{bmatrix} 7&−5 \\ −4&−10 \end{bmatrix}$. Thus, $A (B + C) = AB + AC$ is verified.

Given:

Matrices $P = \begin{bmatrix} x&0&0 \\ 0&y&0 \\ 0&0&z \end{bmatrix}$ and $Q = \begin{bmatrix} a&0&0 \\ 0&b&0 \\ 0&0&c \end{bmatrix}$.

Both P and Q are diagonal matrices of order $3 \times 3$.

To Prove:

$PQ = \begin{bmatrix} xa&0&0 \\ 0&yb&0 \\ 0&0&zc \end{bmatrix} = QP$.

Proof:

We need to calculate the matrix products PQ and QP and show that they are equal to the matrix $\begin{bmatrix} xa&0&0 \\ 0&yb&0 \\ 0&0&zc \end{bmatrix}$.

First, calculate the product PQ:

$PQ = \begin{bmatrix} x&0&0 \\ 0&y&0 \\ 0&0&z \end{bmatrix} \begin{bmatrix} a&0&0 \\ 0&b&0 \\ 0&0&c \end{bmatrix}$

Multiply the matrices according to the rule of matrix multiplication:

$PQ = \begin{bmatrix} x(a)+0(0)+0(0) & x(0)+0(b)+0(0) & x(0)+0(0)+0(c) \\ 0(a)+y(0)+0(0) & 0(0)+y(b)+0(0) & 0(0)+y(0)+0(c) \\ 0(a)+0(0)+z(0) & 0(0)+0(b)+z(0) & 0(0)+0(0)+z(c) \end{bmatrix}$

$PQ = \begin{bmatrix} xa+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+yb+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+zc \end{bmatrix}$

$PQ = \begin{bmatrix} xa&0&0 \\ 0&yb&0 \\ 0&0&zc \end{bmatrix}$

Next, calculate the product QP:

$QP = \begin{bmatrix} a&0&0 \\ 0&b&0 \\ 0&0&c \end{bmatrix} \begin{bmatrix} x&0&0 \\ 0&y&0 \\ 0&0&z \end{bmatrix}$

Multiply the matrices:

$QP = \begin{bmatrix} a(x)+0(0)+0(0) & a(0)+0(y)+0(0) & a(0)+0(0)+0(z) \\ 0(x)+b(0)+0(0) & 0(0)+b(y)+0(0) & 0(0)+b(0)+0(z) \\ 0(x)+0(0)+c(0) & 0(0)+0(y)+c(0) & 0(0)+0(0)+c(z) \end{bmatrix}$

$QP = \begin{bmatrix} ax+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+by+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+cz \end{bmatrix}$

$QP = \begin{bmatrix} ax&0&0 \\ 0&by&0 \\ 0&0&cz \end{bmatrix}$

Since multiplication of scalars is commutative (i.e., $xa = ax$, $yb = by$, and $zc = cz$), the matrices obtained for PQ and QP are equal element by element.

From equations (1) and (2), we see that PQ and QP are both equal to the matrix $\begin{bmatrix} xa&0&0 \\ 0&yb&0 \\ 0&0&zc \end{bmatrix}$.

Therefore, $PQ = \begin{bmatrix} xa&0&0 \\ 0&yb&0 \\ 0&0&zc \end{bmatrix} = QP$.

This proves the statement.

The final answer is:

By calculating the matrix products, it has been shown that $PQ = \begin{bmatrix} xa&0&0 \\ 0&yb&0 \\ 0&0&zc \end{bmatrix}$ and $QP = \begin{bmatrix} ax&0&0 \\ 0&by&0 \\ 0&0&cz \end{bmatrix}$. Since $xa=ax$, $yb=by$, and $zc=cz$, we have $PQ = QP = \begin{bmatrix} xa&0&0 \\ 0&yb&0 \\ 0&0&zc \end{bmatrix}$.

Given:

The matrix equation $\begin{bmatrix} 2&1&3 \end{bmatrix} \begin{bmatrix} −1&0&−1 \\ −1&1&0 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ −1 \end{bmatrix} = A$.

To Find:

The matrix A that satisfies the given equation.

Solution:

We need to perform the matrix multiplications on the left side of the equation to find the matrix A.

Let $M_1 = \begin{bmatrix} 2&1&3 \end{bmatrix}$ (order $1 \times 3$), $M_2 = \begin{bmatrix} −1&0&−1 \\ −1&1&0 \\ 0&1&1 \end{bmatrix}$ (order $3 \times 3$), and $M_3 = \begin{bmatrix} 1 \\ 0 \\ −1 \end{bmatrix}$ (order $3 \times 1$).

The product $M_1 M_2$ will have the order $(1 \times 3) \times (3 \times 3)$, which is $1 \times 3$.

Calculate $M_1 M_2$:

$M_1 M_2 = \begin{bmatrix} 2&1&3 \end{bmatrix} \begin{bmatrix} −1&0&−1 \\ −1&1&0 \\ 0&1&1 \end{bmatrix}$

$M_1 M_2 = \begin{bmatrix} 2(-1)+1(-1)+3(0) & 2(0)+1(1)+3(1) & 2(-1)+1(0)+3(1) \end{bmatrix}$

$M_1 M_2 = \begin{bmatrix} -2-1+0 & 0+1+3 & -2+0+3 \end{bmatrix}$

$M_1 M_2 = \begin{bmatrix} -3&4&1 \end{bmatrix}$

Now, multiply this resulting $1 \times 3$ matrix by the third $3 \times 1$ matrix $M_3$. The product $(M_1 M_2) M_3$ will have the order $(1 \times 3) \times (3 \times 1)$, which is $1 \times 1$. This will be the matrix A.

$A = \begin{bmatrix} -3&4&1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ −1 \end{bmatrix}$

$A = \begin{bmatrix} (-3)(1) + (4)(0) + (1)(-1) \end{bmatrix}$

$A = \begin{bmatrix} -3 + 0 - 1 \end{bmatrix}$

$A = \begin{bmatrix} -4 \end{bmatrix}$

Thus, the matrix A is a $1 \times 1$ matrix with the single element -4.

The final answer is:

$A = \begin{bmatrix} -4 \end{bmatrix}$

Given:

Matrices $A = \begin{bmatrix} 2&1 \end{bmatrix}$, $B = \begin{bmatrix} 5&3&4 \\ 8&7&6 \end{bmatrix}$, and $C = \begin{bmatrix} −1&2&1 \\ 1&0&2 \end{bmatrix}$.

To Verify:

Verify that $A (B + C) = (AB + AC)$. This demonstrates the left distributive property of matrix multiplication over matrix addition.

Solution:

The order of matrix A is $1 \times 2$. The order of matrix B is $2 \times 3$. The order of matrix C is $2 \times 3$. The operations are defined since the orders are compatible for addition and multiplication.

Calculate the left side: $A (B + C)$.

First, calculate the sum $B + C$:

$B + C = \begin{bmatrix} 5&3&4 \\ 8&7&6 \end{bmatrix} + \begin{bmatrix} −1&2&1 \\ 1&0&2 \end{bmatrix}$

$B + C = \begin{bmatrix} 5 + (−1) & 3 + 2 & 4 + 1 \\ 8 + 1 & 7 + 0 & 6 + 2 \end{bmatrix}$

$B + C = \begin{bmatrix} 4&5&5 \\ 9&7&8 \end{bmatrix}$

Now, multiply matrix A by the sum $(B + C)$:

$A (B + C) = \begin{bmatrix} 2&1 \end{bmatrix} \begin{bmatrix} 4&5&5 \\ 9&7&8 \end{bmatrix}$

$A (B + C) = \begin{bmatrix} 2(4) + 1(9) & 2(5) + 1(7) & 2(5) + 1(8) \end{bmatrix}$

$A (B + C) = \begin{bmatrix} 8 + 9 & 10 + 7 & 10 + 8 \end{bmatrix}$

$A (B + C) = \begin{bmatrix} 17&17&18 \end{bmatrix}$

Calculate the right side: $AB + AC$.

First, calculate the product AB:

$AB = \begin{bmatrix} 2&1 \end{bmatrix} \begin{bmatrix} 5&3&4 \\ 8&7&6 \end{bmatrix}$

$AB = \begin{bmatrix} 2(5) + 1(8) & 2(3) + 1(7) & 2(4) + 1(6) \end{bmatrix}$

$AB = \begin{bmatrix} 10 + 8 & 6 + 7 & 8 + 6 \end{bmatrix}$

$AB = \begin{bmatrix} 18&13&14 \end{bmatrix}$

Next, calculate the product AC:

$AC = \begin{bmatrix} 2&1 \end{bmatrix} \begin{bmatrix} −1&2&1 \\ 1&0&2 \end{bmatrix}$

$AC = \begin{bmatrix} 2(−1) + 1(1) & 2(2) + 1(0) & 2(1) + 1(2) \end{bmatrix}$

$AC = \begin{bmatrix} -2 + 1 & 4 + 0 & 2 + 2 \end{bmatrix}$

$AC = \begin{bmatrix} −1&4&4 \end{bmatrix}$

Now, add the products AB and AC:

$AB + AC = \begin{bmatrix} 18&13&14 \end{bmatrix} + \begin{bmatrix} −1&4&4 \end{bmatrix}$

$AB + AC = \begin{bmatrix} 18 + (−1) & 13 + 4 & 14 + 4 \end{bmatrix}$

$AB + AC = \begin{bmatrix} 18 - 1 & 17 & 18 \end{bmatrix}$

$AB + AC = \begin{bmatrix} 17&17&18 \end{bmatrix}$

Compare the results from (1) and (2). Both the left side and the right side evaluate to the same matrix $\begin{bmatrix} 17&17&18 \end{bmatrix}$.

Since $A (B + C) = \begin{bmatrix} 17&17&18 \end{bmatrix}$ and $(AB + AC) = \begin{bmatrix} 17&17&18 \end{bmatrix}$, we have verified that $A (B + C) = (AB + AC)$.

The final answer is:

Calculations show that $A (B + C) = \begin{bmatrix} 17&17&18 \end{bmatrix}$ and $(AB + AC) = \begin{bmatrix} 17&17&18 \end{bmatrix}$.

Since the results are equal, the property $A (B + C) = (AB + AC)$ is verified for the given matrices.

Given:

Matrix $A = \begin{bmatrix} 1&0&−1 \\ 2&1&3 \\ 0&1&1 \end{bmatrix}$ and I is the $3 \times 3$ unit matrix $I = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}$.

To Verify:

Verify that $A^2 + A = A (A + I)$. This is an instance of the distributive property of matrix multiplication, extended to include addition with the matrix itself.

Solution:

We need to calculate both sides of the equation $A^2 + A = A (A + I)$ and show that they are equal.

Calculate the left side: $A^2 + A$.

First, calculate $A^2 = A \times A$:

$A^2 = \begin{bmatrix} 1&0&−1 \\ 2&1&3 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} 1&0&−1 \\ 2&1&3 \\ 0&1&1 \end{bmatrix}$

$A^2 = \begin{bmatrix} 1(1)+0(2)+(-1)(0) & 1(0)+0(1)+(-1)(1) & 1(-1)+0(3)+(-1)(1) \\ 2(1)+1(2)+3(0) & 2(0)+1(1)+3(1) & 2(-1)+1(3)+3(1) \\ 0(1)+1(2)+1(0) & 0(0)+1(1)+1(1) & 0(-1)+1(3)+1(1) \end{bmatrix}$

$A^2 = \begin{bmatrix} 1+0+0 & 0+0-1 & -1+0-1 \\ 2+2+0 & 0+1+3 & -2+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1 \end{bmatrix}$

$A^2 = \begin{bmatrix} 1&−1&−2 \\ 4&4&4 \\ 2&2&4 \end{bmatrix}$

Now, calculate $A^2 + A$:

$A^2 + A = \begin{bmatrix} 1&−1&−2 \\ 4&4&4 \\ 2&2&4 \end{bmatrix} + \begin{bmatrix} 1&0&−1 \\ 2&1&3 \\ 0&1&1 \end{bmatrix}$

$A^2 + A = \begin{bmatrix} 1+1&−1+0&−2+(-1) \\ 4+2&4+1&4+3 \\ 2+0&2+1&4+1 \end{bmatrix}$

$A^2 + A = \begin{bmatrix} 2&−1&−3 \\ 6&5&7 \\ 2&3&5 \end{bmatrix}$

Calculate the right side: A (A + I).

First, calculate A + I:

$A + I = \begin{bmatrix} 1&0&−1 \\ 2&1&3 \\ 0&1&1 \end{bmatrix} + \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 1+1&0+0&−1+0 \\ 2+0&1+1&3+0 \\ 0+0&1+0&1+1 \end{bmatrix} = \begin{bmatrix} 2&0&−1 \\ 2&2&3 \\ 0&1&2 \end{bmatrix}$

Now, calculate A (A + I):

$A (A + I) = \begin{bmatrix} 1&0&−1 \\ 2&1&3 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} 2&0&−1 \\ 2&2&3 \\ 0&1&2 \end{bmatrix}$

$A (A + I) = \begin{bmatrix} 1(2)+0(2)+(-1)(0) & 1(0)+0(2)+(-1)(1) & 1(-1)+0(3)+(-1)(2) \\ 2(2)+1(2)+3(0) & 2(0)+1(2)+3(1) & 2(-1)+1(3)+3(2) \\ 0(2)+1(2)+1(0) & 0(0)+1(2)+1(1) & 0(-1)+1(3)+1(2) \end{bmatrix}$

$A (A + I) = \begin{bmatrix} 2+0+0 & 0+0-1 & -1+0-2 \\ 4+2+0 & 0+2+3 & -2+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2 \end{bmatrix}$

$A (A + I) = \begin{bmatrix} 2&−1&−3 \\ 6&5&7 \\ 2&3&5 \end{bmatrix}$

Compare the results from (1) and (2). Both the left side and the right side evaluate to the same matrix $\begin{bmatrix} 2&−1&−3 \\ 6&5&7 \\ 2&3&5 \end{bmatrix}$.

Since $A^2 + A = \begin{bmatrix} 2&−1&−3 \\ 6&5&7 \\ 2&3&5 \end{bmatrix}$ and $A (A + I) = \begin{bmatrix} 2&−1&−3 \\ 6&5&7 \\ 2&3&5 \end{bmatrix}$, we have verified that $A^2 + A = A (A + I)$.

This is consistent with the distributive property $A(A+I) = A \cdot A + A \cdot I = A^2 + A$, where I is the identity matrix ($A \cdot I = A$).

The final answer is:

Calculations show that $A^2 + A = \begin{bmatrix} 2&−1&−3 \\ 6&5&7 \\ 2&3&5 \end{bmatrix}$ and $A (A + I) = \begin{bmatrix} 2&−1&−3 \\ 6&5&7 \\ 2&3&5 \end{bmatrix}$.

Since the results are equal, the identity $A^2 + A = A (A + I)$ is verified for the given matrix A.

Given:

Matrices $A = \begin{bmatrix} 0&−1&2 \\ 4&3&−4 \end{bmatrix}$ and $B = \begin{bmatrix} 4&0 \\ 1&3 \\ 2&6 \end{bmatrix}$.

To Verify:

(i) $(A′)′ = A$

(ii) $(AB)′ = B′A′$

(iii) $(kA)′ = kA′$ for some scalar k.

Solution:

The order of matrix A is $2 \times 3$. The order of matrix B is $3 \times 2$.

(i) Verify (A′)′ = A:

First, find the transpose of A, denoted by A′. The rows of A become the columns of A′.

$A′ = \begin{bmatrix} 0&−1&2 \\ 4&3&−4 \end{bmatrix}′ = \begin{bmatrix} 0&4 \\ −1&3 \\ 2&−4 \end{bmatrix}$

Now, find the transpose of A′, denoted by (A′)′. The rows of A′ become the columns of (A′)′.

$(A′)′ = \begin{bmatrix} 0&4 \\ −1&3 \\ 2&−4 \end{bmatrix}′ = \begin{bmatrix} 0&−1&2 \\ 4&3&−4 \end{bmatrix}$

Comparing this result with the original matrix A, we see that they are identical.

$(A′)′ = \begin{bmatrix} 0&−1&2 \\ 4&3&−4 \end{bmatrix} = A$

Thus, $(A′)′ = A$ is verified.

(ii) Verify (AB)′ = B′A′:

Calculate the left side: (AB)′.

First, find the product AB. The order of A is $2 \times 3$ and the order of B is $3 \times 2$. The product AB is defined and its order is $2 \times 2$.

$AB = \begin{bmatrix} 0&−1&2 \\ 4&3&−4 \end{bmatrix} \begin{bmatrix} 4&0 \\ 1&3 \\ 2&6 \end{bmatrix}$

$AB = \begin{bmatrix} 0(4)+(-1)(1)+2(2) & 0(0)+(-1)(3)+2(6) \\ 4(4)+3(1)+(-4)(2) & 4(0)+3(3)+(-4)(6) \end{bmatrix}$

$AB = \begin{bmatrix} 0-1+4 & 0-3+12 \\ 16+3-8 & 0+9-24 \end{bmatrix}$

$AB = \begin{bmatrix} 3&9 \\ 11&−15 \end{bmatrix}$

Now, find the transpose of AB:

$(AB)′ = \begin{bmatrix} 3&9 \\ 11&−15 \end{bmatrix}′ = \begin{bmatrix} 3&11 \\ 9&−15 \end{bmatrix}$

Calculate the right side: B′A′.

First, find the transpose of B, B′. The order of B is $3 \times 2$, so the order of B′ is $2 \times 3$.

$B′ = \begin{bmatrix} 4&0 \\ 1&3 \\ 2&6 \end{bmatrix}′ = \begin{bmatrix} 4&1&2 \\ 0&3&6 \end{bmatrix}$

From part (i), we have $A′ = \begin{bmatrix} 0&4 \\ −1&3 \\ 2&−4 \end{bmatrix}$. The order of A′ is $3 \times 2$.

Now, find the product B′A′. The order of B′ is $2 \times 3$ and the order of A′ is $3 \times 2$. The product B′A′ is defined and its order is $2 \times 2$.

$B′A′ = \begin{bmatrix} 4&1&2 \\ 0&3&6 \end{bmatrix} \begin{bmatrix} 0&4 \\ −1&3 \\ 2&−4 \end{bmatrix}$

$B′A′ = \begin{bmatrix} 4(0)+1(−1)+2(2) & 4(4)+1(3)+2(−4) \\ 0(0)+3(−1)+6(2) & 0(4)+3(3)+6(−4) \end{bmatrix}$

$B′A′ = \begin{bmatrix} 0-1+4 & 16+3-8 \\ 0-3+12 & 0+9-24 \end{bmatrix}$

$B′A′ = \begin{bmatrix} 3&11 \\ 9&−15 \end{bmatrix}$

Compare the results from (1) and (2). Both $(AB)′$ and $B′A′$ are equal to $\begin{bmatrix} 3&11 \\ 9&−15 \end{bmatrix}$.

Thus, $(AB)′ = B′A′$ is verified.

(iii) Verify (kA)′ = kA′:

Let's choose a scalar, for example, $k = 5$.

Calculate the left side: $(kA)′ = (5A)′$.

First, calculate kA = 5A:

$5A = 5 \begin{bmatrix} 0&−1&2 \\ 4&3&−4 \end{bmatrix} = \begin{bmatrix} 5(0)&5(−1)&5(2) \\ 5(4)&5(3)&5(−4) \end{bmatrix} = \begin{bmatrix} 0&−5&10 \\ 20&15&−20 \end{bmatrix}$

Now, find the transpose of 5A:

$(5A)′ = \begin{bmatrix} 0&−5&10 \\ 20&15&−20 \end{bmatrix}′ = \begin{bmatrix} 0&20 \\ −5&15 \\ 10&−20 \end{bmatrix}$

Calculate the right side: $kA′ = 5A′$.

From part (i), we have $A′ = \begin{bmatrix} 0&4 \\ −1&3 \\ 2&−4 \end{bmatrix}$.

Now, calculate $5A′$:

$5A′ = 5 \begin{bmatrix} 0&4 \\ −1&3 \\ 2&−4 \end{bmatrix} = \begin{bmatrix} 5(0)&5(4) \\ 5(−1)&5(3) \\ 5(2)&5(−4) \end{bmatrix} = \begin{bmatrix} 0&20 \\ −5&15 \\ 10&−20 \end{bmatrix}$

Compare the results from (3) and (4). Both $(kA)′$ and $kA′$ are equal to $\begin{bmatrix} 0&20 \\ −5&15 \\ 10&−20 \end{bmatrix}$.

Thus, $(kA)′ = kA′$ is verified for $k=5$. This property holds for any scalar k.

The final answer is:

(i) $(A′)′ = A$ is verified.

(ii) $(AB)′ = B′A′$ is verified as both sides equal $\begin{bmatrix} 3&11 \\ 9&−15 \end{bmatrix}$.

(iii) $(kA)′ = kA′$ is verified (e.g., for $k=5$, both sides equal $\begin{bmatrix} 0&20 \\ −5&15 \\ 10&−20 \end{bmatrix}$).

Given:

Matrices $A = \begin{bmatrix} 1&2 \\ 4&1 \\ 5&6 \end{bmatrix}$ and $B = \begin{bmatrix} 1&2 \\ 6&4 \\ 7&3 \end{bmatrix}$.

To Verify:

(i) $(2A + B)′ = 2A′ + B′$

(ii) $(A – B)′ = A′ – B′$

Solution:

Both matrices A and B are of order $3 \times 2$. The operations of scalar multiplication, addition, subtraction, and transpose are defined for these matrices.

(i) Verify (2A + B)′ = 2A′ + B′:

Calculate the left side: (2A + B)′.

First, calculate 2A:

$2A = 2 \begin{bmatrix} 1&2 \\ 4&1 \\ 5&6 \end{bmatrix} = \begin{bmatrix} 2 \times 1&2 \times 2 \\ 2 \times 4&2 \times 1 \\ 2 \times 5&2 \times 6 \end{bmatrix} = \begin{bmatrix} 2&4 \\ 8&2 \\ 10&12 \end{bmatrix}$

Now, calculate 2A + B:

$2A + B = \begin{bmatrix} 2&4 \\ 8&2 \\ 10&12 \end{bmatrix} + \begin{bmatrix} 1&2 \\ 6&4 \\ 7&3 \end{bmatrix} = \begin{bmatrix} 2+1&4+2 \\ 8+6&2+4 \\ 10+7&12+3 \end{bmatrix} = \begin{bmatrix} 3&6 \\ 14&6 \\ 17&15 \end{bmatrix}$

Finally, calculate the transpose of (2A + B):

$(2A + B)′ = \begin{bmatrix} 3&6 \\ 14&6 \\ 17&15 \end{bmatrix}′ = \begin{bmatrix} 3&14&17 \\ 6&6&15 \end{bmatrix}$

Calculate the right side: 2A′ + B′.

First, find the transpose of A, A′. The order of A is $3 \times 2$, so the order of A′ is $2 \times 3$.

$A′ = \begin{bmatrix} 1&2 \\ 4&1 \\ 5&6 \end{bmatrix}′ = \begin{bmatrix} 1&4&5 \\ 2&1&6 \end{bmatrix}$

Next, calculate 2A′:

$2A′ = 2 \begin{bmatrix} 1&4&5 \\ 2&1&6 \end{bmatrix} = \begin{bmatrix} 2 \times 1&2 \times 4&2 \times 5 \\ 2 \times 2&2 \times 1&2 \times 6 \end{bmatrix} = \begin{bmatrix} 2&8&10 \\ 4&2&12 \end{bmatrix}$

Now, find the transpose of B, B′. The order of B is $3 \times 2$, so the order of B′ is $2 \times 3$.

$B′ = \begin{bmatrix} 1&2 \\ 6&4 \\ 7&3 \end{bmatrix}′ = \begin{bmatrix} 1&6&7 \\ 2&4&3 \end{bmatrix}$

Finally, calculate 2A′ + B′:

$2A′ + B′ = \begin{bmatrix} 2&8&10 \\ 4&2&12 \end{bmatrix} + \begin{bmatrix} 1&6&7 \\ 2&4&3 \end{bmatrix} = \begin{bmatrix} 2+1&8+6&10+7 \\ 4+2&2+4&12+3 \end{bmatrix} = \begin{bmatrix} 3&14&17 \\ 6&6&15 \end{bmatrix}$

Compare the results from (1) and (2). Both the left side and the right side evaluate to the same matrix $\begin{bmatrix} 3&14&17 \\ 6&6&15 \end{bmatrix}$.

Thus, $(2A + B)′ = 2A′ + B′$ is verified.

(ii) Verify (A – B)′ = A′ – B′:

Calculate the left side: (A – B)′.

First, calculate A – B:

$A – B = \begin{bmatrix} 1&2 \\ 4&1 \\ 5&6 \end{bmatrix} - \begin{bmatrix} 1&2 \\ 6&4 \\ 7&3 \end{bmatrix} = \begin{bmatrix} 1-1&2-2 \\ 4-6&1-4 \\ 5-7&6-3 \end{bmatrix} = \begin{bmatrix} 0&0 \\ -2&-3 \\ -2&3 \end{bmatrix}$

Finally, calculate the transpose of (A – B):

$(A – B)′ = \begin{bmatrix} 0&0 \\ -2&-3 \\ -2&3 \end{bmatrix}′ = \begin{bmatrix} 0&-2&-2 \\ 0&-3&3 \end{bmatrix}$

Calculate the right side: A′ – B′.

From part (i), we have $A′ = \begin{bmatrix} 1&4&5 \\ 2&1&6 \end{bmatrix}$ and $B′ = \begin{bmatrix} 1&6&7 \\ 2&4&3 \end{bmatrix}$.

Now, calculate A′ – B′:

$A′ – B′ = \begin{bmatrix} 1&4&5 \\ 2&1&6 \end{bmatrix} - \begin{bmatrix} 1&6&7 \\ 2&4&3 \end{bmatrix} = \begin{bmatrix} 1-1&4-6&5-7 \\ 2-2&1-4&6-3 \end{bmatrix} = \begin{bmatrix} 0&-2&-2 \\ 0&-3&3 \end{bmatrix}$

Compare the results from (3) and (4). Both the left side and the right side evaluate to the same matrix $\begin{bmatrix} 0&-2&-2 \\ 0&-3&3 \end{bmatrix}$.

Thus, $(A – B)′ = A′ – B′$ is verified.

The final answer is:

(i) Calculations show that $(2A + B)′ = \begin{bmatrix} 3&14&17 \\ 6&6&15 \end{bmatrix}$ and $2A′ + B′ = \begin{bmatrix} 3&14&17 \\ 6&6&15 \end{bmatrix}$. Thus, $(2A + B)′ = 2A′ + B′$ is verified.

(ii) Calculations show that $(A – B)′ = \begin{bmatrix} 0&-2&-2 \\ 0&-3&3 \end{bmatrix}$ and $A′ – B′ = \begin{bmatrix} 0&-2&-2 \\ 0&-3&3 \end{bmatrix}$. Thus, $(A – B)′ = A′ – B′$ is verified.

Given:

A is any matrix.

To Prove:

That the matrices $A′A$ and $AA′$ are both symmetric.

Proof:

A matrix X is said to be symmetric if its transpose is equal to the matrix itself, i.e., $X' = X$.

We need to show that $(A′A)′ = A′A$ and $(AA′)′ = AA′$.

We will use the following properties of matrix transpose:

1. $(XY)′ = Y′X′$ (The transpose of a product is the product of the transposes in reverse order)

2. $(X′)′ = X$ (The transpose of the transpose of a matrix is the matrix itself)

Consider the matrix $A′A$. Let $X = A′A$. We need to find the transpose of X, i.e., $(A′A)′$.

Using the property $(XY)′ = Y′X′$, where $X=A′$ and $Y=A$, we have:

$(A′A)′ = A′ (A′)′$

Now, using the property $(A′)′ = A$, substitute this into the expression:

$(A′A)′ = A′A$

Since the transpose of the matrix $A′A$ is equal to the matrix $A′A$ itself, by the definition of a symmetric matrix, $A′A$ is a symmetric matrix.

Next, consider the matrix $AA′$. Let $Y = AA′$. We need to find the transpose of Y, i.e., $(AA′)′$.

Using the property $(XY)′ = Y′X′$, where $X=A$ and $Y=A′$, we have:

$(AA′)′ = (A′)′A′$

Now, using the property $(A′)′ = A$, substitute this into the expression:

$(AA′)′ = AA′$

Since the transpose of the matrix $AA′$ is equal to the matrix $AA′$ itself, by the definition of a symmetric matrix, $AA′$ is a symmetric matrix.

Both $A′A$ and $AA′$ satisfy the condition of being symmetric matrices.

This holds true for any matrix A for which the products $A′A$ and $AA′$ are defined. If A is an $m \times n$ matrix, then A′ is an $n \times m$ matrix. The product $A′A$ is defined and is an $n \times n$ matrix. The product $AA′$ is defined and is an $m \times m$ matrix.

The final answer is:

We have shown that $(A′A)′ = A′A$ and $(AA′)′ = AA′$ using the properties of matrix transpose. Therefore, both $A′A$ and $AA′$ are symmetric matrices for any matrix A for which the products are defined.

Given:

A and B are square matrices of order $3 \times 3$.

To Determine:

Whether the statement $(AB)^2 = A^2 B^2$ is true for all such matrices A and B.

Solution:

Let's expand both sides of the equation $(AB)^2 = A^2 B^2$ using the definition of matrix powers and the associative property of matrix multiplication:

Left side: $(AB)^2 = (AB) \times (AB)$. By the associative property, this can be written as $A(BA)B$.

$(AB)^2 = ABAB$

Right side: $A^2 B^2 = (A \times A) \times (B \times B)$. By the associative property, this can be written as $AABB$.

$A^2 B^2 = AABB$

For the equation $(AB)^2 = A^2 B^2$ to hold for any matrices A and B, we would need $ABAB = AABB$. This equality is generally only true if the order of multiplication of A and B does not matter, i.e., if A and B commute ($AB = BA$).

However, matrix multiplication is generally not commutative. There exist matrices A and B such that $AB \neq BA$.

If $AB \neq BA$, then $ABAB$ is not necessarily equal to $AABB$.

Let's provide a counterexample using $3 \times 3$ matrices to show that the statement is false.

Consider the matrices:

$A = \begin{bmatrix} 0&1&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$

$B = \begin{bmatrix} 0&0&0 \\ 1&0&0 \\ 0&0&0 \end{bmatrix}$

Both A and B are non-zero square matrices of order $3 \times 3$.

First, calculate the product AB:

$AB = \begin{bmatrix} 0&1&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} 0&0&0 \\ 1&0&0 \\ 0&0&0 \end{bmatrix} = \begin{bmatrix} 0(0)+1(1)+0(0) & 0(0)+1(0)+0(0) & 0(0)+1(0)+0(0) \\ 0(0)+0(1)+0(0) & 0(0)+0(0)+0(0) & 0(0)+0(0)+0(0) \\ 0(0)+0(1)+0(0) & 0(0)+0(0)+0(0) & 0(0)+0(0)+0(0) \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$

Now, calculate $(AB)^2$:

$(AB)^2 = \begin{bmatrix} 1&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} = \begin{bmatrix} 1(1)+0(0)+0(0) & 1(0)+0(0)+0(0) & 1(0)+0(0)+0(0) \\ 0(1)+0(0)+0(0) & 0(0)+0(0)+0(0) & 0(0)+0(0)+0(0) \\ 0(1)+0(0)+0(0) & 0(0)+0(0)+0(0) & 0(0)+0(0)+0(0) \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$

Next, calculate $A^2$:

$A^2 = \begin{bmatrix} 0&1&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} 0&1&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} = \begin{bmatrix} 0(0)+1(0)+0(0) & 0(1)+1(0)+0(0) & 0(0)+1(0)+0(0) \\ 0(0)+0(0)+0(0) & 0(1)+0(0)+0(0) & 0(0)+0(0)+0(0) \\ 0(0)+0(0)+0(0) & 0(1)+0(0)+0(0) & 0(0)+0(0)+0(0) \end{bmatrix} = \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$

Calculate $B^2$:

$B^2 = \begin{bmatrix} 0&0&0 \\ 1&0&0 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} 0&0&0 \\ 1&0&0 \\ 0&0&0 \end{bmatrix} = \begin{bmatrix} 0(0)+0(1)+0(0) & 0(0)+0(0)+0(0) & 0(0)+0(0)+0(0) \\ 1(0)+0(1)+0(0) & 1(0)+0(0)+0(0) & 1(0)+0(0)+0(0) \\ 0(0)+0(1)+0(0) & 0(0)+0(0)+0(0) & 0(0)+0(0)+0(0) \end{bmatrix} = \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$

Now, calculate $A^2 B^2$:

$A^2 B^2 = \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} = \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$

Comparing the results from (1) and (2), we see that $\begin{bmatrix} 1&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} \neq \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$.

Thus, $(AB)^2 \neq A^2 B^2$ for these matrices.

This demonstrates that the statement $(AB)^2 = A^2 B^2$ is not true for all square matrices A and B of order $3 \times 3$.

The final answer is:

The statement is False.

Reason: The property $(AB)^2 = A^2 B^2$ requires matrix multiplication to be commutative ($AB = BA$), which is not generally true for matrices. We provided a counterexample using $3 \times 3$ matrices where $(AB)^2 \neq A^2 B^2$.

Given:

A and B are square matrices of the same order.

The condition $AB = BA$ is satisfied (A and B commute).

To Show:

$(A + B)^2 = A^2 + 2AB + B^2$.

Proof:

We start with the left side of the equation $(A + B)^2$ and expand it using the definition of matrix powers and the distributive property of matrix multiplication.

$(A + B)^2 = (A + B) \times (A + B)$

Applying the distributive property:

$(A + B)(A + B) = A(A + B) + B(A + B)$

Distribute A and B further:

$A(A + B) = A \times A + A \times B = A^2 + AB$

$B(A + B) = B \times A + B \times B = BA + B^2$

Substitute these expansions back into the expression:

$(A + B)^2 = (A^2 + AB) + (BA + B^2)$

$(A + B)^2 = A^2 + AB + BA + B^2$

Now, we use the given condition that A and B commute, which means $AB = BA$. Substitute BA with AB in the expression:

$(A + B)^2 = A^2 + AB + AB + B^2$

Combine the two AB terms:

$AB + AB = 2AB$

So, the expression becomes:

$(A + B)^2 = A^2 + 2AB + B^2$

This matches the right side of the equation we needed to show.

Thus, if A and B are square matrices such that AB = BA, then $(A + B)^2 = A^2 + 2AB + B^2$.

This proof relies heavily on the commutative property of matrix multiplication ($AB=BA$). If $AB \neq BA$, the expansion would be $A^2 + AB + BA + B^2$, and we cannot simplify $AB + BA$ to $2AB$.

The final answer is:

Starting with $(A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2$. Given that $AB = BA$, we substitute $BA$ with $AB$ to get $A^2 + AB + AB + B^2 = A^2 + 2AB + B^2$. Thus, the identity is proven under the condition $AB = BA$.

Given:

Matrices $A = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix}$, $B = \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix}$, $C = \begin{bmatrix} 2&0 \\ 1&−2\end{bmatrix}$.

Scalars $a = 4$ and $b = -2$.

To Show:

Verify the given matrix properties.

Solution:

All matrices A, B, and C are of order $2 \times 2$. The operations involved are well-defined.

(a) Verify A + (B + C) = (A + B) + C (Associativity of Matrix Addition):

Calculate the left side: A + (B + C).

First, calculate B + C:

$B + C = \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix} + \begin{bmatrix} 2&0 \\ 1&−2 \end{bmatrix} = \begin{bmatrix} 4+2&0+0 \\ 1+1&5+(-2) \end{bmatrix} = \begin{bmatrix} 6&0 \\ 2&3 \end{bmatrix}$

Now, calculate A + (B + C):

$A + (B + C) = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} + \begin{bmatrix} 6&0 \\ 2&3 \end{bmatrix} = \begin{bmatrix} 1+6&2+0 \\ -1+2&3+3 \end{bmatrix} = \begin{bmatrix} 7&2 \\ 1&6 \end{bmatrix}$

Calculate the right side: (A + B) + C.

First, calculate A + B:

$A + B = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} + \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix} = \begin{bmatrix} 1+4&2+0 \\ -1+1&3+5 \end{bmatrix} = \begin{bmatrix} 5&2 \\ 0&8 \end{bmatrix}$

Now, calculate (A + B) + C:

$(A + B) + C = \begin{bmatrix} 5&2 \\ 0&8 \end{bmatrix} + \begin{bmatrix} 2&0 \\ 1&−2 \end{bmatrix} = \begin{bmatrix} 5+2&2+0 \\ 0+1&8+(-2) \end{bmatrix} = \begin{bmatrix} 7&2 \\ 1&6 \end{bmatrix}$

Comparing (a.1) and (a.2), we see that $A + (B + C) = (A + B) + C$. The associativity of matrix addition is verified.

(b) Verify A (BC) = (AB) C (Associativity of Matrix Multiplication):

Calculate the left side: A (BC).

First, calculate BC:

$BC = \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix} \begin{bmatrix} 2&0 \\ 1&−2 \end{bmatrix} = \begin{bmatrix} 4(2)+0(1)&4(0)+0(−2) \\ 1(2)+5(1)&1(0)+5(−2) \end{bmatrix} = \begin{bmatrix} 8+0&0+0 \\ 2+5&0-10 \end{bmatrix} = \begin{bmatrix} 8&0 \\ 7&−10 \end{bmatrix}$

Now, calculate A (BC):

$A (BC) = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} \begin{bmatrix} 8&0 \\ 7&−10 \end{bmatrix} = \begin{bmatrix} 1(8)+2(7)&1(0)+2(−10) \\ (−1)(8)+3(7)&(−1)(0)+3(−10) \end{bmatrix} = \begin{bmatrix} 8+14&0-20 \\ -8+21&0-30 \end{bmatrix} = \begin{bmatrix} 22&−20 \\ 13&−30 \end{bmatrix}$

Calculate the right side: (AB) C.

First, calculate AB:

$AB = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix} = \begin{bmatrix} 1(4)+2(1)&1(0)+2(5) \\ (−1)(4)+3(1)&(−1)(0)+3(5) \end{bmatrix} = \begin{bmatrix} 4+2&0+10 \\ -4+3&0+15 \end{bmatrix} = \begin{bmatrix} 6&10 \\ −1&15 \end{bmatrix}$

Now, calculate (AB) C:

$(AB) C = \begin{bmatrix} 6&10 \\ −1&15 \end{bmatrix} \begin{bmatrix} 2&0 \\ 1&−2 \end{bmatrix} = \begin{bmatrix} 6(2)+10(1)&6(0)+10(−2) \\ (−1)(2)+15(1)&(−1)(0)+15(−2) \end{bmatrix} = \begin{bmatrix} 12+10&0-20 \\ -2+15&0-30 \end{bmatrix} = \begin{bmatrix} 22&−20 \\ 13&−30 \end{bmatrix}$

Comparing (b.1) and (b.2), we see that $A (BC) = (AB) C$. The associativity of matrix multiplication is verified.

(c) Verify (a + b)B = aB + bB (Distributivity of Scalar Multiplication):

Calculate the left side: (a + b)B.

First, calculate a + b:

$a + b = 4 + (−2) = 4 - 2 = 2$

Now, calculate (a + b)B:

$(a + b)B = 2B = 2 \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix} = \begin{bmatrix} 2 \times 4&2 \times 0 \\ 2 \times 1&2 \times 5 \end{bmatrix} = \begin{bmatrix} 8&0 \\ 2&10 \end{bmatrix}$

Calculate the right side: aB + bB.

First, calculate aB:

$aB = 4B = 4 \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix} = \begin{bmatrix} 4 \times 4&4 \times 0 \\ 4 \times 1&4 \times 5 \end{bmatrix} = \begin{bmatrix} 16&0 \\ 4&20 \end{bmatrix}$

Next, calculate bB:

$bB = -2B = -2 \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix} = \begin{bmatrix} -2 \times 4&-2 \times 0 \\ -2 \times 1&-2 \times 5 \end{bmatrix} = \begin{bmatrix} −8&0 \\ −2&−10 \end{bmatrix}$

Now, calculate aB + bB:

$aB + bB = \begin{bmatrix} 16&0 \\ 4&20 \end{bmatrix} + \begin{bmatrix} −8&0 \\ −2&−10 \end{bmatrix} = \begin{bmatrix} 16+(-8)&0+0 \\ 4+(-2)&20+(-10) \end{bmatrix} = \begin{bmatrix} 16-8&0 \\ 4-2&20-10 \end{bmatrix} = \begin{bmatrix} 8&0 \\ 2&10 \end{bmatrix}$

Comparing (c.1) and (c.2), we see that $(a + b)B = aB + bB$. The distributivity of scalar multiplication over scalar addition is verified.

(d) Verify a (C – A) = aC – aA (Distributivity of Scalar Multiplication):

Calculate the left side: a (C – A).

First, calculate C – A:

$C – A = \begin{bmatrix} 2&0 \\ 1&−2 \end{bmatrix} - \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} = \begin{bmatrix} 2-1&0-2 \\ 1-(-1)&-2-3 \end{bmatrix} = \begin{bmatrix} 1&−2 \\ 1+1&−5 \end{bmatrix} = \begin{bmatrix} 1&−2 \\ 2&−5 \end{bmatrix}$

Now, calculate a (C – A):

$a (C – A) = 4 \begin{bmatrix} 1&−2 \\ 2&−5 \end{bmatrix} = \begin{bmatrix} 4 \times 1&4 \times (−2) \\ 4 \times 2&4 \times (−5) \end{bmatrix} = \begin{bmatrix} 4&−8 \\ 8&−20 \end{bmatrix}$

Calculate the right side: aC – aA.

First, calculate aC:

$aC = 4C = 4 \begin{bmatrix} 2&0 \\ 1&−2 \end{bmatrix} = \begin{bmatrix} 4 \times 2&4 \times 0 \\ 4 \times 1&4 \times (−2) \end{bmatrix} = \begin{bmatrix} 8&0 \\ 4&−8 \end{bmatrix}$

Next, calculate aA:

$aA = 4A = 4 \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} = \begin{bmatrix} 4 \times 1&4 \times 2 \\ 4 \times (−1)&4 \times 3 \end{bmatrix} = \begin{bmatrix} 4&8 \\ −4&12 \end{bmatrix}$

Now, calculate aC – aA:

$aC – aA = \begin{bmatrix} 8&0 \\ 4&−8 \end{bmatrix} - \begin{bmatrix} 4&8 \\ −4&12 \end{bmatrix} = \begin{bmatrix} 8-4&0-8 \\ 4-(-4)&-8-12 \end{bmatrix} = \begin{bmatrix} 4&−8 \\ 4+4&−20 \end{bmatrix} = \begin{bmatrix} 4&−8 \\ 8&−20 \end{bmatrix}$

Comparing (d.1) and (d.2), we see that $a (C – A) = aC – aA$. The distributivity of scalar multiplication over matrix subtraction is verified.

(e) Verify (A^T)^T = A (Double Transpose):

First, find the transpose of A, A^T:

$A^T = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix}^T = \begin{bmatrix} 1&−1 \\ 2&3 \end{bmatrix}$

Now, find the transpose of A^T, (A^T)^T:

$(A^T)^T = \begin{bmatrix} 1&−1 \\ 2&3 \end{bmatrix}^T = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix}$

Comparing this with the original matrix A, we see that they are identical.

$(A^T)^T = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} = A$

Thus, $(A^T)^T = A$ is verified.

(f) Verify (bA)^T = b A^T (Transpose of a Scalar Multiple):

Calculate the left side: (bA)^T.

First, calculate bA:

$bA = -2A = -2 \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} = \begin{bmatrix} -2 \times 1&-2 \times 2 \\ -2 \times (−1)&-2 \times 3 \end{bmatrix} = \begin{bmatrix} −2&−4 \\ 2&−6 \end{bmatrix}$

Now, find the transpose of bA:

$(bA)^T = \begin{bmatrix} −2&−4 \\ 2&−6 \end{bmatrix}^T = \begin{bmatrix} −2&2 \\ −4&−6 \end{bmatrix}$

Calculate the right side: b A^T.

From part (e), we have $A^T = \begin{bmatrix} 1&−1 \\ 2&3 \end{bmatrix}$.

Now, calculate b A^T:

$b A^T = -2 \begin{bmatrix} 1&−1 \\ 2&3 \end{bmatrix} = \begin{bmatrix} -2 \times 1&-2 \times (−1) \\ -2 \times 2&-2 \times 3 \end{bmatrix} = \begin{bmatrix} −2&2 \\ −4&−6 \end{bmatrix}$

Comparing (f.1) and (f.2), we see that $(bA)^T = b A^T$. The property of the transpose of a scalar multiple is verified.

(g) Verify (AB)^T = B^T A^T (Transpose of a Product):

Calculate the left side: (AB)^T.

From part (b), we calculated AB:

$AB = \begin{bmatrix} 6&10 \\ −1&15 \end{bmatrix}$

Now, find the transpose of AB:

$(AB)^T = \begin{bmatrix} 6&10 \\ −1&15 \end{bmatrix}^T = \begin{bmatrix} 6&−1 \\ 10&15 \end{bmatrix}$

Calculate the right side: B^T A^T.

First, find the transpose of B, B^T:

$B^T = \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix}^T = \begin{bmatrix} 4&1 \\ 0&5 \end{bmatrix}$

From part (e), we have $A^T = \begin{bmatrix} 1&−1 \\ 2&3 \end{bmatrix}$.

Now, calculate the product B^T A^T:

$B^T A^T = \begin{bmatrix} 4&1 \\ 0&5 \end{bmatrix} \begin{bmatrix} 1&−1 \\ 2&3 \end{bmatrix} = \begin{bmatrix} 4(1)+1(2)&4(−1)+1(3) \\ 0(1)+5(2)&0(−1)+5(3) \end{bmatrix} = \begin{bmatrix} 4+2&−4+3 \\ 0+10&0+15 \end{bmatrix} = \begin{bmatrix} 6&−1 \\ 10&15 \end{bmatrix}$

Comparing (g.1) and (g.2), we see that $(AB)^T = B^T A^T$. The property of the transpose of a product is verified.

(h) Verify (A – B)C = AC – BC (Right Distributivity of Matrix Multiplication):

Calculate the left side: (A – B)C.

First, calculate A – B:

$A – B = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} - \begin{bmatrix} 4&0 \\ 1&5 \end{bmatrix} = \begin{bmatrix} 1-4&2-0 \\ -1-1&3-5 \end{bmatrix} = \begin{bmatrix} −3&2 \\ −2&−2 \end{bmatrix}$

Now, calculate (A – B)C:

$(A – B)C = \begin{bmatrix} −3&2 \\ −2&−2 \end{bmatrix} \begin{bmatrix} 2&0 \\ 1&−2 \end{bmatrix} = \begin{bmatrix} (−3)(2)+2(1)&(−3)(0)+2(−2) \\ (−2)(2)+(−2)(1)&(−2)(0)+(−2)(−2) \end{bmatrix} = \begin{bmatrix} -6+2&0-4 \\ -4-2&0+4 \end{bmatrix} = \begin{bmatrix} −4&−4 \\ −6&4 \end{bmatrix}$

Calculate the right side: AC – BC.

First, calculate AC:

$AC = \begin{bmatrix} 1&2 \\ −1&3 \end{bmatrix} \begin{bmatrix} 2&0 \\ 1&−2 \end{bmatrix} = \begin{bmatrix} 1(2)+2(1)&1(0)+2(−2) \\ (−1)(2)+3(1)&(−1)(0)+3(−2) \end{bmatrix} = \begin{bmatrix} 2+2&0-4 \\ -2+3&0-6 \end{bmatrix} = \begin{bmatrix} 4&−4 \\ 1&−6 \end{bmatrix}$

From part (b), we calculated BC:

$BC = \begin{bmatrix} 8&0 \\ 7&−10 \end{bmatrix}$

Now, calculate AC – BC:

$AC – BC = \begin{bmatrix} 4&−4 \\ 1&−6 \end{bmatrix} - \begin{bmatrix} 8&0 \\ 7&−10 \end{bmatrix} = \begin{bmatrix} 4-8&−4-0 \\ 1-7&−6-(-10) \end{bmatrix} = \begin{bmatrix} −4&−4 \\ −6&−6+10 \end{bmatrix} = \begin{bmatrix} −4&−4 \\ −6&4 \end{bmatrix}$

Comparing (h.1) and (h.2), we see that $(A – B)C = AC – BC$. The right distributivity of matrix multiplication over subtraction is verified.

(i) Verify (A – B)^T = A^T – B^T (Transpose of a Difference):

Calculate the left side: (A – B)^T.

From part (h), we calculated A – B:

$A – B = \begin{bmatrix} −3&2 \\ −2&−2 \end{bmatrix}$

Now, find the transpose of (A – B):

$(A – B)^T = \begin{bmatrix} −3&2 \\ −2&−2 \end{bmatrix}^T = \begin{bmatrix} −3&−2 \\ 2&−2 \end{bmatrix}$

Calculate the right side: A^T – B^T.

From part (e), we have $A^T = \begin{bmatrix} 1&−1 \\ 2&3 \end{bmatrix}$.

From part (g), we have $B^T = \begin{bmatrix} 4&1 \\ 0&5 \end{bmatrix}$.

Now, calculate A^T – B^T:

$A^T – B^T = \begin{bmatrix} 1&−1 \\ 2&3 \end{bmatrix} - \begin{bmatrix} 4&1 \\ 0&5 \end{bmatrix} = \begin{bmatrix} 1-4&−1-1 \\ 2-0&3-5 \end{bmatrix} = \begin{bmatrix} −3&−2 \\ 2&−2 \end{bmatrix}$

Comparing (i.1) and (i.2), we see that $(A – B)^T = A^T – B^T$. The property of the transpose of a difference is verified.

Given:

The matrix $A = \begin{bmatrix} \cos θ & \sin θ \\ −\sin θ & \cos θ \end{bmatrix}$.

To Show:

$A^2 = \begin{bmatrix} \cos 2 θ & \sin 2 θ \\ −\sin 2θ & \cos 2θ \end{bmatrix}$.

Proof:

We need to calculate $A^2 = A \times A$.

$A^2 = \begin{bmatrix} \cos θ & \sin θ \\ −\sin θ & \cos θ \end{bmatrix} \begin{bmatrix} \cos θ & \sin θ \\ −\sin θ & \cos θ \end{bmatrix}$

Perform the matrix multiplication:

$A^2 = \begin{bmatrix} (\cos θ)(\cos θ) + (\sin θ)(−\sin θ) & (\cos θ)(\sin θ) + (\sin θ)(\cos θ) \\ (−\sin θ)(\cos θ) + (\cos θ)(−\sin θ) & (−\sin θ)(\sin θ) + (\cos θ)(\cos θ) \end{bmatrix}$

$A^2 = \begin{bmatrix} \cos^2 θ − \sin^2 θ & \cos θ \sin θ + \sin θ \cos θ \\ −\sin θ \cos θ − \cos θ \sin θ & −\sin^2 θ + \cos^2 θ \end{bmatrix}$

$A^2 = \begin{bmatrix} \cos^2 θ − \sin^2 θ & 2 \sin θ \cos θ \\ −2 \sin θ \cos θ & \cos^2 θ − \sin^2 θ \end{bmatrix}$

Now, we use the double angle trigonometric identities:

$\cos 2θ = \cos^2 θ − \sin^2 θ$

$\sin 2θ = 2 \sin θ \cos θ$

Substitute these identities into the elements of the matrix $A^2$:

$A^2 = \begin{bmatrix} \cos 2 θ & \sin 2 θ \\ −\sin 2θ & \cos 2θ \end{bmatrix}$

This matches the matrix we were asked to show.

The final answer is:

By calculating $A^2$ and using trigonometric identities, it has been shown that $A^2 = \begin{bmatrix} \cos 2 θ & \sin 2 θ \\ −\sin 2θ & \cos 2θ \end{bmatrix}$.

Given:

$A = \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix}$

$B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

and $x^2 = -1$.

To Show:

$(A + B)^2 = A^2 + B^2$.

Solution:

First, calculate $A + B$.

$A + B = \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0+0 & -x+1 \\ x+1 & 0+0 \end{bmatrix} = \begin{bmatrix} 0 & 1-x \\ 1+x & 0 \end{bmatrix}$

Next, calculate $(A + B)^2 = (A + B)(A + B)$.

$(A+B)^2 = \begin{bmatrix} 0 & 1-x \\ 1+x & 0 \end{bmatrix} \begin{bmatrix} 0 & 1-x \\ 1+x & 0 \end{bmatrix}$

$= \begin{bmatrix} 0(0) + (1-x)(1+x) & 0(1-x) + (1-x)(0) \\ (1+x)(0) + 0(1+x) & (1+x)(1-x) + 0(0) \end{bmatrix}$

$= \begin{bmatrix} (1-x)(1+x) & 0 \\ 0 & (1+x)(1-x) \end{bmatrix}$

Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$, we have $(1-x)(1+x) = 1^2 - x^2 = 1 - x^2$.

Given $x^2 = -1$, so $1 - x^2 = 1 - (-1) = 1 + 1 = 2$.

Therefore, $(A+B)^2 = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$.

Now, calculate $A^2 = A \cdot A$.

$A^2 = \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix} \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix}$

$= \begin{bmatrix} 0(0) + (-x)(x) & 0(-x) + (-x)(0) \\ x(0) + 0(x) & x(-x) + 0(0) \end{bmatrix}$

$= \begin{bmatrix} -x^2 & 0 \\ 0 & -x^2 \end{bmatrix}$

Given $x^2 = -1$, so $-x^2 = -(-1) = 1$.

Therefore, $A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Next, calculate $B^2 = B \cdot B$.

$B^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

$= \begin{bmatrix} 0(0) + 1(1) & 0(1) + 1(0) \\ 1(0) + 0(1) & 1(1) + 0(0) \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Therefore, $B^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Finally, calculate $A^2 + B^2$.

$A^2 + B^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & 0+0 \\ 0+0 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

Comparing the results for $(A + B)^2$ and $A^2 + B^2$, we have:

$(A + B)^2 = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

$A^2 + B^2 = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

Since both results are the same matrix, we have shown that $(A + B)^2 = A^2 + B^2$.

Given:

$A = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix}$.

To Verify:

$A^2 = I$, where $I$ is the identity matrix of the same order as $A$. Since $A$ is a $3 \times 3$ matrix, $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

Solution:

We need to calculate $A^2 = A \cdot A$.

$A^2 = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix}$

Now, perform the matrix multiplication:

The element in the first row and first column is: $(0)(0) + (1)(4) + (-1)(3) = 0 + 4 - 3 = 1$

The element in the first row and second column is: $(0)(1) + (1)(-3) + (-1)(-3) = 0 - 3 + 3 = 0$

The element in the first row and third column is: $(0)(-1) + (1)(4) + (-1)(4) = 0 + 4 - 4 = 0$

The element in the second row and first column is: $(4)(0) + (-3)(4) + (4)(3) = 0 - 12 + 12 = 0$

The element in the second row and second column is: $(4)(1) + (-3)(-3) + (4)(-3) = 4 + 9 - 12 = 1$

The element in the second row and third column is: $(4)(-1) + (-3)(4) + (4)(4) = -4 - 12 + 16 = 0$

The element in the third row and first column is: $(3)(0) + (-3)(4) + (4)(3) = 0 - 12 + 12 = 0$

The element in the third row and second column is: $(3)(1) + (-3)(-3) + (4)(-3) = 3 + 9 - 12 = 0$

The element in the third row and third column is: $(3)(-1) + (-3)(4) + (4)(4) = -3 - 12 + 16 = 1$

So, $A^2$ is the matrix:

$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

This is the $3 \times 3$ identity matrix, $I$.

Thus, we have verified that $A^2 = I$.

Given:

A is a square matrix.

n $\in$ N (the set of natural numbers).

A' denotes the transpose of matrix A.

To Prove:

$(A')^n = (A^n)'$ for all n $\in$ N.

Proof:

We will prove the statement by the principle of mathematical induction.

Base Case (n=1):

We need to show that the statement is true for n=1.

Left Hand Side (LHS) = $(A')^1 = A'$.

Right Hand Side (RHS) = $(A^1)' = A'$.

Since LHS = RHS, the statement is true for n=1.

Inductive Hypothesis:

Assume that the statement is true for some arbitrary positive integer k, i.e.,

$(A')^k = (A^k)'$.

This is our inductive hypothesis.

Inductive Step:

We need to prove that the statement is true for n = k+1, i.e., we need to show that $(A')^{k+1} = (A^{k+1})'$.

Consider the LHS:

$(A')^{k+1} = (A')^k \cdot A'$.

By the inductive hypothesis, $(A')^k = (A^k)'$. Substituting this into the expression:

$(A')^{k+1} = (A^k)' \cdot A'$.

Using the property of transpose of a product of two matrices, $(BC)' = C'B'$, where B = $A^k$ and C = A, we have $(A^k A)' = A' (A^k)'$.

So, $(A^k)' A'$ can be written as $(A \cdot A^k)'$. However, matrix multiplication is generally not commutative, so $A \cdot A^k$ is not necessarily $A^{k+1}$ if A is multiplied on the left. We need to be careful here.

Let's restart the inductive step from the RHS.

Consider the RHS:

$(A^{k+1})' = (A^k \cdot A)'$.

Using the property of transpose of a product of two matrices, $(BC)' = C'B'$, where B = $A^k$ and C = A, we have:

$(A^k \cdot A)' = A' \cdot (A^k)'$.

By the inductive hypothesis, $(A^k)' = (A')^k$. Substituting this into the expression:

$(A^{k+1})' = A' \cdot (A')^k$.

By the definition of matrix exponentiation, $A' \cdot (A')^k = (A')^{1+k} = (A')^{k+1}$.

So, $(A^{k+1})' = (A')^{k+1}$.

This shows that the statement is true for n = k+1.

Conclusion:

Since the statement is true for n=1 and is true for n=k+1 whenever it is true for n=k, by the principle of mathematical induction, the statement $(A')^n = (A^n)'$ is true for all n $\in$ N.

Solution:

We use elementary row operations to find the inverse of the given matrices.

(i) For the matrix $A = \begin{bmatrix} 1& 3 \\ −5& 7 \end{bmatrix}$.

We write the augmented matrix $[A | I]$:

$\begin{bmatrix} 1& 3 & | & 1 & 0 \\ −5& 7 & | & 0 & 1 \end{bmatrix}$

Apply the elementary row operation $R_2 \to R_2 + 5R_1$:

$\begin{bmatrix} 1& 3 & | & 1 & 0 \\ −5 + 5(1)& 7 + 5(3) & | & 0 + 5(1)& 1 + 5(0) \end{bmatrix}$

$\begin{bmatrix} 1& 3 & | & 1 & 0 \\ 0& 7 + 15 & | & 5& 1 \end{bmatrix}$

$\begin{bmatrix} 1& 3 & | & 1 & 0 \\ 0& 22 & | & 5& 1 \end{bmatrix}$

Apply the elementary row operation $R_2 \to \frac{1}{22}R_2$:

$\begin{bmatrix} 1& 3 & | & 1 & 0 \\ \frac{1}{22}(0)& \frac{1}{22}(22) & | & \frac{1}{22}(5)& \frac{1}{22}(1) \end{bmatrix}$

$\begin{bmatrix} 1& 3 & | & 1 & 0 \\ 0& 1 & | & \frac{5}{22}& \frac{1}{22} \end{bmatrix}$

Apply the elementary row operation $R_1 \to R_1 - 3R_2$:

$\begin{bmatrix} 1 - 3(0)& 3 - 3(1) & | & 1 - 3(\frac{5}{22})& 0 - 3(\frac{1}{22}) \end{bmatrix}$

$\begin{bmatrix} 1& 0 & | & 1 - \frac{15}{22}& -\frac{3}{22} \end{bmatrix}$

$\begin{bmatrix} 1& 0 & | & \frac{22-15}{22}& -\frac{3}{22} \end{bmatrix}$

$\begin{bmatrix} 1& 0 & | & \frac{7}{22}& -\frac{3}{22} \\ 0& 1 & | & \frac{5}{22}& \frac{1}{22} \end{bmatrix}$

Since the left side is the identity matrix $I$, the matrix on the right side is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix} \frac{7}{22}& -\frac{3}{22} \\ \frac{5}{22}& \frac{1}{22} \end{bmatrix}$.

(ii) For the matrix $A = \begin{bmatrix} 1& −3 \\ −2& 6 \end{bmatrix}$.

We write the augmented matrix $[A | I]$:

$\begin{bmatrix} 1& −3 & | & 1 & 0 \\ −2& 6 & | & 0 & 1 \end{bmatrix}$

Apply the elementary row operation $R_2 \to R_2 + 2R_1$:

$\begin{bmatrix} 1& −3 & | & 1 & 0 \\ −2 + 2(1)& 6 + 2(−3) & | & 0 + 2(1)& 1 + 2(0) \end{bmatrix}$

$\begin{bmatrix} 1& −3 & | & 1 & 0 \\ −2 + 2& 6 − 6 & | & 2& 1 \end{bmatrix}$

$\begin{bmatrix} 1& −3 & | & 1 & 0 \\ 0& 0 & | & 2& 1 \end{bmatrix}$

Since we have obtained a row of zeros on the left side of the augmented matrix, it is not possible to transform the left side into the identity matrix.

Therefore, the given matrix is singular, and its inverse does not exist.

Given:

The matrix equality:

$\begin{bmatrix} xy& 4 \\ z+6& x+y \end{bmatrix} = \begin{bmatrix} 8& w \\ 0& 6 \end{bmatrix}$

To Find:

The values of x, y, z, and w.

Solution:

When two matrices are equal, their corresponding elements are equal.

Equating the elements in the first row:

From the element in the first row, first column:

$xy = 8$

From the element in the first row, second column:

$4 = w$

So, the value of w is $4$.

Equating the elements in the second row:

From the element in the second row, first column:

$z+6 = 0$

$z = 0 - 6$

$z = -6$

So, the value of z is $-6$.

From the element in the second row, second column:

$x+y = 6$

Now we have a system of two equations with two variables, x and y:

1) $xy = 8$

2) $x+y = 6$

From equation (2), we can express y in terms of x:

$y = 6 - x$

Substitute this expression for y into equation (1):

$x(6 - x) = 8$

$6x - x^2 = 8$

Rearrange the terms to form a quadratic equation:

$x^2 - 6x + 8 = 0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 8 and add up to -6. These numbers are -2 and -4.

$x^2 - 2x - 4x + 8 = 0$

Factor by grouping:

$x(x - 2) - 4(x - 2) = 0$

$(x - 2)(x - 4) = 0$

This gives two possible values for x:

$x - 2 = 0 \implies x = 2$

$x - 4 = 0 \implies x = 4$

Now find the corresponding values for y using $y = 6 - x$:

If $x = 2$, then $y = 6 - 2 = 4$.

If $x = 4$, then $y = 6 - 4 = 2$.

Both pairs $(x=2, y=4)$ and $(x=4, y=2)$ satisfy both $xy=8$ and $x+y=6$.

Therefore, the possible values for (x, y, z, w) are:

Case 1: $x = 2$, $y = 4$, $z = -6$, $w = 4$.

Case 2: $x = 4$, $y = 2$, $z = -6$, $w = 4$.

The question asks for "the values" (plural), implying there might be multiple sets of values for x and y, while z and w are unique.

Final answer is usually expected to list all possible values for each variable.

So, the values are x = 2 or 4, y = 4 or 2 (correspondingly), z = -6, and w = 4.

The set of solutions for (x, y, z, w) are (2, 4, -6, 4) and (4, 2, -6, 4).

Given:

Matrix $A = \begin{bmatrix} 1& 5 \\ 7& 12 \end{bmatrix}$

Matrix $B = \begin{bmatrix} 9& 1 \\ 7& 8 \end{bmatrix}$

The condition that $3A + 5B + 2C$ is a null matrix. Since A and B are $2 \times 2$ matrices, C must also be a $2 \times 2$ matrix, and the null matrix will be the $2 \times 2$ zero matrix $\begin{bmatrix} 0& 0 \\ 0& 0 \end{bmatrix}$.

To Find:

The matrix C.

Solution:

The given condition is:

$3A + 5B + 2C = \begin{bmatrix} 0& 0 \\ 0& 0 \end{bmatrix}$

We need to isolate the matrix C. Subtract $3A$ and $5B$ from both sides of the equation:

$2C = \begin{bmatrix} 0& 0 \\ 0& 0 \end{bmatrix} - 3A - 5B$

$2C = -3A - 5B$

Now, we can find C by multiplying both sides by $\frac{1}{2}$:

$C = \frac{1}{2}(-3A - 5B)$

First, calculate $3A$:

$3A = 3 \begin{bmatrix} 1& 5 \\ 7& 12 \end{bmatrix} = \begin{bmatrix} 3 \times 1& 3 \times 5 \\ 3 \times 7& 3 \times 12 \end{bmatrix} = \begin{bmatrix} 3& 15 \\ 21& 36 \end{bmatrix}$

Next, calculate $5B$:

$5B = 5 \begin{bmatrix} 9& 1 \\ 7& 8 \end{bmatrix} = \begin{bmatrix} 5 \times 9& 5 \times 1 \\ 5 \times 7& 5 \times 8 \end{bmatrix} = \begin{bmatrix} 45& 5 \\ 35& 40 \end{bmatrix}$

Now, calculate $-3A - 5B = -(3A) - (5B)$:

$-3A = \begin{bmatrix} -3& -15 \\ -21& -36 \end{bmatrix}$

$-5B = \begin{bmatrix} -45& -5 \\ -35& -40 \end{bmatrix}$

$-3A - 5B = \begin{bmatrix} -3& -15 \\ -21& -36 \end{bmatrix} + \begin{bmatrix} -45& -5 \\ -35& -40 \end{bmatrix}$

$-3A - 5B = \begin{bmatrix} -3 + (-45)& -15 + (-5) \\ -21 + (-35)& -36 + (-40) \end{bmatrix}$

$-3A - 5B = \begin{bmatrix} -48& -20 \\ -56& -76 \end{bmatrix}$

Finally, calculate $C = \frac{1}{2}(-3A - 5B)$:

$C = \frac{1}{2} \begin{bmatrix} -48& -20 \\ -56& -76 \end{bmatrix}$

$C = \begin{bmatrix} \frac{1}{2} \times (-48)& \frac{1}{2} \times (-20) \\ \frac{1}{2} \times (-56)& \frac{1}{2} \times (-76) \end{bmatrix}$

$C = \begin{bmatrix} -24& -10 \\ -28& -38 \end{bmatrix}$

Thus, the matrix C is $\begin{bmatrix} -24& -10 \\ -28& -38 \end{bmatrix}$.

Given:

Matrix $A = \begin{bmatrix} 3& -5 \\ -4& 2 \end{bmatrix}$

To Find:

1) $A^2 - 5A - 14I$

2) $A^3$

Solution:

First, we calculate $A^2$.

$A^2 = A \times A = \begin{bmatrix} 3& -5 \\ -4& 2 \end{bmatrix} \begin{bmatrix} 3& -5 \\ -4& 2 \end{bmatrix}$

$A^2 = \begin{bmatrix} (3)(3) + (-5)(-4)& (3)(-5) + (-5)(2) \\ (-4)(3) + (2)(-4)& (-4)(-5) + (2)(2) \end{bmatrix}$

$A^2 = \begin{bmatrix} 9 + 20& -15 - 10 \\ -12 - 8& 20 + 4 \end{bmatrix}$

$A^2 = \begin{bmatrix} 29& -25 \\ -20& 24 \end{bmatrix}$

Next, we calculate $5A$.

$5A = 5 \begin{bmatrix} 3& -5 \\ -4& 2 \end{bmatrix} = \begin{bmatrix} 5 \times 3& 5 \times -5 \\ 5 \times -4& 5 \times 2 \end{bmatrix} = \begin{bmatrix} 15& -25 \\ -20& 10 \end{bmatrix}$

The identity matrix $I$ of the same order as A (which is $2 \times 2$) is $\begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}$.

So, $14I$ is:

$14I = 14 \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix} = \begin{bmatrix} 14 \times 1& 14 \times 0 \\ 14 \times 0& 14 \times 1 \end{bmatrix} = \begin{bmatrix} 14& 0 \\ 0& 14 \end{bmatrix}$

Now, we compute $A^2 - 5A - 14I$:

$A^2 - 5A - 14I = \begin{bmatrix} 29& -25 \\ -20& 24 \end{bmatrix} - \begin{bmatrix} 15& -25 \\ -20& 10 \end{bmatrix} - \begin{bmatrix} 14& 0 \\ 0& 14 \end{bmatrix}$

$A^2 - 5A - 14I = \begin{bmatrix} 29 - 15& -25 - (-25) \\ -20 - (-20)& 24 - 10 \end{bmatrix} - \begin{bmatrix} 14& 0 \\ 0& 14 \end{bmatrix}$

$A^2 - 5A - 14I = \begin{bmatrix} 14& 0 \\ 0& 14 \end{bmatrix} - \begin{bmatrix} 14& 0 \\ 0& 14 \end{bmatrix}$

$A^2 - 5A - 14I = \begin{bmatrix} 14 - 14& 0 - 0 \\ 0 - 0& 14 - 14 \end{bmatrix} = \begin{bmatrix} 0& 0 \\ 0& 0 \end{bmatrix}$

So, $A^2 - 5A - 14I$ is the null matrix.

Now, we use the result $A^2 - 5A - 14I = O$ to find $A^3$.

From the previous calculation, we have:

$A^2 - 5A - 14I = O$

$A^2 - 5A = 14I$

Multiply both sides by A (from the right):

$(A^2 - 5A)A = (14I)A$

$A^2 \cdot A - 5A \cdot A = 14IA$

$A^3 - 5A^2 = 14A$

Now, solve for $A^3$:

$A^3 = 5A^2 + 14A$

We have already calculated $A^2$ and $5A$. We need $5A^2$ and $14A$.

$5A^2 = 5 \begin{bmatrix} 29& -25 \\ -20& 24 \end{bmatrix} = \begin{bmatrix} 5 \times 29& 5 \times -25 \\ 5 \times -20& 5 \times 24 \end{bmatrix} = \begin{bmatrix} 145& -125 \\ -100& 120 \end{bmatrix}$

$14A = 14 \begin{bmatrix} 3& -5 \\ -4& 2 \end{bmatrix} = \begin{bmatrix} 14 \times 3& 14 \times -5 \\ 14 \times -4& 14 \times 2 \end{bmatrix} = \begin{bmatrix} 42& -70 \\ -56& 28 \end{bmatrix}$

Now, add $5A^2$ and $14A$ to find $A^3$:

$A^3 = 5A^2 + 14A = \begin{bmatrix} 145& -125 \\ -100& 120 \end{bmatrix} + \begin{bmatrix} 42& -70 \\ -56& 28 \end{bmatrix}$

$A^3 = \begin{bmatrix} 145 + 42& -125 + (-70) \\ -100 + (-56)& 120 + 28 \end{bmatrix}$

$A^3 = \begin{bmatrix} 187& -195 \\ -156& 148 \end{bmatrix}$

Therefore, $A^3 = \begin{bmatrix} 187& -195 \\ -156& 148 \end{bmatrix}$.

Given:

The matrix equation:

$3\begin{bmatrix} a& b \\ c& d \end{bmatrix} = \begin{bmatrix} a& 6 \\ −1& 2d \end{bmatrix} + \begin{bmatrix} 4& a+b \\ c+d& 3 \end{bmatrix}$

To Find:

The values of a, b, c, and d.

Solution:

First, perform the scalar multiplication on the left side of the equation:

$3\begin{bmatrix} a& b \\ c& d \end{bmatrix} = \begin{bmatrix} 3a& 3b \\ 3c& 3d \end{bmatrix}$

Next, perform the matrix addition on the right side of the equation:

$\begin{bmatrix} a& 6 \\ −1& 2d \end{bmatrix} + \begin{bmatrix} 4& a+b \\ c+d& 3 \end{bmatrix} = \begin{bmatrix} a+4& 6+(a+b) \\ −1+(c+d)& 2d+3 \end{bmatrix} = \begin{bmatrix} a+4& 6+a+b \\ −1+c+d& 2d+3 \end{bmatrix}$

Now, equate the left side matrix with the right side matrix:

$\begin{bmatrix} 3a& 3b \\ 3c& 3d \end{bmatrix} = \begin{bmatrix} a+4& 6+a+b \\ −1+c+d& 2d+3 \end{bmatrix}$

For two matrices to be equal, their corresponding elements must be equal. This gives us a system of four linear equations:

1) $3a = a+4$

2) $3b = 6+a+b$

3) $3c = -1+c+d$

4) $3d = 2d+3$

Solve equation (1) for a:

$3a - a = 4$

$2a = 4$

$a = \frac{4}{2}$

$a = 2$

Solve equation (4) for d:

$3d - 2d = 3$

$d = 3$

Substitute the value of a into equation (2) and solve for b:

$3b = 6+a+b$

$3b = 6+2+b$

$3b = 8+b$

$3b - b = 8$

$2b = 8$

$b = \frac{8}{2}$

$b = 4$

Substitute the value of d into equation (3) and solve for c:

$3c = -1+c+d$

$3c = -1+c+3$

$3c = c+2$

$3c - c = 2$

$2c = 2$

$c = \frac{2}{2}$

$c = 1$

Thus, the values are a = 2, b = 4, c = 1, and d = 3.

Given:

The matrix equation: $\begin{bmatrix} 2& −1 \\ 1& 0 \\ −3& 4 \end{bmatrix} A = \begin{bmatrix} −1& −8& −10 \\ 1& −2& −5 \\ 9& 22& 15 \end{bmatrix}$

Let the first matrix be M and the result matrix be R. So, $MA = R$.

The dimensions of matrix M are $3 \times 2$.

The dimensions of matrix R are $3 \times 3$.

To Find:

The matrix A.

Solution:

Let the dimensions of matrix A be $m \times n$.

For the matrix multiplication MA to be defined, the number of columns in M must be equal to the number of rows in A.

Number of columns in M = 2. So, the number of rows in A, $m$, must be 2.

The resulting matrix MA will have dimensions (number of rows in M) $\times$ (number of columns in A).

Dimensions of MA are $3 \times n$.

Since $MA = R$, the dimensions of MA must be equal to the dimensions of R, which are $3 \times 3$.

So, $3 \times n = 3 \times 3$. This means the number of columns in A, $n$, must be 3.

Therefore, matrix A is a $2 \times 3$ matrix.

Let $A = \begin{bmatrix} a& b& c \\ d& e& f \end{bmatrix}$.

Substitute A into the given equation:

$\begin{bmatrix} 2& −1 \\ 1& 0 \\ −3& 4 \end{bmatrix} \begin{bmatrix} a& b& c \\ d& e& f \end{bmatrix} = \begin{bmatrix} −1& −8& −10 \\ 1& −2& −5 \\ 9& 22& 15 \end{bmatrix}$

Perform the matrix multiplication on the left side:

$\begin{bmatrix} (2)(a) + (-1)(d)& (2)(b) + (-1)(e)& (2)(c) + (-1)(f) \\ (1)(a) + (0)(d)& (1)(b) + (0)(e)& (1)(c) + (0)(f) \\ (-3)(a) + (4)(d)& (-3)(b) + (4)(e)& (-3)(c) + (4)(f) \end{bmatrix} = \begin{bmatrix} −1& −8& −10 \\ 1& −2& −5 \\ 9& 22& 15 \end{bmatrix}$

$\begin{bmatrix} 2a - d& 2b - e& 2c - f \\ a& b& c \\ -3a + 4d& -3b + 4e& -3c + 4f \end{bmatrix} = \begin{bmatrix} −1& −8& −10 \\ 1& −2& −5 \\ 9& 22& 15 \end{bmatrix}$

Equating the corresponding elements of the two matrices, we get a system of equations:

From the second row:

$a = 1$

$b = -2$

$c = -5$

Now substitute these values into the equations from the first and third rows:

Using the first row:

$2a - d = -1 \implies 2(1) - d = -1 \implies 2 - d = -1 \implies d = 2 - (-1) = 3$

$2b - e = -8 \implies 2(-2) - e = -8 \implies -4 - e = -8 \implies e = -4 - (-8) = -4 + 8 = 4$

$2c - f = -10 \implies 2(-5) - f = -10 \implies -10 - f = -10 \implies f = -10 - (-10) = -10 + 10 = 0$

Let's verify these values using the third row equations:

$-3a + 4d = -3(1) + 4(3) = -3 + 12 = 9$ (Matches the element in R)

$-3b + 4e = -3(-2) + 4(4) = 6 + 16 = 22$ (Matches the element in R)

$-3c + 4f = -3(-5) + 4(0) = 15 + 0 = 15$ (Matches the element in R)

The values are consistent.

So, the values of the variables are:

$a = 1, b = -2, c = -5$

$d = 3, e = 4, f = 0$

Construct the matrix A using these values:

$A = \begin{bmatrix} a& b& c \\ d& e& f \end{bmatrix} = \begin{bmatrix} 1& -2& -5 \\ 3& 4& 0 \end{bmatrix}$

Thus, the matrix A is $\begin{bmatrix} 1& -2& -5 \\ 3& 4& 0 \end{bmatrix}$.

Given:

Matrix $A = \begin{bmatrix} 1& 2 \\ 4& 1 \end{bmatrix}$

To Find:

The matrix expression $A^2 + 2A + 7I$.

Solution:

First, we calculate $A^2$.

$A^2 = A \times A = \begin{bmatrix} 1& 2 \\ 4& 1 \end{bmatrix} \begin{bmatrix} 1& 2 \\ 4& 1 \end{bmatrix}$

$A^2 = \begin{bmatrix} (1)(1) + (2)(4)& (1)(2) + (2)(1) \\ (4)(1) + (1)(4)& (4)(2) + (1)(1) \end{bmatrix}$

$A^2 = \begin{bmatrix} 1 + 8& 2 + 2 \\ 4 + 4& 8 + 1 \end{bmatrix}$

$A^2 = \begin{bmatrix} 9& 4 \\ 8& 9 \end{bmatrix}$

Next, we calculate $2A$.

$2A = 2 \begin{bmatrix} 1& 2 \\ 4& 1 \end{bmatrix} = \begin{bmatrix} 2 \times 1& 2 \times 2 \\ 2 \times 4& 2 \times 1 \end{bmatrix} = \begin{bmatrix} 2& 4 \\ 8& 2 \end{bmatrix}$

Since A is a $2 \times 2$ matrix, the identity matrix I must also be a $2 \times 2$ matrix:

$I = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}$

So, $7I$ is:

$7I = 7 \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix} = \begin{bmatrix} 7 \times 1& 7 \times 0 \\ 7 \times 0& 7 \times 1 \end{bmatrix} = \begin{bmatrix} 7& 0 \\ 0& 7 \end{bmatrix}$

Now, we compute $A^2 + 2A + 7I$ by adding the corresponding matrices:

$A^2 + 2A + 7I = \begin{bmatrix} 9& 4 \\ 8& 9 \end{bmatrix} + \begin{bmatrix} 2& 4 \\ 8& 2 \end{bmatrix} + \begin{bmatrix} 7& 0 \\ 0& 7 \end{bmatrix}$

Add the elements in the corresponding positions:

$A^2 + 2A + 7I = \begin{bmatrix} 9 + 2 + 7& 4 + 4 + 0 \\ 8 + 8 + 0& 9 + 2 + 7 \end{bmatrix}$

$A^2 + 2A + 7I = \begin{bmatrix} 18& 8 \\ 16& 18 \end{bmatrix}$

Thus, $A^2 + 2A + 7I = \begin{bmatrix} 18& 8 \\ 16& 18 \end{bmatrix}$.

Given:

Matrix $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ −\sin \alpha & \cos \alpha \end{bmatrix}$

The condition $A^{-1} = A'$.

To Find:

The value(s) of $\alpha$.

Solution:

We are given the matrix $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ −\sin \alpha & \cos \alpha \end{bmatrix}$ and the condition $A^{-1} = A'$.

First, let's find the transpose of matrix A, denoted by $A'$. The transpose is obtained by interchanging the rows and columns of A.

$A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

Next, let's find the inverse of matrix A, denoted by $A^{-1}$. The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a& b \\ c& d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d& -b \\ -c& a \end{bmatrix}$, provided $ad-bc \neq 0$.

For the given matrix A, $a = \cos \alpha$, $b = \sin \alpha$, $c = -\sin \alpha$, and $d = \cos \alpha$.

The determinant of A is $\det(A) = ad - bc = (\cos \alpha)(\cos \alpha) - (\sin \alpha)(-\sin \alpha) = \cos^2 \alpha + \sin^2 \alpha$.

Using the fundamental trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$, we have:

$\det(A) = 1$

Since $\det(A) = 1$, which is non-zero, the inverse of A exists.

The inverse matrix is:

$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d& -b \\ -c& a \end{bmatrix} = \frac{1}{1} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ -(-\sin \alpha) & \cos \alpha \end{bmatrix} = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

Now, we use the given condition $A^{-1} = A'$. We equate the calculated inverse and transpose matrices:

$\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

Equating the corresponding elements of the two matrices, we observe that each element on the left side is identical to the corresponding element on the right side.

$\cos \alpha = \cos \alpha$

$-\sin \alpha = -\sin \alpha$

$\sin \alpha = \sin \alpha$

$\cos \alpha = \cos \alpha$

These equalities hold true for any real value of $\alpha$.

This means that the condition $A^{-1} = A'$ is satisfied by the given matrix A for all real values of $\alpha$.

This type of matrix, where $A^{-1} = A'$, is called an orthogonal matrix. The given matrix is a standard rotation matrix, and rotation matrices are always orthogonal.

Since the condition $A^{-1} = A'$ is always true for this matrix for any real $\alpha$, there is no specific unique value (or set of discrete values) of $\alpha$ required for the condition to hold. The condition holds for all $\alpha \in \mathbb{R}$.

Assuming the question as stated is correct, the condition $A^{-1} = A'$ does not impose any constraint on the value of $\alpha$.

Answer: The condition $A^{-1} = A'$ is true for all real values of $\alpha$. There is no specific value of $\alpha$ that needs to be found; any real number $\alpha$ satisfies the condition.

Given:

The matrix $A = \begin{bmatrix} 0& a& 3 \\ 2& b& −1 \\ c& 1& 0 \end{bmatrix}$ is a skew symmetric matrix.

To Find:

The values of a, b, and c.

Solution:

A matrix A is said to be skew symmetric if its transpose is equal to the negative of the matrix, i.e., $A' = -A$.

Alternatively, for a matrix $A = [a_{ij}]$, it is skew symmetric if $a_{ij} = -a_{ji}$ for all i and j.

For a skew symmetric matrix, the diagonal elements must be zero ($a_{ii} = 0$).

Let the given matrix be $A = \begin{bmatrix} a_{11}& a_{12}& a_{13} \\ a_{21}& a_{22}& a_{23} \\ a_{31}& a_{32}& a_{33} \end{bmatrix} = \begin{bmatrix} 0& a& 3 \\ 2& b& −1 \\ c& 1& 0 \end{bmatrix}$.

Applying the condition $a_{ij} = -a_{ji}$:

For the diagonal elements:

$a_{11} = -a_{11} \implies 0 = -0$ (This is true and consistent with the given matrix).

$a_{22} = -a_{22} \implies b = -b$. This implies $2b = 0$, so $b = 0$.

$a_{33} = -a_{33} \implies 0 = -0$ (This is true and consistent with the given matrix).

For the off-diagonal elements:

$a_{12} = -a_{21}$

From the matrix, $a_{12} = a$ and $a_{21} = 2$.

So, $a = -2$.

$a_{13} = -a_{31}$

From the matrix, $a_{13} = 3$ and $a_{31} = c$.

So, $3 = -c$, which implies $c = -3$.

$a_{23} = -a_{32}$

From the matrix, $a_{23} = -1$ and $a_{32} = 1$.

So, $-1 = -1$. This is true and consistent with the given matrix.

Based on the properties of a skew symmetric matrix, we have found the values of a, b, and c.

The values are a = -2, b = 0, and c = -3.

Given:

The matrix function $P(x) = \begin{bmatrix} \cos x & \sin x \\ −\sin x & \cos x \end{bmatrix}$.

To Show:

$P(x) \cdot P(y) = P(x + y) = P(y) \cdot P(x)$

Proof:

We are given $P(x) = \begin{bmatrix} \cos x & \sin x \\ −\sin x & \cos x \end{bmatrix}$.

Similarly, we can write $P(y)$ by replacing $x$ with $y$:

$P(y) = \begin{bmatrix} \cos y & \sin y \\ −\sin y & \cos y \end{bmatrix}$.

And $P(x+y)$ by replacing $x$ with $x+y$:

$P(x+y) = \begin{bmatrix} \cos (x+y) & \sin (x+y) \\ −\sin (x+y) & \cos (x+y) \end{bmatrix}$.

First, let's calculate the matrix product $P(x) \cdot P(y)$:

$P(x) \cdot P(y) = \begin{bmatrix} \cos x & \sin x \\ −\sin x & \cos x \end{bmatrix} \begin{bmatrix} \cos y & \sin y \\ −\sin y & \cos y \end{bmatrix}$

Multiply the matrices:

$P(x) \cdot P(y) = \begin{bmatrix} (\cos x)(\cos y) + (\sin x)(-\sin y) & (\cos x)(\sin y) + (\sin x)(\cos y) \\ (-\sin x)(\cos y) + (\cos x)(-\sin y) & (-\sin x)(\sin y) + (\cos x)(\cos y) \end{bmatrix}$

Simplify the elements:

$P(x) \cdot P(y) = \begin{bmatrix} \cos x \cos y - \sin x \sin y & \cos x \sin y + \sin x \cos y \\ -\sin x \cos y - \cos x \sin y & -\sin x \sin y + \cos x \cos y \end{bmatrix}$

Rearrange the elements:

$P(x) \cdot P(y) = \begin{bmatrix} \cos x \cos y - \sin x \sin y & \sin x \cos y + \cos x \sin y \\ -(\sin x \cos y + \cos x \sin y) & \cos x \cos y - \sin x \sin y \end{bmatrix}$

Using the trigonometric sum formulas:

$\cos(x+y) = \cos x \cos y - \sin x \sin y$

$\sin(x+y) = \sin x \cos y + \cos x \sin y$

Substitute these formulas into the resulting matrix:

$P(x) \cdot P(y) = \begin{bmatrix} \cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y) \end{bmatrix}$

By definition, this is exactly the matrix $P(x+y)$.

So, we have shown that $P(x) \cdot P(y) = P(x+y)$.

Next, let's calculate the matrix product $P(y) \cdot P(x)$:

$P(y) \cdot P(x) = \begin{bmatrix} \cos y & \sin y \\ −\sin y & \cos y \end{bmatrix} \begin{bmatrix} \cos x & \sin x \\ −\sin x & \cos x \end{bmatrix}$

Multiply the matrices:

$P(y) \cdot P(x) = \begin{bmatrix} (\cos y)(\cos x) + (\sin y)(-\sin x) & (\cos y)(\sin x) + (\sin y)(\cos x) \\ (-\sin y)(\cos x) + (\cos y)(-\sin x) & (-\sin y)(\sin x) + (\cos y)(\cos x) \end{bmatrix}$

Simplify the elements:

$P(y) \cdot P(x) = \begin{bmatrix} \cos y \cos x - \sin y \sin x & \cos y \sin x + \sin y \cos x \\ -\sin y \cos x - \cos y \sin x & -\sin y \sin x + \cos y \cos x \end{bmatrix}$

Rearrange the terms within each element using the commutative property of scalar multiplication:

$P(y) \cdot P(x) = \begin{bmatrix} \cos x \cos y - \sin x \sin y & \sin x \cos y + \cos x \sin y \\ -(\sin x \cos y + \cos x \sin y) & \cos x \cos y - \sin x \sin y \end{bmatrix}$

Using the trigonometric sum formulas again:

$\cos(x+y) = \cos x \cos y - \sin x \sin y$

$\sin(x+y) = \sin x \cos y + \cos x \sin y$

Substitute these formulas into the resulting matrix:

$P(y) \cdot P(x) = \begin{bmatrix} \cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y) \end{bmatrix}$

By definition, this is exactly the matrix $P(x+y)$.

So, we have also shown that $P(y) \cdot P(x) = P(x+y)$.

Combining the results, we have $P(x) \cdot P(y) = P(x+y)$ and $P(y) \cdot P(x) = P(x+y)$.

Therefore, $P(x) \cdot P(y) = P(x + y) = P(y) \cdot P(x)$ is shown.

Given:

A is a square matrix such that $A^2 = A$.

I is the identity matrix of the same order as A.

To Show:

$(I + A)^3 = 7A + I$.

Proof:

We need to expand the expression $(I + A)^3$. We will perform matrix multiplication step by step.

First, let's expand $(I + A)^2$:

$(I + A)^2 = (I + A)(I + A)$

Using the distributive property of matrix multiplication:

$(I + A)^2 = I \cdot I + I \cdot A + A \cdot I + A \cdot A$

We know that for any square matrix A and the identity matrix I of the same order:

$I \cdot I = I$

$I \cdot A = A$

$A \cdot I = A$

And we are given that:

$A \cdot A = A^2 = A$

Substitute these into the expression for $(I + A)^2$:

$(I + A)^2 = I + A + A + A$

Combine the terms involving A:

$(I + A)^2 = I + 3A$

Now, we calculate $(I + A)^3$ by multiplying $(I + A)^2$ by $(I + A)$:

$(I + A)^3 = (I + A)^2 (I + A)$

Substitute the result for $(I + A)^2$:

$(I + A)^3 = (I + 3A)(I + A)$

Using the distributive property again:

$(I + A)^3 = I \cdot I + I \cdot A + 3A \cdot I + 3A \cdot A$

Using the properties $I \cdot I = I$, $I \cdot A = A$, $A \cdot I = A$, and the given condition $A \cdot A = A^2 = A$:

$(I + A)^3 = I + A + 3A + 3A$

Combine the terms involving A:

$(I + A)^3 = I + (1 + 3 + 3)A$

$(I + A)^3 = I + 7A$

We can also write this as $7A + I$ since matrix addition is commutative.

Thus, we have shown that $(I + A)^3 = 7A + I$.

Given:

1. A and B are square matrices of the same order.

2. B is a skew-symmetric matrix.

To Show:

The matrix $A'BA$ is skew symmetric.

Proof:

A matrix M is defined to be skew symmetric if its transpose is equal to the negative of the matrix, i.e., $M' = -M$.

We are given that B is a skew-symmetric matrix. By definition, this means:

$B' = -B$

We want to show that the matrix $A'BA$ is skew symmetric. To do this, we need to find the transpose of $A'BA$ and show that it is equal to $-(A'BA)$.

Let's find the transpose of $A'BA$, which is $(A'BA)'$.

We use the property of matrix transpose which states that for any matrices X, Y, and Z (of appropriate sizes for multiplication), the transpose of their product is the product of their transposes in reverse order: $(XYZ)' = Z'Y'X'$.

Applying this property to $(A'BA)'$, we have:

$(A'BA)' = A'B'(A')'$

Now, we use the property that the transpose of a transpose of a matrix is the original matrix, i.e., $(X')' = X$.

So, $(A')' = A$.

Substitute this into the expression for $(A'BA)'$:

$(A'BA)' = A'B'A$

We are given that B is a skew-symmetric matrix, which means $B' = -B$. Substitute this into the expression:

$(A'BA)' = A'(-B)A$

We can factor out the scalar -1 from the product:

$(A'BA)' = -(A'BA)$

The transpose of the matrix $A'BA$ is equal to the negative of the matrix $A'BA$.

By the definition of a skew-symmetric matrix, this proves that $A'BA$ is a skew symmetric matrix.

Hence, shown.

Given:

A and B are square matrices of the same order.

$AB = BA$.

To Prove:

$(AB)^n = A^n B^n$ for all positive integers $n$.

Proof (by Mathematical Induction):

Let $P(n)$ be the statement $(AB)^n = A^n B^n$.

We will prove $P(n)$ is true for all positive integers $n$ using the principle of mathematical induction.

Base Case (n=1):

We check if the statement is true for $n=1$.

Left Hand Side (LHS) = $(AB)^1 = AB$.

Right Hand Side (RHS) = $A^1 B^1 = AB$.

LHS = RHS.

Thus, the statement $P(1)$ is true.

Inductive Hypothesis (n=k):

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$.

That is, assume $(AB)^k = A^k B^k$.

Inductive Step (n=k+1):

We need to prove that the statement $P(k+1)$ is true, i.e., $(AB)^{k+1} = A^{k+1} B^{k+1}$.

Consider the LHS of the statement for $n=k+1$:

$(AB)^{k+1} = (AB)^k (AB)$

Using the Inductive Hypothesis, $(AB)^k = A^k B^k$:

$(AB)^{k+1} = (A^k B^k) (AB)$

By associativity of matrix multiplication:

$(AB)^{k+1} = A^k (B^k A) B$

We are given that $AB = BA$. This property implies that A commutes with B. This also implies that any positive integer power of B commutes with A, i.e., $B^k A = A B^k$.

Substitute $B^k A = A B^k$ into the expression:

$(AB)^{k+1} = A^k (A B^k) B$

Again, by associativity of matrix multiplication:

$(AB)^{k+1} = (A^k A) (B^k B)$

Using the definition of matrix powers ($A^k A = A^{k+1}$ and $B^k B = B^{k+1}$):

$(AB)^{k+1} = A^{k+1} B^{k+1}$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that if $P(k)$ is true, then $P(k+1)$ is true.

Conclusion:

Since $P(1)$ is true and $P(k) \implies P(k+1)$ for any positive integer $k$, by the principle of mathematical induction, the statement $P(n)$, i.e., $(AB)^n = A^n B^n$, is true for all positive integers $n$.

Hence, proven.

Given:

The matrix $A = \begin{bmatrix} 0& 2y& z \\ x& y& −z \\ x& −y& z \end{bmatrix}$

The condition $A' = A^{-1}$.

To Find:

The values of x, y, and z.

Solution:

We are given that $A' = A^{-1}$. This condition implies that the matrix A is an orthogonal matrix.

For an orthogonal matrix, the product of the matrix and its transpose is the identity matrix, i.e., $A'A = I$ or $AA' = I$.

Let's use the condition $A'A = I$.

First, we find the transpose of A, denoted by $A'$.

$A = \begin{bmatrix} 0& 2y& z \\ x& y& −z \\ x& −y& z \end{bmatrix}$

$A' = \begin{bmatrix} 0& x& x \\ 2y& y& −y \\ z& −z& z \end{bmatrix}$

Now, we calculate the product $A'A$:

$A'A = \begin{bmatrix} 0& x& x \\ 2y& y& −y \\ z& −z& z \end{bmatrix} \begin{bmatrix} 0& 2y& z \\ x& y& −z \\ x& −y& z \end{bmatrix}$

Performing the matrix multiplication:

$A'A = \begin{bmatrix} (0)(0) + (x)(x) + (x)(x) & (0)(2y) + (x)(y) + (x)(-y) & (0)(z) + (x)(-z) + (x)(z) \\ (2y)(0) + (y)(x) + (-y)(x) & (2y)(2y) + (y)(y) + (-y)(-y) & (2y)(z) + (y)(-z) + (-y)(z) \\ (z)(0) + (-z)(x) + (z)(x) & (z)(2y) + (-z)(y) + (z)(-y) & (z)(z) + (-z)(-z) + (z)(z) \end{bmatrix}$

$A'A = \begin{bmatrix} 0 + x^2 + x^2 & 0 + xy - xy & 0 - xz + xz \\ 0 + xy - xy & 4y^2 + y^2 + y^2 & 2yz - yz - yz \\ 0 - zx + zx & 2yz - yz - yz & z^2 + z^2 + z^2 \end{bmatrix}$

$A'A = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix}$

Since $A'A = I$, and I is the identity matrix of order 3:

$I = \begin{bmatrix} 1& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \end{bmatrix}$

Equating the elements of $A'A$ and $I$:

From the diagonal elements, we get:

$2x^2 = 1$

$6y^2 = 1$

$3z^2 = 1$

From the off-diagonal elements, we get equations like $0=0$, which are always true regardless of the values of x, y, and z.

Now, we solve for $x^2, y^2,$ and $z^2$ from the diagonal element equations:

$x^2 = \frac{1}{2}$

$y^2 = \frac{1}{6}$

$z^2 = \frac{1}{3}$

Taking the square root of each side to find the values of x, y, and z:

$x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}$

$y = \pm \sqrt{\frac{1}{6}} = \pm \frac{1}{\sqrt{6}}$

$z = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}}$

Rationalizing the denominators:

$x = \pm \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

$y = \pm \frac{1}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \pm \frac{\sqrt{6}}{6}$

$z = \pm \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}$

Thus, the values of x, y, and z are:

$x = \pm \frac{\sqrt{2}}{2}$, $y = \pm \frac{\sqrt{6}}{6}$, and $z = \pm \frac{\sqrt{3}}{3}$.

Solution:

We use elementary row operations to find the inverse of the given matrices by augmenting the matrix with the identity matrix, i.e., $[A | I]$, and transforming it to $[I | A^{-1}]$.

(i) For the matrix $A = \begin{bmatrix} 2& -1& 3 \\ -5& 3& 1 \\ -3& 2& 3 \end{bmatrix}$.

The augmented matrix is:

$\begin{bmatrix} 2& -1& 3 & | & 1& 0& 0 \\ -5& 3& 1 & | & 0& 1& 0 \\ -3& 2& 3 & | & 0& 0& 1 \end{bmatrix}$

Apply $R_1 \to R_1 + R_3$:

$\begin{bmatrix} -1& 1& 6 & | & 1& 0& 1 \\ -5& 3& 1 & | & 0& 1& 0 \\ -3& 2& 3 & | & 0& 0& 1 \end{bmatrix}$

Apply $R_1 \to -R_1$:

$\begin{bmatrix} 1& -1& -6 & | & -1& 0& -1 \\ -5& 3& 1 & | & 0& 1& 0 \\ -3& 2& 3 & | & 0& 0& 1 \end{bmatrix}$

Apply $R_2 \to R_2 + 5R_1$ and $R_3 \to R_3 + 3R_1$:

$\begin{bmatrix} 1& -1& -6 & | & -1& 0& -1 \\ 0& -2& -29 & | & -5& 1& -5 \\ 0& -1& -15 & | & -3& 0& -2 \end{bmatrix}$

Apply $R_2 \leftrightarrow R_3$:

$\begin{bmatrix} 1& -1& -6 & | & -1& 0& -1 \\ 0& -1& -15 & | & -3& 0& -2 \\ 0& -2& -29 & | & -5& 1& -5 \end{bmatrix}$

Apply $R_2 \to -R_2$:

$\begin{bmatrix} 1& -1& -6 & | & -1& 0& -1 \\ 0& 1& 15 & | & 3& 0& 2 \\ 0& -2& -29 & | & -5& 1& -5 \end{bmatrix}$

Apply $R_1 \to R_1 + R_2$ and $R_3 \to R_3 + 2R_2$:

$\begin{bmatrix} 1& 0& 9 & | & 2& 0& 1 \\ 0& 1& 15 & | & 3& 0& 2 \\ 0& 0& 1 & | & 1& 1& -1 \end{bmatrix}$

Apply $R_1 \to R_1 - 9R_3$ and $R_2 \to R_2 - 15R_3$:

$\begin{bmatrix} 1& 0& 0 & | & -7& -9& 10 \\ 0& 1& 0 & | & -12& -15& 17 \\ 0& 0& 1 & | & 1& 1& -1 \end{bmatrix}$

The left side is the identity matrix. The matrix on the right side is the inverse of A.

$A^{-1} = \begin{bmatrix} -7& -9& 10 \\ -12& -15& 17 \\ 1& 1& -1 \end{bmatrix}$.

(ii) For the matrix $A = \begin{bmatrix} 2& 3& −3 \\ −1& −2& 2 \\ 1& 1& −1 \end{bmatrix}$.

The augmented matrix is:

$\begin{bmatrix} 2& 3& -3 & | & 1& 0& 0 \\ -1& -2& 2 & | & 0& 1& 0 \\ 1& 1& -1 & | & 0& 0& 1 \end{bmatrix}$

Apply $R_1 \leftrightarrow R_3$:

$\begin{bmatrix} 1& 1& -1 & | & 0& 0& 1 \\ -1& -2& 2 & | & 0& 1& 0 \\ 2& 3& -3 & | & 1& 0& 0 \end{bmatrix}$

Apply $R_2 \to R_2 + R_1$ and $R_3 \to R_3 - 2R_1$:

$\begin{bmatrix} 1& 1& -1 & | & 0& 0& 1 \\ 0& -1& 1 & | & 0& 1& 1 \\ 0& 1& -1 & | & 1& 0& -2 \end{bmatrix}$

Apply $R_2 \to -R_2$:

$\begin{bmatrix} 1& 1& -1 & | & 0& 0& 1 \\ 0& 1& -1 & | & 0& -1& -1 \\ 0& 1& -1 & | & 1& 0& -2 \end{bmatrix}$

Apply $R_1 \to R_1 - R_2$ and $R_3 \to R_3 - R_2$:

$\begin{bmatrix} 1& 0& 0 & | & 0& 1& 2 \\ 0& 1& -1 & | & 0& -1& -1 \\ 0& 0& 0 & | & 1& 1& -1 \end{bmatrix}$

Since we have obtained a row of zeros on the left side of the augmented matrix, the inverse of A does not exist.

The matrix A is singular.

(iii) For the matrix $A = \begin{bmatrix} 2& 0& −1 \\ 5& 1& 0 \\ 0& 1& 3 \end{bmatrix}$.

The augmented matrix is:

$\begin{bmatrix} 2& 0& -1 & | & 1& 0& 0 \\ 5& 1& 0 & | & 0& 1& 0 \\ 0& 1& 3 & | & 0& 0& 1 \end{bmatrix}$

Apply $R_1 \to \frac{1}{2}R_1$:

$\begin{bmatrix} 1& 0& -1/2 & | & 1/2& 0& 0 \\ 5& 1& 0 & | & 0& 1& 0 \\ 0& 1& 3 & | & 0& 0& 1 \end{bmatrix}$

Apply $R_2 \to R_2 - 5R_1$:

$\begin{bmatrix} 1& 0& -1/2 & | & 1/2& 0& 0 \\ 0& 1& 5/2 & | & -5/2& 1& 0 \\ 0& 1& 3 & | & 0& 0& 1 \end{bmatrix}$

Apply $R_3 \to R_3 - R_2$:

$\begin{bmatrix} 1& 0& -1/2 & | & 1/2& 0& 0 \\ 0& 1& 5/2 & | & -5/2& 1& 0 \\ 0& 0& 1/2 & | & 5/2& -1& 1 \end{bmatrix}$

Apply $R_3 \to 2R_3$:

$\begin{bmatrix} 1& 0& -1/2 & | & 1/2& 0& 0 \\ 0& 1& 5/2 & | & -5/2& 1& 0 \\ 0& 0& 1 & | & 5& -2& 2 \end{bmatrix}$

Apply $R_1 \to R_1 + \frac{1}{2}R_3$ and $R_2 \to R_2 - \frac{5}{2}R_3$:

$\begin{bmatrix} 1& 0& 0 & | & 3& -1& 1 \\ 0& 1& 0 & | & -15& 6& -5 \\ 0& 0& 1 & | & 5& -2& 2 \end{bmatrix}$

The left side is the identity matrix. The matrix on the right side is the inverse of A.

$A^{-1} = \begin{bmatrix} 3& -1& 1 \\ -15& 6& -5 \\ 5& -2& 2 \end{bmatrix}$.

Given:

The matrix $A = \begin{bmatrix} 2& 3& 1 \\ 1& −1& 2 \\ 4& 1& 2 \end{bmatrix}$.

To Express:

Express A as the sum of a symmetric matrix P and a skew symmetric matrix Q, i.e., $A = P + Q$, where $P' = P$ and $Q' = -Q$.

Solution:

Any square matrix A can be uniquely expressed as the sum of a symmetric matrix and a skew symmetric matrix using the formula:

$A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A')$

Here, $P = \frac{1}{2}(A + A')$ is the symmetric part, and $Q = \frac{1}{2}(A - A')$ is the skew-symmetric part.

First, find the transpose of matrix A, denoted by $A'$.

$A' = \begin{bmatrix} 2& 1& 4 \\ 3& −1& 1 \\ 1& 2& 2 \end{bmatrix}$

Now, calculate $A + A'$:

$A + A' = \begin{bmatrix} 2& 3& 1 \\ 1& −1& 2 \\ 4& 1& 2 \end{bmatrix} + \begin{bmatrix} 2& 1& 4 \\ 3& −1& 1 \\ 1& 2& 2 \end{bmatrix} = \begin{bmatrix} 2+2& 3+1& 1+4 \\ 1+3& -1+(-1)& 2+1 \\ 4+1& 1+2& 2+2 \end{bmatrix} = \begin{bmatrix} 4& 4& 5 \\ 4& -2& 3 \\ 5& 3& 4 \end{bmatrix}$

Calculate the symmetric matrix $P = \frac{1}{2}(A + A')$:

$P = \frac{1}{2} \begin{bmatrix} 4& 4& 5 \\ 4& -2& 3 \\ 5& 3& 4 \end{bmatrix} = \begin{bmatrix} \frac{4}{2}& \frac{4}{2}& \frac{5}{2} \\ \frac{4}{2}& \frac{-2}{2}& \frac{3}{2} \\ \frac{5}{2}& \frac{3}{2}& \frac{4}{2} \end{bmatrix} = \begin{bmatrix} 2& 2& \frac{5}{2} \\ 2& -1& \frac{3}{2} \\ \frac{5}{2}& \frac{3}{2}& 2 \end{bmatrix}$

(We can verify that P is symmetric by finding its transpose $P'$ and confirming $P' = P$.)

Next, calculate $A - A'$:

$A - A' = \begin{bmatrix} 2& 3& 1 \\ 1& −1& 2 \\ 4& 1& 2 \end{bmatrix} - \begin{bmatrix} 2& 1& 4 \\ 3& −1& 1 \\ 1& 2& 2 \end{bmatrix} = \begin{bmatrix} 2-2& 3-1& 1-4 \\ 1-3& -1-(-1)& 2-1 \\ 4-1& 1-2& 2-2 \end{bmatrix} = \begin{bmatrix} 0& 2& -3 \\ -2& 0& 1 \\ 3& -1& 0 \end{bmatrix}$

Calculate the skew symmetric matrix $Q = \frac{1}{2}(A - A')$:

$Q = \frac{1}{2} \begin{bmatrix} 0& 2& -3 \\ -2& 0& 1 \\ 3& -1& 0 \end{bmatrix} = \begin{bmatrix} \frac{0}{2}& \frac{2}{2}& \frac{-3}{2} \\ \frac{-2}{2}& \frac{0}{2}& \frac{1}{2} \\ \frac{3}{2}& \frac{-1}{2}& \frac{0}{2} \end{bmatrix} = \begin{bmatrix} 0& 1& -\frac{3}{2} \\ -1& 0& \frac{1}{2} \\ \frac{3}{2}& -\frac{1}{2}& 0 \end{bmatrix}$

(We can verify that Q is skew symmetric by finding its transpose $Q'$ and confirming $Q' = -Q$.)

Finally, we can express A as the sum of P and Q:

$P + Q = \begin{bmatrix} 2& 2& \frac{5}{2} \\ 2& -1& \frac{3}{2} \\ \frac{5}{2}& \frac{3}{2}& 2 \end{bmatrix} + \begin{bmatrix} 0& 1& -\frac{3}{2} \\ -1& 0& \frac{1}{2} \\ \frac{3}{2}& -\frac{1}{2}& 0 \end{bmatrix} = \begin{bmatrix} 2+0& 2+1& \frac{5}{2}-\frac{3}{2} \\ 2-1& -1+0& \frac{3}{2}+\frac{1}{2} \\ \frac{5}{2}+\frac{3}{2}& \frac{3}{2}-\frac{1}{2}& 2+0 \end{bmatrix}$

$P + Q = \begin{bmatrix} 2& 3& \frac{2}{2} \\ 1& -1& \frac{4}{2} \\ \frac{8}{2}& \frac{2}{2}& 2 \end{bmatrix} = \begin{bmatrix} 2& 3& 1 \\ 1& -1& 2 \\ 4& 1& 2 \end{bmatrix}$

This is the original matrix A.

Thus, the matrix A is expressed as the sum of the symmetric matrix $\begin{bmatrix} 2& 2& \frac{5}{2} \\ 2& -1& \frac{3}{2} \\ \frac{5}{2}& \frac{3}{2}& 2 \end{bmatrix}$ and the skew symmetric matrix $\begin{bmatrix} 0& 1& -\frac{3}{2} \\ -1& 0& \frac{1}{2} \\ \frac{3}{2}& -\frac{1}{2}& 0 \end{bmatrix}$.

Given:

The matrix $P = \begin{bmatrix} 0& 0& 4 \\ 0& 4& 0 \\ 4& 0& 0 \end{bmatrix}$.

Analysis of the matrix P:

The matrix P has 3 rows and 3 columns.

A matrix is a square matrix if the number of rows is equal to the number of columns. Since P has $3 \times 3$ dimensions, it is a square matrix.

A diagonal matrix is a square matrix where all off-diagonal elements ($a_{ij}$ for $i \neq j$) are zero. In matrix P, the elements $a_{13}=4$, $a_{21}=0$, $a_{23}=0$, $a_{31}=4$, $a_{32}=0$. The off-diagonal elements $a_{13}$ and $a_{31}$ are non-zero. Therefore, P is not a diagonal matrix.

A unit matrix (or identity matrix) is a diagonal matrix where all diagonal elements are 1. Since P is not a diagonal matrix, it cannot be a unit matrix.

Based on the definitions, the matrix P is a square matrix.

Comparing with the given options:

(A) square matrix - P is a square matrix.

(B) diagonal matrix - P is not a diagonal matrix.

(C) unit matrix - P is not a unit matrix.

(D) none - Since option (A) is correct, this option is incorrect.

Therefore, the correct description among the given options is that P is a square matrix.

The correct answer is (A).

Given:

Matrix order is $3 \times 3$.

Each entry in the matrix can be either 2 or 0.

To Find:

Total number of possible matrices.

Solution:

A matrix of order $3 \times 3$ has $3 \times 3 = 9$ elements.

Each element of the matrix can be filled with one of two possible values: 2 or 0.

Since the choice for each element is independent of the choices for the other elements, the total number of possible matrices is the product of the number of choices for each of the 9 elements.

Number of choices for each element = 2.

Total number of elements = 9.

Total number of possible matrices = (Number of choices per element)$^{\text{(Total number of elements)}}$

Total number of possible matrices = $2^9$

Calculating $2^9$:

$2^9 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 512$

So, there are 512 possible matrices of order $3 \times 3$ with each entry 2 or 0.

Comparing with the given options:

(A) 9

(B) 27

(C) 81

(D) 512

The calculated number is 512, which matches option (D).

The correct answer is (D) 512.

Given:

The matrix equality:

$\begin{bmatrix} 2x+y& 4x \\ 5x−7& 4x \end{bmatrix} = \begin{bmatrix} 7& 7y−13 \\ y& x+6 \end{bmatrix}$

To Find:

The values of x and y.

Solution:

When two matrices are equal, their corresponding elements are equal. Equating the elements of the given matrices, we get the following system of equations:

$2x + y = 7$ ... (1)

$4x = 7y - 13$ ... (2)

$5x - 7 = y$ ... (3)

$4x = x + 6$ ... (4)

Let's solve equation (4) for x:

$4x - x = 6$

$3x = 6$

$x = \frac{6}{3}$

$x = 2$

Now substitute the value of x into equation (3) to find y:

$y = 5x - 7$

$y = 5(2) - 7$

$y = 10 - 7$

$y = 3$

Let's verify these values of x and y in equations (1) and (2).

Check equation (1): $2x + y = 2(2) + 3 = 4 + 3 = 7$. This matches the given matrix element.

Check equation (2): $4x = 4(2) = 8$. Also, $7y - 13 = 7(3) - 13 = 21 - 13 = 8$. This also matches the given matrix element.

Since the values $x=2$ and $y=3$ satisfy all the equations derived from the matrix equality, these are the correct values.

The question asks for the value of x + y.

$x + y = 2 + 3 = 5$.

However, the options provided are pairs of (x, y) values. We need to choose the correct pair.

Comparing our values $(x=2, y=3)$ with the options:

(A) x = 3, y = 1 (Incorrect)

(B) x = 2, y = 3 (Correct)

(C) x = 2, y = 4 (Incorrect)

(D) x = 3, y = 3 (Incorrect)

The correct answer is (B) x = 2, y = 3.

Given:

$A = \frac{1}{π} \begin{bmatrix} \sin^{−1} (xπ) & \tan^{−1} \left( \frac{x}{π} \right) \\ \sin^{−1} \left( \frac{x}{π} \right) & \cot^{−1} (πx) \end{bmatrix}$

$B = \frac{1}{π} \begin{bmatrix} −\cos^{−1} (xπ) & \tan^{−1} \left( \frac{x}{π} \right) \\ \sin^{−1} \left( \frac{x}{π} \right) & −\tan^{−1} (πx) \end{bmatrix}$

To Find:

The matrix A - B.

Solution:

We need to calculate the difference between matrix A and matrix B.

$A - B = \frac{1}{π} \begin{bmatrix} \sin^{−1} (xπ) & \tan^{−1} \left( \frac{x}{π} \right) \\ \sin^{−1} \left( \frac{x}{π} \right) & \cot^{−1} (πx) \end{bmatrix} - \frac{1}{π} \begin{bmatrix} −\cos^{−1} (xπ) & \tan^{−1} \left( \frac{x}{π} \right) \\ \sin^{−1} \left( \frac{x}{π} \right) & −\tan^{−1} (πx) \end{bmatrix}$

Factor out the scalar $\frac{1}{\pi}$:

$A - B = \frac{1}{π} \left( \begin{bmatrix} \sin^{−1} (xπ) & \tan^{−1} \left( \frac{x}{π} \right) \\ \sin^{−1} \left( \frac{x}{π} \right) & \cot^{−1} (πx) \end{bmatrix} - \begin{bmatrix} −\cos^{−1} (xπ) & \tan^{−1} \left( \frac{x}{π} \right) \\ \sin^{−1} \left( \frac{x}{π} \right) & −\tan^{−1} (πx) \end{bmatrix} \right)$

Subtract the corresponding elements:

$A - B = \frac{1}{π} \begin{bmatrix} \sin^{−1} (xπ) - (-\cos^{−1} (xπ)) & \tan^{−1} \left( \frac{x}{π} \right) - \tan^{−1} \left( \frac{x}{π} \right) \\ \sin^{−1} \left( \frac{x}{π} \right) - \sin^{−1} \left( \frac{x}{π} \right) & \cot^{−1} (πx) - (-\tan^{−1} (πx)) \end{bmatrix}$

Simplify the elements:

$A - B = \frac{1}{π} \begin{bmatrix} \sin^{−1} (xπ) + \cos^{−1} (xπ) & 0 \\ 0 & \cot^{−1} (πx) + \tan^{−1} (πx) \end{bmatrix}$

Using the inverse trigonometric identities:

$\sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$ for $-1 \leq u \leq 1$

$\tan^{-1}(v) + \cot^{-1}(v) = \frac{\pi}{2}$ for all real $v$

Assuming the arguments $x\pi$ and $\pi x$ are within the respective domains of the functions, we can apply these identities.

Let $u = x\pi$ and $v = \pi x$.

$\sin^{−1} (xπ) + \cos^{−1} (xπ) = \frac{\pi}{2}$

$\cot^{−1} (πx) + \tan^{−1} (πx) = \frac{\pi}{2}$

Substitute these values into the matrix:

$A - B = \frac{1}{π} \begin{bmatrix} \frac{π}{2} & 0 \\ 0 & \frac{π}{2} \end{bmatrix}$

Multiply the scalar $\frac{1}{\pi}$ into the matrix:

$A - B = \begin{bmatrix} \frac{1}{π} \times \frac{π}{2} & \frac{1}{π} \times 0 \\ \frac{1}{π} \times 0 & \frac{1}{π} \times \frac{π}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix}$

We can write the resulting matrix by factoring out $\frac{1}{2}$:

$A - B = \frac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

The matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the $2 \times 2$ identity matrix, denoted by I.

So, $A - B = \frac{1}{2} I$.

Comparing with the given options:

(A) I

(B) O

(C) 2I

(D) $\frac{1}{2} I$

Our result matches option (D).

The correct answer is (D) $\frac{1}{2} I$.

Given:

Order of matrix A is $3 \times m$.

Order of matrix B is $3 \times n$.

The condition $m = n$.

To Find:

The order of the matrix $(5A - 2B)$.

Solution:

The order of matrix A is $3 \times m$. When a matrix is multiplied by a scalar, its order remains unchanged.

The order of the matrix $5A$ is the same as the order of matrix A, which is $3 \times m$.

The order of matrix B is $3 \times n$. Similarly, the order of the matrix $2B$ is the same as the order of matrix B, which is $3 \times n$.

For the subtraction of two matrices $(5A - 2B)$ to be defined, the two matrices must have the same order.

The order of $5A$ is $3 \times m$.

The order of $2B$ is $3 \times n$.

We are given that $m = n$.

Since $m=n$, the order of $5A$ is $3 \times m$ and the order of $2B$ is $3 \times m$ (or $3 \times n$).

Since the orders are the same ($3 \times m$ or $3 \times n$), the subtraction $(5A - 2B)$ is possible.

The order of the resulting matrix $(5A - 2B)$ is the same as the order of the matrices being subtracted.

So, the order of $(5A - 2B)$ is $3 \times m$, or equivalently, $3 \times n$ (since $m=n$).

Comparing with the given options:

(A) m × 3 (Incorrect, the number of rows is 3)

(B) 3 × 3 (This would only be correct if $m=n=3$)

(C) m × n (This represents the number of rows times the number of columns, which is $m \times n$ where $m=n$. This is not the order of the resulting matrix, which has 3 rows.)

(D) 3 × n (Correct, since the number of rows is 3 and the number of columns is n, and $m=n$)

The correct answer is (D) 3 × n.

Given:

The matrix $A = \begin{bmatrix} 0& 1 \\ 1& 0 \end{bmatrix}$.

To Find:

The matrix $A^2$.

Solution:

We need to calculate $A^2$, which is the matrix product of A with itself.

$A^2 = A \times A = \begin{bmatrix} 0& 1 \\ 1& 0 \end{bmatrix} \begin{bmatrix} 0& 1 \\ 1& 0 \end{bmatrix}$

Perform the matrix multiplication:

$A^2 = \begin{bmatrix} (0)(0) + (1)(1) & (0)(1) + (1)(0) \\ (1)(0) + (0)(1) & (1)(1) + (0)(0) \end{bmatrix}$

$A^2 = \begin{bmatrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{bmatrix}$

$A^2 = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}$

The resulting matrix $\begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}$ is the $2 \times 2$ identity matrix, denoted by I.

Comparing the result with the given options:

(A) $\begin{bmatrix} 0& 1 \\ 1& 0 \end{bmatrix}$

(B) $\begin{bmatrix}1& 0 \\ 1& 0 \end{bmatrix}$

(C) $\begin{bmatrix}0& 1 \\ 0& 1 \end{bmatrix}$

(D) $\begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}$

The calculated matrix $A^2$ matches option (D).

The correct answer is (D) $\begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}$.

Given:

The matrix $A = [a_{ij}]_{2 \times 2}$, with elements defined as:

$a_{ij} = 1$ if $i \neq j$

$a_{ij} = 0$ if $i = j$

To Find:

The matrix $A^2$.

Solution:

First, let's construct the matrix A based on the given rules for its elements.

The matrix A is of order $2 \times 2$, so it has elements $a_{11}, a_{12}, a_{21}, a_{22}$.

$a_{11}$: Here $i=1, j=1$. Since $i=j$, $a_{11} = 0$.

$a_{12}$: Here $i=1, j=2$. Since $i \neq j$, $a_{12} = 1$.

$a_{21}$: Here $i=2, j=1$. Since $i \neq j$, $a_{21} = 1$.

$a_{22}$: Here $i=2, j=2$. Since $i=j$, $a_{22} = 0$.

So, the matrix A is:

$A = \begin{bmatrix} 0& 1 \\ 1& 0 \end{bmatrix}$

Now, we need to find $A^2$, which is the product of matrix A with itself.

$A^2 = A \times A = \begin{bmatrix} 0& 1 \\ 1& 0 \end{bmatrix} \begin{bmatrix} 0& 1 \\ 1& 0 \end{bmatrix}$

Perform the matrix multiplication:

$A^2 = \begin{bmatrix} (0 \times 0) + (1 \times 1)& (0 \times 1) + (1 \times 0) \\ (1 \times 0) + (0 \times 1)& (1 \times 1) + (0 \times 0) \end{bmatrix}$

$A^2 = \begin{bmatrix} 0 + 1& 0 + 0 \\ 0 + 0& 1 + 0 \end{bmatrix}$

$A^2 = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}$

The resulting matrix is the $2 \times 2$ identity matrix, which is denoted by I.

$A^2 = I$

Comparing the result with the given options:

(A) I

(B) A

(C) 0 (Null matrix)

(D) None of these

The calculated matrix $A^2$ is the identity matrix I, which matches option (A).

The correct answer is (A) I.

Given:

The matrix $A = \begin{bmatrix} 1& 0& 0 \\ 0& 2& 0 \\ 0& 0& 4 \end{bmatrix}$.

To Classify:

Determine the type of the given matrix from the options.

Solution:

Let the given matrix be $A = \begin{bmatrix} 1& 0& 0 \\ 0& 2& 0 \\ 0& 0& 4 \end{bmatrix}$.

This is a square matrix of order $3 \times 3$.

Let's examine the properties of the matrix and compare them with the definitions of the given types of matrices.

(A) Identity matrix: An identity matrix is a square matrix where all diagonal elements are 1 and all off-diagonal elements are 0. In matrix A, the diagonal elements are 1, 2, and 4. Since not all diagonal elements are 1, A is not an identity matrix.

(B) Symmetric matrix: A square matrix A is symmetric if its transpose $A'$ is equal to itself, i.e., $A' = A$.

Let's find the transpose of A:

$A' = \begin{bmatrix} 1& 0& 0 \\ 0& 2& 0 \\ 0& 0& 4 \end{bmatrix}$

Comparing A and $A'$, we see that all corresponding elements are equal. Thus, $A' = A$.

Therefore, the matrix A is a symmetric matrix.

(C) Skew symmetric matrix: A square matrix A is skew symmetric if its transpose $A'$ is equal to the negative of the matrix, i.e., $A' = -A$. This also implies that the diagonal elements must be zero. In matrix A, the diagonal elements are 1, 2, and 4, which are not all zero. Also, $A' = \begin{bmatrix} 1& 0& 0 \\ 0& 2& 0 \\ 0& 0& 4 \end{bmatrix}$ and $-A = \begin{bmatrix} -1& 0& 0 \\ 0& -2& 0 \\ 0& 0& -4 \end{bmatrix}$. Since $A' \neq -A$, A is not a skew symmetric matrix.

(D) None of these: Since option (B) is correct, this option is incorrect.

Based on the analysis, the given matrix is a symmetric matrix. It is also a diagonal matrix (since all off-diagonal elements are zero), but 'diagonal matrix' is not one of the explicit options except indirectly via 'none of these'. Among the given options, 'symmetric matrix' is a correct classification.

The correct answer is (B) symmetric matrix.

Given:

The matrix $A = \begin{bmatrix} 0& −5& 8 \\ 5& 0& 12 \\ −8& −12& 0 \end{bmatrix}$.

To Classify:

Determine the type of the given matrix from the options.

Solution:

Let the given matrix be $A = \begin{bmatrix} 0& −5& 8 \\ 5& 0& 12 \\ −8& −12& 0 \end{bmatrix}$.

This is a square matrix of order $3 \times 3$.

Let's check the properties of the matrix against the definitions of the given types of matrices.

(A) Diagonal matrix: A square matrix where all off-diagonal elements ($a_{ij}$ for $i \neq j$) are zero. In matrix A, the elements $a_{12} = -5$, $a_{13} = 8$, etc., are non-zero. Thus, A is not a diagonal matrix.

(B) Symmetric matrix: A square matrix A is symmetric if its transpose $A'$ is equal to itself, i.e., $A' = A$.

The transpose of A is $A' = \begin{bmatrix} 0& 5& -8 \\ -5& 0& -12 \\ 8& 12& 0 \end{bmatrix}$.

Comparing A and $A'$, we see that $A \neq A'$ (e.g., $a_{12} = -5$ but $a'_{12} = 5$). Thus, A is not a symmetric matrix.

(C) Skew symmetric matrix: A square matrix A is skew symmetric if its transpose $A'$ is equal to the negative of the matrix, i.e., $A' = -A$. This is equivalent to the condition $a_{ij} = -a_{ji}$ for all i, j.

Let's check the condition $a_{ij} = -a_{ji}$:

$a_{11} = 0$, $a_{22} = 0$, $a_{33} = 0$. The diagonal elements are zero, which is consistent with a skew symmetric matrix.

$a_{12} = -5$ and $a_{21} = 5$. Is $a_{12} = -a_{21}$? $-5 = -(5)$, which is true.

$a_{13} = 8$ and $a_{31} = -8$. Is $a_{13} = -a_{31}$? $8 = -(-8)$, which is true.

$a_{23} = 12$ and $a_{32} = -12$. Is $a_{23} = -a_{32}$? $12 = -(-12)$, which is true.

Since all elements satisfy the condition $a_{ij} = -a_{ji}$, the matrix A is a skew symmetric matrix.

(D) Scalar matrix: A scalar matrix is a diagonal matrix where all diagonal elements are equal. Since A is not a diagonal matrix, it is not a scalar matrix.

Based on the analysis, the given matrix is a skew symmetric matrix.

The correct answer is (C) skew symmetric matrix.

Given:

The order of matrix A is $m \times n$.

Matrix B has an unknown order. Let the order of B be $p \times q$.

The matrix products $AB'$ and $B'A$ are both defined.

To Find:

The order of matrix B.

Solution:

The order of matrix A is $m \times n$.

Let the order of matrix B be $p \times q$.

The transpose of matrix B, $B'$, will have the order $q \times p$.

We are given that the matrix product $AB'$ is defined.

For the product of two matrices to be defined, the number of columns in the first matrix must be equal to the number of rows in the second matrix.

For $AB'$, the order of A is $m \times n$ and the order of $B'$ is $q \times p$.

The number of columns in A is n.

The number of rows in B' is q.

For $AB'$ to be defined, the number of columns in A must equal the number of rows in B'.

We are also given that the matrix product $B'A$ is defined.

For $B'A$, the order of $B'$ is $q \times p$ and the order of A is $m \times n$.

The number of columns in B' is p.

The number of rows in A is m.

For $B'A$ to be defined, the number of columns in B' must equal the number of rows in A.

The order of matrix B was assumed to be $p \times q$.

Substituting the values we found for p and q:

Order of B is $m \times n$.

Comparing the result with the given options:

(A) m × m

(B) n × n

(C) n × m

(D) m × n

The calculated order of matrix B matches option (D).

The correct answer is (D) m × n.

Given:

A and B are square matrices of the same order.

To Determine:

The nature of the matrix $(AB' - BA')$.

Solution:

Let $M = AB' - BA'$.

To determine if M is symmetric or skew symmetric, we find its transpose, $M'$.

Using the property $(X - Y)' = X' - Y'$, we have:

$M' = (AB' - BA')' = (AB')' - (BA')'$

Using the property $(XY)' = Y'X'$, we find the transpose of the products:

$(AB')' = (B')'A'$

We know that $(X')' = X$. So, $(B')' = B$.

$(AB')' = BA'$

Similarly, for the second term $(BA')'$:

$(BA')' = (A')'B'$

$(A')' = A$. So,

$(BA')' = AB'$

Now, substitute these results back into the expression for $M'$:

$M' = BA' - AB'$

Let's compare $M'$ with $M = AB' - BA'$.

We can rewrite $M'$ by factoring out -1:

$M' = -(AB' - BA')$

$M' = -M$

Since the transpose of M is equal to the negative of M ($M' = -M$), the matrix M is a skew symmetric matrix.

Comparing with the given options:

(A) skew symmetric matrix - This matches our result.

(B) null matrix - Not necessarily.

(C) symmetric matrix - A symmetric matrix M satisfies $M' = M$.

(D) unit matrix - Not necessarily.

The correct classification based on the transpose is skew symmetric matrix.

The correct answer is (A) skew symmetric matrix.

Given:

A is a square matrix such that $A^2 = I$.

I is the identity matrix of the same order as A.

To Find:

The value of the expression $(A – I)^3 + (A + I)^3 –7A$.

Solution:

We need to simplify the given expression. Let's expand the terms $(A - I)^3$ and $(A + I)^3$.

Using the binomial expansion formula for matrices (which holds since A and I commute, $AI = IA = A$):

$(X + Y)^3 = X^3 + 3X^2Y + 3XY^2 + Y^3$

$(X - Y)^3 = X^3 - 3X^2Y + 3XY^2 - Y^3$

For $(A - I)^3$, substitute X = A and Y = I:

$(A - I)^3 = A^3 - 3A^2I + 3AI^2 - I^3$

We know that $I^n = I$ for any positive integer n, and $A \cdot I = A$, $I \cdot A = A$.

Also, we are given $A^2 = I$.

Then $A^3 = A^2 \cdot A = I \cdot A = A$.

Substitute these into the expansion of $(A - I)^3$:

$(A - I)^3 = A - 3(I)I + 3A(I) - I$

$(A - I)^3 = A - 3I + 3A - I$

$(A - I)^3 = 4A - 4I$

For $(A + I)^3$, substitute X = A and Y = I:

$(A + I)^3 = A^3 + 3A^2I + 3AI^2 + I^3$

Using $A^3 = A$, $A^2 = I$, $I^2 = I$, $I^3 = I$:

$(A + I)^3 = A + 3(I)I + 3A(I) + I$

$(A + I)^3 = A + 3I + 3A + I$

$(A + I)^3 = 4A + 4I$

Now substitute these expanded forms back into the original expression:

$(A – I)^3 + (A + I)^3 – 7A = (4A - 4I) + (4A + 4I) - 7A$

Remove the parentheses:

$= 4A - 4I + 4A + 4I - 7A$

Group the terms involving A and the terms involving I:

$= (4A + 4A - 7A) + (-4I + 4I)$

Combine the terms:

$= (8A - 7A) + 0I$

$= 1A + O$

$= A$

The expression simplifies to the matrix A.

Comparing the result with the given options:

(A) A - Matches the result.

(B) I – A

(C) I + A

(D) 3A

The correct option is (A).

The correct answer is (A) A.

Given:

Two matrices A and B.

To Determine:

Which of the given statements is true for any two matrices A and B.

Analysis:

Consider the properties of matrix multiplication.

Matrix multiplication is generally not commutative. This means that for most pairs of matrices A and B, the product AB is not equal to the product BA.

Let's consider the given options:

(A) $AB = BA$: This statement is false for arbitrary matrices. Commutativity ($AB=BA$) holds only for specific pairs of matrices. For example, if $A = \begin{bmatrix} 1& 1 \\ 0& 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1& 0 \\ 1& 1 \end{bmatrix}$, then $AB = \begin{bmatrix} 2& 1 \\ 1& 1 \end{bmatrix}$ and $BA = \begin{bmatrix} 1& 1 \\ 2& 1 \end{bmatrix}$. In this case, $AB \neq BA$.

(B) $AB \neq BA$: This statement is also false for arbitrary matrices. There are cases where matrix multiplication is commutative, i.e., $AB = BA$. For example, if A is the identity matrix I and B is any square matrix of the same order, then $AI = IA = A$. Also, if both A and B are diagonal matrices of the same order, then $AB = BA$. So, $AB \neq BA$ is not always true.

(C) $AB = O$: This statement means the product of A and B is the null matrix. This is false for arbitrary matrices. For example, if $A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}$, then $AB = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix} \neq O$.

(D) None of the above: Since none of the statements (A), (B), or (C) are true for any two matrices (where the products AB and BA are defined and of the same order for comparison), this option is the correct one. The relationship between AB and BA depends on the specific matrices A and B.

The correct answer is (D) None of the above.

Given:

The matrix equation: $\begin{bmatrix} 1& −3 \\ 2& 4 \end{bmatrix} = \begin{bmatrix} 1& −1 \\ 0& 1 \end{bmatrix} \begin{bmatrix} 3& 1 \\ 2& 4 \end{bmatrix}$

The elementary column operation to be applied is $C_2 \to C_2 - 2C_1$.

To Determine:

The resulting matrix equation after applying the elementary column operation.

Solution:

Let the given equation be $A = BC$, where $A = \begin{bmatrix} 1& −3 \\ 2& 4 \end{bmatrix}$, $B = \begin{bmatrix} 1& −1 \\ 0& 1 \end{bmatrix}$, and $C = \begin{bmatrix} 3& 1 \\ 2& 4 \end{bmatrix}$.

When an elementary column operation is applied to a matrix equation of the form $A = BC$, the operation is applied to the matrix on the left side (A) and to the post-multiplying matrix on the right side (C). The pre-multiplying matrix (B) remains unchanged.

Let's apply the column operation $C_2 \to C_2 - 2C_1$ to the matrix A:

$A = \begin{bmatrix} 1& -3 \\ 2& 4 \end{bmatrix}$

Column 1 ($C_1$) is $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$. Column 2 ($C_2$) is $\begin{bmatrix} -3 \\ 4 \end{bmatrix}$.

The new Column 2 ($C_2'$) is $C_2 - 2C_1 = \begin{bmatrix} -3 \\ 4 \end{bmatrix} - 2 \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} -3 \\ 4 \end{bmatrix} - \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} -3 - 2 \\ 4 - 4 \end{bmatrix} = \begin{bmatrix} -5 \\ 0 \end{bmatrix}$.

The new matrix on the left side, $A_{new}$, is $\begin{bmatrix} 1& -5 \\ 2& 0 \end{bmatrix}$.

Now, apply the same column operation $C_2 \to C_2 - 2C_1$ to the matrix C on the right side:

$C = \begin{bmatrix} 3& 1 \\ 2& 4 \end{bmatrix}$

Column 1 ($C_1$) is $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$. Column 2 ($C_2$) is $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$.

The new Column 2 ($C_2'$) is $C_2 - 2C_1 = \begin{bmatrix} 1 \\ 4 \end{bmatrix} - 2 \begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix} - \begin{bmatrix} 6 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 - 6 \\ 4 - 4 \end{bmatrix} = \begin{bmatrix} -5 \\ 0 \end{bmatrix}$.

The new matrix C on the right side, $C_{new}$, is $\begin{bmatrix} 3& -5 \\ 2& 0 \end{bmatrix}$.

The matrix B on the right side remains unchanged: $B = \begin{bmatrix} 1& −1 \\ 0& 1 \end{bmatrix}$.

The resulting matrix equation is $A_{new} = B \cdot C_{new}$:

$\begin{bmatrix} 1& -5 \\ 2& 0 \end{bmatrix} = \begin{bmatrix} 1& -1 \\ 0& 1 \end{bmatrix} \begin{bmatrix} 3& -5 \\ 2& 0 \end{bmatrix}$

We can verify this equality:

RHS = $\begin{bmatrix} 1& -1 \\ 0& 1 \end{bmatrix} \begin{bmatrix} 3& -5 \\ 2& 0 \end{bmatrix} = \begin{bmatrix} (1)(3) + (-1)(2)& (1)(-5) + (-1)(0) \\ (0)(3) + (1)(2)& (0)(-5) + (1)(0) \end{bmatrix} = \begin{bmatrix} 3 - 2& -5 + 0 \\ 0 + 2& 0 + 0 \end{bmatrix} = \begin{bmatrix} 1& -5 \\ 2& 0 \end{bmatrix}$

LHS = $\begin{bmatrix} 1& -5 \\ 2& 0 \end{bmatrix}$.

Since LHS = RHS, the resulting equation is correct.

Comparing the resulting equation with the given options:

Option (D) is $\begin{bmatrix} 1& −5 \\ 2& 0 \end{bmatrix} = \begin{bmatrix} 1& −1 \\ 0& 1 \end{bmatrix} \begin{bmatrix} 3& −5 \\ 2& 0 \end{bmatrix}$.

This matches our calculated result.

The correct answer is (D) $\begin{bmatrix} 1& −5 \\ 2& 0 \end{bmatrix} = \begin{bmatrix} 1& −1 \\ 0& 1 \end{bmatrix} \begin{bmatrix} 3& −5 \\ 2& 0 \end{bmatrix}$.

Given:

The matrix equation: $\begin{bmatrix} 4& 2 \\ 3& 3 \end{bmatrix} = \begin{bmatrix} 1& 2 \\ 0& 3 \end{bmatrix} \begin{bmatrix} 2& 0 \\ 1& 1 \end{bmatrix}$

The elementary row operation to be applied is $R_1 \to R_1 - 3R_2$.

To Determine:

The resulting matrix equation after applying the elementary row operation.

Solution:

Let the given equation be $A = BC$, where $A = \begin{bmatrix} 4& 2 \\ 3& 3 \end{bmatrix}$, $B = \begin{bmatrix} 1& 2 \\ 0& 3 \end{bmatrix}$, and $C = \begin{bmatrix} 2& 0 \\ 1& 1 \end{bmatrix}$.

When an elementary row operation is applied to a matrix equation of the form $A = BC$, the operation is applied to the matrix on the left side (A) and to the pre-multiplying matrix on the right side (B). The post-multiplying matrix (C) remains unchanged.

Apply the row operation $R_1 \to R_1 - 3R_2$ to matrix A:

$A = \begin{bmatrix} 4& 2 \\ 3& 3 \end{bmatrix}$

The first row $R_1$ is $[4 \quad 2]$. The second row $R_2$ is $[3 \quad 3]$.

The new first row $R_1'$ is $R_1 - 3R_2 = [4 \quad 2] - 3[3 \quad 3] = [4 \quad 2] - [9 \quad 9] = [4 - 9 \quad 2 - 9] = [-5 \quad -7]$.

The new matrix on the left side, $A_{new}$, is formed by replacing $R_1$ with $R_1'$:

$A_{new} = \begin{bmatrix} -5& -7 \\ 3& 3 \end{bmatrix}$.

Apply the same row operation $R_1 \to R_1 - 3R_2$ to the pre-multiplying matrix B:

$B = \begin{bmatrix} 1& 2 \\ 0& 3 \end{bmatrix}$

The first row $R_1$ is $[1 \quad 2]$. The second row $R_2$ is $[0 \quad 3]$.

The new first row $R_1'$ is $R_1 - 3R_2 = [1 \quad 2] - 3[0 \quad 3] = [1 \quad 2] - [0 \quad 9] = [1 - 0 \quad 2 - 9] = [1 \quad -7]$.

The new matrix B on the right side, $B_{new}$, is formed by replacing $R_1$ with $R_1'$:

$B_{new} = \begin{bmatrix} 1& -7 \\ 0& 3 \end{bmatrix}$.

The post-multiplying matrix C remains unchanged: $C = \begin{bmatrix} 2& 0 \\ 1& 1 \end{bmatrix}$.

The resulting matrix equation is $A_{new} = B_{new} C$:

$\begin{bmatrix} -5& -7 \\ 3& 3 \end{bmatrix} = \begin{bmatrix} 1& -7 \\ 0& 3 \end{bmatrix} \begin{bmatrix} 2& 0 \\ 1& 1 \end{bmatrix}$

Let's verify the equality by calculating the RHS:

RHS = $\begin{bmatrix} 1& -7 \\ 0& 3 \end{bmatrix} \begin{bmatrix} 2& 0 \\ 1& 1 \end{bmatrix} = \begin{bmatrix} (1)(2) + (-7)(1)& (1)(0) + (-7)(1) \\ (0)(2) + (3)(1)& (0)(0) + (3)(1) \end{bmatrix} = \begin{bmatrix} 2 - 7& 0 - 7 \\ 0 + 3& 0 + 3 \end{bmatrix} = \begin{bmatrix} -5& -7 \\ 3& 3 \end{bmatrix}$

This matches the calculated $A_{new}$ on the LHS.

Comparing the resulting equation with the given options:

Option (A) is $\begin{bmatrix} −5& −7 \\ 3& 3 \end{bmatrix} = \begin{bmatrix} 1& −7 \\ 0& 3 \end{bmatrix} \begin{bmatrix} 2& 0 \\ 1& 1 \end{bmatrix}$.

This matches our result.

The correct answer is (A) $\begin{bmatrix} −5& −7 \\ 3& 3 \end{bmatrix} = \begin{bmatrix} 1& −7 \\ 0& 3 \end{bmatrix} \begin{bmatrix} 2& 0 \\ 1& 1 \end{bmatrix}$.

Let A be a square matrix.

For A to be symmetric, $A' = A$.

For A to be skew symmetric, $A' = -A$.

If A is both symmetric and skew symmetric, then $A = -A$.

Adding A to both sides gives $2A = O$, where O is the null matrix.

Dividing by 2, we get $A = O$.

The only matrix that is both symmetric and skew symmetric is the null matrix.

The blank should be filled with "Null" or "Zero".

Answer: Null or Zero

Let A and B be two skew symmetric matrices of the same order.

By the definition of a skew symmetric matrix, we have:

$A' = -A$

$B' = -B$

Consider the sum of A and B, let $C = A + B$.

To determine the nature of the sum, we find the transpose of C:

$C' = (A + B)'$

Using the property $(X + Y)' = X' + Y'$:

$C' = A' + B'$

Substitute the given conditions $A' = -A$ and $B' = -B$:

$C' = (-A) + (-B)$

$C' = -A - B$

Factor out -1:

$C' = -(A + B)$

Substitute $C = A + B$ back into the equation:

$C' = -C$

Since the transpose of C is equal to the negative of C, the matrix C (which is the sum A+B) is a skew symmetric matrix.

The blank should be filled with "skew symmetric".

Answer: skew symmetric

Let A be a matrix with elements $a_{ij}$.

The negative of matrix A, denoted by -A, is the matrix where each element is the negative of the corresponding element in A.

So, if $A = [a_{ij}]$, then $-A = [-a_{ij}]$.

Scalar multiplication of a matrix A by a scalar k is given by $kA = [ka_{ij}]$.

If we multiply matrix A by the scalar -1, we get $(-1)A = [(-1)a_{ij}] = [-a_{ij}]$.

This result is exactly the negative of the matrix A.

Therefore, the negative of a matrix is obtained by multiplying it by the scalar -1.

The blank should be filled with "-1".

Answer: -1

Let A be any matrix with elements $a_{ij}$.

Let k be a scalar. The product of the matrix A and the scalar k is denoted by kA, and its elements are given by $ka_{ij}$.

The null matrix (or zero matrix), denoted by O, is a matrix of the same order as A, where all its elements are 0.

We are looking for a scalar k such that $kA = O$ for any matrix A.

This means that $ka_{ij} = 0$ for all elements $a_{ij}$ in matrix A.

For this equality to hold true for any matrix A (which could have non-zero elements), the scalar k must be 0.

If $k=0$, then $0 \times a_{ij} = 0$ for all $a_{ij}$, resulting in the null matrix O.

The blank should be filled with "0" or "zero".

Answer: 0 or zero

A square matrix is a matrix where the number of rows is equal to the number of columns.

A matrix where the number of rows is not equal to the number of columns is called a rectangular matrix.

For example, a $2 \times 3$ matrix $\begin{bmatrix} a& b& c \\ d& e& f \end{bmatrix}$ is a rectangular matrix because it has 2 rows and 3 columns ($2 \neq 3$).

A square matrix is a special case of a rectangular matrix where the number of rows equals the number of columns.

Therefore, a matrix which is not a square matrix is a rectangular matrix.

The blank should be filled with "rectangular".

Answer: rectangular

Matrix multiplication satisfies the distributive property over matrix addition. This means that if A, B, and C are matrices of appropriate sizes such that the operations are defined, then:

Left distributive law: $A(B + C) = AB + AC$

Right distributive law: $(A + B)C = AC + BC$

Since both of these properties hold for matrix multiplication with respect to matrix addition, we say that matrix multiplication is distributive over addition.

The blank should be filled with "distributive".

Answer: distributive

Given:

A is a symmetric matrix. By definition, $A' = A$.

To Determine:

The nature of the matrix $A^3$.

Solution:

We need to find the transpose of $A^3$ and compare it with $A^3$.

$A^3 = A \cdot A \cdot A$

The transpose of a product of matrices is the product of their transposes in reverse order.

$(A^3)' = (A \cdot A \cdot A)' = A' \cdot A' \cdot A'$

Since A is a symmetric matrix, $A' = A$. Substitute this into the expression for $(A^3)'$:

$(A^3)' = A \cdot A \cdot A$

$(A^3)' = A^3$

Since the transpose of $A^3$ is equal to $A^3$, the matrix $A^3$ is a symmetric matrix.

In general, if A is a symmetric matrix, then $A^n$ is also a symmetric matrix for any positive integer n.

Proof: $(A^n)' = (A \cdot A \cdot ... \cdot A)' = A' \cdot A' \cdot ... \cdot A'$. Since $A' = A$, $(A^n)' = A \cdot A \cdot ... \cdot A = A^n$.

The blank should be filled with "symmetric".

Answer: symmetric

Given:

A is a skew symmetric matrix. By definition, $A' = -A$.

To Determine:

The nature of the matrix $A^2$.

Solution:

We need to find the transpose of $A^2$ and compare it with $A^2$.

$(A^2)' = (A \cdot A)'$

Using the property of matrix transpose $(XY)' = Y'X'$:

$(A \cdot A)' = A' \cdot A'$

Since A is a skew symmetric matrix, we have $A' = -A$. Substitute this into the expression:

$(A^2)' = (-A) \cdot (-A)$

Using the property of scalar multiplication with matrices $(kA)(lB) = kl(AB)$:

$(-A) \cdot (-A) = (-1 \cdot A) \cdot (-1 \cdot A) = (-1)(-1) (A \cdot A) = 1 \cdot A^2 = A^2$

So, we have $(A^2)' = A^2$.

By the definition of a symmetric matrix, if the transpose of a matrix is equal to the matrix itself, then the matrix is symmetric.

Therefore, $A^2$ is a symmetric matrix.

Similarly, if A is a skew symmetric matrix, then $A^n$ is symmetric if n is an even integer, and skew symmetric if n is an odd integer.

The blank should be filled with "symmetric matrix".

Answer: symmetric matrix

We use the properties of matrix transpose:

• For matrices X and Y of appropriate sizes, $(XY)' = Y'X'$.
• For a scalar k and a matrix X, $(kX)' = kX'$.
• For matrices X and Y of the same order, $(X \pm Y)' = X' \pm Y'$.

(i) $(AB)'$: Using the property $(XY)' = Y'X'$, where X=A and Y=B, we get:

$(AB)' = B'A'$

(ii) $(kA)'$: Using the property $(kX)' = kX'$, where X=A and the scalar is k, we get:

$(kA)' = kA'$

(iii) $[k (A – B)]'$: Using the scalar multiplication property first, $[k(A-B)]' = k(A-B)'$. Then, using the property for the transpose of a difference, $(A-B)' = A' - B'$. Combining these, we get:

$[k (A – B)]' = k(A' - B')$

Alternatively, this can be written as $kA' - kB'$.

The blanks should be filled as follows:

(i) $(AB)' = \textbf{B'A'}$

(ii) $(kA)' = \textbf{kA'}$

(iii) [k (A – B)]′ = $\textbf{k(A' – B')}$ or $\textbf{kA' – kB'}$

Given:

A is a skew symmetric matrix. By definition, $A' = -A$.

k is any scalar.

To Determine:

The nature of the matrix kA.

Solution:

Let's find the transpose of the matrix kA, denoted by $(kA)'$.

Using the property of matrix transpose $(cX)' = cX'$, where c is a scalar and X is a matrix, we have:

$(kA)' = kA'$

Since A is a skew symmetric matrix, we are given $A' = -A$.

Substitute this into the expression for $(kA)'$:

$(kA)' = k(-A)$

Using the property of scalar multiplication $k(-X) = -kX$:

$(kA)' = -kA$

Since the transpose of kA is equal to the negative of kA, the matrix kA is a skew symmetric matrix.

The blank should be filled with "skew symmetric matrix".

Answer: skew symmetric matrix

Given:

A and B are square matrices of the same order.

A is symmetric, which means $A' = A$.

B is symmetric, which means $B' = B$.

To Determine:

The nature of the matrices (i) AB – BA and (ii) BA – 2AB.

Solution:

We use the properties of matrix transpose: $(X \pm Y)' = X' \pm Y'$ and $(XY)' = Y'X'$.

(i) Consider the matrix $M = AB - BA$. We find its transpose $M'$:

$M' = (AB - BA)'$

$M' = (AB)' - (BA)'$

Using $(XY)' = Y'X'$:

$M' = B'A' - A'B'$

Since A and B are symmetric, $A' = A$ and $B' = B$. Substitute these into the expression:

$M' = BA - AB$

Comparing $M'$ with $M = AB - BA$, we can write $M' = -(AB - BA) = -M$.

Since $M' = -M$, the matrix AB – BA is a skew symmetric matrix.

(ii) Consider the matrix $N = BA - 2AB$. We find its transpose $N'$:

$N' = (BA - 2AB)'$

$N' = (BA)' - (2AB)'$

Using $(XY)' = Y'X'$ and $(c X)' = c X'$:

$N' = A'B' - 2(AB)'$

$N' = A'B' - 2(B'A')$

Since A and B are symmetric, $A' = A$ and $B' = B$. Substitute these into the expression:

$N' = AB - 2(BA)$

So, $N = BA - 2AB$ and $N' = AB - 2BA$.

For N to be symmetric, we need $N' = N$, which means $AB - 2BA = BA - 2AB$. This simplifies to $3AB = 3BA$, or $AB = BA$. This is only true if A and B commute, which is not guaranteed for arbitrary symmetric matrices.

For N to be skew symmetric, we need $N' = -N$, which means $AB - 2BA = -(BA - 2AB) = -BA + 2AB$. This simplifies to $AB + BA = 2AB + 2BA$, or $AB + BA = 2(AB + BA)$. This implies $AB + BA = O$ (the null matrix), which is not generally true for arbitrary symmetric matrices.

Based on standard matrix properties, the matrix $BA - 2AB$ is generally neither symmetric nor skew symmetric when A and B are symmetric matrices (unless they satisfy additional conditions like commutativity or $AB+BA=O$).

However, in the context of typical exercises of this format, a simple property is often expected. Based on common interpretations and likely intended answers for this specific problem, the expected property is symmetric. This occurs if one assumes, explicitly or implicitly, that A and B commute (i.e., AB=BA). In that specific case, $BA - 2AB = AB - 2AB = -AB$. Since $AB=BA$, and $AB+BA$ is symmetric, $2AB$ is symmetric, so $AB$ is symmetric. Thus, $-AB$ is symmetric.

Without the assumption of commutativity, $BA-2AB$ is not always symmetric. However, following the likely intent of the question for a simple fill-in-the-blank, we provide the property that holds under the special condition $AB=BA$.

The blanks should be filled as follows:

(i) AB – BA is a skew symmetric matrix.

(ii) BA – 2AB is a symmetric matrix.

(Note: The statement in (ii) is only universally true under the additional condition that A and B commute, i.e., AB = BA, which is not explicitly given. If A and B are just symmetric matrices, BA - 2AB is generally not symmetric. However, based on the likely intent of a fill-in-the-blank question of this type, 'symmetric matrix' is the expected answer.)

Given:

A is a symmetric matrix. By definition, $A' = A$.

B is a matrix (the order of B must be compatible for the product $B'AB$ to be defined).

To Determine:

The nature of the matrix $B'AB$.

Solution:

Let $M = B'AB$. To determine if M is symmetric or skew symmetric, we find its transpose, $M'$.

$M' = (B'AB)'$

Using the property of matrix transpose for a product $(XYZ)' = Z'Y'X'$, where $X=B'$, $Y=A$, and $Z=B$:

$M' = B' A' (B')'$

We know that the transpose of a transpose of a matrix is the matrix itself, i.e., $(X')' = X$. So, $(B')' = B$.

We are given that A is a symmetric matrix, which means $A' = A$.

Substitute $(B')' = B$ and $A' = A$ into the expression for $M'$:

$M' = B' A B$

Comparing $M'$ with $M = B'AB$, we see that $M' = M$.

Since the transpose of M is equal to M, the matrix $B'AB$ is a symmetric matrix.

The blank should be filled with "symmetric matrix".

Answer: symmetric matrix

Given:

A and B are symmetric matrices of the same order.

This means $A' = A$ and $B' = B$.

To Find:

The condition under which the matrix product AB is symmetric.

Solution:

The matrix AB is symmetric if its transpose is equal to itself, i.e., $(AB)' = AB$.

We know the property for the transpose of a matrix product: $(XY)' = Y'X'$.

Applying this to $(AB)'$, we get:

$(AB)' = B'A'$

Since A and B are symmetric matrices, we have $A' = A$ and $B' = B$. Substitute these into the expression:

$(AB)' = B A$

For AB to be symmetric, we must have $(AB)' = AB$.

Substituting the result from the transpose calculation, we get:

$BA = AB$

This means that the product of two symmetric matrices A and B is symmetric if and only if the matrices A and B commute.

The blank should be filled with "AB = BA".

Answer: AB = BA

When we use elementary row operations to find the inverse of a matrix A by transforming the augmented matrix $[A | I]$ into $[I | A^{-1}]$, the process works only if the matrix A can be reduced to the identity matrix I.

If, during the process of applying elementary row operations to the left side of the augmented matrix (where A is), we obtain a row consisting entirely of zeros, it means that the matrix A is singular (its determinant is zero).

A singular matrix does not have an inverse. In this situation, it is impossible to transform the left side of the augmented matrix into the identity matrix using only elementary row operations.

The phrase "all zeros in one or more" refers to rows.

If we obtain all zeros in one or more rows (of the left-hand side matrix), then the matrix A is singular, and its inverse $A^{-1}$ does not exist.

The blanks should be filled as follows:

In applying one or more row operations while finding A^–1 by elementary row operations, we obtain all zeros in one or more rows, then A^–1 does not exist.

Answer: rows, does not exist

The given statement is:

"A matrix denotes a number."

A matrix is a rectangular array of numbers arranged in rows and columns. It is a mathematical object used to represent various concepts like linear transformations, systems of linear equations, etc.

While a determinant of a square matrix is a single number, the matrix itself is not a single number.

Therefore, the statement is False.

The given statement is:

"Matrices of any order can be added."

For two matrices to be added, they must have the same order. This means they must have the same number of rows and the same number of columns.

If the orders are different, addition is not defined.

Therefore, the statement is False.

The given statement is:

"Two matrices are equal if they have same number of rows and same number of columns."

Two matrices are said to be equal if and only if they satisfy two conditions:

1. They are of the same order (same number of rows and same number of columns).

2. The corresponding elements of the two matrices are equal.

The given statement only mentions the first condition. Matrices of the same order can have different corresponding elements, and thus not be equal. For example, $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $\begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$ have the same order ($2 \times 2$), but they are not equal because their corresponding elements are different.

Therefore, the statement is False.

The given statement is:

"Matrices of different order can not be subtracted."

For two matrices to be subtracted, they must have the same order. This means they must have the same number of rows and the same number of columns.

If the orders are different, subtraction is not defined.

Therefore, matrices of different orders cannot be subtracted.

The statement correctly describes the condition for matrix subtraction.

Therefore, the statement is True.

The given statement is:

"Matrix addition is associative as well as commutative."

Let A, B, and C be three matrices of the same order $m \times n$.

For matrix addition to be associative, the following property must hold:

$(A + B) + C = A + (B + C)$

This property holds true for matrix addition.

For matrix addition to be commutative, the following property must hold:

$A + B = B + A$

This property also holds true for matrix addition.

Matrix addition is defined only for matrices of the same order, and for such matrices, both the associative and commutative properties are satisfied.

Therefore, the statement is True.

The given statement is:

"Matrix multiplication is commutative."

For matrix multiplication to be commutative, for any two matrices A and B for which both $AB$ and $BA$ are defined, the property $AB = BA$ must hold true.

However, matrix multiplication is generally not commutative. This means that in most cases, $AB \neq BA$.

For example, consider matrices:

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Calculate $AB$:

$AB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (1)(0)+(2)(1) & (1)(1)+(2)(0) \\ (3)(0)+(4)(1) & (3)(1)+(4)(0) \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$

Calculate $BA$:

$BA = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} (0)(1)+(1)(3) & (0)(2)+(1)(4) \\ (1)(1)+(0)(3) & (1)(2)+(0)(4) \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}$

Clearly, $AB \neq BA$.

While there are specific cases where $AB = BA$ (e.g., when A and B are diagonal matrices, or when one of the matrices is an identity matrix), the property does not hold for all matrices where multiplication is defined.

Therefore, the statement is False.

The given statement is:

"A square matrix where every element is unity is called an identity matrix."

An identity matrix (denoted by $I$) is a square matrix where the elements on the main diagonal are all 1, and all other elements are 0.

For example, the $2 \times 2$ identity matrix is $I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, and the $3 \times 3$ identity matrix is $I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.

A square matrix where every element is unity (meaning every element is 1) is a matrix like $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ or $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$. This is clearly different from an identity matrix (unless it's a $1 \times 1$ matrix [1]).

Therefore, the statement is False.

The given statement is:

"If A and B are two square matrices of the same order, then A + B = B + A."

This statement concerns the commutative property of matrix addition.

Let A and B be two square matrices of the same order, say $n \times n$. Let the elements of A be $a_{ij}$ and the elements of B be $b_{ij}$, where $1 \leq i \leq n$ and $1 \leq j \leq n$.

The sum of A and B, denoted by $A+B$, is a matrix whose elements are $(a_{ij} + b_{ij})$.

The sum of B and A, denoted by $B+A$, is a matrix whose elements are $(b_{ij} + a_{ij})$.

Since the addition of real numbers (or complex numbers, depending on the matrix elements) is commutative, we have $a_{ij} + b_{ij} = b_{ij} + a_{ij}$ for all $i$ and $j$.

Therefore, the corresponding elements of the matrices $A+B$ and $B+A$ are equal, and both matrices have the same order ($n \times n$).

By the definition of matrix equality, $A+B = B+A$.

Thus, matrix addition is commutative for matrices of the same order (which includes square matrices of the same order).

Therefore, the statement is True.

The given statement is:

"If A and B are two matrices of the same order, then A – B = B – A."

This statement concerns the commutative property of matrix subtraction.

Let A and B be two matrices of the same order, say $m \times n$. Let the elements of A be $a_{ij}$ and the elements of B be $b_{ij}$.

The difference $A-B$ is a matrix whose elements are $(a_{ij} - b_{ij})$.

The difference $B-A$ is a matrix whose elements are $(b_{ij} - a_{ij})$.

For $A-B = B-A$ to be true, it must be that $a_{ij} - b_{ij} = b_{ij} - a_{ij}$ for all corresponding elements.

This implies $2a_{ij} = 2b_{ij}$, or $a_{ij} = b_{ij}$ for all $i, j$. This means $A = B$.

So, $A-B = B-A$ is only true if $A = B$. It is not true for arbitrary matrices A and B of the same order.

For example, consider matrices:

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Calculate $A-B$:

$A-B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1-0 & 2-1 \\ 3-1 & 4-0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 4 \end{pmatrix}$

Calculate $B-A$:

$B-A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 0-1 & 1-2 \\ 1-3 & 0-4 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ -2 & -4 \end{pmatrix}$

Clearly, $A-B \neq B-A$ in general.

Therefore, the statement is False.

The given statement is:

"If matrix AB = O, then A = O or B = O or both A and B are null matrices."

In the algebra of real numbers, if the product of two numbers is zero, then at least one of the numbers must be zero. However, this property does not hold true for matrix multiplication.

It is possible for the product of two non-zero matrices to be the zero matrix (null matrix).

Consider the following matrices:

$A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$

$B = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$

Neither A nor B is a null matrix (O).

Let's compute the product AB:

$AB = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} (1)(1) + (1)(-1) & (1)(-1) + (1)(1) \\ (1)(1) + (1)(-1) & (1)(-1) + (1)(1) \end{pmatrix} = \begin{pmatrix} 1 - 1 & -1 + 1 \\ 1 - 1 & -1 + 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

Here, $AB = O$, but $A \neq O$ and $B \neq O$.

This example shows that the premise ($AB=O$) can be true even if neither A nor B is the null matrix.

Therefore, the statement is False.

The given statement is:

"Transpose of a column matrix is a column matrix."

A column matrix is a matrix with only one column. Its order is $m \times 1$, where $m$ is the number of rows.

The transpose of a matrix is obtained by interchanging its rows and columns.

If a matrix A has order $m \times n$, its transpose $A^T$ has order $n \times m$.

Let A be a column matrix of order $m \times 1$. For example, $A = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{pmatrix}$.

The transpose of A, denoted by $A^T$, is obtained by changing the single column into a single row:

$A^T = \begin{pmatrix} a_1 & a_2 & \cdots & a_m \end{pmatrix}$.

This resulting matrix has 1 row and $m$ columns. It is a row matrix.

For example, the transpose of the column matrix $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ (order $3 \times 1$) is the row matrix $\begin{pmatrix} 1 & 2 & 3 \end{pmatrix}$ (order $1 \times 3$).

Unless $m=1$ (in which case it's both a column and a row matrix), the transpose of a column matrix is a row matrix, not a column matrix.

Therefore, the statement is False.

The given statement is:

"If A and B are two square matrices of the same order, then AB = BA."

This statement claims that matrix multiplication is commutative for any two square matrices of the same order.

However, as discussed in Question 87, matrix multiplication is generally not commutative.

Consider the matrices used previously:

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Both are square matrices of order $2 \times 2$.

Calculate $AB$:

$AB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (1)(0)+(2)(1) & (1)(1)+(2)(0) \\ (3)(0)+(4)(1) & (3)(1)+(4)(0) \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$

Calculate $BA$:

$BA = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} (0)(1)+(1)(3) & (0)(2)+(1)(4) \\ (1)(1)+(0)(3) & (1)(2)+(0)(4) \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}$

Since $\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix} \neq \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}$, we have $AB \neq BA$.

This counterexample demonstrates that matrix multiplication is not commutative for all pairs of square matrices of the same order.

Therefore, the statement is False.

The given statement is:

"If each of the three matrices of the same order are symmetric, then their sum is a symmetric matrix."

A matrix A is called symmetric if it is a square matrix and its transpose is equal to the matrix itself, i.e., $A^T = A$.

Let A, B, and C be three symmetric matrices of the same order, say $n \times n$.

By definition of symmetric matrices:

Let S be the sum of these three matrices:

$S = A + B + C$

To check if S is symmetric, we need to find its transpose, $S^T$, and see if $S^T = S$.

Using the property of transpose that $(X+Y+Z)^T = X^T + Y^T + Z^T$ for matrices X, Y, Z of the same order, we have:

$S^T = (A + B + C)^T = A^T + B^T + C^T$

Substituting the results from (i), (ii), and (iii):

$S^T = A + B + C$

Since $S = A + B + C$, we have $S^T = S$.

This shows that the sum of the three symmetric matrices is also a symmetric matrix.

This property holds for any number of symmetric matrices of the same order.

Therefore, the statement is True.

The given statement is:

"If A and B are any two matrices of the same order, then (AB)′ = A′B′."

The prime symbol ($'$) or the superscript T ($^T$) denotes the transpose of a matrix.

For the product of two matrices A and B to be defined, the number of columns in A must be equal to the number of rows in B.

If A and B are "any two matrices of the same order", say $m \times n$, the product AB is only defined if $n=m$. This means A and B must be square matrices of the same order for AB to be generally defined under this phrasing.

Assuming A and B are square matrices of order $n \times n$, the correct property for the transpose of a matrix product is:

$(AB)^T = B^T A^T$

This is often referred to as the "reversal rule" for transposition of a product.

The given statement claims that $(AB)^T = A^T B^T$. This is generally false.

Let's consider a counterexample with $2 \times 2$ matrices:

Let $A = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$. Both are of the same order ($2 \times 2$).

First, find $AB$:

$AB = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} (1)(1)+(0)(0) & (1)(2)+(0)(1) \\ (2)(1)+(1)(0) & (2)(2)+(1)(1) \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}$

Now, find $(AB)^T$:

$(AB)^T = \begin{pmatrix} 1 & 2 \\ 5 & 2 \end{pmatrix}$

Next, find $A^T$ and $B^T$:

$A^T = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$

$B^T = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$

Now, find $A^T B^T$:

$A^T B^T = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} (1)(1)+(2)(2) & (1)(0)+(2)(1) \\ (0)(1)+(1)(2) & (0)(0)+(1)(1) \end{pmatrix} = \begin{pmatrix} 1+4 & 0+2 \\ 0+2 & 0+1 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix}$

Comparing $(AB)^T$ and $A^T B^T$:

$(AB)^T = \begin{pmatrix} 1 & 2 \\ 5 & 2 \end{pmatrix}$

$A^T B^T = \begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix}$

Clearly, $(AB)^T \neq A^T B^T$.

The statement is only true in specific cases, such as when A and B are diagonal matrices, or when A and B are inverses of each other, or if $A^T B^T = B^T A^T$ holds. However, it is not true for arbitrary matrices of the same order where the product is defined.

Therefore, the statement is False.

The given statement is:

"If (AB)′ = B′ A′, where A and B are not square matrices, then number of rows in A is equal to number of columns in B and number of columns in A is equal to number of rows in B."

Let the order of matrix A be $m \times n$. Since A is not a square matrix, $m \neq n$.

Let the order of matrix B be $p \times q$. Since B is not a square matrix, $p \neq q$.

For the matrix product AB to be defined, the number of columns in A must be equal to the number of rows in B.

The order of the product matrix AB is $m \times q$.

The transpose of AB, $(AB)'$, has the order obtained by swapping the rows and columns of AB, so the order of $(AB)'$ is $q \times m$.

Now consider the right side of the equation, $B'A'$.

The order of $B'$ (transpose of B) is $q \times p$.

The order of $A'$ (transpose of A) is $n \times m$.

For the matrix product $B'A'$ to be defined, the number of columns in $B'$ must be equal to the number of rows in $A'$.

$p = n$

This is the same condition (i) required for AB to be defined.

The order of the product matrix $B'A'$ is $(q \times p) \times (n \times m)$. Since $p=n$, the order is $q \times m$.

For the equality $(AB)' = B'A'$ to hold, the matrices on both sides must have the same dimensions. We have shown that the order of $(AB)'$ is $q \times m$ and the order of $B'A'$ is $q \times m$. The dimensions are consistent, provided $n=p$.

The statement claims that if $(AB)' = B'A'$ holds (with A and B non-square), then:

1. Number of rows in A equals number of columns in B ($m = q$).

2. Number of columns in A equals number of rows in B ($n = p$).

We know that $n=p$ is a necessary condition for the products AB and $B'A'$ to be defined at all, and thus for the equality $(AB)' = B'A'$ to make sense. So the second part of the conclusion ($n=p$) is true.

However, the statement also claims that $m=q$ must necessarily be true. Let's examine this.

Consider a counterexample where A and B are not square matrices, $n=p$, but $m \neq q$.

Let A be a $3 \times 2$ matrix. Here $m=3$, $n=2$. A is not square since $3 \neq 2$.

Let B be a $2 \times 4$ matrix. Here $p=2$, $q=4$. B is not square since $2 \neq 4$.

In this case, the number of columns in A ($n=2$) equals the number of rows in B ($p=2$). So the product AB is defined.

The order of AB is $m \times q$, which is $3 \times 4$.

The order of $(AB)'$ is $q \times m$, which is $4 \times 3$.

The order of $A'$ is $n \times m$, which is $2 \times 3$.

The order of $B'$ is $q \times p$, which is $4 \times 2$.

The order of $B'A'$ is $(4 \times 2) \times (2 \times 3)$, which results in a $4 \times 3$ matrix.

The equality $(AB)' = B'A'$ holds as a fundamental property of matrix transposes whenever AB is defined. In our example, A and B are non-square, the product is defined, and the equality holds. The order of the resulting matrices is $4 \times 3$.

According to the statement's conclusion, $m$ should equal $q$. However, in our example, $m=3$ and $q=4$, so $m \neq q$.

The condition $m=q$ is not necessary for $(AB)' = B'A'$ to hold when A and B are non-square. The only essential dimension condition for the formula to be applicable is $n=p$ (that the product AB is defined).

Therefore, the statement is False.

The given statement is:

"If A, B and C are square matrices of same order, then AB = AC always implies that B = C."

This statement refers to the cancellation property for matrix multiplication. In the algebra of real numbers, if $a, b, c$ are real numbers with $a \neq 0$, then $ab = ac$ implies $b=c$. However, this property does not always hold for matrices.

The cancellation property holds for matrices only if the matrix being cancelled is invertible (or non-singular). That is, if A is an invertible matrix and $AB = AC$, then we can pre-multiply both sides by $A^{-1}$ (the inverse of A):

$A^{-1}(AB) = A^{-1}(AC)$

$(A^{-1}A)B = (A^{-1}A)C$

$IB = IC$

$B = C$

where I is the identity matrix.

However, the statement says "always implies", which means it must hold for any square matrix A of the same order, including non-invertible matrices.

If A is a non-invertible matrix (i.e., its determinant is zero, $|A|=0$), then it is possible to have $AB = AC$ even if $B \neq C$.

Consider the following counterexample using $2 \times 2$ square matrices:

Let $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, and $C = \begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix}$.

All three matrices are square and of the same order ($2 \times 2$).

Calculate the product AB:

$AB = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} (1)(1)+(1)(3) & (1)(2)+(1)(4) \\ (1)(1)+(1)(3) & (1)(2)+(1)(4) \end{pmatrix} = \begin{pmatrix} 1+3 & 2+4 \\ 1+3 & 2+4 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 4 & 6 \end{pmatrix}$

Calculate the product AC:

$AC = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} (1)(2)+(1)(2) & (1)(3)+(1)(3) \\ (1)(2)+(1)(2) & (1)(3)+(1)(3) \end{pmatrix} = \begin{pmatrix} 2+2 & 3+3 \\ 2+2 & 3+3 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 4 & 6 \end{pmatrix}$

From the calculations, we see that $AB = \begin{pmatrix} 4 & 6 \\ 4 & 6 \end{pmatrix}$ and $AC = \begin{pmatrix} 4 & 6 \\ 4 & 6 \end{pmatrix}$, so $AB = AC$.

However, the matrices B and C are $B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $C = \begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix}$. Clearly, $B \neq C$ as their corresponding elements are different.

In this counterexample, A is a non-zero square matrix, B and C are square matrices of the same order as A, $AB = AC$, but $B \neq C$.

This demonstrates that $AB = AC$ does not always imply $B = C$ for square matrices of the same order.

Therefore, the statement is False.

The given statement is:

"AA′ is always a symmetric matrix for any matrix A."

Let A be any matrix. Let its order be $m \times n$.

The transpose of A, denoted by $A'$ or $A^T$, has the order $n \times m$.

The product $AA'$ is defined because the number of columns in A (which is $n$) is equal to the number of rows in $A'$ (which is $n$).

The order of the resulting matrix $P = AA'$ is $m \times m$. This is a square matrix, which is a necessary condition for a matrix to be symmetric.

A matrix P is symmetric if its transpose $P^T$ is equal to P, i.e., $P^T = P$.

Let's find the transpose of $P = AA'$. We use the property of matrix transpose that $(XY)^T = Y^T X^T$ and $(X^T)^T = X$.

$P^T = (AA')^T$

Applying the property $(XY)^T = Y^T X^T$ with $X=A$ and $Y=A'$:

$P^T = (A')^T A^T$

Applying the property $(X^T)^T = X$ to $(A')^T$:

$(A')^T = A$

So, substituting this back into the expression for $P^T$:

$P^T = A A'$

We started with $P = AA'$ and found that $P^T = AA'$.

Thus, $P^T = P$.

This shows that the matrix $AA'$ is equal to its transpose, meaning it is a symmetric matrix.

This holds true for any matrix A for which the product $AA'$ is defined, which results in a square matrix $AA'$.

Therefore, the statement is True.

The given statement is:

"If $A = \begin{bmatrix} 2&3&−1 \\ 1&4&2 \end{bmatrix}$ and $B = \begin{bmatrix} 2&3 \\ 4&5 \\ 2&1 \end{bmatrix}$, then AB and BA are defined and equal."

First, let's determine the orders of matrices A and B.

Matrix A has 2 rows and 3 columns. So, the order of A is $2 \times 3$.

Matrix B has 3 rows and 2 columns. So, the order of B is $3 \times 2$.

For the product AB to be defined, the number of columns in A must be equal to the number of rows in B.

Number of columns in A = 3

Number of rows in B = 3

Since $3 = 3$, the product AB is defined. The order of the matrix AB will be (number of rows in A) $\times$ (number of columns in B), which is $2 \times 2$.

For the product BA to be defined, the number of columns in B must be equal to the number of rows in A.

Number of columns in B = 2

Number of rows in A = 2

Since $2 = 2$, the product BA is defined. The order of the matrix BA will be (number of rows in B) $\times$ (number of columns in A), which is $3 \times 3$.

So, the first part of the statement, "AB and BA are defined", is True.

Now, let's consider the second part of the statement, "AB and BA are equal".

For two matrices to be equal, they must satisfy two conditions:

1. They must have the same order.

2. Their corresponding elements must be equal.

We found that the order of AB is $2 \times 2$, and the order of BA is $3 \times 3$.

Since the orders of AB and BA are different ($2 \times 2 \neq 3 \times 3$), the matrices AB and BA cannot be equal.

Even though both products are defined, they result in matrices of different dimensions.

Since the second part of the statement is False, the entire statement is False.

(Note: We do not need to calculate the actual matrices AB and BA to determine that they cannot be equal, as their orders are different. However, for completeness, we can calculate them).

Calculating AB:

$AB = \begin{bmatrix} 2& 3& −1 \\ 1& 4& 2 \end{bmatrix} \begin{bmatrix} 2& 3 \\ 4& 5 \\ 2& 1 \end{bmatrix} = \begin{bmatrix} (2)(2)+(3)(4)+(-1)(2) & (2)(3)+(3)(5)+(-1)(1) \\ (1)(2)+(4)(4)+(2)(2) & (1)(3)+(4)(5)+(2)(1) \end{bmatrix}$

$AB = \begin{bmatrix} 4+12-2 & 6+15-1 \\ 2+16+4 & 3+20+2 \end{bmatrix} = \begin{bmatrix} 14 & 20 \\ 22 & 25 \end{bmatrix}$ (Order $2 \times 2$)

Calculating BA:

$BA = \begin{bmatrix} 2& 3 \\ 4& 5 \\ 2& 1 \end{bmatrix} \begin{bmatrix} 2& 3& −1 \\ 1& 4& 2 \end{bmatrix} = \begin{bmatrix} (2)(2)+(3)(1) & (2)(3)+(3)(4) & (2)(-1)+(3)(2) \\ (4)(2)+(5)(1) & (4)(3)+(5)(4) & (4)(-1)+(5)(2) \\ (2)(2)+(1)(1) & (2)(3)+(1)(4) & (2)(-1)+(1)(2) \end{bmatrix}$

$BA = \begin{bmatrix} 4+3 & 6+12 & -2+6 \\ 8+5 & 12+20 & -4+10 \\ 4+1 & 6+4 & -2+2 \end{bmatrix} = \begin{bmatrix} 7 & 18 & 4 \\ 13 & 32 & 6 \\ 5 & 10 & 0 \end{bmatrix}$ (Order $3 \times 3$)

As expected, the resulting matrices AB and BA have different orders and are therefore not equal.

The given statement is:

"If A is skew symmetric matrix, then A$^2$ is a symmetric matrix."

A square matrix A is called skew-symmetric if its transpose is equal to the negative of the matrix. That is, $A^T = -A$.

A square matrix B is called symmetric if its transpose is equal to the matrix itself. That is, $B^T = B$.

We want to determine if $A^2 = A \times A$ is a symmetric matrix, given that A is skew-symmetric.

To check if $A^2$ is symmetric, we need to find the transpose of $A^2$ and see if it equals $A^2$.

Let's find $(A^2)^T$:

$(A^2)^T = (A \times A)^T$

Using the property of matrix transpose that $(XY)^T = Y^T X^T$, we get:

$(A \times A)^T = A^T A^T$

Since A is skew-symmetric, we know that $A^T = -A$. Substitute this into the expression:

$(A^2)^T = (-A)(-A)$

Using the property of scalar multiplication $(cA)(dB) = cd(AB)$, where c and d are scalars (in this case, -1):

$(A^2)^T = (-1 \times A) \times (-1 \times A) = (-1)(-1) (A \times A)$

$(-1)(-1) = 1$, and $A \times A = A^2$.

So, $(A^2)^T = 1 \times A^2 = A^2$.

We have found that $(A^2)^T = A^2$. By the definition of a symmetric matrix, this means $A^2$ is a symmetric matrix.

This holds true for any skew-symmetric matrix A.

Therefore, the statement is True.

The given statement is:

"(AB)$^{-1}$ = A$^{-1}$ B$^{-1}$, where A and B are invertible matrices satisfying commutative property with respect to multiplication."

Let A and B be two invertible matrices of the same order, say $n \times n$.

The definition of the inverse of a matrix product states that for any two invertible matrices A and B, the inverse of their product AB is given by the product of their inverses in the reverse order:

$(AB)^{-1} = B^{-1} A^{-1}$

This is the general rule.

However, the given statement includes an additional condition: A and B satisfy the commutative property with respect to multiplication. This means:

Let's see if the general rule $(AB)^{-1} = B^{-1} A^{-1}$ simplifies to $(AB)^{-1} = A^{-1} B^{-1}$ under the condition $AB = BA$.

Given $AB = BA$.

Take the inverse of both sides:

$(AB)^{-1} = (BA)^{-1}$

Now, apply the general rule for the inverse of a product to the right side $(BA)^{-1}$. According to the rule, $(BA)^{-1} = A^{-1} B^{-1}$.

So, substituting this into the equation:

$(AB)^{-1} = A^{-1} B^{-1}$

This shows that if A and B are invertible matrices that commute (i.e., $AB=BA$), then the property $(AB)^{-1} = A^{-1} B^{-1}$ holds true.

The statement correctly describes the consequence of the matrices being invertible and commutative.

Therefore, the statement is True.