Chapter 3 Matrices

Representation as a 3x2 Matrix:

The given information can be represented in the form of a 3 × 2 matrix, where the rows correspond to the factories (I, II, III) and the columns correspond to the number of men and women workers, respectively.

Let A be the required 3 × 2 matrix:

$A = \begin{bmatrix} 30 & 25 \\ 25 & 31 \\ 27 & 26 \end{bmatrix}$

Here, the first column represents the number of Men Workers and the second column represents the number of Women Workers. The first, second, and third rows represent factories I, II, and III respectively.

Interpretation of the Entry in the Third Row and Second Column:

The entry in the third row and second column of the matrix A is $A_{32}$.

Looking at the matrix:

$A = \begin{bmatrix} 30 & 25 \\ 25 & 31 \\ 27 & \mathbf{26} \end{bmatrix}$

The entry is 26.

The third row corresponds to Factory III.

The second column corresponds to Women Workers.

Therefore, the entry in the third row and second column (which is 26) represents the number of women workers in Factory III.

We know that if a matrix is of order $m \times n$, it has $m \times n$ elements.

We are given that the matrix has 8 elements.

Let the order of the matrix be $m \times n$. Then the number of elements is $m \times n = 8$.

We need to find all possible pairs of positive integers $(m, n)$ whose product is 8.

The pairs of factors of 8 are:

(1, 8) such that $1 \times 8 = 8$

(8, 1) such that $8 \times 1 = 8$

(2, 4) such that $2 \times 4 = 8$

(4, 2) such that $4 \times 2 = 8$

Thus, the possible orders of a matrix having 8 elements are:

$\mathbf{1 \times 8}$, $\mathbf{8 \times 1}$, $\mathbf{2 \times 4}$, and $\mathbf{4 \times 2}$.

We need to construct a 3 × 2 matrix A. A general 3 × 2 matrix is given by:

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}$

The elements $a_{ij}$ are given by the formula $a_{ij} = \frac{1}{2} |i - 2j|$, where $i$ is the row number ($i=1, 2, 3$) and $j$ is the column number ($j=1, 2$).

We calculate each element:

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

$a_{12} = \frac{1}{2} |1 - 2(2)| = \frac{1}{2} |1 - 4| = \frac{1}{2} |-3| = \frac{1}{2}(3) = \frac{3}{2}$

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

$a_{22} = \frac{1}{2} |2 - 2(2)| = \frac{1}{2} |2 - 4| = \frac{1}{2} |-2| = \frac{1}{2}(2) = 1$

$a_{31} = \frac{1}{2} |3 - 2(1)| = \frac{1}{2} |3 - 2| = \frac{1}{2} |1| = \frac{1}{2}(1) = \frac{1}{2}$

$a_{32} = \frac{1}{2} |3 - 2(2)| = \frac{1}{2} |3 - 4| = \frac{1}{2} |-1| = \frac{1}{2}(1) = \frac{1}{2}$

Substituting these values into the matrix structure, we get the required matrix:

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

The following matrix equation is given:

$\begin{bmatrix}x+3 & z+4 & 2y-7 \\ -6 & a-1 & 0 \\ b-3 & -21 & 0 \end{bmatrix} = \begin{bmatrix}0 & 6 & 3y − 2 \\ −6 & −3 & 2c + 2 \\ 2b + 4 & −21 & 0 \end{bmatrix}$

To Find:

The values of the variables a, b, c, x, y, and z.

Solution:

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

By comparing the corresponding elements of the two given matrices, we get the following equations:

Equating element (1,1):

$x + 3 = 0$

$x = -3$

Equating element (1,2):

$z + 4 = 6$

$z = 6 - 4$

$z = 2$

Equating element (1,3):

$2y - 7 = 3y - 2$

$-7 + 2 = 3y - 2y$

$-5 = y$

$y = -5$

Equating element (2,2):

$a - 1 = -3$

$a = -3 + 1$

$a = -2$

Equating element (2,3):

$0 = 2c + 2$

$-2 = 2c$

$c = \frac{-2}{2}$

$c = -1$

Equating element (3,1):

$b - 3 = 2b + 4$

$-3 - 4 = 2b - b$

$-7 = b$

$b = -7$

(Note: Equating elements (2,1), (3,2), and (3,3) gives $-6=-6$, $-21=-21$, and $0=0$, which confirms consistency but does not yield new information about the variables).

The values of the variables are:

a = -2

b = -7

c = -1

x = -3

y = -5

z = 2

Given:

The matrix equation:

$\begin{bmatrix}2a + b&a − 2b \\ 5c − d&4c + 3d \end{bmatrix} = \begin{bmatrix}4&−3 \\ 11&24 \end{bmatrix}$

To Find:

The values of a, b, c, and d.

Solution:

By the definition of equality of matrices, the corresponding elements must be equal. Comparing the elements, we get the following system of linear equations:

Solving equations (i) and (ii) for a and b:

Multiply equation (i) by 2:

Add equation (ii) and (v):

$(a - 2b) + (4a + 2b) = -3 + 8$

$5a = 5$

$a = 1$

Substitute $a = 1$ into equation (i):

$2(1) + b = 4$

$2 + b = 4$

$b = 4 - 2 = 2$

So, $\mathbf{a = 1}$ and $\mathbf{b = 2}$.

Solving equations (iii) and (iv) for c and d:

Multiply equation (iii) by 3:

Add equation (iv) and (vi):

$(4c + 3d) + (15c - 3d) = 24 + 33$

$19c = 57$

$c = \frac{57}{19} = 3$

Substitute $c = 3$ into equation (iii):

$5(3) - d = 11$

$15 - d = 11$

$d = 15 - 11 = 4$

So, $\mathbf{c = 3}$ and $\mathbf{d = 4}$.

The required values are:

a = 1, b = 2, c = 3, d = 4

The given matrix is $A = \begin{bmatrix}2&5&19&-7 \\ 35&-2& \frac{5}{2} &12 \\ \sqrt{3}&1&-5&17 \end{bmatrix}$.

(i) The order of the matrix:

The matrix A has 3 rows and 4 columns.

Therefore, the order of the matrix A is 3 × 4.

(ii) The number of elements:

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) × (Number of columns) = 3 × 4 = 12.

(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃:

The element $a_{ij}$ is the element in the $i$-th row and $j$-th column.

$a_{13}$ is the element in the 1st row and 3rd column, which is 19.

$a_{21}$ is the element in the 2nd row and 1st column, which is 35.

$a_{33}$ is the element in the 3rd row and 3rd column, which is -5.

$a_{24}$ is the element in the 2nd row and 4th column, which is 12.

$a_{23}$ is the element in the 2nd row and 3rd column, which is $\frac{5}{2}$.

We know that if a matrix is of order $m \times n$, it has $m \times n$ elements.

Case 1: Matrix has 24 elements

Let the order of the matrix be $m \times n$. Then the number of elements is $m \times n = 24$.

We need to find all ordered pairs of positive integers $(m, n)$ whose product is 24.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

The possible pairs $(m, n)$ are:

(1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4).

Therefore, the possible orders of a matrix having 24 elements are:

$1 \times 24, 24 \times 1, 2 \times 12, 12 \times 2, 3 \times 8, 8 \times 3, 4 \times 6, 6 \times 4$.

Case 2: Matrix has 13 elements

Let the order of the matrix be $m \times n$. Then the number of elements is $m \times n = 13$.

Since 13 is a prime number, its only factors are 1 and 13.

We need to find all ordered pairs of positive integers $(m, n)$ whose product is 13.

The possible pairs $(m, n)$ are:

(1, 13), (13, 1).

Therefore, the possible orders of a matrix having 13 elements are:

$1 \times 13, 13 \times 1$.

We know that if a matrix is of order $m \times n$, it has $m \times n$ elements.

Case 1: Matrix has 18 elements

Let the order of the matrix be $m \times n$. Then the number of elements is $m \times n = 18$.

We need to find all ordered pairs of positive integers $(m, n)$ whose product is 18.

The factors of 18 are 1, 2, 3, 6, 9, and 18.

The possible pairs $(m, n)$ are:

(1, 18), (18, 1), (2, 9), (9, 2), (3, 6), (6, 3).

Therefore, the possible orders of a matrix having 18 elements are:

$1 \times 18, 18 \times 1, 2 \times 9, 9 \times 2, 3 \times 6, 6 \times 3$.

Case 2: Matrix has 5 elements

Let the order of the matrix be $m \times n$. Then the number of elements is $m \times n = 5$.

Since 5 is a prime number, its only factors are 1 and 5.

We need to find all ordered pairs of positive integers $(m, n)$ whose product is 5.

The possible pairs $(m, n)$ are:

(1, 5), (5, 1).

Therefore, the possible orders of a matrix having 5 elements are:

$1 \times 5, 5 \times 1$.

A general 2 × 2 matrix is given by $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$. We need to find the elements $a_{ij}$ using the given formulas for each case.

(i) $a_{ij} = \frac{(i + j)^2}{2}$

Calculate each element:

$a_{11} = \frac{(1 + 1)^2}{2} = \frac{2^2}{2} = \frac{4}{2} = 2$

$a_{12} = \frac{(1 + 2)^2}{2} = \frac{3^2}{2} = \frac{9}{2}$

$a_{21} = \frac{(2 + 1)^2}{2} = \frac{3^2}{2} = \frac{9}{2}$

$a_{22} = \frac{(2 + 2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8$

The constructed matrix is:

$A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}$

(ii) $a_{ij} = \frac{i}{j}$

Calculate each element:

$a_{11} = \frac{1}{1} = 1$

$a_{12} = \frac{1}{2}$

$a_{21} = \frac{2}{1} = 2$

$a_{22} = \frac{2}{2} = 1$

The constructed matrix is:

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

(iii) $a_{ij} = \frac{(i + 2j)^2}{2}$

Calculate each element:

$a_{11} = \frac{(1 + 2(1))^2}{2} = \frac{(1 + 2)^2}{2} = \frac{3^2}{2} = \frac{9}{2}$

$a_{12} = \frac{(1 + 2(2))^2}{2} = \frac{(1 + 4)^2}{2} = \frac{5^2}{2} = \frac{25}{2}$

$a_{21} = \frac{(2 + 2(1))^2}{2} = \frac{(2 + 2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8$

$a_{22} = \frac{(2 + 2(2))^2}{2} = \frac{(2 + 4)^2}{2} = \frac{6^2}{2} = \frac{36}{2} = 18$

The constructed matrix is:

$A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}$

A general 3 × 4 matrix is given by $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$. We need to find the elements $a_{ij}$ using the given formulas for each case, where $i$ is the row number ($i=1, 2, 3$) and $j$ is the column number ($j=1, 2, 3, 4$).

(i) $a_{ij} = \frac{1}{2} |-3i + j|$

Calculate each element:

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

$a_{12} = \frac{1}{2} |-3(1) + 2| = \frac{1}{2} |-1| = \frac{1}{2}$

$a_{13} = \frac{1}{2} |-3(1) + 3| = \frac{1}{2} |0| = 0$

$a_{14} = \frac{1}{2} |-3(1) + 4| = \frac{1}{2} |1| = \frac{1}{2}$

$a_{21} = \frac{1}{2} |-3(2) + 1| = \frac{1}{2} |-5| = \frac{5}{2}$

$a_{22} = \frac{1}{2} |-3(2) + 2| = \frac{1}{2} |-4| = 2$

$a_{23} = \frac{1}{2} |-3(2) + 3| = \frac{1}{2} |-3| = \frac{3}{2}$

$a_{24} = \frac{1}{2} |-3(2) + 4| = \frac{1}{2} |-2| = 1$

$a_{31} = \frac{1}{2} |-3(3) + 1| = \frac{1}{2} |-8| = 4$

$a_{32} = \frac{1}{2} |-3(3) + 2| = \frac{1}{2} |-7| = \frac{7}{2}$

$a_{33} = \frac{1}{2} |-3(3) + 3| = \frac{1}{2} |-6| = 3$

$a_{34} = \frac{1}{2} |-3(3) + 4| = \frac{1}{2} |-5| = \frac{5}{2}$

The constructed matrix is:

$A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}$

(ii) $a_{ij} = 2i - j$

Calculate each element:

$a_{11} = 2(1) - 1 = 1$

$a_{12} = 2(1) - 2 = 0$

$a_{13} = 2(1) - 3 = -1$

$a_{14} = 2(1) - 4 = -2$

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

$a_{22} = 2(2) - 2 = 2$

$a_{23} = 2(2) - 3 = 1$

$a_{24} = 2(2) - 4 = 0$

$a_{31} = 2(3) - 1 = 5$

$a_{32} = 2(3) - 2 = 4$

$a_{33} = 2(3) - 3 = 3$

$a_{34} = 2(3) - 4 = 2$

The constructed matrix is:

$A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}$

By the definition of equality of matrices, the corresponding elements must be equal.

(i) $\begin{bmatrix}4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix}y & z \\ 1 & 5 \end{bmatrix}$

Comparing the corresponding elements:

$4 = y \implies \mathbf{y = 4}$

$3 = z \implies \mathbf{z = 3}$

$x = 1 \implies \mathbf{x = 1}$

$5 = 5$ (This is consistent)

So, the values are $x=1, y=4, z=3$.

(ii) $\begin{bmatrix}x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix}6&2\\5&8 \end{bmatrix}$

Comparing the corresponding elements:

From equation (2):

$z = 5 - 5 = 0$

From equation (1):

$y = 6 - x$

Substitute this expression for $y$ into equation (3):

$x(6 - x) = 8$

$6x - x^2 = 8$

$x^2 - 6x + 8 = 0$

Factor the quadratic equation:

$(x - 2)(x - 4) = 0$

This gives two possible values for $x$: $x = 2$ or $x = 4$.

If $x = 2$, then $y = 6 - 2 = 4$.

If $x = 4$, then $y = 6 - 4 = 2$.

So, there are two possible sets of solutions:

$x = 2, y = 4, z = 0$ or $x = 4, y = 2, z = 0$.

(iii) $\begin{bmatrix}x + y + z\\x + z\\y + z \end{bmatrix} = \begin{bmatrix}9\\5\\7 \end{bmatrix}$

Comparing the corresponding elements:

Substitute equation (B) into equation (A):

$(x + z) + y = 9$

$5 + y = 9$

$y = 9 - 5 = 4$

Substitute $y = 4$ into equation (C):

$4 + z = 7$

$z = 7 - 4 = 3$

Substitute $z = 3$ into equation (B):

$x + 3 = 5$

$x = 5 - 3 = 2$

So, the values are $\mathbf{x = 2, y = 4, z = 3}$.

Given:

The matrix equation:

$\begin{bmatrix}a− b&2a + c \\ 2a − b&3c + d \end{bmatrix} = \begin{bmatrix}−1&5\\0&13 \end{bmatrix}$

To Find:

The values of a, b, c, and d.

Solution:

By the definition of equality of matrices, the corresponding elements must be equal. Comparing the elements, we get the following system of linear equations:

Solving equations (i) and (iii) for a and b:

From equation (iii):

$2a = b$

Substitute $b = 2a$ into equation (i):

$a - (2a) = -1$

$-a = -1$

$a = 1$

Now substitute $a=1$ back into $b=2a$:

$b = 2(1) = 2$

So, $\mathbf{a = 1}$ and $\mathbf{b = 2}$.

Solving for c using equation (ii):

Substitute $a=1$ into equation (ii):

$2(1) + c = 5$

$2 + c = 5$

$c = 5 - 2 = 3$

So, $\mathbf{c = 3}$.

Solving for d using equation (iv):

Substitute $c=3$ into equation (iv):

$3(3) + d = 13$

$9 + d = 13$

$d = 13 - 9 = 4$

So, $\mathbf{d = 4}$.

The required values are:

a = 1, b = 2, c = 3, d = 4

By definition, a square matrix is a matrix in which the number of rows is equal to the number of columns.

The given matrix A = [a_ij]_{m × n} has $m$ rows and $n$ columns.

For A to be a square matrix, the number of rows ($m$) must be equal to the number of columns ($n$).

So, the condition for A to be a square matrix is m = n.

Comparing this with the given options:

(A) m < n (Rectangular matrix, more columns than rows)

(B) m > n (Rectangular matrix, more rows than columns)

(C) m = n (Square matrix)

(D) None of these

Therefore, the correct option is (C).

Given:

Two matrices that are supposed to be equal:

Matrix 1: $\begin{bmatrix}3x + 7& 5 \\ y + 1& 2-3x \end{bmatrix}$

Matrix 2: $\begin{bmatrix}0& y − 2 \\ 8& 4 \end{bmatrix}$

To Find:

The values of x and y that make the two matrices equal.

Solution:

For two matrices to be equal, their corresponding elements must be equal. We set up equations by equating the corresponding elements:

Equating element (1,1):

Equating element (1,2):

Equating element (2,1):

Equating element (2,2):

Now, we solve these equations:

From equation (i):

$3x = -7 \implies x = -\frac{7}{3}$

From equation (ii):

$y = 5 + 2 = 7$

From equation (iii):

$y = 8 - 1 = 7$

(The value $y=7$ is consistent from equations (ii) and (iii)).

From equation (iv):

$-3x = 4 - 2$

$-3x = 2 \implies x = -\frac{2}{3}$

Comparing the values obtained for x:

From equation (i), we get $x = -\frac{7}{3}$.

From equation (iv), we get $x = -\frac{2}{3}$.

Since the value of $x$ obtained from equation (i) is different from the value of $x$ obtained from equation (iv) ($-\frac{7}{3} \neq -\frac{2}{3}$), there is a contradiction. It is impossible for both equations to be true simultaneously for a single value of $x$.

Therefore, there are no values of x and y for which the given matrices are equal.

The correct option is (B) not possible to find.

A matrix of order 3 × 3 has $3 \times 3 = 9$ elements.

Each element (entry) of the matrix can be either 0 or 1.

So, for each of the 9 positions in the matrix, there are 2 possible choices (0 or 1).

Since the choice for each element is independent, the total number of possible matrices can be found using the multiplication principle of counting.

Total number of possible matrices = (Number of choices for 1st element) × (Number of choices for 2nd element) × ... × (Number of choices for 9th element)

Total number of possible matrices = $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9$.

Calculating $2^9$:

$2^9 = 512$.

Therefore, there are 512 possible matrices of order 3 × 3 with each entry being either 0 or 1.

Comparing this with the given options, the correct option is (D).

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

Matrix B = $\begin{bmatrix}2&\sqrt{5}&1\\−2&3&\frac{1}{2} \end{bmatrix}$

To Find:

The sum of the matrices A + B.

Solution:

Matrix addition is defined for matrices of the same order. Both matrix A and matrix B have the order 2 × 3. Therefore, their sum A + B is defined.

To find the sum A + B, we add the corresponding elements of matrix A and matrix B.

$A + B = \begin{bmatrix}\sqrt{3}&1&−1 \\ 2&3&0 \end{bmatrix} + \begin{bmatrix}2&\sqrt{5}&1\\−2&3&\frac{1}{2} \end{bmatrix}$

$A + B = \begin{bmatrix}(\sqrt{3} + 2) & (1 + \sqrt{5}) & (-1 + 1) \\ (2 + (-2)) & (3 + 3) & (0 + \frac{1}{2}) \end{bmatrix}$

Simplifying the elements:

$A + B = \begin{bmatrix} 2 + \sqrt{3} & 1 + \sqrt{5} & 0 \\ 0 & 6 & \frac{1}{2} \end{bmatrix}$

Therefore, the sum of the matrices is:

$A + B = \mathbf{\begin{bmatrix} 2 + \sqrt{3} & 1 + \sqrt{5} & 0 \\ 0 & 6 & \frac{1}{2} \end{bmatrix}}$

Given:

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

Matrix B = $\begin{bmatrix}3&−1&3\\−1&0&2 \end{bmatrix}$

To Find:

The matrix 2A - B.

Solution:

First, we calculate the scalar multiple 2A.

$2A = 2 \times \begin{bmatrix}1&2&3\\2&3&1 \end{bmatrix} = \begin{bmatrix}2(1)&2(2)&2(3)\\2(2)&2(3)&2(1) \end{bmatrix} = \begin{bmatrix}2&4&6\\4&6&2 \end{bmatrix}$

Next, we subtract matrix B from 2A. Since both 2A and B are of the same order (2 × 3), subtraction is possible.

$2A - B = \begin{bmatrix}2&4&6\\4&6&2 \end{bmatrix} - \begin{bmatrix}3&−1&3\\−1&0&2 \end{bmatrix}$

We subtract the corresponding elements:

$2A - B = \begin{bmatrix}(2 - 3) & (4 - (-1)) & (6 - 3) \\ (4 - (-1)) & (6 - 0) & (2 - 2) \end{bmatrix}$

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

$2A - B = \begin{bmatrix}-1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}$

Therefore, the resulting matrix is:

$\mathbf{2A - B = \begin{bmatrix}-1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}}$

Given:

Matrix A = $\begin{bmatrix}8&0\\4&−2\\3&6 \end{bmatrix}$

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

The equation: $2A + 3X = 5B$

To Find:

The matrix X.

Solution:

We start with the given equation:

$2A + 3X = 5B$

To find X, we first isolate 3X:

$3X = 5B - 2A$

Now, we calculate the matrices 5B and 2A.

$5B = 5 \begin{bmatrix}2&−2\\4&2\\−5&1 \end{bmatrix} = \begin{bmatrix}5(2)&5(−2)\\5(4)&5(2)\\5(−5)&5(1) \end{bmatrix} = \begin{bmatrix}10&−10\\20&10\\−25&5 \end{bmatrix}$

$2A = 2 \begin{bmatrix}8&0\\4&−2\\3&6 \end{bmatrix} = \begin{bmatrix}2(8)&2(0)\\2(4)&2(−2)\\2(3)&2(6) \end{bmatrix} = \begin{bmatrix}16&0\\8&−4\\6&12 \end{bmatrix}$

Next, we calculate $5B - 2A$:

$5B - 2A = \begin{bmatrix}10&−10\\20&10\\−25&5 \end{bmatrix} - \begin{bmatrix}16&0\\8&−4\\6&12 \end{bmatrix}$

$5B - 2A = \begin{bmatrix}(10 - 16) & (-10 - 0) \\ (20 - 8) & (10 - (-4)) \\ (-25 - 6) & (5 - 12) \end{bmatrix}$

$5B - 2A = \begin{bmatrix}-6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}$

Since $3X = 5B - 2A$, we have:

$3X = \begin{bmatrix}-6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}$

To find X, we multiply the resulting matrix by $\frac{1}{3}$:

$X = \frac{1}{3} \begin{bmatrix}-6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}$

$X = \begin{bmatrix} \frac{1}{3}(-6) & \frac{1}{3}(-10) \\ \frac{1}{3}(12) & \frac{1}{3}(14) \\ \frac{1}{3}(-31) & \frac{1}{3}(-7) \end{bmatrix}$

$X = \begin{bmatrix} -2 & -\frac{10}{3} \\ 4 & \frac{14}{3} \\ -\frac{31}{3} & -\frac{7}{3} \end{bmatrix}$

Therefore, the matrix X is:

$\mathbf{X = \begin{bmatrix} -2 & -\frac{10}{3} \\ 4 & \frac{14}{3} \\ -\frac{31}{3} & -\frac{7}{3} \end{bmatrix}}$

Given:

We are given two matrix equations:

To Find:

The matrices X and Y.

Solution:

We can solve this system of matrix equations similar to how we solve a system of linear equations.

Step 1: Find X by adding equations (i) and (ii).

$(X + Y) + (X - Y) = \begin{bmatrix}5&2\\0&9 \end{bmatrix} + \begin{bmatrix}3&6\\0&−1 \end{bmatrix}$

$2X = \begin{bmatrix} 5+3 & 2+6 \\ 0+0 & 9+(-1) \end{bmatrix}$

$2X = \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}$

To find X, multiply by $\frac{1}{2}$:

$X = \frac{1}{2} \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}(8) & \frac{1}{2}(8) \\ \frac{1}{2}(0) & \frac{1}{2}(8) \end{bmatrix}$

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

Step 2: Find Y by subtracting equation (ii) from equation (i).

$(X + Y) - (X - Y) = \begin{bmatrix}5&2\\0&9 \end{bmatrix} - \begin{bmatrix}3&6\\0&−1 \end{bmatrix}$

$X + Y - X + Y = \begin{bmatrix} 5-3 & 2-6 \\ 0-0 & 9-(-1) \end{bmatrix}$

$2Y = \begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}$

To find Y, multiply by $\frac{1}{2}$:

$Y = \frac{1}{2} \begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}(2) & \frac{1}{2}(-4) \\ \frac{1}{2}(0) & \frac{1}{2}(10) \end{bmatrix}$

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

Therefore, the required matrices are:

$\mathbf{X = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}}$

$\mathbf{Y = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}}$

Given:

The matrix equation:

$2\begin{bmatrix}x&5\\7&y−3 \end{bmatrix} + \begin{bmatrix}3&−4\\1&2 \end{bmatrix} = \begin{bmatrix}7&6\\15&14 \end{bmatrix}$

To Find:

The values of x and y.

Solution:

First, perform the scalar multiplication:

$2\begin{bmatrix}x&5\\7&y−3 \end{bmatrix} = \begin{bmatrix}2(x)&2(5)\\2(7)&2(y−3) \end{bmatrix} = \begin{bmatrix}2x&10\\14&2y−6 \end{bmatrix}$

Now substitute this back into the equation:

$\begin{bmatrix}2x&10\\14&2y−6 \end{bmatrix} + \begin{bmatrix}3&−4\\1&2 \end{bmatrix} = \begin{bmatrix}7&6\\15&14 \end{bmatrix}$

Perform the matrix addition on the left side:

$\begin{bmatrix} 2x + 3 & 10 + (-4) \\ 14 + 1 & (2y - 6) + 2 \end{bmatrix} = \begin{bmatrix}7&6\\15&14 \end{bmatrix}$

$\begin{bmatrix} 2x + 3 & 6 \\ 15 & 2y - 4 \end{bmatrix} = \begin{bmatrix}7&6\\15&14 \end{bmatrix}$

By the equality of matrices, the corresponding elements must be equal. Equating the elements:

Equating element (1,1):

$2x + 3 = 7$

$2x = 7 - 3$

$2x = 4$

$x = \frac{4}{2} = 2$

Equating element (2,2):

$2y - 4 = 14$

$2y = 14 + 4$

$2y = 18$

$y = \frac{18}{2} = 9$

(Equating the elements (1,2) and (2,1) gives $6=6$ and $15=15$, which confirms consistency.)

Therefore, the values are:

$\mathbf{x = 2}$ and $\mathbf{y = 9}$.

Given:

Matrix A represents the September sales (in $\textsf{₹}$) for Ramkishan (row 1) and Gurcharan Singh (row 2) for Basmati (col 1), Permal (col 2), and Naura (col 3) rice varieties:

$A = \begin{bmatrix}10000 & 20000 & 30000 \\ 50000 & 30000 & 10000 \end{bmatrix}$

Matrix B represents the October sales (in $\textsf{₹}$) with the same structure:

$B = \begin{bmatrix}5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix}$

(i) Combined sales in September and October:

The combined sales are represented by the sum of matrix A and matrix B (A + B).

$A + B = \begin{bmatrix}10000 & 20000 & 30000 \\ 50000 & 30000 & 10000 \end{bmatrix} + \begin{bmatrix}5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix}$

$A + B = \begin{bmatrix} (10000 + 5000) & (20000 + 10000) & (30000 + 6000) \\ (50000 + 20000) & (30000 + 10000) & (10000 + 10000) \end{bmatrix}$

$A + B = \begin{bmatrix} 15000 & 30000 & 36000 \\ 70000 & 40000 & 20000 \end{bmatrix}$

This matrix represents the combined sales in $\textsf{₹}$ for each farmer and each variety over the two months. For example, Ramkishan's combined sale of Basmati is $\textsf{₹}$15,000, Permal is $\textsf{₹}$30,000, and Naura is $\textsf{₹}$36,000.

(ii) Decrease in sales from September to October:

The decrease in sales is represented by the difference between matrix A and matrix B (A - B).

$A - B = \begin{bmatrix}10000 & 20000 & 30000 \\ 50000 & 30000 & 10000 \end{bmatrix} - \begin{bmatrix}5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix}$

$A - B = \begin{bmatrix} (10000 - 5000) & (20000 - 10000) & (30000 - 6000) \\ (50000 - 20000) & (30000 - 10000) & (10000 - 10000) \end{bmatrix}$

$A - B = \begin{bmatrix} 5000 & 10000 & 24000 \\ 30000 & 20000 & 0 \end{bmatrix}$

This matrix represents the decrease in sales in $\textsf{₹}$ from September to October for each farmer and each variety. For example, the decrease in Ramkishan's sale of Naura rice was $\textsf{₹}$24,000.

(iii) Profit in October:

Both farmers receive a 2% profit on gross sales in October. The profit is calculated as 2% of the October sales (matrix B).

Profit = 2% of B = $0.02 \times B$.

$0.02 \times B = 0.02 \times \begin{bmatrix}5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix}$

$0.02 \times B = \begin{bmatrix} 0.02(5000) & 0.02(10000) & 0.02(6000) \\ 0.02(20000) & 0.02(10000) & 0.02(10000) \end{bmatrix}$

$0.02 \times B = \begin{bmatrix} 100 & 200 & 120 \\ 400 & 200 & 200 \end{bmatrix}$

This matrix represents the profit in $\textsf{₹}$ for each farmer and each variety in October. For example, Ramkishan's profit on Basmati rice in October was $\textsf{₹}$100, and Gurcharan Singh's profit on Permal rice was $\textsf{₹}$200.

Given:

Matrix A = $\begin{bmatrix}6&9\\2&3 \end{bmatrix}$

Matrix B = $\begin{bmatrix}2&6&0\\7&9&8 \end{bmatrix}$

To Find:

The product matrix AB.

Solution:

First, we check if the matrix product AB is defined.

The order of matrix A is 2 × 2.

The order of matrix B is 2 × 3.

For the product AB to be defined, the number of columns in matrix A must be equal to the number of rows in matrix B.

Number of columns in A = 2.

Number of rows in B = 2.

Since the number of columns in A is equal to the number of rows in B, the product AB is defined.

The order of the resulting matrix AB will be (Number of rows in A) × (Number of columns in B), which is 2 × 3.

Let $AB = C = \begin{bmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \end{bmatrix}$.

We calculate each element $c_{ij}$ by multiplying the $i$-th row of A by the $j$-th column of B:

$c_{11} = (\text{1st row of A}) \times (\text{1st column of B}) = (6)(2) + (9)(7) = 12 + 63 = 75$

$c_{12} = (\text{1st row of A}) \times (\text{2nd column of B}) = (6)(6) + (9)(9) = 36 + 81 = 117$

$c_{13} = (\text{1st row of A}) \times (\text{3rd column of B}) = (6)(0) + (9)(8) = 0 + 72 = 72$

$c_{21} = (\text{2nd row of A}) \times (\text{1st column of B}) = (2)(2) + (3)(7) = 4 + 21 = 25$

$c_{22} = (\text{2nd row of A}) \times (\text{2nd column of B}) = (2)(6) + (3)(9) = 12 + 27 = 39$

$c_{23} = (\text{2nd row of A}) \times (\text{3rd column of B}) = (2)(0) + (3)(8) = 0 + 24 = 24$

Substituting these values back into the matrix C:

$AB = \begin{bmatrix} 75 & 117 & 72 \\ 25 & 39 & 24 \end{bmatrix}$

Therefore, the product matrix is:

$\mathbf{AB = \begin{bmatrix} 75 & 117 & 72 \\ 25 & 39 & 24 \end{bmatrix}}$

Given:

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

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

To Find:

1. The product matrix AB.

2. The product matrix BA.

3. Show that AB ≠ BA.

Solution:

1. Calculation of AB:

The order of A is 2 × 3.

The order of B is 3 × 2.

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 × 2.

$AB = \begin{bmatrix}1&−2&3\\−4&2&5 \end{bmatrix} \begin{bmatrix}2&3\\4&5\\2&1 \end{bmatrix}$

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

$AB = \begin{bmatrix} 2 - 8 + 6 & 3 - 10 + 3 \\ -8 + 8 + 10 & -12 + 10 + 5 \end{bmatrix}$

$AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$

So, $\mathbf{AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}}$.

2. Calculation of BA:

The order of B is 3 × 2.

The order of A is 2 × 3.

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 × 3.

$BA = \begin{bmatrix}2&3\\4&5\\2&1 \end{bmatrix} \begin{bmatrix}1&−2&3\\−4&2&5 \end{bmatrix}$

$BA = \begin{bmatrix} (2)(1) + (3)(-4) & (2)(-2) + (3)(2) & (2)(3) + (3)(5) \\ (4)(1) + (5)(-4) & (4)(-2) + (5)(2) & (4)(3) + (5)(5) \\ (2)(1) + (1)(-4) & (2)(-2) + (1)(2) & (2)(3) + (1)(5) \end{bmatrix}$

$BA = \begin{bmatrix} 2 - 12 & -4 + 6 & 6 + 15 \\ 4 - 20 & -8 + 10 & 12 + 25 \\ 2 - 4 & -4 + 2 & 6 + 5 \end{bmatrix}$

$BA = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}$

So, $\mathbf{BA = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}}$.

3. Show that AB ≠ BA:

We found that $AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$, which is a 2 × 2 matrix.

We found that $BA = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}$, which is a 3 × 3 matrix.

Since the matrices AB and BA have different orders, they cannot be equal.

Therefore, AB ≠ BA.

Given:

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

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

Calculation of AB:

Both A and B are 2 × 2 matrices, so the product AB is defined and will be a 2 × 2 matrix.

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

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

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

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

Calculation of BA:

The product BA is also defined and will be a 2 × 2 matrix.

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

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

$BA = \begin{bmatrix} 0 + 0 & 0 - 1 \\ 1 + 0 & 0 + 0 \end{bmatrix}$

$BA = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Conclusion:

Comparing AB and BA:

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

$BA = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Since the corresponding elements are not all equal (e.g., the element in the first row, second column is 1 in AB but -1 in BA), we have:

AB ≠ BA.

This example demonstrates that matrix multiplication is not commutative in general.

Given:

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

Matrix B = $\begin{bmatrix}3&5\\0&0 \end{bmatrix}$

To Find:

The product matrix AB.

Solution:

First, check if the matrix product AB is defined.

The order of matrix A is 2 × 2.

The order of matrix B is 2 × 2.

Since the number of columns in A (2) is equal to the number of rows in B (2), the product AB is defined.

The order of the resulting matrix AB will be 2 × 2.

Now, we compute the product AB:

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

$AB = \begin{bmatrix} (0)(3) + (-1)(0) & (0)(5) + (-1)(0) \\ (0)(3) + (2)(0) & (0)(5) + (2)(0) \end{bmatrix}$

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

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

Therefore, the product matrix is the zero matrix:

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

Note: This example shows that the product of two non-zero matrices can be a zero matrix. This is different from the multiplication of real numbers, where if the product is zero, at least one of the numbers must be zero.

Given:

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

To Find:

1. A(BC)

2. (AB)C

3. Show that (AB)C = A(BC)

Solution:

Step 1: Calculate BC

The order of B is 3 × 2 and the order of C is 2 × 4. The product BC is defined and its order is 3 × 4.

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

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

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

$BC = \begin{bmatrix} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{bmatrix}$

Step 2: Calculate A(BC)

The order of A is 3 × 3 and the order of BC is 3 × 4. The product A(BC) is defined and its order is 3 × 4.

$A(BC) = \begin{bmatrix}1&1&−1\\2&0&3\\3&−1&2 \end{bmatrix} \begin{bmatrix} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{bmatrix}$

$A(BC) = \begin{bmatrix} (1)(7)+(1)(4)+(-1)(7) & (1)(2)+(1)(0)+(-1)(-2) & (1)(-3)+(1)(-4)+(-1)(-11) & (1)(-1)+(1)(2)+(-1)(8) \\ (2)(7)+(0)(4)+(3)(7) & (2)(2)+(0)(0)+(3)(-2) & (2)(-3)+(0)(-4)+(3)(-11) & (2)(-1)+(0)(2)+(3)(8) \\ (3)(7)+(-1)(4)+(2)(7) & (3)(2)+(-1)(0)+(2)(-2) & (3)(-3)+(-1)(-4)+(2)(-11) & (3)(-1)+(-1)(2)+(2)(8) \end{bmatrix}$

$A(BC) = \begin{bmatrix} 7+4-7 & 2+0+2 & -3-4+11 & -1+2-8 \\ 14+0+21 & 4+0-6 & -6+0-33 & -2+0+24 \\ 21-4+14 & 6+0-4 & -9+4-22 & -3-2+16 \end{bmatrix}$

$A(BC) = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}$

Step 3: Calculate AB

The order of A is 3 × 3 and the order of B is 3 × 2. The product AB is defined and its order is 3 × 2.

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

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

$AB = \begin{bmatrix} 1+0+1 & 3+2-4 \\ 2+0-3 & 6+0+12 \\ 3+0-2 & 9-2+8 \end{bmatrix}$

$AB = \begin{bmatrix} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{bmatrix}$

Step 4: Calculate (AB)C

The order of AB is 3 × 2 and the order of C is 2 × 4. The product (AB)C is defined and its order is 3 × 4.

$(AB)C = \begin{bmatrix} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{bmatrix} \begin{bmatrix}1&2&3&-4\\2&0&−2&1 \end{bmatrix}$

$(AB)C = \begin{bmatrix} (2)(1)+(1)(2) & (2)(2)+(1)(0) & (2)(3)+(1)(-2) & (2)(-4)+(1)(1) \\ (-1)(1)+(18)(2) & (-1)(2)+(18)(0) & (-1)(3)+(18)(-2) & (-1)(-4)+(18)(1) \\ (1)(1)+(15)(2) & (1)(2)+(15)(0) & (1)(3)+(15)(-2) & (1)(-4)+(15)(1) \end{bmatrix}$

$(AB)C = \begin{bmatrix} 2+2 & 4+0 & 6-2 & -8+1 \\ -1+36 & -2+0 & -3-36 & 4+18 \\ 1+30 & 2+0 & 3-30 & -4+15 \end{bmatrix}$

$(AB)C = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}$

Step 5: Compare A(BC) and (AB)C

We found:

$A(BC) = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}$

$(AB)C = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}$

Since the resulting matrices are identical, we have shown that (AB)C = A(BC). This demonstrates the associative property of matrix multiplication.

Given:

$A = \begin{bmatrix}0&6&7\\−6&0&8\\7&−8&0 \end{bmatrix}$, $B = \begin{bmatrix}0&1&1\\1&0&2\\1&2&0 \end{bmatrix}$, $C = \begin{bmatrix}2\\−2\\3 \end{bmatrix}$

Calculations:

1. Calculate AC:

The order of A is 3 × 3 and the order of C is 3 × 1. The product AC is defined and its order is 3 × 1.

$AC = \begin{bmatrix}0&6&7\\−6&0&8\\7&−8&0 \end{bmatrix} \begin{bmatrix}2\\−2\\3 \end{bmatrix} = \begin{bmatrix} (0)(2) + (6)(-2) + (7)(3) \\ (-6)(2) + (0)(-2) + (8)(3) \\ (7)(2) + (-8)(-2) + (0)(3) \end{bmatrix}$

$AC = \begin{bmatrix} 0 - 12 + 21 \\ -12 + 0 + 24 \\ 14 + 16 + 0 \end{bmatrix} = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix}$

2. Calculate BC:

The order of B is 3 × 3 and the order of C is 3 × 1. The product BC is defined and its order is 3 × 1.

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

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

3. Calculate A + B:

Both A and B are 3 × 3 matrices, so their sum is defined.

$A + B = \begin{bmatrix}0&6&7\\−6&0&8\\7&−8&0 \end{bmatrix} + \begin{bmatrix}0&1&1\\1&0&2\\1&2&0 \end{bmatrix} = \begin{bmatrix} 0+0 & 6+1 & 7+1 \\ -6+1 & 0+0 & 8+2 \\ 7+1 & -8+2 & 0+0 \end{bmatrix}$

$A + B = \begin{bmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0 \end{bmatrix}$

4. Calculate (A + B)C:

The order of (A + B) is 3 × 3 and the order of C is 3 × 1. The product (A + B)C is defined and its order is 3 × 1.

$(A + B)C = \begin{bmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0 \end{bmatrix} \begin{bmatrix}2\\−2\\3 \end{bmatrix} = \begin{bmatrix} (0)(2) + (7)(-2) + (8)(3) \\ (-5)(2) + (0)(-2) + (10)(3) \\ (8)(2) + (-6)(-2) + (0)(3) \end{bmatrix}$

$(A + B)C = \begin{bmatrix} 0 - 14 + 24 \\ -10 + 0 + 30 \\ 16 + 12 + 0 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$

Verification:

We need to verify that (A + B)C = AC + BC.

Calculate AC + BC:

$AC + BC = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix} + \begin{bmatrix} 1 \\ 8 \\ -2 \end{bmatrix} = \begin{bmatrix} 9+1 \\ 12+8 \\ 30+(-2) \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$

Comparing the results:

$(A + B)C = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$

$AC + BC = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$

Since both results are identical, we have verified that (A + B)C = AC + BC. This demonstrates the distributive property of matrix multiplication over matrix addition.

Given:

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

$I$ is the identity matrix of order 3, $I = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$.

$O$ is the zero matrix of order 3, $O = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0 \end{bmatrix}$.

To Show:

$A^3 – 23A – 40 I = O$

Proof:

Step 1: Calculate A²

$A^2 = A \times A = \begin{bmatrix}1&2&3\\3&−2&1\\4&2&1 \end{bmatrix} \begin{bmatrix}1&2&3\\3&−2&1\\4&2&1 \end{bmatrix}$

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

$A^2 = \begin{bmatrix} 1+6+12 & 2-4+6 & 3+2+3 \\ 3-6+4 & 6+4+2 & 9-2+1 \\ 4+6+4 & 8-4+2 & 12+2+1 \end{bmatrix}$

$A^2 = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix}$

Step 2: Calculate A³

$A^3 = A^2 \times A = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix} \begin{bmatrix}1&2&3\\3&−2&1\\4&2&1 \end{bmatrix}$

$A^3 = \begin{bmatrix} 19(1)+4(3)+8(4) & 19(2)+4(-2)+8(2) & 19(3)+4(1)+8(1) \\ 1(1)+12(3)+8(4) & 1(2)+12(-2)+8(2) & 1(3)+12(1)+8(1) \\ 14(1)+6(3)+15(4) & 14(2)+6(-2)+15(2) & 14(3)+6(1)+15(1) \end{bmatrix}$

$A^3 = \begin{bmatrix} 19+12+32 & 38-8+16 & 57+4+8 \\ 1+36+32 & 2-24+16 & 3+12+8 \\ 14+18+60 & 28-12+30 & 42+6+15 \end{bmatrix}$

$A^3 = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix}$

Step 3: Calculate 23A

$23A = 23 \begin{bmatrix}1&2&3\\3&−2&1\\4&2&1 \end{bmatrix} = \begin{bmatrix} 23(1) & 23(2) & 23(3) \\ 23(3) & 23(-2) & 23(1) \\ 23(4) & 23(2) & 23(1) \end{bmatrix} = \begin{bmatrix} 23 & 46 & 69 \\ 69 & -46 & 23 \\ 92 & 46 & 23 \end{bmatrix}$

Step 4: Calculate 40I

$40I = 40 \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix} 40 & 0 & 0 \\ 0 & 40 & 0 \\ 0 & 0 & 40 \end{bmatrix}$

Step 5: Calculate A³ – 23A – 40 I

$A^3 - 23A - 40I = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix} - \begin{bmatrix} 23 & 46 & 69 \\ 69 & -46 & 23 \\ 92 & 46 & 23 \end{bmatrix} - \begin{bmatrix} 40 & 0 & 0 \\ 0 & 40 & 0 \\ 0 & 0 & 40 \end{bmatrix}$

$A^3 - 23A - 40I = \begin{bmatrix} (63 - 23 - 40) & (46 - 46 - 0) & (69 - 69 - 0) \\ (69 - 69 - 0) & (-6 - (-46) - 40) & (23 - 23 - 0) \\ (92 - 92 - 0) & (46 - 46 - 0) & (63 - 23 - 40) \end{bmatrix}$

$A^3 - 23A - 40I = \begin{bmatrix} (40 - 40) & (0 - 0) & (0 - 0) \\ (0 - 0) & (-6 + 46 - 40) & (0 - 0) \\ (0 - 0) & (0 - 0) & (40 - 40) \end{bmatrix}$

$A^3 - 23A - 40I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & (40 - 40) & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Therefore, $A^3 - 23A - 40I = O$.

Hence Proved.

Given:

Matrix A representing the cost per contact (in paise):

$A = \begin{bmatrix}40 \\ 100 \\ 50 \end{bmatrix} \begin{matrix} \text{Telephone} \\ \text{Housecall} \\ \text{Letter} \end{matrix}$

Matrix B representing the number of contacts in cities X and Y:

$B = \begin{matrix} \text{City X} \\ \text{City Y} \end{matrix} \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{matrix} \text{(Telephone} & \text{Housecall} & \text{Letter)} \end{matrix}$

To Find:

The total amount spent by the group in City X and City Y.

Solution:

The total amount spent in each city can be calculated by multiplying the number of contacts matrix (B) by the cost per contact matrix (A).

The order of B is 2 × 3 and the order of A is 3 × 1. The product BA is defined and will have the order 2 × 1.

The resulting matrix BA will represent the total cost (in paise) for each city:

$BA = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix}$

Calculate the elements of the product matrix:

Row 1 (City X Cost): $(1000)(40) + (500)(100) + (5000)(50) = 40000 + 50000 + 250000 = 340000$ paise.

Row 2 (City Y Cost): $(3000)(40) + (1000)(100) + (10000)(50) = 120000 + 100000 + 500000 = 720000$ paise.

So, the total cost matrix (in paise) is:

$BA = \begin{bmatrix} 340000 \\ 720000 \end{bmatrix} \begin{matrix} \text{Cost in City X} \\ \text{Cost in City Y} \end{matrix}$

To express the cost in Rupees, we divide by 100:

Cost in City X = $\frac{340000}{100} = \textsf{₹} 3400$.

Cost in City Y = $\frac{720000}{100} = \textsf{₹} 7200$.

Therefore, the total amount spent by the group is $\textsf{₹} 3400$ in city X and $\textsf{₹} 7200$ in city Y.

Given:

The matrices are:

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

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

$C = \begin{bmatrix}−2&5\\3&4 \end{bmatrix}$

(i) Find A + B

Matrix addition is performed by adding corresponding elements.

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

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

$= \begin{bmatrix}3 & 7 \\ 1 & 7 \end{bmatrix}$

(ii) Find A – B

Matrix subtraction is performed by subtracting corresponding elements.

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

$= \begin{bmatrix}2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix}$

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

$= \begin{bmatrix}1 & 1 \\ 5 & -3 \end{bmatrix}$

(iii) Find 3A – C

First, perform scalar multiplication of 3 with matrix A.

$3A = 3 \begin{bmatrix}2&4\\3&2 \end{bmatrix} = \begin{bmatrix}3 \times 2 & 3 \times 4 \\ 3 \times 3 & 3 \times 2 \end{bmatrix} = \begin{bmatrix}6 & 12 \\ 9 & 6 \end{bmatrix}$

Now, subtract matrix C from 3A.

$3A - C = \begin{bmatrix}6 & 12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix}−2&5\\3&4 \end{bmatrix}$

$= \begin{bmatrix}6-(-2) & 12-5 \\ 9-3 & 6-4 \end{bmatrix}$

$= \begin{bmatrix}6+2 & 7 \\ 6 & 2 \end{bmatrix}$

$= \begin{bmatrix}8 & 7 \\ 6 & 2 \end{bmatrix}$

(iv) Find AB

Matrix multiplication AB is performed by multiplying the rows of A by the columns of B.

$AB = \begin{bmatrix}2&4\\3&2 \end{bmatrix} \begin{bmatrix}1&3\\−2&5 \end{bmatrix}$

The element in the $i$-th row and $j$-th column of the product is the dot product of the $i$-th row of A and the $j$-th column of B.

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

$= \begin{bmatrix}2-8 & 6+20 \\ 3-4 & 9+10 \end{bmatrix}$

$= \begin{bmatrix}-6 & 26 \\ -1 & 19 \end{bmatrix}$

(v) Find BA

Matrix multiplication BA is performed by multiplying the rows of B by the columns of A.

$BA = \begin{bmatrix}1&3\\−2&5 \end{bmatrix} \begin{bmatrix}2&4\\3&2 \end{bmatrix}$

The element in the $i$-th row and $j$-th column of the product is the dot product of the $i$-th row of B and the $j$-th column of A.

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

$= \begin{bmatrix}2+9 & 4+6 \\ -4+15 & -8+10 \end{bmatrix}$

$= \begin{bmatrix}11 & 10 \\ 11 & 2 \end{bmatrix}$

Solution:

We compute the sum of the matrices by adding their corresponding elements.

(i)

$\begin{bmatrix}a&b\\−b&a \end{bmatrix} + \begin{bmatrix}a&b\\b&a \end{bmatrix}$

$= \begin{bmatrix}a+a & b+b \\ −b+b & a+a \end{bmatrix}$

$= \begin{bmatrix}2a & 2b \\ 0 & 2a \end{bmatrix}$

(ii)

$\begin{bmatrix}a^2 + b^2&b^2 + c^2\\a^2 + c^2&a^2 + b^2 \end{bmatrix} + \begin{bmatrix}2ab&2bc\\−2ac&−2ab \end{bmatrix}$

$= \begin{bmatrix}a^2 + b^2 + 2ab & b^2 + c^2 + 2bc \\ a^2 + c^2 - 2ac & a^2 + b^2 - 2ab \end{bmatrix}$

Using the algebraic identities $(x+y)^2 = x^2 + 2xy + y^2$ and $(x-y)^2 = x^2 - 2xy + y^2$:

$= \begin{bmatrix}(a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2 \end{bmatrix}$

(iii)

$\begin{bmatrix}−1&4&−6\\8&5&16\\2&8&5 \end{bmatrix} + \begin{bmatrix}12&7&6\\8&0&5\\3&2&4 \end{bmatrix}$

$= \begin{bmatrix}−1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix}$

$= \begin{bmatrix}11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}$

(iv)

$\begin{bmatrix}\cos^2x&\sin^2x\\\sin^2x&\cos^2x \end{bmatrix} + \begin{bmatrix}\sin^2x&\cos^2x \\ \cos^2x&\sin^2x \end{bmatrix}$

$= \begin{bmatrix}\cos^2x + \sin^2x & \sin^2x + \cos^2x \\ \sin^2x + \cos^2x & \cos^2x + \sin^2x \end{bmatrix}$

Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$:

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

Solution:

We compute the indicated matrix products using the rules of matrix multiplication.

(i)

$\begin{bmatrix}a&b\\−b&a \end{bmatrix} \begin{bmatrix}a&−b\\b&a \end{bmatrix}$

$= \begin{bmatrix} (a)(a)+(b)(b) & (a)(-b)+(b)(a) \\ (-b)(a)+(a)(b) & (-b)(-b)+(a)(a) \end{bmatrix}$

$= \begin{bmatrix} a^2+b^2 & -ab+ab \\ -ab+ab & b^2+a^2 \end{bmatrix}$

$= \begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix}$

(ii)

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

The first matrix is 3x1 and the second is 1x3. The resulting matrix will be 3x3.

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

$= \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$

(iii)

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

The first matrix is 2x2 and the second is 2x3. The resulting matrix will be 2x3.

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

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

$= \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$

(iv)

$\begin{bmatrix}2&3&4\\3&4&5\\4&5&6 \end{bmatrix} \begin{bmatrix}1&−3&5\\0&2&4\\3&0&5 \end{bmatrix}$

Both matrices are 3x3. The resulting matrix will be 3x3.

$= \begin{bmatrix} (2)(1)+(3)(0)+(4)(3) & (2)(-3)+(3)(2)+(4)(0) & (2)(5)+(3)(4)+(4)(5) \\ (3)(1)+(4)(0)+(5)(3) & (3)(-3)+(4)(2)+(5)(0) & (3)(5)+(4)(4)+(5)(5) \\ (4)(1)+(5)(0)+(6)(3) & (4)(-3)+(5)(2)+(6)(0) & (4)(5)+(5)(4)+(6)(5) \end{bmatrix}$

$= \begin{bmatrix} 2+0+12 & -6+6+0 & 10+12+20 \\ 3+0+15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30 \end{bmatrix}$

$= \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}$

(v)

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

The first matrix is 3x2 and the second is 2x3. The resulting matrix will be 3x3.

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

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

$= \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix}$

(vi)

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

The first matrix is 2x3 and the second is 3x2. The resulting matrix will be 2x2.

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

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

$= \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$

Given:

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

$B = \begin{bmatrix}3&−1&2\\4&2&5\\2&0&3 \end{bmatrix}$

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

Compute (A + B):

We add the corresponding elements of matrices A and B.

$A + B = \begin{bmatrix}1&2&−3\\5&0&2\\1&−1&1 \end{bmatrix} + \begin{bmatrix}3&−1&2\\4&2&5\\2&0&3 \end{bmatrix}$

$= \begin{bmatrix}1+3 & 2+(-1) & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix}$

$= \begin{bmatrix}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix}$

Compute (B - C):

We subtract the corresponding elements of matrix C from matrix B.

$B - C = \begin{bmatrix}3&−1&2\\4&2&5\\2&0&3 \end{bmatrix} - \begin{bmatrix}4&1&2\\0&3&2\\1&−2&3 \end{bmatrix}$

$= \begin{bmatrix}3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{bmatrix}$

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

Verify that A + (B – C) = (A + B) – C:

First, compute the left-hand side (LHS): A + (B - C)

$A + (B - C) = \begin{bmatrix}1&2&−3\\5&0&2\\1&−1&1 \end{bmatrix} + \begin{bmatrix}-1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}$

$= \begin{bmatrix}1+(-1) & 2+(-2) & -3+0 \\ 5+4 & 0+(-1) & 2+3 \\ 1+1 & -1+2 & 1+0 \end{bmatrix}$

$= \begin{bmatrix}0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$

Next, compute the right-hand side (RHS): (A + B) - C

$(A + B) - C = \begin{bmatrix}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} - \begin{bmatrix}4&1&2\\0&3&2\\1&−2&3 \end{bmatrix}$

$= \begin{bmatrix}4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1-(-2) & 4-3 \end{bmatrix}$

$= \begin{bmatrix}0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$

Since the LHS = $\begin{bmatrix}0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$ and the RHS = $\begin{bmatrix}0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$, we have LHS = RHS.

Therefore, A + (B – C) = (A + B) – C is verified.

Given:

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

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

To Find: $3A - 5B$

Solution:

First, compute the scalar multiple $3A$:

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

$= \begin{bmatrix}3 \times \frac{2}{3}&3 \times 1&3 \times \frac{5}{3}\\3 \times \frac{1}{3}&3 \times \frac{2}{3}&3 \times \frac{4}{3}\\ 3 \times \frac{7}{3}&3 \times 2&3 \times \frac{2}{3} \end{bmatrix}$

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

Next, compute the scalar multiple $5B$:

$5B = 5 \times \begin{bmatrix}\frac{2}{5}&\frac{3}{5}&1\\\frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\\frac{7}{5}&\frac{6}{5}&\frac{2}{5} \end{bmatrix}$

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

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

Now, compute $3A - 5B$ by subtracting the corresponding elements:

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

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

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

Therefore, $3A - 5B$ is the zero matrix of order 3x3.

Solution:

We are asked to simplify the expression:

$\cos \theta\begin{bmatrix}\cos\theta&\sin\theta\\− \sin\theta&\cos\theta \end{bmatrix} + \sin \theta \begin{bmatrix}\sin\theta&−\cos\theta\\\cos\theta&\sin\theta \end{bmatrix}$

First, perform the scalar multiplication for each term:

$\cos \theta \begin{bmatrix}\cos\theta&\sin\theta\\− \sin\theta&\cos\theta \end{bmatrix} = \begin{bmatrix}\cos \theta \times \cos\theta&\cos \theta \times \sin\theta\\\cos \theta \times (− \sin\theta)&\cos \theta \times \cos\theta \end{bmatrix} = \begin{bmatrix}\cos^2\theta&\cos\theta\sin\theta\\− \sin\theta\cos\theta&\cos^2\theta \end{bmatrix}$

$\sin \theta \begin{bmatrix}\sin\theta&−\cos\theta\\\cos\theta&\sin\theta \end{bmatrix} = \begin{bmatrix}\sin \theta \times \sin\theta&\sin \theta \times (−\cos\theta)\\\sin \theta \times \cos\theta&\sin \theta \times \sin\theta \end{bmatrix} = \begin{bmatrix}\sin^2\theta&−\sin\theta\cos\theta\\\sin\theta\cos\theta&\sin^2\theta \end{bmatrix}$

Now, add the two resulting matrices:

$\begin{bmatrix}\cos^2\theta&\cos\theta\sin\theta\\− \sin\theta\cos\theta&\cos^2\theta \end{bmatrix} + \begin{bmatrix}\sin^2\theta&−\sin\theta\cos\theta\\\sin\theta\cos\theta&\sin^2\theta \end{bmatrix}$

$= \begin{bmatrix}\cos^2\theta + \sin^2\theta & \cos\theta\sin\theta - \sin\theta\cos\theta \\ − \sin\theta\cos\theta + \sin\theta\cos\theta & \cos^2\theta + \sin^2\theta \end{bmatrix}$

Using the trigonometric identity $\cos^2\theta + \sin^2\theta = 1$ and simplifying the off-diagonal terms:

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

The simplified result is the identity matrix of order 2.

Solution:

(i)

Given:

To Find: Matrices X and Y

Solution:

Add equations (1) and (2):

$(X + Y) + (X - Y) = \begin{bmatrix}7&0\\2&5 \end{bmatrix} + \begin{bmatrix}3&0\\0&3 \end{bmatrix}$

$2X = \begin{bmatrix}7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix}$

$2X = \begin{bmatrix}10 & 0 \\ 2 & 8 \end{bmatrix}$

$X = \frac{1}{2} \begin{bmatrix}10 & 0 \\ 2 & 8 \end{bmatrix}$

$X = \begin{bmatrix} \frac{1}{2}(10) & \frac{1}{2}(0) \\ \frac{1}{2}(2) & \frac{1}{2}(8) \end{bmatrix}$

$\mathbf{X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}}$

Subtract equation (2) from equation (1):

$(X + Y) - (X - Y) = \begin{bmatrix}7&0\\2&5 \end{bmatrix} - \begin{bmatrix}3&0\\0&3 \end{bmatrix}$

$2Y = \begin{bmatrix}7-3 & 0-0 \\ 2-0 & 5-3 \end{bmatrix}$

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

$Y = \frac{1}{2} \begin{bmatrix}4 & 0 \\ 2 & 2 \end{bmatrix}$

$Y = \begin{bmatrix} \frac{1}{2}(4) & \frac{1}{2}(0) \\ \frac{1}{2}(2) & \frac{1}{2}(2) \end{bmatrix}$

$\mathbf{Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}}$

(ii)

Given:

To Find: Matrices X and Y

Solution:

Multiply equation (3) by 3 and equation (4) by 2:

Subtract equation (6) from equation (5):

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

$5Y = \begin{bmatrix}6-4 & 9-(-4) \\ 12-(-2) & 0-10 \end{bmatrix}$

$5Y = \begin{bmatrix}2 & 13 \\ 14 & -10 \end{bmatrix}$

$Y = \frac{1}{5} \begin{bmatrix}2 & 13 \\ 14 & -10 \end{bmatrix}$

$\mathbf{Y = \begin{bmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}}$

Multiply equation (3) by 2 and equation (4) by 3:

Subtract equation (7) from equation (8):

$(9X + 6Y) - (4X + 6Y) = \begin{bmatrix}6&−6\\−3&15 \end{bmatrix} - \begin{bmatrix}4&6\\8&0 \end{bmatrix}$

$5X = \begin{bmatrix}6-4 & -6-6 \\ -3-8 & 15-0 \end{bmatrix}$

$5X = \begin{bmatrix}2 & -12 \\ -11 & 15 \end{bmatrix}$

$X = \frac{1}{5} \begin{bmatrix}2 & -12 \\ -11 & 15 \end{bmatrix}$

$\mathbf{X = \begin{bmatrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \end{bmatrix}}$

Given:

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

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

To Find: Matrix X

Solution:

We have the equation:

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

To find X, we first isolate 2X by subtracting Y from both sides:

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

Substitute the given matrix Y into the equation:

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

Perform the matrix subtraction by subtracting corresponding elements:

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

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

Now, divide the resulting matrix by 2 (or multiply by $\frac{1}{2}$) to find X:

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

$X = \begin{bmatrix}\frac{1}{2}(-2) & \frac{1}{2}(-2) \\ \frac{1}{2}(-4) & \frac{1}{2}(-2) \end{bmatrix}$

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

Therefore, the matrix $\mathbf{X = \begin{bmatrix}-1 & -1 \\ -2 & -1 \end{bmatrix}}$.

Given:

The matrix equation is:

$2\begin{bmatrix}1&3\\0&x \end{bmatrix} + \begin{bmatrix}y&0\\1&2 \end{bmatrix} = \begin{bmatrix}5&6\\1&8 \end{bmatrix}$

To Find: The values of $x$ and $y$.

Solution:

First, perform the scalar multiplication on the first term:

$2\begin{bmatrix}1&3\\0&x \end{bmatrix} = \begin{bmatrix}2 \times 1 & 2 \times 3 \\ 2 \times 0 & 2 \times x \end{bmatrix} = \begin{bmatrix}2 & 6 \\ 0 & 2x \end{bmatrix}$

Now substitute this back into the original equation:

$\begin{bmatrix}2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix}y&0\\1&2 \end{bmatrix} = \begin{bmatrix}5&6\\1&8 \end{bmatrix}$

Perform the matrix addition on the left side:

$\begin{bmatrix}2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix} = \begin{bmatrix}5&6\\1&8 \end{bmatrix}$

$\begin{bmatrix}2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix}5&6\\1&8 \end{bmatrix}$

For two matrices to be equal, their corresponding elements must be equal. Therefore, we can equate the corresponding elements:

Element (1,1): $2 + y = 5$

Element (1,2): $6 = 6$ (This is consistent)

Element (2,1): $1 = 1$ (This is consistent)

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

Now, solve the equations for $y$ and $x$.

From the equation $2 + y = 5$:

$y = 5 - 2$

$y = 3$

From the equation $2x + 2 = 8$:

$2x = 8 - 2$

$2x = 6$

$x = \frac{6}{2}$

$x = 3$

Thus, the values are $\mathbf{x = 3}$ and $\mathbf{y = 3}$.

Given:

The matrix equation is:

$2\begin{bmatrix}x&z\\y&t \end{bmatrix} + 3\begin{bmatrix}1&−1\\0&2 \end{bmatrix} = 3\begin{bmatrix}3&5\\4&6 \end{bmatrix}$

To Find: The values of $x, y, z,$ and $t$.

Solution:

First, perform the scalar multiplications on both sides of the equation:

$2\begin{bmatrix}x&z\\y&t \end{bmatrix} = \begin{bmatrix}2x & 2z \\ 2y & 2t \end{bmatrix}$

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

$3\begin{bmatrix}3&5\\4&6 \end{bmatrix} = \begin{bmatrix}3 \times 3 & 3 \times 5 \\ 3 \times 4 & 3 \times 6 \end{bmatrix} = \begin{bmatrix}9 & 15 \\ 12 & 18 \end{bmatrix}$

Substitute these results back into the original equation:

$\begin{bmatrix}2x & 2z \\ 2y & 2t \end{bmatrix} + \begin{bmatrix}3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix}9 & 15 \\ 12 & 18 \end{bmatrix}$

Perform the matrix addition on the left side:

$\begin{bmatrix}2x+3 & 2z+(-3) \\ 2y+0 & 2t+6 \end{bmatrix} = \begin{bmatrix}9 & 15 \\ 12 & 18 \end{bmatrix}$

$\begin{bmatrix}2x+3 & 2z-3 \\ 2y & 2t+6 \end{bmatrix} = \begin{bmatrix}9 & 15 \\ 12 & 18 \end{bmatrix}$

For the matrices to be equal, their corresponding elements must be equal. Equating the corresponding elements gives us four equations:

$2x + 3 = 9$

$2z - 3 = 15$

$2y = 12$

$2t + 6 = 18$

Now, solve each equation:

For $x$:

$2x = 9 - 3$

$2x = 6$

$x = \frac{6}{2} = 3$

For $z$:

$2z = 15 + 3$

$2z = 18$

$z = \frac{18}{2} = 9$

For $y$:

$y = \frac{12}{2} = 6$

For $t$:

$2t = 18 - 6$

$2t = 12$

$t = \frac{12}{2} = 6$

Therefore, the values are $\mathbf{x = 3}$, $\mathbf{y = 6}$, $\mathbf{z = 9}$, and $\mathbf{t = 6}$.

Given:

The matrix equation is:

$x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix}−1\\1 \end{bmatrix} = \begin{bmatrix}10\\5 \end{bmatrix}$

To Find: The values of $x$ and $y$.

Solution:

First, perform the scalar multiplication on the terms on the left side:

$x\begin{bmatrix}2\\3 \end{bmatrix} = \begin{bmatrix}2x\\3x \end{bmatrix}$

$y\begin{bmatrix}−1\\1 \end{bmatrix} = \begin{bmatrix}−y\\y \end{bmatrix}$

Substitute these back into the original equation:

$\begin{bmatrix}2x\\3x \end{bmatrix} + \begin{bmatrix}−y\\y \end{bmatrix} = \begin{bmatrix}10\\5 \end{bmatrix}$

Perform the matrix addition on the left side:

$\begin{bmatrix}2x + (-y)\\3x + y \end{bmatrix} = \begin{bmatrix}10\\5 \end{bmatrix}$

$\begin{bmatrix}2x - y\\3x + y \end{bmatrix} = \begin{bmatrix}10\\5 \end{bmatrix}$

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

We can solve this system. Adding equation (i) and equation (ii):

$(2x - y) + (3x + y) = 10 + 5$

$5x = 15$

$x = \frac{15}{5}$

$x = 3$

Substitute the value of $x = 3$ into equation (i):

$2(3) - y = 10$

$6 - y = 10$

$-y = 10 - 6$

$-y = 4$

$y = -4$

Thus, the values are $\mathbf{x = 3}$ and $\mathbf{y = -4}$.

Given:

The matrix equation is:

$3\begin{bmatrix}x&y\\z&w \end{bmatrix} = \begin{bmatrix}x&6\\−1&2w \end{bmatrix} + \begin{bmatrix}4&x+y\\z+w&3 \end{bmatrix}$

To Find: The values of $x, y, z,$ and $w$.

Solution:

First, perform the scalar multiplication on the left side:

$3\begin{bmatrix}x&y\\z&w \end{bmatrix} = \begin{bmatrix}3x & 3y \\ 3z & 3w \end{bmatrix}$

Next, perform the matrix addition on the right side:

$\begin{bmatrix}x&6\\−1&2w \end{bmatrix} + \begin{bmatrix}4&x+y\\z+w&3 \end{bmatrix} = \begin{bmatrix}x+4 & 6+(x+y) \\ -1+(z+w) & 2w+3 \end{bmatrix} = \begin{bmatrix}x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix}$

Now, equate the resulting matrices from the left and right sides:

$\begin{bmatrix}3x & 3y \\ 3z & 3w \end{bmatrix} = \begin{bmatrix}x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix}$

For the matrices to be equal, their corresponding elements must be equal. Equating the corresponding elements gives us four equations:

Now, solve each equation:

From equation (i):

$3x - x = 4$

$2x = 4$

$x = \frac{4}{2} = 2$

From equation (iv):

$3w - 2w = 3$

$w = 3$

Substitute the value of $x=2$ into equation (ii):

$3y = 2 + y + 6$

$3y = y + 8$

$3y - y = 8$

$2y = 8$

$y = \frac{8}{2} = 4$

Substitute the value of $w=3$ into equation (iii):

$3z = z + 3 - 1$

$3z = z + 2$

$3z - z = 2$

$2z = 2$

$z = \frac{2}{2} = 1$

Therefore, the values are $\mathbf{x = 2}$, $\mathbf{y = 4}$, $\mathbf{z = 1}$, and $\mathbf{w = 3}$.

Given:

$F(x) = \begin{bmatrix}\cos x&−\sin x&0\\\sin x&\cos x&0\\0&0&1 \end{bmatrix}$

To Prove: $F(x) F(y) = F(x + y)$

Proof:

First, let's write down the matrix $F(y)$ by replacing $x$ with $y$ in the definition of $F(x)$:

$F(y) = \begin{bmatrix}\cos y&−\sin y&0\\\sin y&\cos y&0\\0&0&1 \end{bmatrix}$

Now, let's compute the product $F(x) F(y)$ using matrix multiplication:

$F(x) F(y) = \begin{bmatrix}\cos x&−\sin x&0\\\sin x&\cos x&0\\0&0&1 \end{bmatrix} \begin{bmatrix}\cos y&−\sin y&0\\\sin y&\cos y&0\\0&0&1 \end{bmatrix}$

Multiply the rows of the first matrix by the columns of the second matrix:

$= \begin{bmatrix} (\cos x)(\cos y)+(-\sin x)(\sin y)+(0)(0) & (\cos x)(-\sin y)+(-\sin x)(\cos y)+(0)(0) & (\cos x)(0)+(-\sin x)(0)+(0)(1) \\ (\sin x)(\cos y)+(\cos x)(\sin y)+(0)(0) & (\sin x)(-\sin y)+(\cos x)(\cos y)+(0)(0) & (\sin x)(0)+(\cos x)(0)+(0)(1) \\ (0)(\cos y)+(0)(\sin y)+(1)(0) & (0)(-\sin y)+(0)(\cos y)+(1)(0) & (0)(0)+(0)(0)+(1)(1) \end{bmatrix}$

Simplify each element:

$= \begin{bmatrix} \cos x \cos y - \sin x \sin y & -(\cos x \sin y + \sin x \cos y) & 0 \\ \sin x \cos y + \cos x \sin y & \cos x \cos y - \sin x \sin y & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Now, we use the trigonometric angle addition identities:

$\cos(A+B) = \cos A \cos B - \sin A \sin B$

$\sin(A+B) = \sin A \cos B + \cos A \sin B$

Applying these identities to the elements of the resulting matrix:

$F(x) F(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Now, let's find the matrix $F(x + y)$ by replacing $x$ with $(x+y)$ in the definition of $F(x)$:

$F(x+y) = \begin{bmatrix}\cos(x+y)&−\sin(x+y)&0\\\sin(x+y)&\cos(x+y)&0\\0&0&1 \end{bmatrix}$

Comparing the result of $F(x) F(y)$ with $F(x+y)$, we see that they are identical.

Therefore, $\mathbf{F(x) F(y) = F(x + y)}$.

Hence Proved.

Solution:

We need to show that matrix multiplication is generally not commutative (i.e., AB ≠ BA).

(i)

To Show: $\begin{bmatrix}5&−1\\6&7 \end{bmatrix} \begin{bmatrix}2&1\\3&4 \end{bmatrix} ≠ \begin{bmatrix}2&1\\3&4 \end{bmatrix} \begin{bmatrix}5&−1\\6&7 \end{bmatrix}$

Calculate the Left Hand Side (LHS):

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

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

$= \begin{bmatrix} 10-3 & 5-4 \\ 12+21 & 6+28 \end{bmatrix}$

$= \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$

Calculate the Right Hand Side (RHS):

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

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

$= \begin{bmatrix} 10+6 & -2+7 \\ 15+24 & -3+28 \end{bmatrix}$

$= \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$

Comparison:

Since LHS = $\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$ and RHS = $\begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$, we see that LHS ≠ RHS.

Thus, $\begin{bmatrix}5&−1\\6&7 \end{bmatrix} \begin{bmatrix}2&1\\3&4 \end{bmatrix} ≠ \begin{bmatrix}2&1\\3&4 \end{bmatrix} \begin{bmatrix}5&−1\\6&7 \end{bmatrix}$ is shown.

(ii)

To Show: $\begin{bmatrix}1&2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix}−1&1&0\\0&−1&1\\2&3&4 \end{bmatrix} ≠ \begin{bmatrix}−1&1&0\\0&−1&1\\2&3&4 \end{bmatrix} \begin{bmatrix}1&2&3\\0&1&0\\1&1&0 \end{bmatrix}$

Calculate the Left Hand Side (LHS):

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

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

$= \begin{bmatrix} -1+0+6 & 1-2+9 & 0+2+12 \\ 0+0+0 & 0-1+0 & 0+1+0 \\ -1+0+0 & 1-1+0 & 0+1+0 \end{bmatrix}$

$= \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$

Calculate the Right Hand Side (RHS):

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

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

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

$= \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$

Comparison:

Since LHS = $\begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$ and RHS = $\begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$, we see that LHS ≠ RHS.

Thus, $\begin{bmatrix}1&2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix}−1&1&0\\0&−1&1\\2&3&4 \end{bmatrix} ≠ \begin{bmatrix}−1&1&0\\0&−1&1\\2&3&4 \end{bmatrix} \begin{bmatrix}1&2&3\\0&1&0\\1&1&0 \end{bmatrix}$ is shown.

Given:

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

The identity matrix $I = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$

To Find: The value of the matrix expression $A^2 - 5A + 6I$.

Solution:

First, we need to calculate $A^2$.

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

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

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

$A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$

Next, calculate $5A$.

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

$5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}$

Next, calculate $6I$.

$6I = 6 \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix} 6 \times 1 & 6 \times 0 & 6 \times 0 \\ 6 \times 0 & 6 \times 1 & 6 \times 0 \\ 6 \times 0 & 6 \times 0 & 6 \times 1 \end{bmatrix}$

$6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

Finally, compute $A^2 - 5A + 6I$.

$A^2 - 5A + 6I = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

$= \begin{bmatrix} (5-10+6) & (-1-0+0) & (2-5+0) \\ (9-10+0) & (-2-5+6) & (5-15+0) \\ (0-5+0) & (-1-(-5)+0) & (-2-0+6) \end{bmatrix}$

$= \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$

Therefore, $A^2 - 5A + 6I = \mathbf{\begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}}$.

Given:

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

The identity matrix $I = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$

The zero matrix $O = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0 \end{bmatrix}$

To Prove: $A^3 - 6A^2 + 7A + 2I = O$

Proof:

First, calculate $A^2$.

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

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

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

$A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}$

Next, calculate $A^3$.

$A^3 = A^2 \times A = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

$= \begin{bmatrix} (5)(1)+(0)(0)+(8)(2) & (5)(0)+(0)(2)+(8)(0) & (5)(2)+(0)(1)+(8)(3) \\ (2)(1)+(4)(0)+(5)(2) & (2)(0)+(4)(2)+(5)(0) & (2)(2)+(4)(1)+(5)(3) \\ (8)(1)+(0)(0)+(13)(2) & (8)(0)+(0)(2)+(13)(0) & (8)(2)+(0)(1)+(13)(3) \end{bmatrix}$

$= \begin{bmatrix} 5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39 \end{bmatrix}$

$A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}$

Now, calculate the terms $6A^2$, $7A$, and $2I$.

$6A^2 = 6 \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} = \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix}$

$7A = 7 \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix} = \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix}$

$2I = 2 \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

Finally, compute the expression $A^3 - 6A^2 + 7A + 2I$.

$A^3 - 6A^2 + 7A + 2I = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} + \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

$= \begin{bmatrix} (21-30+7+2) & (0-0+0+0) & (34-48+14+0) \\ (12-12+0+0) & (8-24+14+2) & (23-30+7+0) \\ (34-48+14+0) & (0-0+0+0) & (55-78+21+2) \end{bmatrix}$

$= \begin{bmatrix} (30-30) & 0 & (48-48) \\ 0 & (24-24) & (30-30) \\ (48-48) & 0 & (78-78) \end{bmatrix}$

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

$= O$

Thus, $\mathbf{A^3 - 6A^2 + 7A + 2I = O}$ is proved.

Given:

$A = \begin{bmatrix}3&−2\\4&−2 \end{bmatrix}$

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

The equation $A^2 = kA - 2I$

To Find: The value of $k$.

Solution:

First, calculate $A^2$.

$A^2 = A \times A = \begin{bmatrix}3&−2\\4&−2 \end{bmatrix} \begin{bmatrix}3&−2\\4&−2 \end{bmatrix}$

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

$= \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix}$

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

Next, calculate $kA$ and $2I$.

$kA = k \begin{bmatrix}3&−2\\4&−2 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}$

$2I = 2 \begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

Now substitute these into the given equation $A^2 = kA - 2I$:

$\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

Perform the matrix subtraction on the right side:

$\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k-0 \\ 4k-0 & -2k-2 \end{bmatrix}$

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

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

Element (1,1): $1 = 3k - 2$

Element (1,2): $-2 = -2k$

Element (2,1): $4 = 4k$

Element (2,2): $-4 = -2k - 2$

Solve any one of these equations for $k$. Let's use the simplest one, from element (2,1):

$4 = 4k$

$k = \frac{4}{4}$

$k = 1$

We can check this with the other equations:

From (1,1): $1 = 3(1) - 2 \implies 1 = 3 - 2 \implies 1 = 1$ (True)

From (1,2): $-2 = -2(1) \implies -2 = -2$ (True)

From (2,2): $-4 = -2(1) - 2 \implies -4 = -2 - 2 \implies -4 = -4$ (True)

All equations give $k=1$.

Therefore, the value of $\mathbf{k = 1}$.

Given:

$A = \begin{bmatrix}0&−\tan \frac{\alpha}{2}\\\tan \frac{\alpha}{2}&0 \end{bmatrix}$

$I = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$ (Identity matrix of order 2)

To Show: $I + A = (I – A) \begin{bmatrix}\cos \alpha&−\sin \alpha \\ \sin \alpha&\cos \alpha \end{bmatrix}$

Proof:

Let $t = \tan \frac{\alpha}{2}$. Then the matrix A becomes:

$A = \begin{bmatrix}0&-t\\t&0 \end{bmatrix}$

Calculate the Left Hand Side (LHS):

$LHS = I + A = \begin{bmatrix}1&0\\0&1 \end{bmatrix} + \begin{bmatrix}0&-t\\t&0 \end{bmatrix}$

$LHS = \begin{bmatrix}1+0 & 0+(-t) \\ 0+t & 1+0 \end{bmatrix} = \begin{bmatrix}1 & -t \\ t & 1 \end{bmatrix}$

Calculate the terms for the Right Hand Side (RHS):

First, calculate $I - A$:

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

$I - A = \begin{bmatrix}1-0 & 0-(-t) \\ 0-t & 1-0 \end{bmatrix} = \begin{bmatrix}1 & t \\ -t & 1 \end{bmatrix}$

Next, express $\cos \alpha$ and $\sin \alpha$ in terms of $t = \tan \frac{\alpha}{2}$ using the half-angle tangent formulas:

$\cos \alpha = \frac{1 - \tan^2 \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} = \frac{1 - t^2}{1 + t^2}$

$\sin \alpha = \frac{2 \tan \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} = \frac{2t}{1 + t^2}$

Substitute these into the rotation matrix:

$\begin{bmatrix}\cos \alpha&−\sin \alpha \\ \sin \alpha&\cos \alpha \end{bmatrix} = \begin{bmatrix}\frac{1 - t^2}{1 + t^2} & -\frac{2t}{1 + t^2} \\ \frac{2t}{1 + t^2} & \frac{1 - t^2}{1 + t^2} \end{bmatrix}$

Now, calculate the RHS:

$RHS = (I - A) \begin{bmatrix}\cos \alpha&−\sin \alpha \\ \sin \alpha&\cos \alpha \end{bmatrix}$

$RHS = \begin{bmatrix}1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix}\frac{1 - t^2}{1 + t^2} & -\frac{2t}{1 + t^2} \\ \frac{2t}{1 + t^2} & \frac{1 - t^2}{1 + t^2} \end{bmatrix}$

Factor out the common term $\frac{1}{1+t^2}$:

$RHS = \frac{1}{1+t^2} \begin{bmatrix}1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix}1 - t^2 & -2t \\ 2t & 1 - t^2 \end{bmatrix}$

Perform the matrix multiplication:

$= \frac{1}{1+t^2} \begin{bmatrix} (1)(1-t^2)+(t)(2t) & (1)(-2t)+(t)(1-t^2) \\ (-t)(1-t^2)+(1)(2t) & (-t)(-2t)+(1)(1-t^2) \end{bmatrix}$

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

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

Factor out $(1+t^2)$ from the elements where possible:

$= \frac{1}{1+t^2} \begin{bmatrix} 1+t^2 & -t(1+t^2) \\ t(1+t^2) & 1+t^2 \end{bmatrix}$

Multiply the scalar $\frac{1}{1+t^2}$ into the matrix:

$RHS = \begin{bmatrix} \frac{1+t^2}{1+t^2} & \frac{-t(1+t^2)}{1+t^2} \\ \frac{t(1+t^2)}{1+t^2} & \frac{1+t^2}{1+t^2} \end{bmatrix}$

$RHS = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$

Compare LHS and RHS:

LHS = $\begin{bmatrix}1 & -t \\ t & 1 \end{bmatrix}$

RHS = $\begin{bmatrix}1 & -t \\ t & 1 \end{bmatrix}$

Since LHS = RHS, we have shown that:

$I + A = (I – A) \begin{bmatrix}\cos \alpha&−\sin \alpha \\ \sin \alpha&\cos \alpha \end{bmatrix}$

Substituting back $t = \tan \frac{\alpha}{2}$ gives the final result in terms of $\alpha$.

Hence Proved.

Given:

Total amount to be invested = $\textsf{₹} 30,000$.

Investment options:

Type 1 Bond: Pays 5% interest per year.

Type 2 Bond: Pays 7% interest per year.

To Find:

How to divide $\textsf{₹} 30,000$ among the two types of bonds using matrix multiplication to obtain an annual total interest of:

(a) $\textsf{₹} 1800$

(b) $\textsf{₹} 2000$

Solution:

Let the amount invested in the first bond (5% interest) be $\textsf{₹} x$.

Then, the amount invested in the second bond (7% interest) will be $\textsf{₹} (30000 - x)$.

We can represent the amounts invested as a 1x2 row matrix (Investment Matrix):

$A = \begin{bmatrix} x & (30000 - x) \end{bmatrix}$

The annual interest rates can be represented as a 2x1 column matrix (Interest Rate Matrix):

$R = \begin{bmatrix} 5\% \\ 7\% \end{bmatrix} = \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix}$

The total annual interest is obtained by the matrix product $A \times R$.

Total Interest = $A \times R = \begin{bmatrix} x & (30000 - x) \end{bmatrix} \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix}$

Total Interest = $[x(0.05) + (30000 - x)(0.07)]$

(a) Target annual interest = $\textsf{₹} 1800$

Set the total interest matrix equal to the target interest:

$[x(0.05) + (30000 - x)(0.07)] = [1800]$

Equating the elements gives the equation:

$0.05x + 0.07(30000 - x) = 1800$

$0.05x + 2100 - 0.07x = 1800$

$-0.02x = 1800 - 2100$

$-0.02x = -300$

$x = \frac{-300}{-0.02} = \frac{300}{0.02} = 15000$

Amount in the first bond (5% interest) = $\textsf{₹} x = \textsf{₹} 15,000$.

Amount in the second bond (7% interest) = $\textsf{₹} (30000 - x) = \textsf{₹} (30000 - 15000) = \textsf{₹} 15,000$.

Conclusion for (a): To obtain an annual interest of $\textsf{₹} 1800$, invest $\mathbf{\textsf{₹} 15,000}$ in the 5% bond and $\mathbf{\textsf{₹} 15,000}$ in the 7% bond.

(b) Target annual interest = $\textsf{₹} 2000$

Set the total interest matrix equal to the target interest:

$[x(0.05) + (30000 - x)(0.07)] = [2000]$

Equating the elements gives the equation:

$0.05x + 0.07(30000 - x) = 2000$

$0.05x + 2100 - 0.07x = 2000$

$-0.02x = 2000 - 2100$

$-0.02x = -100$

$x = \frac{-100}{-0.02} = \frac{100}{0.02} = 5000$

Amount in the first bond (5% interest) = $\textsf{₹} x = \textsf{₹} 5,000$.

Amount in the second bond (7% interest) = $\textsf{₹} (30000 - x) = \textsf{₹} (30000 - 5000) = \textsf{₹} 25,000$.

Conclusion for (b): To obtain an annual interest of $\textsf{₹} 2000$, invest $\mathbf{\textsf{₹} 5,000}$ in the 5% bond and $\mathbf{\textsf{₹} 25,000}$ in the 7% bond.

Given:

Number of Chemistry books = 10 dozen = $10 \times 12 = 120$

Number of Physics books = 8 dozen = $8 \times 12 = 96$

Number of Economics books = 10 dozen = $10 \times 12 = 120$

Selling price of one Chemistry book = $\textsf{₹} 80$

Selling price of one Physics book = $\textsf{₹} 60$

Selling price of one Economics book = $\textsf{₹} 40$

To Find: The total amount the bookshop will receive from selling all the books using matrix algebra.

Solution:

We can represent the number of books of each type as a 1x3 row matrix (Quantity Matrix):

$Q = \begin{bmatrix} 120 & 96 & 120 \end{bmatrix}$

We can represent the selling price of each book as a 3x1 column matrix (Price Matrix):

$P = \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}$

The total amount received by the bookshop is the product of the Quantity Matrix and the Price Matrix, $Q \times P$.

Total Amount = $Q \times P = \begin{bmatrix} 120 & 96 & 120 \end{bmatrix} \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}$

Total Amount = $[(120 \times 80) + (96 \times 60) + (120 \times 40)] $

Total Amount = $[9600 + 5760 + 4800]$

Total Amount = $[20160]$

The total amount the bookshop will receive is $\mathbf{\textsf{₹} 20,160}$.

Given:

Orders of the matrices:

X: $2 \times n$

Y: $3 \times k$

Z: $2 \times p$

W: $n \times 3$

P: $p \times k$

To Find: The restrictions on $n, k,$ and $p$ such that the matrix expression $PY + WY$ is defined.

Solution:

For the matrix product $PY$ to be defined, the number of columns in matrix P must be equal to the number of rows in matrix Y.

Order of P = $p \times k$

Order of Y = $3 \times k$

Number of columns in P = $k$

Number of rows in Y = $3$

Therefore, for $PY$ to be defined, we must have $\mathbf{k = 3}$.

If $k=3$, the order of the resulting matrix $PY$ is (rows of P) $\times$ (columns of Y) = $p \times k = p \times 3$.

For the matrix product $WY$ to be defined, the number of columns in matrix W must be equal to the number of rows in matrix Y.

Order of W = $n \times 3$

Order of Y = $3 \times k$

Number of columns in W = $3$

Number of rows in Y = $3$

Since $3=3$, the product $WY$ is always defined (given the structure). The order of the resulting matrix $WY$ is (rows of W) $\times$ (columns of Y) = $n \times k$.

Since we already established $k=3$, the order of $WY$ is $n \times 3$.

For the matrix addition $PY + WY$ to be defined, the matrices $PY$ and $WY$ must have the same order.

Order of $PY = p \times 3$

Order of $WY = n \times 3$

For the orders to be the same, the number of rows must be equal and the number of columns must be equal.

Number of columns: $3 = 3$ (This is consistent).

Number of rows: $p = n$. Therefore, we must have $\mathbf{p = n}$.

Conclusion:

The restrictions required for $PY + WY$ to be defined are:

1. $k = 3$ (for $PY$ to be defined)

2. $p = n$ (for $PY + WY$ to be defined)

Comparing these restrictions with the given options:

(A) $k = 3, p = n$ - Matches our findings.

(B) $k$ is arbitrary, $p = 2$ - Incorrect.

(C) $p$ is arbitrary, $k = 3$ - Incorrect.

(D) $k = 2, p = 3$ - Incorrect.

Therefore, the correct option is (A) k = 3, p = n.

Given:

Orders of the matrices:

X: $2 \times n$

Y: $3 \times k$

Z: $2 \times p$

W: $n \times 3$

P: $p \times k$

Condition: $n = p$

To Find: The order of the matrix $7X - 5Z$.

Solution:

We need to determine the order of the matrix resulting from the expression $7X - 5Z$.

First, consider the matrix $X$. Its order is given as $2 \times n$.

Scalar multiplication of a matrix by a constant (like 7) does not change its order.

So, the order of the matrix $7X$ is also $2 \times n$.

Next, consider the matrix $Z$. Its order is given as $2 \times p$.

Scalar multiplication of a matrix by a constant (like 5) does not change its order.

So, the order of the matrix $5Z$ is also $2 \times p$.

For the subtraction $7X - 5Z$ to be defined, the matrices $7X$ and $5Z$ must have the same order.

Order of $7X = 2 \times n$

Order of $5Z = 2 \times p$

For these orders to be the same, the number of rows must be equal ($2=2$) and the number of columns must be equal ($n=p$).

The problem statement gives the condition that $\mathbf{n = p}$. Therefore, the subtraction is defined.

When two matrices of the same order are subtracted, the resulting matrix has the same order.

Since $n=p$, both $7X$ and $5Z$ have the order $2 \times n$ (or equivalently, $2 \times p$).

Thus, the order of the resulting matrix $7X - 5Z$ is $\mathbf{2 \times n}$.

Comparing this with the given options:

(A) $p \times 2$

(B) $2 \times n$

(C) $n \times 3$

(D) $p \times n$

The correct option is (B) 2 × n.

Given:

$A = \begin{bmatrix}3&\sqrt{3}&2\\4&2&0 \end{bmatrix}$

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

(i) Verify that (A′)′ = A

First, find the transpose of A, denoted by A′.

To find the transpose, we interchange the rows and columns of A.

$A' = \begin{bmatrix}3&4\\\sqrt{3}&2\\2&0 \end{bmatrix}$

Now, find the transpose of A′, denoted by (A′)′.

To find the transpose of A′, we interchange the rows and columns of A′.

$(A')' = \begin{bmatrix}3&\sqrt{3}&2\\4&2&0 \end{bmatrix}$

Comparing the result with the original matrix A:

$(A')' = \begin{bmatrix}3&\sqrt{3}&2\\4&2&0 \end{bmatrix} = A$

Thus, (A′)′ = A is verified.

(ii) Verify that (A + B)′ = A′ + B′

First, calculate the sum A + B.

$A + B = \begin{bmatrix}3&\sqrt{3}&2\\4&2&0 \end{bmatrix} + \begin{bmatrix}2&−1&2\\1&2&4 \end{bmatrix}$

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

$A + B = \begin{bmatrix}5 & \sqrt{3}-1 & 4 \\ 5 & 4 & 4 \end{bmatrix}$

Now, find the transpose of (A + B), which is the Left Hand Side (LHS).

LHS = $(A + B)' = \begin{bmatrix}5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{bmatrix}$

Next, find the transposes A′ and B′.

$A' = \begin{bmatrix}3&4\\\sqrt{3}&2\\2&0 \end{bmatrix}$

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

Now, calculate the sum A′ + B′, which is the Right Hand Side (RHS).

RHS = $A' + B' = \begin{bmatrix}3&4\\\sqrt{3}&2\\2&0 \end{bmatrix} + \begin{bmatrix}2&1\\−1&2\\2&4 \end{bmatrix}$

RHS = $A' + B' = \begin{bmatrix}3+2 & 4+1 \\ \sqrt{3}+(-1) & 2+2 \\ 2+2 & 0+4 \end{bmatrix}$

RHS = $A' + B' = \begin{bmatrix}5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{bmatrix}$

Comparing LHS and RHS:

LHS = $\begin{bmatrix}5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{bmatrix}$ and RHS = $\begin{bmatrix}5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{bmatrix}$

Since LHS = RHS, (A + B)′ = A′ + B′ is verified.

(iii) Verify that (kB)′ = kB′, where k is any constant.

First, calculate the scalar multiple kB.

$kB = k \begin{bmatrix}2&−1&2\\1&2&4 \end{bmatrix} = \begin{bmatrix}k(2) & k(-1) & k(2) \\ k(1) & k(2) & k(4) \end{bmatrix}$

$kB = \begin{bmatrix}2k & -k & 2k \\ k & 2k & 4k \end{bmatrix}$

Now, find the transpose of (kB), which is the Left Hand Side (LHS).

LHS = $(kB)' = \begin{bmatrix}2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}$

Next, find the transpose B′.

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

Now, calculate the scalar multiple kB′, which is the Right Hand Side (RHS).

RHS = $kB' = k \begin{bmatrix}2&1\\−1&2\\2&4 \end{bmatrix} = \begin{bmatrix}k(2) & k(1) \\ k(-1) & k(2) \\ k(2) & k(4) \end{bmatrix}$

RHS = $kB' = \begin{bmatrix}2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}$

Comparing LHS and RHS:

LHS = $\begin{bmatrix}2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}$ and RHS = $\begin{bmatrix}2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}$

Since LHS = RHS, (kB)′ = kB′ is verified.

Given:

$A = \begin{bmatrix}−2\\4\\5 \end{bmatrix}$

$B = \begin{bmatrix}1&3&−6 \end{bmatrix}$

To Verify: $(AB)' = B'A'$

Verification:

Step 1: Calculate the product AB (LHS).

Matrix A has order $3 \times 1$ and matrix B has order $1 \times 3$. The product AB will have order $3 \times 3$.

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

$= \begin{bmatrix} (-2)(1) & (-2)(3) & (-2)(-6) \\ (4)(1) & (4)(3) & (4)(-6) \\ (5)(1) & (5)(3) & (5)(-6) \end{bmatrix}$

$AB = \begin{bmatrix} -2 & -6 & 12 \\ 4 & 12 & -24 \\ 5 & 15 & -30 \end{bmatrix}$

Step 2: Calculate the transpose of AB, (AB)'.

Interchange the rows and columns of AB.

LHS = $(AB)' = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$

Step 3: Calculate the transposes A' and B' (RHS).

Transpose of A:

$A' = \begin{bmatrix} -2 & 4 & 5 \end{bmatrix}$ (Order $1 \times 3$)

Transpose of B:

$B' = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix}$ (Order $3 \times 1$)

Step 4: Calculate the product B'A'.

Matrix B' has order $3 \times 1$ and matrix A' has order $1 \times 3$. The product B'A' will have order $3 \times 3$.

RHS = $B'A' = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix} \begin{bmatrix} -2 & 4 & 5 \end{bmatrix}$

$= \begin{bmatrix} (1)(-2) & (1)(4) & (1)(5) \\ (3)(-2) & (3)(4) & (3)(5) \\ (-6)(-2) & (-6)(4) & (-6)(5) \end{bmatrix}$

$B'A' = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$

Step 5: Compare LHS and RHS.

LHS = $(AB)' = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$

RHS = $B'A' = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$

Since LHS = RHS, the property (AB)’ = B’A’ is verified.

Given:

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

To Find: Express matrix B as the sum of a symmetric matrix and a skew-symmetric matrix.

Solution:

Any square matrix can be expressed as the sum of a symmetric matrix P and a skew-symmetric matrix Q, where:

$P = \frac{1}{2}(B + B')$ (Symmetric part)

$Q = \frac{1}{2}(B - B')$ (Skew-symmetric part)

Step 1: Find the transpose of B, denoted by B'.

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

Step 2: Calculate the symmetric part P.

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

$B + B' = \begin{bmatrix}2+2 & -2-1 & -4+1 \\ -1-2 & 3+3 & 4-2 \\ 1-4 & -2+4 & -3-3 \end{bmatrix} = \begin{bmatrix}4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix}$

$P = \frac{1}{2}(B + B') = \frac{1}{2}\begin{bmatrix}4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix}$

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

Check if P is symmetric: $P' = \begin{bmatrix}2 & -3/2 & -3/2 \\ -3/2 & 3 & 1 \\ -3/2 & 1 & -3 \end{bmatrix}$. Since $P' = P$, P is symmetric.

Step 3: Calculate the skew-symmetric part Q.

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

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

$Q = \frac{1}{2}(B - B') = \frac{1}{2}\begin{bmatrix}0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix}$

$Q = \begin{bmatrix}0 & -1/2 & -5/2 \\ 1/2 & 0 & 3 \\ 5/2 & -3 & 0 \end{bmatrix}$

Check if Q is skew-symmetric: $Q' = \begin{bmatrix}0 & 1/2 & 5/2 \\ -1/2 & 0 & -3 \\ -5/2 & 3 & 0 \end{bmatrix}$. Since $Q' = -Q$, Q is skew-symmetric.

Step 4: Express B as the sum P + Q.

$P + Q = \begin{bmatrix}2 & -3/2 & -3/2 \\ -3/2 & 3 & 1 \\ -3/2 & 1 & -3 \end{bmatrix} + \begin{bmatrix}0 & -1/2 & -5/2 \\ 1/2 & 0 & 3 \\ 5/2 & -3 & 0 \end{bmatrix}$

$P + Q = \begin{bmatrix}2+0 & -3/2-1/2 & -3/2-5/2 \\ -3/2+1/2 & 3+0 & 1+3 \\ -3/2+5/2 & 1-3 & -3+0 \end{bmatrix}$

$P + Q = \begin{bmatrix}2 & -4/2 & -8/2 \\ -2/2 & 3 & 4 \\ 2/2 & -2 & -3 \end{bmatrix} = \begin{bmatrix}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = B$

Thus, the matrix B is expressed as the sum of a symmetric matrix P and a skew-symmetric matrix Q:

$B = \underbrace{\begin{bmatrix}2 & -3/2 & -3/2 \\ -3/2 & 3 & 1 \\ -3/2 & 1 & -3 \end{bmatrix}}_{\text{Symmetric (P)}} + \underbrace{\begin{bmatrix}0 & -1/2 & -5/2 \\ 1/2 & 0 & 3 \\ 5/2 & -3 & 0 \end{bmatrix}}_{\text{Skew-symmetric (Q)}}$

Solution:

The transpose of a matrix is obtained by interchanging its rows and columns. If $A = [a_{ij}]_{m \times n}$, then its transpose $A' = [a_{ji}]_{n \times m}$.

(i)

Let $A = \begin{bmatrix}5\\\frac{1}{2}\\−1 \end{bmatrix}$.

This is a $3 \times 1$ matrix. Its transpose, $A'$, will be a $1 \times 3$ matrix.

The column $\begin{bmatrix}5\\\frac{1}{2}\\−1 \end{bmatrix}$ becomes the row $\begin{bmatrix}5 & \frac{1}{2} & -1 \end{bmatrix}$.

Therefore, $A' = \mathbf{\begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}}$.

(ii)

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

This is a $2 \times 2$ matrix. Its transpose, $B'$, will be a $2 \times 2$ matrix.

Row 1 $\begin{bmatrix}1 & -1\end{bmatrix}$ becomes Column 1 $\begin{bmatrix}1 \\ -1\end{bmatrix}$.

Row 2 $\begin{bmatrix}2 & 3\end{bmatrix}$ becomes Column 2 $\begin{bmatrix}2 \\ 3\end{bmatrix}$.

Therefore, $B' = \mathbf{\begin{bmatrix}1 & 2 \\ -1 & 3 \end{bmatrix}}$.

(iii)

Let $C = \begin{bmatrix}−1&5&6\\\sqrt{3}&5&6\\2&3&−1 \end{bmatrix}$.

This is a $3 \times 3$ matrix. Its transpose, $C'$, will be a $3 \times 3$ matrix.

Row 1 $\begin{bmatrix}-1 & 5 & 6\end{bmatrix}$ becomes Column 1 $\begin{bmatrix}-1 \\ 5 \\ 6\end{bmatrix}$.

Row 2 $\begin{bmatrix}\sqrt{3} & 5 & 6\end{bmatrix}$ becomes Column 2 $\begin{bmatrix}\sqrt{3} \\ 5 \\ 6\end{bmatrix}$.

Row 3 $\begin{bmatrix}2 & 3 & -1\end{bmatrix}$ becomes Column 3 $\begin{bmatrix}2 \\ 3 \\ -1\end{bmatrix}$.

Therefore, $C' = \mathbf{\begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}}$.

Given:

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

$B = \begin{bmatrix}−4&1&−5\\1&2&0\\1&3&1 \end{bmatrix}$

First, let's find the transposes of A and B:

$A' = \begin{bmatrix}−1&5&−2\\2&7&1\\3&9&1 \end{bmatrix}$

$B' = \begin{bmatrix}−4&1&1\\1&2&3\\−5&0&1 \end{bmatrix}$

(i) Verify that (A + B)′ = A′ + B′

Calculate the Left Hand Side (LHS): (A + B)′

First, find A + B:

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

$= \begin{bmatrix}−1+(-4) & 2+1 & 3+(-5) \\ 5+1 & 7+2 & 9+0 \\ -2+1 & 1+3 & 1+1 \end{bmatrix} = \begin{bmatrix}−5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}$

Now, find the transpose of (A + B):

LHS = $(A + B)' = \begin{bmatrix}−5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$

Calculate the Right Hand Side (RHS): A′ + B′

$A' + B' = \begin{bmatrix}−1&5&−2\\2&7&1\\3&9&1 \end{bmatrix} + \begin{bmatrix}−4&1&1\\1&2&3\\−5&0&1 \end{bmatrix}$

$= \begin{bmatrix}−1+(-4) & 5+1 & -2+1 \\ 2+1 & 7+2 & 1+3 \\ 3+(-5) & 9+0 & 1+1 \end{bmatrix} = \begin{bmatrix}−5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$

RHS = $\begin{bmatrix}−5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$

Compare LHS and RHS:

Since LHS = RHS, the property (A + B)′ = A′ + B′ is verified.

(ii) Verify that (A – B)′ = A′ – B′

Calculate the Left Hand Side (LHS): (A - B)′

First, find A - B:

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

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

Now, find the transpose of (A - B):

LHS = $(A - B)' = \begin{bmatrix}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$

Calculate the Right Hand Side (RHS): A′ - B′

$A' - B' = \begin{bmatrix}−1&5&−2\\2&7&1\\3&9&1 \end{bmatrix} - \begin{bmatrix}−4&1&1\\1&2&3\\−5&0&1 \end{bmatrix}$

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

RHS = $\begin{bmatrix}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$

Compare LHS and RHS:

Since LHS = RHS, the property (A – B)′ = A′ – B′ is verified.

Given:

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

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

To perform the operations, we first need to find matrix A and the transpose of B (B').

Finding A from A':

$A = (A')' = \left(\begin{bmatrix}3&4\\−1&2\\0&1 \end{bmatrix}\right)' = \begin{bmatrix}3&−1&0\\4&2&1 \end{bmatrix}$

Finding B':

$B' = \left(\begin{bmatrix}−1&2&1\\1&2&3 \end{bmatrix}\right)' = \begin{bmatrix}−1&1\\2&2\\1&3 \end{bmatrix}$

(i) Verify that (A + B)′ = A′ + B′

Calculate the Left Hand Side (LHS): (A + B)′

First, find A + B:

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

$= \begin{bmatrix}3+(-1) & -1+2 & 0+1 \\ 4+1 & 2+2 & 1+3 \end{bmatrix} = \begin{bmatrix}2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}$

Now, find the transpose of (A + B):

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

Calculate the Right Hand Side (RHS): A′ + B′

$A' + B' = \begin{bmatrix}3&4\\−1&2\\0&1 \end{bmatrix} + \begin{bmatrix}−1&1\\2&2\\1&3 \end{bmatrix}$

$= \begin{bmatrix}3+(-1) & 4+1 \\ -1+2 & 2+2 \\ 0+1 & 1+3 \end{bmatrix} = \begin{bmatrix}2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$

RHS = $\begin{bmatrix}2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$

Compare LHS and RHS:

Since LHS = RHS, the property (A + B)′ = A′ + B′ is verified.

(ii) Verify that (A – B)′ = A′ – B′

Calculate the Left Hand Side (LHS): (A - B)′

First, find A - B:

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

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

Now, find the transpose of (A - B):

LHS = $(A - B)' = \begin{bmatrix}4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$

Calculate the Right Hand Side (RHS): A′ - B′

$A' - B' = \begin{bmatrix}3&4\\−1&2\\0&1 \end{bmatrix} - \begin{bmatrix}−1&1\\2&2\\1&3 \end{bmatrix}$

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

RHS = $\begin{bmatrix}4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$

Compare LHS and RHS:

Since LHS = RHS, the property (A – B)′ = A′ – B′ is verified.

Given:

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

$B = \begin{bmatrix}−1&0\\1&2 \end{bmatrix}$

To Find: The value of $(A + 2B)’$.

Solution:

We can use the property of transposes: $(X + Y)' = X' + Y'$ and $(kX)' = kX'$ for any scalar $k$.

Therefore, $(A + 2B)' = A' + (2B)'$.

Also, $(2B)' = 2B'$.

So, $(A + 2B)' = A' + 2B'$.

First, find the transpose of B, denoted by B'.

$B' = \begin{bmatrix}−1&0\\1&2 \end{bmatrix}' = \begin{bmatrix}−1&1\\0&2 \end{bmatrix}$

Next, calculate $2B'$.

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

Now, calculate $A' + 2B'$.

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

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

$(A + 2B)' = \begin{bmatrix}−4 & 5 \\ 1 & 6 \end{bmatrix}$

Alternate Solution:

First, find A from A'.

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

Next, calculate 2B.

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

Now, calculate A + 2B.

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

Finally, find the transpose of (A + 2B).

$(A + 2B)' = \begin{bmatrix}−4 & 1 \\ 5 & 6 \end{bmatrix}' = \begin{bmatrix}−4 & 5 \\ 1 & 6 \end{bmatrix}$

Both methods yield the same result: $\mathbf{(A + 2B)' = \begin{bmatrix}−4 & 5 \\ 1 & 6 \end{bmatrix}}$.

We need to verify the property $(AB)' = B'A'$ for the given matrices.

(i)

Given: $A = \begin{bmatrix}1\\−4\\3 \end{bmatrix}$, $B = \begin{bmatrix}−1&2&1 \end{bmatrix}$

Calculate the Left Hand Side (LHS): (AB)′

First, find the product AB. (A is $3 \times 1$, B is $1 \times 3$, so AB is $3 \times 3$)

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

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

Now, find the transpose of AB:

LHS = $(AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Calculate the Right Hand Side (RHS): B′A′

First, find the transposes A' and B':

$A' = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$

$B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$

Now, find the product B'A'. (B' is $3 \times 1$, A' is $1 \times 3$, so B'A' is $3 \times 3$)

$B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} (-1)(1) & (-1)(-4) & (-1)(3) \\ (2)(1) & (2)(-4) & (2)(3) \\ (1)(1) & (1)(-4) & (1)(3) \end{bmatrix}$

$B'A' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

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

Compare LHS and RHS:

Since LHS = RHS, the property (AB)′ = B′A′ is verified for (i).

(ii)

Given: $A = \begin{bmatrix}0\\1\\2 \end{bmatrix}$, $B = \begin{bmatrix}1&5&7 \end{bmatrix}$

Calculate the Left Hand Side (LHS): (AB)′

First, find the product AB. (A is $3 \times 1$, B is $1 \times 3$, so AB is $3 \times 3$)

$AB = \begin{bmatrix}0\\1\\2 \end{bmatrix} \begin{bmatrix}1&5&7 \end{bmatrix} = \begin{bmatrix} (0)(1) & (0)(5) & (0)(7) \\ (1)(1) & (1)(5) & (1)(7) \\ (2)(1) & (2)(5) & (2)(7) \end{bmatrix}$

$AB = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}$

Now, find the transpose of AB:

LHS = $(AB)' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

Calculate the Right Hand Side (RHS): B′A′

First, find the transposes A' and B':

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

$B' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}$

Now, find the product B'A'. (B' is $3 \times 1$, A' is $1 \times 3$, so B'A' is $3 \times 3$)

$B'A' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} (1)(0) & (1)(1) & (1)(2) \\ (5)(0) & (5)(1) & (5)(2) \\ (7)(0) & (7)(1) & (7)(2) \end{bmatrix}$

$B'A' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

RHS = $\begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

Compare LHS and RHS:

Since LHS = RHS, the property (AB)′ = B′A′ is verified for (ii).

We need to verify the property $A'A = I$, where I is the identity matrix of order 2, $I = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$.

(i)

Given: $A = \begin{bmatrix}\cos\alpha&\sin\alpha\\−\sin\alpha&\cos\alpha \end{bmatrix}$

Step 1: Find the transpose A'.

$A' = \begin{bmatrix}\cos\alpha&−\sin\alpha\\\sin\alpha&\cos\alpha \end{bmatrix}$

Step 2: Calculate the product A'A.

$A'A = \begin{bmatrix}\cos\alpha&−\sin\alpha\\\sin\alpha&\cos\alpha \end{bmatrix} \begin{bmatrix}\cos\alpha&\sin\alpha\\−\sin\alpha&\cos\alpha \end{bmatrix}$

$= \begin{bmatrix} (\cos\alpha)(\cos\alpha) + (-\sin\alpha)(-\sin\alpha) & (\cos\alpha)(\sin\alpha) + (-\sin\alpha)(\cos\alpha) \\ (\sin\alpha)(\cos\alpha) + (\cos\alpha)(-\sin\alpha) & (\sin\alpha)(\sin\alpha) + (\cos\alpha)(\cos\alpha) \end{bmatrix}$

$= \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & \cos\alpha\sin\alpha - \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha - \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha \end{bmatrix}$

Using the trigonometric identity $\sin^2\alpha + \cos^2\alpha = 1$:

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

Step 3: Compare with I.

Since $A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$, the property A'A = I is verified for (i).

(ii)

Given: $A = \begin{bmatrix}\sin\alpha&\cos\alpha \\ −\cos\alpha&\sin\alpha \end{bmatrix}$

Step 1: Find the transpose A'.

$A' = \begin{bmatrix}\sin\alpha&−\cos\alpha\\\cos\alpha&\sin\alpha \end{bmatrix}$

Step 2: Calculate the product A'A.

$A'A = \begin{bmatrix}\sin\alpha&−\cos\alpha\\\cos\alpha&\sin\alpha \end{bmatrix} \begin{bmatrix}\sin\alpha&\cos\alpha \\ −\cos\alpha&\sin\alpha \end{bmatrix}$

$= \begin{bmatrix} (\sin\alpha)(\sin\alpha) + (-\cos\alpha)(-\cos\alpha) & (\sin\alpha)(\cos\alpha) + (-\cos\alpha)(\sin\alpha) \\ (\cos\alpha)(\sin\alpha) + (\sin\alpha)(-\cos\alpha) & (\cos\alpha)(\cos\alpha) + (\sin\alpha)(\sin\alpha) \end{bmatrix}$

$= \begin{bmatrix} \sin^2\alpha + \cos^2\alpha & \sin\alpha\cos\alpha - \cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha - \sin\alpha\cos\alpha & \cos^2\alpha + \sin^2\alpha \end{bmatrix}$

Using the trigonometric identity $\sin^2\alpha + \cos^2\alpha = 1$:

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

Step 3: Compare with I.

Since $A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$, the property A'A = I is verified for (ii).

Definitions:

A square matrix A is called symmetric if its transpose $A'$ is equal to the matrix itself, i.e., $A' = A$.

A square matrix A is called skew-symmetric if its transpose $A'$ is equal to the negative of the matrix itself, i.e., $A' = -A$.

(i) Show that the matrix A = $\begin{bmatrix}1&−1&5\\−1&2&1\\5&1&3 \end{bmatrix}$ is a symmetric matrix.

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

Step 1: Find the transpose of A, denoted by A'.

To find the transpose, we interchange the rows and columns of A.

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

Step 2: Compare A' with A.

We have $A = \begin{bmatrix}1&−1&5\\−1&2&1\\5&1&3 \end{bmatrix}$ and $A' = \begin{bmatrix}1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$.

Since $A' = A$, the matrix A is a symmetric matrix.

Hence Shown.

(ii) Show that the matrix A = $\begin{bmatrix}0&1&−1\\−1&0&1\\1&−1&0 \end{bmatrix}$ is a skew symmetric matrix.

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

Step 1: Find the transpose of A, denoted by A'.

To find the transpose, we interchange the rows and columns of A.

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

Step 2: Find the negative of A, denoted by -A.

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

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

Step 3: Compare A' with -A.

We have $A' = \begin{bmatrix}0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$ and $-A = \begin{bmatrix}0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$.

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

Hence Shown.

Given:

The matrix $A = \begin{bmatrix}1&5\\6&7 \end{bmatrix}$

Step 1: Find the transpose of A, denoted by A'.

$A' = \begin{bmatrix}1&5\\6&7 \end{bmatrix}' = \begin{bmatrix}1 & 6 \\ 5 & 7 \end{bmatrix}$

(i) Verify that (A + A′) is a symmetric matrix

Step 2: Calculate the sum A + A'.

Let $P = A + A'$.

$P = \begin{bmatrix}1&5\\6&7 \end{bmatrix} + \begin{bmatrix}1 & 6 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix}1+1 & 5+6 \\ 6+5 & 7+7 \end{bmatrix}$

$P = \begin{bmatrix}2 & 11 \\ 11 & 14 \end{bmatrix}$

Step 3: Find the transpose of P, denoted by P'.

$P' = \begin{bmatrix}2 & 11 \\ 11 & 14 \end{bmatrix}' = \begin{bmatrix}2 & 11 \\ 11 & 14 \end{bmatrix}$

Step 4: Check if P is symmetric.

A matrix is symmetric if $P' = P$.

Comparing P and P', we see that $P' = P = \begin{bmatrix}2 & 11 \\ 11 & 14 \end{bmatrix}$.

Therefore, (A + A′) is a symmetric matrix. Verified.

(ii) Verify that (A – A′) is a skew symmetric matrix

Step 2: Calculate the difference A - A'.

Let $Q = A - A'$.

$Q = \begin{bmatrix}1&5\\6&7 \end{bmatrix} - \begin{bmatrix}1 & 6 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix}1-1 & 5-6 \\ 6-5 & 7-7 \end{bmatrix}$

$Q = \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}$

Step 3: Find the transpose of Q, denoted by Q'.

$Q' = \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}' = \begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}$

Step 4: Find the negative of Q, denoted by -Q.

$-Q = -1 \times \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}$

Step 5: Check if Q is skew-symmetric.

A matrix is skew-symmetric if $Q' = -Q$.

Comparing Q' and -Q, we see that $Q' = -Q = \begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}$.

Therefore, (A – A′) is a skew-symmetric matrix. Verified.

Given:

The matrix $A = \begin{bmatrix}0&a&b\\−a&0&c\\−b&−c&0 \end{bmatrix}$

To Find:

(i) $\frac{1}{2} (A + A’)$

(ii) $\frac{1}{2} (A - A’)$

Solution:

Step 1: Find the transpose of A, denoted by A'.

$A' = \begin{bmatrix}0&a&b\\−a&0&c\\−b&−c&0 \end{bmatrix}' = \begin{bmatrix}0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}$

Step 2: Calculate $\frac{1}{2} (A + A’)$.

First, find the sum $A + A'$.

$A + A' = \begin{bmatrix}0&a&b\\−a&0&c\\−b&−c&0 \end{bmatrix} + \begin{bmatrix}0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}$

$A + A' = \begin{bmatrix}0+0 & a+(-a) & b+(-b) \\ -a+a & 0+0 & c+(-c) \\ -b+b & -c+c & 0+0 \end{bmatrix} = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Now, multiply by $\frac{1}{2}$.

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

Step 3: Calculate $\frac{1}{2} (A - A’)$.

First, find the difference $A - A'$.

$A - A' = \begin{bmatrix}0&a&b\\−a&0&c\\−b&−c&0 \end{bmatrix} - \begin{bmatrix}0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}$

$A - A' = \begin{bmatrix}0-0 & a-(-a) & b-(-b) \\ -a-a & 0-0 & c-(-c) \\ -b-b & -c-c & 0-0 \end{bmatrix} = \begin{bmatrix}0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}$

Now, multiply by $\frac{1}{2}$.

$\frac{1}{2} (A - A') = \frac{1}{2} \begin{bmatrix}0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}$

$\frac{1}{2} (A - A') = \begin{bmatrix}\frac{1}{2}(0) & \frac{1}{2}(2a) & \frac{1}{2}(2b) \\ \frac{1}{2}(-2a) & \frac{1}{2}(0) & \frac{1}{2}(2c) \\ \frac{1}{2}(-2b) & \frac{1}{2}(-2c) & \frac{1}{2}(0) \end{bmatrix} = \mathbf{\begin{bmatrix}0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}}$

(Note: This is equal to the original matrix A, as A itself is a skew-symmetric matrix).

Any square matrix A can be expressed as the sum of a symmetric matrix $P = \frac{1}{2}(A + A')$ and a skew-symmetric matrix $Q = \frac{1}{2}(A - A')$.

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

First, find the transpose A':

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

Calculate the symmetric part P:

$P = \frac{1}{2}(A + A') = \frac{1}{2} \left( \begin{bmatrix}3&5\\1&−1 \end{bmatrix} + \begin{bmatrix}3&1\\5&−1 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix}3+3 & 5+1 \\ 1+5 & -1-1 \end{bmatrix}$

$P = \frac{1}{2} \begin{bmatrix}6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix}3 & 3 \\ 3 & -1 \end{bmatrix}$

Calculate the skew-symmetric part Q:

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

$Q = \frac{1}{2} \begin{bmatrix}0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix}0 & 2 \\ -2 & 0 \end{bmatrix}$

Verify the sum P + Q:

$P + Q = \begin{bmatrix}3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix}0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix}3+0 & 3+2 \\ 3-2 & -1+0 \end{bmatrix} = \begin{bmatrix}3 & 5 \\ 1 & -1 \end{bmatrix} = A$

Thus, $A = \underbrace{\begin{bmatrix}3 & 3 \\ 3 & -1 \end{bmatrix}}_{\text{Symmetric}} + \underbrace{\begin{bmatrix}0 & 2 \\ -2 & 0 \end{bmatrix}}_{\text{Skew-symmetric}}$.

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

First, find the transpose A':

$A' = \begin{bmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$ (Note: A is symmetric, so A' = A)

Calculate the symmetric part P:

$P = \frac{1}{2}(A + A') = \frac{1}{2}(A + A) = \frac{1}{2}(2A) = A$

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

Calculate the skew-symmetric part Q:

$Q = \frac{1}{2}(A - A') = \frac{1}{2}(A - A) = \frac{1}{2}(O)$, where O is the zero matrix.

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

Verify the sum P + Q:

$P + Q = \begin{bmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} + \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = A$

Thus, $A = \underbrace{\begin{bmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}}_{\text{Symmetric}} + \underbrace{\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}}_{\text{Skew-symmetric}}$.

(iii) $A = \begin{bmatrix}3&3&−1\\−2&−2&1\\−4&−5&2 \end{bmatrix}$

First, find the transpose A':

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

Calculate the symmetric part P:

$P = \frac{1}{2}(A + A') = \frac{1}{2} \left( \begin{bmatrix}3&3&−1\\−2&−2&1\\−4&−5&2 \end{bmatrix} + \begin{bmatrix}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \right)$

$P = \frac{1}{2} \begin{bmatrix}3+3 & 3-2 & -1-4 \\ -2+3 & -2-2 & 1-5 \\ -4-1 & -5+1 & 2+2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix}$

$P = \begin{bmatrix}3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix}$

Calculate the skew-symmetric part Q:

$Q = \frac{1}{2}(A - A') = \frac{1}{2} \left( \begin{bmatrix}3&3&−1\\−2&−2&1\\−4&−5&2 \end{bmatrix} - \begin{bmatrix}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \right)$

$Q = \frac{1}{2} \begin{bmatrix}3-3 & 3-(-2) & -1-(-4) \\ -2-3 & -2-(-2) & 1-(-5) \\ -4-(-1) & -5-1 & 2-2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix}$

$Q = \begin{bmatrix}0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{bmatrix}$

Verify the sum P + Q:

$P + Q = \begin{bmatrix}3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} + \begin{bmatrix}0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{bmatrix} = \begin{bmatrix}3 & 6/2 & -2/2 \\ -4/2 & -2 & 1 \\ -8/2 & -5 & 2 \end{bmatrix} = \begin{bmatrix}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = A$

Thus, $A = \underbrace{\begin{bmatrix}3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix}}_{\text{Symmetric}} + \underbrace{\begin{bmatrix}0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{bmatrix}}_{\text{Skew-symmetric}}$.

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

First, find the transpose A':

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

Calculate the symmetric part P:

$P = \frac{1}{2}(A + A') = \frac{1}{2} \left( \begin{bmatrix}1&5\\−1&2 \end{bmatrix} + \begin{bmatrix}1&−1\\5&2 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix}1+1 & 5-1 \\ -1+5 & 2+2 \end{bmatrix}$

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

Calculate the skew-symmetric part Q:

$Q = \frac{1}{2}(A - A') = \frac{1}{2} \left( \begin{bmatrix}1&5\\−1&2 \end{bmatrix} - \begin{bmatrix}1&−1\\5&2 \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix}1-1 & 5-(-1) \\ -1-5 & 2-2 \end{bmatrix}$

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

Verify the sum P + Q:

$P + Q = \begin{bmatrix}1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix}0 & 3 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix}1+0 & 2+3 \\ 2-3 & 2+0 \end{bmatrix} = \begin{bmatrix}1 & 5 \\ -1 & 2 \end{bmatrix} = A$

Thus, $A = \underbrace{\begin{bmatrix}1 & 2 \\ 2 & 2 \end{bmatrix}}_{\text{Symmetric}} + \underbrace{\begin{bmatrix}0 & 3 \\ -3 & 0 \end{bmatrix}}_{\text{Skew-symmetric}}$.

Given:

A and B are symmetric matrices of the same order.

This implies that $A' = A$ and $B' = B$.

To Determine: The nature of the matrix $AB - BA$.

Solution:

Let $P = AB - BA$.

To determine if P is symmetric or skew-symmetric, we need to find its transpose, P'.

$P' = (AB - BA)'$

Using the property of transpose $(X - Y)' = X' - Y'$:

$P' = (AB)' - (BA)'$

Using the reversal law for transpose of a product $(XY)' = Y'X'$:

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

Now, substitute the given conditions $A' = A$ and $B' = B$ into the expression for P':

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

$P' = BA - AB$

Now, compare P' with P:

$P = AB - BA$

$P' = BA - AB$

We can rewrite P' by factoring out -1:

$P' = -(AB - BA)$

Since $P = AB - BA$, we have:

$P' = -P$

By definition, a matrix M is skew-symmetric if $M' = -M$.

Since $P' = -P$, the matrix $P = AB - BA$ is a skew-symmetric matrix.

Therefore, the correct option is (A).

Given:

$A = \begin{bmatrix}\cos\alpha&−\sin\alpha\\\sin\alpha&\cos\alpha \end{bmatrix}$

$I = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$ (Identity matrix of order 2)

Condition: $A + A’ = I$

To Find: The value of $\alpha$.

Solution:

Step 1: Find the transpose of A, denoted by A'.

$A' = \begin{bmatrix}\cos\alpha&−\sin\alpha\\\sin\alpha&\cos\alpha \end{bmatrix}' = \begin{bmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

Step 2: Substitute A, A', and I into the given condition $A + A’ = I$.

$\begin{bmatrix}\cos\alpha&−\sin\alpha\\\sin\alpha&\cos\alpha \end{bmatrix} + \begin{bmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Step 3: Perform the matrix addition on the left side.

$\begin{bmatrix}\cos\alpha + \cos\alpha & -\sin\alpha + \sin\alpha \\ \sin\alpha + (-\sin\alpha) & \cos\alpha + \cos\alpha \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

$\begin{bmatrix}2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Step 4: Equate the corresponding elements of the matrices.

From the equality of matrices, we get:

$2\cos\alpha = 1$

$0 = 0$

Step 5: Solve for $\cos\alpha$.

$\cos\alpha = \frac{1}{2}$

Step 6: Find the value of $\alpha$ from the given options that satisfies $\cos\alpha = \frac{1}{2}$.

(A) $\alpha = \frac{\pi}{6} \implies \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$

(B) $\alpha = \frac{\pi}{3} \implies \cos(\frac{\pi}{3}) = \frac{1}{2}$

(C) $\alpha = \pi \implies \cos(\pi) = -1$

(D) $\alpha = \frac{3\pi}{2} \implies \cos(\frac{3\pi}{2}) = 0$

The value $\alpha = \frac{\pi}{3}$ satisfies the condition $\cos\alpha = \frac{1}{2}$.

Therefore, the correct option is (B) $\frac{π}{3}$.

Given:

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

To Find: The inverse of matrix A, denoted by $A^{-1}$, using elementary operations.

Solution:

We use elementary row operations to find the inverse. We start with the augmented matrix $[A | I]$, where I is the identity matrix of the same order as A.

$[A | I] = \left[ \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 2 & -1 & 0 & 1 \end{array} \right]$

Our goal is to transform the matrix A on the left side into the identity matrix I. The matrix on the right side will then become $A^{-1}$.

Step 1: Apply the row operation $R_2 \rightarrow R_2 - 2R_1$ to make the element in the second row, first column zero.

$R_2 = [2 \quad -1 \quad | \quad 0 \quad 1]$

$2R_1 = [2 \times 1 \quad 2 \times 2 \quad | \quad 2 \times 1 \quad 2 \times 0] = [2 \quad 4 \quad | \quad 2 \quad 0]$

$R_2 - 2R_1 = [2-2 \quad -1-4 \quad | \quad 0-2 \quad 1-0] = [0 \quad -5 \quad | \quad -2 \quad 1]$

The matrix becomes:

$\left[ \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -5 & -2 & 1 \end{array} \right]$

Step 2: Apply the row operation $R_2 \rightarrow -\frac{1}{5}R_2$ to make the element in the second row, second column one.

$-\frac{1}{5} R_2 = [-\frac{1}{5}(0) \quad -\frac{1}{5}(-5) \quad | \quad -\frac{1}{5}(-2) \quad -\frac{1}{5}(1)] = [0 \quad 1 \quad | \quad \frac{2}{5} \quad -\frac{1}{5}]$

The matrix becomes:

$\left[ \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{2}{5} & -\frac{1}{5} \end{array} \right]$

Step 3: Apply the row operation $R_1 \rightarrow R_1 - 2R_2$ to make the element in the first row, second column zero.

$R_1 = [1 \quad 2 \quad | \quad 1 \quad 0]$

$2R_2 = [2 \times 0 \quad 2 \times 1 \quad | \quad 2 \times \frac{2}{5} \quad 2 \times (-\frac{1}{5})] = [0 \quad 2 \quad | \quad \frac{4}{5} \quad -\frac{2}{5}]$

$R_1 - 2R_2 = [1-0 \quad 2-2 \quad | \quad 1-\frac{4}{5} \quad 0-(-\frac{2}{5})] = [1 \quad 0 \quad | \quad \frac{5-4}{5} \quad \frac{2}{5}] = [1 \quad 0 \quad | \quad \frac{1}{5} \quad \frac{2}{5}]$

The matrix becomes:

$\left[ \begin{array}{cc|cc} 1 & 0 & \frac{1}{5} & \frac{2}{5} \\ 0 & 1 & \frac{2}{5} & -\frac{1}{5} \end{array} \right]$

The left side is now the identity matrix I. The right side is the inverse matrix $A^{-1}$.

Therefore, the inverse of A is:

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

Given:

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

To Find: The inverse of matrix A, denoted by $A^{-1}$, using elementary operations.

Solution:

We use elementary row operations to find the inverse. We start with the augmented matrix $[A | I]$, where I is the identity matrix of order 3.

$[A | I] = \left[ \begin{array}{ccc|ccc} 0 & 1 & 2 & 1 & 0 & 0 \\ 1 & 2 & 3 & 0 & 1 & 0 \\ 3 & 1 & 1 & 0 & 0 & 1 \end{array} \right]$

Our goal is to transform the matrix A on the left side into the identity matrix I using row operations. The matrix on the right side will then become $A^{-1}$.

Apply $R_1 \leftrightarrow R_2$ to get a non-zero element in the top-left position.

$\left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 3 & 1 & 1 & 0 & 0 & 1 \end{array} \right]$

Apply $R_3 \rightarrow R_3 - 3R_1$ to make the element in the third row, first column zero.

$\left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 1 - 3(2) & 1 - 3(3) & 0 - 3(0) & 0 - 3(1) & 1 - 3(0) \end{array} \right]$

$\left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & -5 & -8 & 0 & -3 & 1 \end{array} \right]$

Apply $R_1 \rightarrow R_1 - 2R_2$ to make the element in the first row, second column zero.

$\left[ \begin{array}{ccc|ccc} 1 - 2(0) & 2 - 2(1) & 3 - 2(2) & 0 - 2(1) & 1 - 2(0) & 0 - 2(0) \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & -5 & -8 & 0 & -3 & 1 \end{array} \right]$

$\left[ \begin{array}{ccc|ccc} 1 & 0 & -1 & -2 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & -5 & -8 & 0 & -3 & 1 \end{array} \right]$

Apply $R_3 \rightarrow R_3 + 5R_2$ to make the element in the third row, second column zero.

$\left[ \begin{array}{ccc|ccc} 1 & 0 & -1 & -2 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 + 5(0) & -5 + 5(1) & -8 + 5(2) & 0 + 5(1) & -3 + 5(0) & 1 + 5(0) \end{array} \right]$

$\left[ \begin{array}{ccc|ccc} 1 & 0 & -1 & -2 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 2 & 5 & -3 & 1 \end{array} \right]$

Apply $R_3 \rightarrow \frac{1}{2}R_3$ to make the element in the third row, third column one.

$\left[ \begin{array}{ccc|ccc} 1 & 0 & -1 & -2 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right]$

Apply $R_1 \rightarrow R_1 + R_3$ to make the element in the first row, third column zero.

$\left[ \begin{array}{ccc|ccc} 1+0 & 0+0 & -1+1 & -2+\frac{5}{2} & 1+(-\frac{3}{2}) & 0+\frac{1}{2} \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right]$

$\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right]$

Apply $R_2 \rightarrow R_2 - 2R_3$ to make the element in the second row, third column zero.

$\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 - 2(0) & 1 - 2(0) & 2 - 2(1) & 1 - 2(\frac{5}{2}) & 0 - 2(-\frac{3}{2}) & 0 - 2(\frac{1}{2}) \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right]$

$\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 & 1 - 5 & 0 + 3 & 0 - 1 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right]$

$\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 & -4 & 3 & -1 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right]$

The left side is now the identity matrix I. The right side is the inverse matrix $A^{-1}$.

Therefore, the inverse of A is:

$\mathbf{A^{-1} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}}$

Given:

The matrix $P = \begin{bmatrix}10&−2\\−5&1 \end{bmatrix}$

To Find: The inverse of matrix P, denoted by $P^{-1}$, if it exists, using elementary operations.

Solution:

We use elementary row operations to find the inverse. We start with the augmented matrix $[P | I]$, where I is the identity matrix of order 2.

$[P | I] = \left[ \begin{array}{cc|cc} 10 & -2 & 1 & 0 \\ -5 & 1 & 0 & 1 \end{array} \right]$

Our goal is to transform the matrix P on the left side into the identity matrix I. If successful, the matrix on the right side will become $P^{-1}$.

Apply the row operation $R_1 \rightarrow \frac{1}{10}R_1$ to make the element in the first row, first column one.

$\left[ \begin{array}{cc|cc} \frac{10}{10} & \frac{-2}{10} & \frac{1}{10} & \frac{0}{10} \\ -5 & 1 & 0 & 1 \end{array} \right] = \left[ \begin{array}{cc|cc} 1 & -\frac{1}{5} & \frac{1}{10} & 0 \\ -5 & 1 & 0 & 1 \end{array} \right]$

Apply the row operation $R_2 \rightarrow R_2 + 5R_1$ to make the element in the second row, first column zero.

$\left[ \begin{array}{cc|cc} 1 & -\frac{1}{5} & \frac{1}{10} & 0 \\ -5 + 5(1) & 1 + 5(-\frac{1}{5}) & 0 + 5(\frac{1}{10}) & 1 + 5(0) \end{array} \right]$

$\left[ \begin{array}{cc|cc} 1 & -\frac{1}{5} & \frac{1}{10} & 0 \\ -5 + 5 & 1 - 1 & 0 + \frac{5}{10} & 1 + 0 \end{array} \right]$

$\left[ \begin{array}{cc|cc} 1 & -\frac{1}{5} & \frac{1}{10} & 0 \\ 0 & 0 & \frac{1}{2} & 1 \end{array} \right]$

At this step, we obtained a row of all zeros ($[0 \quad 0]$) in the left-hand matrix (the part corresponding to A). When this happens during the application of elementary row operations to find the inverse, it means that the original matrix is singular (non-invertible).

Therefore, the inverse of the matrix P does not exist.

(We can also check the determinant: $\det(P) = (10)(1) - (-2)(-5) = 10 - 10 = 0$. Since the determinant is 0, the matrix is singular and has no inverse).

Given:

Let the given matrix be $A = \begin{bmatrix}1&−1\\2&3 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

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

Our goal is to transform the matrix on the left-hand side (LHS) into the identity matrix $I$ by applying elementary row operations. The same operations will be applied simultaneously to the matrix $I$ on the right-hand side (RHS).

Apply the operation $R_2 \to R_2 - 2R_1$:

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

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

Apply the operation $R_2 \to \frac{1}{5}R_2$:

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

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

Apply the operation $R_1 \to R_1 + R_2$:

$\begin{bmatrix}1+0&−1+1\\0&1 \end{bmatrix} = \begin{bmatrix}1 + (-\frac{2}{5})&0 + \frac{1}{5}\\-\frac{2}{5}&\frac{1}{5} \end{bmatrix} A$

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

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

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Our goal is to transform the matrix on the left-hand side (LHS) into the identity matrix $I$ by applying elementary column operations. The same operations will be applied simultaneously to the matrix $I$ on the right-hand side (RHS).

Apply the operation $C_2 \to C_2 + C_1$:

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

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

Apply the operation $C_1 \to C_1 - 2C_2$:

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

$\begin{bmatrix}1&0\\-8&5 \end{bmatrix} = A \begin{bmatrix}-1&1\\-2&1 \end{bmatrix}$

Apply the operation $C_2 \to \frac{1}{5}C_2$:

$\begin{bmatrix}1&\frac{1}{5}(0)\\-8&\frac{1}{5}(5) \end{bmatrix} = A \begin{bmatrix}-1&\frac{1}{5}(1)\\-2&\frac{1}{5}(1) \end{bmatrix}$

$\begin{bmatrix}1&0\\-8&1 \end{bmatrix} = A \begin{bmatrix}-1&\frac{1}{5}\\-2&\frac{1}{5} \end{bmatrix}$

Apply the operation $C_1 \to C_1 + 8C_2$:

$\begin{bmatrix}1 + 8(0)&0\\-8 + 8(1)&1 \end{bmatrix} = A \begin{bmatrix}-1 + 8(\frac{1}{5})&\frac{1}{5}\\-2 + 8(\frac{1}{5})&\frac{1}{5} \end{bmatrix}$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}\frac{-5+8}{5}&\frac{1}{5}\\ \frac{-10+8}{5}&\frac{1}{5} \end{bmatrix}$

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

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

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}2&1\\1&1 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

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

Apply the operation $R_1 \leftrightarrow R_2$:

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

Apply the operation $R_2 \to R_2 - 2R_1$:

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

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

Apply the operation $R_2 \to -R_2$:

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

Apply the operation $R_1 \to R_1 - R_2$:

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}1&-1\\-1&2 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Apply the operation $C_1 \to C_1 - C_2$:

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

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

Apply the operation $C_2 \to C_2 - C_1$:

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}1&-1\\-1&2 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}1&3\\2&7 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

$\begin{bmatrix}1&3\\2&7 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix} A$

Apply the operation $R_2 \to R_2 - 2R_1$:

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

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

Apply the operation $R_1 \to R_1 - 3R_2$:

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

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

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}7&-3\\-2&1 \end{bmatrix} A$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}7&-3\\-2&1 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

$\begin{bmatrix}1&3\\2&7 \end{bmatrix} = A \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Apply the operation $C_2 \to C_2 - 3C_1$:

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

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

Apply the operation $C_1 \to C_1 - 2C_2$:

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

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

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}7&-3\\-2&1 \end{bmatrix}$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}7&-3\\-2&1 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}2&3\\5&7 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

$\begin{bmatrix}2&3\\5&7 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix} A$

Apply the operation $R_1 \to \frac{1}{2} R_1$:

$\begin{bmatrix}1&\frac{3}{2}\\5&7 \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\0&1 \end{bmatrix} A$

Apply the operation $R_2 \to R_2 - 5R_1$:

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

$\begin{bmatrix}1&\frac{3}{2}\\0&7 - \frac{15}{2} \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\-\frac{5}{2}&1 \end{bmatrix} A$

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

Apply the operation $R_2 \to -2R_2$:

$\begin{bmatrix}1&\frac{3}{2}\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\(-2)(-\frac{5}{2})&(-2)(1) \end{bmatrix} A$

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

Apply the operation $R_1 \to R_1 - \frac{3}{2}R_2$:

$\begin{bmatrix}1 - \frac{3}{2}(0)&\frac{3}{2} - \frac{3}{2}(1)\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{2} - \frac{3}{2}(5)&0 - \frac{3}{2}(-2)\\5&-2 \end{bmatrix} A$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{2} - \frac{15}{2}&3\\5&-2 \end{bmatrix} A$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}-\frac{14}{2}&3\\5&-2 \end{bmatrix} A$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}-7&3\\5&-2 \end{bmatrix} A$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

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

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

$\begin{bmatrix}2&3\\5&7 \end{bmatrix} = A \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Apply the operation $C_1 \to \frac{1}{2} C_1$:

$\begin{bmatrix}1&3\\\frac{5}{2}&7 \end{bmatrix} = A \begin{bmatrix}\frac{1}{2}&0\\0&1 \end{bmatrix}$

Apply the operation $C_2 \to C_2 - 3C_1$:

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

$\begin{bmatrix}1&0\\\frac{5}{2}&7 - \frac{15}{2} \end{bmatrix} = A \begin{bmatrix}\frac{1}{2}&-\frac{3}{2}\\0&1 \end{bmatrix}$

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

Apply the operation $C_2 \to -2C_2$:

$\begin{bmatrix}1&0\\\frac{5}{2}&(-2)(-\frac{1}{2}) \end{bmatrix} = A \begin{bmatrix}\frac{1}{2}&(-2)(-\frac{3}{2})\\0&(-2)(1) \end{bmatrix}$

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

Apply the operation $C_1 \to C_1 - \frac{5}{2}C_2$:

$\begin{bmatrix}1 - \frac{5}{2}(0)&0\\\frac{5}{2} - \frac{5}{2}(1)&1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{2} - \frac{5}{2}(3)&3\\0 - \frac{5}{2}(-2)&-2 \end{bmatrix}$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{2} - \frac{15}{2}&3\\0 + 5&-2 \end{bmatrix}$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}-\frac{14}{2}&3\\5&-2 \end{bmatrix}$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}-7&3\\5&-2 \end{bmatrix}$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

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

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}2&1\\7&4 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

$\begin{bmatrix}2&1\\7&4 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix} A$

Apply the operation $R_1 \to \frac{1}{2} R_1$:

$\begin{bmatrix}1&\frac{1}{2}\\7&4 \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\0&1 \end{bmatrix} A$

Apply the operation $R_2 \to R_2 - 7R_1$:

$\begin{bmatrix}1&\frac{1}{2}\\7 - 7(1)&4 - 7(\frac{1}{2}) \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\0 - 7(\frac{1}{2})&1 - 7(0) \end{bmatrix} A$

$\begin{bmatrix}1&\frac{1}{2}\\0&4 - \frac{7}{2} \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\-\frac{7}{2}&1 \end{bmatrix} A$

$\begin{bmatrix}1&\frac{1}{2}\\0&\frac{1}{2} \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\-\frac{7}{2}&1 \end{bmatrix} A$

Apply the operation $R_2 \to 2R_2$:

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

$\begin{bmatrix}1&\frac{1}{2}\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\-7&2 \end{bmatrix} A$

Apply the operation $R_1 \to R_1 - \frac{1}{2}R_2$:

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

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{2} + \frac{7}{2}&-1\\-7&2 \end{bmatrix} A$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{8}{2}&-1\\-7&2 \end{bmatrix} A$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}4&-1\\-7&2 \end{bmatrix} A$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}4&-1\\-7&2 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

$\begin{bmatrix}2&1\\7&4 \end{bmatrix} = A \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Apply the operation $C_1 \to \frac{1}{2} C_1$:

$\begin{bmatrix}1&1\\\frac{7}{2}&4 \end{bmatrix} = A \begin{bmatrix}\frac{1}{2}&0\\0&1 \end{bmatrix}$

Apply the operation $C_2 \to C_2 - C_1$:

$\begin{bmatrix}1&1 - 1\\\frac{7}{2}&4 - \frac{7}{2} \end{bmatrix} = A \begin{bmatrix}\frac{1}{2}&0 - \frac{1}{2}\\0&1 - 0 \end{bmatrix}$

$\begin{bmatrix}1&0\\\frac{7}{2}&\frac{1}{2} \end{bmatrix} = A \begin{bmatrix}\frac{1}{2}&-\frac{1}{2}\\0&1 \end{bmatrix}$

Apply the operation $C_2 \to 2C_2$:

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

$\begin{bmatrix}1&0\\\frac{7}{2}&1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{2}&-1\\0&2 \end{bmatrix}$

Apply the operation $C_1 \to C_1 - \frac{7}{2}C_2$:

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

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{2} + \frac{7}{2}&-1\\0 - 7&2 \end{bmatrix}$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}\frac{8}{2}&-1\\-7&2 \end{bmatrix}$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}4&-1\\-7&2 \end{bmatrix}$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}4&-1\\-7&2 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}2&5\\1&3 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

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

Apply the operation $R_1 \leftrightarrow R_2$:

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

Apply the operation $R_2 \to R_2 - 2R_1$:

$\begin{bmatrix}1&3\\2 - 2(1)&5 - 2(3) \end{bmatrix} = \begin{bmatrix}0&1\\1 - 2(0)&0 - 2(1) \end{bmatrix} A$

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

Apply the operation $R_2 \to -R_2$:

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

Apply the operation $R_1 \to R_1 - 3R_2$:

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

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}3&-5\\-1&2 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Apply the operation $C_2 \to C_2 - 2C_1$:

$\begin{bmatrix}2&5 - 2(2)\\1&3 - 2(1) \end{bmatrix} = A \begin{bmatrix}1&0 - 2(1)\\0&1 - 2(0) \end{bmatrix}$

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

Apply the operation $C_1 \to C_1 - C_2$:

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

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

Apply the operation $C_2 \to C_2 - C_1$:

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}3&-5\\-1&2 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}3&1\\5&2 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

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

Apply the operation $R_1 \to \frac{1}{3} R_1$:

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

Apply the operation $R_2 \to R_2 - 5R_1$:

$\begin{bmatrix}1&\frac{1}{3}\\5 - 5(1)&2 - 5(\frac{1}{3}) \end{bmatrix} = \begin{bmatrix}\frac{1}{3}&0\\0 - 5(\frac{1}{3})&1 - 5(0) \end{bmatrix} A$

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

$\begin{bmatrix}1&\frac{1}{3}\\0&\frac{1}{3} \end{bmatrix} = \begin{bmatrix}\frac{1}{3}&0\\-\frac{5}{3}&1 \end{bmatrix} A$

Apply the operation $R_2 \to 3R_2$:

$\begin{bmatrix}1&\frac{1}{3}\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{3}&0\\3(-\frac{5}{3})&3(1) \end{bmatrix} A$

$\begin{bmatrix}1&\frac{1}{3}\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{3}&0\\-5&3 \end{bmatrix} A$

Apply the operation $R_1 \to R_1 - \frac{1}{3}R_2$:

$\begin{bmatrix}1 - \frac{1}{3}(0)&\frac{1}{3} - \frac{1}{3}(1)\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{3} - \frac{1}{3}(-5)&0 - \frac{1}{3}(3)\\-5&3 \end{bmatrix} A$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{3} + \frac{5}{3}&-1\\-5&3 \end{bmatrix} A$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{6}{3}&-1\\-5&3 \end{bmatrix} A$

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}2&-1\\-5&3 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Apply the operation $C_1 \to \frac{1}{3} C_1$:

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

Apply the operation $C_2 \to C_2 - C_1$:

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

$\begin{bmatrix}1&0\\\frac{5}{3}&\frac{1}{3} \end{bmatrix} = A \begin{bmatrix}\frac{1}{3}&-\frac{1}{3}\\0&1 \end{bmatrix}$

Apply the operation $C_2 \to 3C_2$:

$\begin{bmatrix}1&0\\\frac{5}{3}&3(\frac{1}{3}) \end{bmatrix} = A \begin{bmatrix}\frac{1}{3}&3(-\frac{1}{3})\\0&3(1) \end{bmatrix}$

$\begin{bmatrix}1&0\\\frac{5}{3}&1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{3}&-1\\0&3 \end{bmatrix}$

Apply the operation $C_1 \to C_1 - \frac{5}{3}C_2$:

$\begin{bmatrix}1 - \frac{5}{3}(0)&0\\\frac{5}{3} - \frac{5}{3}(1)&1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{3} - \frac{5}{3}(-1)&-1\\0 - \frac{5}{3}(3)&3 \end{bmatrix}$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{3} + \frac{5}{3}&-1\\-5&3 \end{bmatrix}$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}\frac{6}{3}&-1\\-5&3 \end{bmatrix}$

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}2&-1\\-5&3 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}4&5\\3&4 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

$\begin{bmatrix}4&5\\3&4 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix} A$

Apply the operation $R_1 \to R_1 - R_2$:

$\begin{bmatrix}4-3&5-4\\3&4 \end{bmatrix} = \begin{bmatrix}1-0&0-1\\0&1 \end{bmatrix} A$

$\begin{bmatrix}1&1\\3&4 \end{bmatrix} = \begin{bmatrix}1&-1\\0&1 \end{bmatrix} A$

Apply the operation $R_2 \to R_2 - 3R_1$:

$\begin{bmatrix}1&1\\3 - 3(1)&4 - 3(1) \end{bmatrix} = \begin{bmatrix}1&-1\\0 - 3(1)&1 - 3(-1) \end{bmatrix} A$

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

$\begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}1&-1\\-3&4 \end{bmatrix} A$

Apply the operation $R_1 \to R_1 - R_2$:

$\begin{bmatrix}1 - 0&1 - 1\\0&1 \end{bmatrix} = \begin{bmatrix}1 - (-3)&-1 - 4\\-3&4 \end{bmatrix} A$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}1 + 3&-5\\-3&4 \end{bmatrix} A$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}4&-5\\-3&4 \end{bmatrix} A$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}4&-5\\-3&4 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

$\begin{bmatrix}4&5\\3&4 \end{bmatrix} = A \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Apply the operation $C_1 \to C_1 - C_2$:

$\begin{bmatrix}4-5&5\\3-4&4 \end{bmatrix} = A \begin{bmatrix}1-0&0\\0-1&1 \end{bmatrix}$

$\begin{bmatrix}-1&5\\-1&4 \end{bmatrix} = A \begin{bmatrix}1&0\\-1&1 \end{bmatrix}$

Apply the operation $C_1 \to -C_1$:

$\begin{bmatrix}1&5\\1&4 \end{bmatrix} = A \begin{bmatrix}-1&0\\1&1 \end{bmatrix}$

Apply the operation $C_2 \to C_2 - 5C_1$:

$\begin{bmatrix}1&5 - 5(1)\\1&4 - 5(1) \end{bmatrix} = A \begin{bmatrix}-1&0 - 5(-1)\\1&1 - 5(1) \end{bmatrix}$

$\begin{bmatrix}1&0\\1&-1 \end{bmatrix} = A \begin{bmatrix}-1&5\\1&-4 \end{bmatrix}$

Apply the operation $C_2 \to -C_2$:

$\begin{bmatrix}1&0\\1&1 \end{bmatrix} = A \begin{bmatrix}-1&-5\\1&4 \end{bmatrix}$

Apply the operation $C_1 \to C_1 - C_2$:

$\begin{bmatrix}1 - 0&0\\1 - 1&1 \end{bmatrix} = A \begin{bmatrix}-1 - (-5)&-5\\1 - 4&4 \end{bmatrix}$

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}4&-5\\-3&4 \end{bmatrix}$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}4&-5\\-3&4 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}3&10\\2&7 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

$\begin{bmatrix}3&10\\2&7 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix} A$

Apply the operation $R_1 \to R_1 - R_2$:

$\begin{bmatrix}3-2&10-7\\2&7 \end{bmatrix} = \begin{bmatrix}1-0&0-1\\0&1 \end{bmatrix} A$

$\begin{bmatrix}1&3\\2&7 \end{bmatrix} = \begin{bmatrix}1&-1\\0&1 \end{bmatrix} A$

Apply the operation $R_2 \to R_2 - 2R_1$:

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

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

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

Apply the operation $R_1 \to R_1 - 3R_2$:

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

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

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}7&-10\\-2&3 \end{bmatrix} A$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}7&-10\\-2&3 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

$\begin{bmatrix}3&10\\2&7 \end{bmatrix} = A \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Apply the operation $C_2 \to C_2 - 3C_1$:

$\begin{bmatrix}3&10 - 3(3)\\2&7 - 3(2) \end{bmatrix} = A \begin{bmatrix}1&0 - 3(1)\\0&1 - 3(0) \end{bmatrix}$

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

Apply the operation $C_1 \to C_1 - 2C_2$:

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

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

$\begin{bmatrix}1&1\\0&1 \end{bmatrix} = A \begin{bmatrix}7&-3\\-2&1 \end{bmatrix}$

Apply the operation $C_2 \to C_2 - C_1$:

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

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = A \begin{bmatrix}7&-10\\-2&3 \end{bmatrix}$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}7&-10\\-2&3 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}3&−1\\−4&2 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

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

Apply the operation $R_1 \to \frac{1}{3} R_1$:

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

Apply the operation $R_2 \to R_2 + 4R_1$:

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

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

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

Apply the operation $R_2 \to \frac{3}{2}R_2$:

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

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

Apply the operation $R_1 \to R_1 + \frac{1}{3}R_2$:

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

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{3} + \frac{2}{3}&\frac{1}{2}\\2&\frac{3}{2} \end{bmatrix} A$

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}1&\frac{1}{2}\\2&\frac{3}{2} \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Apply the operation $C_1 \to \frac{1}{3} C_1$:

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

Apply the operation $C_2 \to C_2 + C_1$:

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

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

Apply the operation $C_2 \to \frac{3}{2}C_2$:

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

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

Apply the operation $C_1 \to C_1 + \frac{4}{3}C_2$:

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

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}1&\frac{1}{2}\\2&\frac{3}{2} \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}2&−6\\1&−2 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

$\begin{bmatrix}2&−6\\1&−2 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix} A$

Apply the operation $R_1 \leftrightarrow R_2$:

$\begin{bmatrix}1&−2\\2&−6 \end{bmatrix} = \begin{bmatrix}0&1\\1&0 \end{bmatrix} A$

Apply the operation $R_2 \to R_2 - 2R_1$:

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

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

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

Apply the operation $R_2 \to -\frac{1}{2}R_2$:

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

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

Apply the operation $R_1 \to R_1 + 2R_2$:

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}-1&3\\-\frac{1}{2}&1 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

$\begin{bmatrix}2&−6\\1&−2 \end{bmatrix} = A \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Apply the operation $C_1 \to \frac{1}{2} C_1$:

$\begin{bmatrix}1&−6\\\frac{1}{2}&−2 \end{bmatrix} = A \begin{bmatrix}\frac{1}{2}&0\\0&1 \end{bmatrix}$

Apply the operation $C_2 \to C_2 + 6C_1$:

$\begin{bmatrix}1&−6 + 6(1)\\\frac{1}{2}&−2 + 6(\frac{1}{2}) \end{bmatrix} = A \begin{bmatrix}\frac{1}{2}&0 + 6(\frac{1}{2})\\0&1 + 6(0) \end{bmatrix}$

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

Apply the operation $C_1 \to C_1 - \frac{1}{2}C_2$:

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

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}-1&3\\-\frac{1}{2}&1 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}6&−3\\−2&1 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations, if it exists.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

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

Apply the operation $R_1 \to \frac{1}{6} R_1$:

$\begin{bmatrix}1&−\frac{3}{6}\\−2&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{6}&0\\0&1 \end{bmatrix} A$

$\begin{bmatrix}1&−\frac{1}{2}\\−2&1 \end{bmatrix} = \begin{bmatrix}\frac{1}{6}&0\\0&1 \end{bmatrix} A$

Apply the operation $R_2 \to R_2 + 2R_1$:

$\begin{bmatrix}1&−\frac{1}{2}\\−2 + 2(1)&1 + 2(−\frac{1}{2}) \end{bmatrix} = \begin{bmatrix}\frac{1}{6}&0\\0 + 2(\frac{1}{6})&1 + 2(0) \end{bmatrix} A$

$\begin{bmatrix}1&−\frac{1}{2}\\0&1 - 1 \end{bmatrix} = \begin{bmatrix}\frac{1}{6}&0\\\frac{2}{6}&1 \end{bmatrix} A$

$\begin{bmatrix}1&−\frac{1}{2}\\0&0 \end{bmatrix} = \begin{bmatrix}\frac{1}{6}&0\\\frac{1}{3}&1 \end{bmatrix} A$

In the above equation, we can see that the second row of the matrix on the LHS consists entirely of zeros. Therefore, we cannot proceed further to obtain the identity matrix on the LHS.

Hence, the inverse of matrix $A$ does not exist.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Apply the operation $C_1 \to \frac{1}{6} C_1$:

$\begin{bmatrix}1&−3\\−\frac{2}{6}&1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{6}&0\\0&1 \end{bmatrix}$

$\begin{bmatrix}1&−3\\−\frac{1}{3}&1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{6}&0\\0&1 \end{bmatrix}$

Apply the operation $C_2 \to C_2 + 3C_1$:

$\begin{bmatrix}1&−3 + 3(1)\\−\frac{1}{3}&1 + 3(−\frac{1}{3}) \end{bmatrix} = A \begin{bmatrix}\frac{1}{6}&0 + 3(\frac{1}{6})\\0&1 + 3(0) \end{bmatrix}$

$\begin{bmatrix}1&0\\−\frac{1}{3}&1 - 1 \end{bmatrix} = A \begin{bmatrix}\frac{1}{6}&\frac{3}{6}\\0&1 \end{bmatrix}$

$\begin{bmatrix}1&0\\−\frac{1}{3}&0 \end{bmatrix} = A \begin{bmatrix}\frac{1}{6}&\frac{1}{2}\\0&1 \end{bmatrix}$

In the above equation, we can see that the second column of the matrix on the LHS consists entirely of zeros. Therefore, we cannot proceed further to obtain the identity matrix on the LHS.

Hence, the inverse of matrix $A$ does not exist.

Given:

Let the given matrix be $A = \begin{bmatrix}2&−3\\−1&2 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

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

Apply the operation $R_1 \leftrightarrow R_2$:

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

Apply the operation $R_1 \to -R_1$:

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

Apply the operation $R_2 \to R_2 - 2R_1$:

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

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

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

Apply the operation $R_1 \to R_1 + 2R_2$:

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

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}2&3\\1&2 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Apply the operation $C_1 \to C_1 + C_2$:

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

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

Apply the operation $C_1 \to -C_1$:

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

Apply the operation $C_2 \to C_2 + 3C_1$:

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

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

Apply the operation $C_2 \to -C_2$:

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

Apply the operation $C_1 \to C_1 + C_2$:

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

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}2&3\\1&2 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}2&1\\4&2 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations, if it exists.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of the same order as $A$.

$\begin{bmatrix}2&1\\4&2 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix} A$

Apply the operation $R_1 \to \frac{1}{2} R_1$:

$\begin{bmatrix}1&\frac{1}{2}\\4&2 \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\0&1 \end{bmatrix} A$

Apply the operation $R_2 \to R_2 - 4R_1$:

$\begin{bmatrix}1&\frac{1}{2}\\4 - 4(1)&2 - 4(\frac{1}{2}) \end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\0 - 4(\frac{1}{2})&1 - 4(0) \end{bmatrix} A$

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

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

In the above equation, we can see that the second row of the matrix on the LHS consists entirely of zeros. Therefore, we cannot proceed further to obtain the identity matrix on the LHS.

Hence, the inverse of matrix $A$ does not exist.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

$\begin{bmatrix}2&1\\4&2 \end{bmatrix} = A \begin{bmatrix}1&0\\0&1 \end{bmatrix}$

Apply the operation $C_1 \to \frac{1}{2} C_1$:

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

Apply the operation $C_2 \to C_2 - C_1$:

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

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

In the above equation, we can see that the second column of the matrix on the LHS consists entirely of zeros. Therefore, we cannot proceed further to obtain the identity matrix on the LHS.

Hence, the inverse of matrix $A$ does not exist.

(Note: A matrix inverse exists if and only if its determinant is non-zero. Here, det(A) = $2 \times 2 - 1 \times 4 = 4 - 4 = 0$.)

Given:

Let the given matrix be $A = \begin{bmatrix}2&−3&3\\2&2&3\\3&−2&2 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of order 3.

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

Apply $R_1 \to \frac{1}{2}R_1$:

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

Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$:

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

Apply $R_2 \to \frac{1}{5}R_2$:

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

Apply $R_1 \to R_1 + \frac{3}{2}R_2$ and $R_3 \to R_3 - \frac{5}{2}R_2$:

$\begin{bmatrix}1&0&3/2\\0&1&0\\0&0&−5/2 \end{bmatrix} = \begin{bmatrix}1/5&3/10&0\\-1/5&1/5&0\\-1&−1/2&1 \end{bmatrix} A$

Apply $R_3 \to -\frac{2}{5}R_3$:

$\begin{bmatrix}1&0&3/2\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix}1/5&3/10&0\\-1/5&1/5&0\\2/5&1/5&−2/5 \end{bmatrix} A$

Apply $R_1 \to R_1 - \frac{3}{2}R_3$:

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}-2/5&0&3/5\\-1/5&1/5&0\\2/5&1/5&−2/5 \end{bmatrix} = \frac{1}{5} \begin{bmatrix}-2&0&3\\-1&1&0\\2&1&-2 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Apply $C_1 \to \frac{1}{2}C_1$:

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

Apply $C_2 \to C_2 + 3C_1$ and $C_3 \to C_3 - 3C_1$:

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

Apply $C_2 \to \frac{1}{5}C_2$:

$\begin{bmatrix}1&0&0\\1&1&0\\3/2&1/2&−5/2 \end{bmatrix} = A \begin{bmatrix}1/2&3/10&-3/2\\0&1/5&0\\0&0&1 \end{bmatrix}$

Apply $C_1 \to C_1 - C_2$:

$\begin{bmatrix}1&0&0\\0&1&0\\1&1/2&−5/2 \end{bmatrix} = A \begin{bmatrix}1/5&3/10&-3/2\\-1/5&1/5&0\\0&0&1 \end{bmatrix}$

Apply $C_3 \to -\frac{2}{5}C_3$:

$\begin{bmatrix}1&0&0\\0&1&0\\1&1/2&1 \end{bmatrix} = A \begin{bmatrix}1/5&3/10&3/5\\-1/5&1/5&0\\0&0&-2/5 \end{bmatrix}$

Apply $C_1 \to C_1 - C_3$ and $C_2 \to C_2 - \frac{1}{2}C_3$:

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

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}-2/5&0&3/5\\-1/5&1/5&0\\2/5&1/5&−2/5 \end{bmatrix} = \frac{1}{5} \begin{bmatrix}-2&0&3\\-1&1&0\\2&1&-2 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}1&3&−2\\−3&0&−5\\2&5&0 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of order 3.

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

Apply $R_2 \to R_2 + 3R_1$ and $R_3 \to R_3 - 2R_1$:

$\begin{bmatrix}1&3&−2\\0&9&−11\\0&-1&4 \end{bmatrix} = \begin{bmatrix}1&0&0\\3&1&0\\-2&0&1 \end{bmatrix} A$

Apply $R_2 \leftrightarrow R_3$:

$\begin{bmatrix}1&3&−2\\0&-1&4\\0&9&−11 \end{bmatrix} = \begin{bmatrix}1&0&0\\-2&0&1\\3&1&0 \end{bmatrix} A$

Apply $R_2 \to -R_2$:

$\begin{bmatrix}1&3&−2\\0&1&-4\\0&9&−11 \end{bmatrix} = \begin{bmatrix}1&0&0\\2&0&-1\\3&1&0 \end{bmatrix} A$

Apply $R_1 \to R_1 - 3R_2$ and $R_3 \to R_3 - 9R_2$:

$\begin{bmatrix}1&0&10\\0&1&-4\\0&0&25 \end{bmatrix} = \begin{bmatrix}-5&0&3\\2&0&-1\\-15&1&9 \end{bmatrix} A$

Apply $R_3 \to \frac{1}{25}R_3$:

$\begin{bmatrix}1&0&10\\0&1&-4\\0&0&1 \end{bmatrix} = \begin{bmatrix}-5&0&3\\2&0&-1\\-15/25&1/25&9/25 \end{bmatrix} A$

$\begin{bmatrix}1&0&10\\0&1&-4\\0&0&1 \end{bmatrix} = \begin{bmatrix}-5&0&3\\2&0&-1\\-3/5&1/25&9/25 \end{bmatrix} A$

Apply $R_1 \to R_1 - 10R_3$ and $R_2 \to R_2 + 4R_3$:

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix}-5 - 10(-3/5)&0 - 10(1/25)&3 - 10(9/25)\\2 + 4(-3/5)&0 + 4(1/25)&-1 + 4(9/25)\\-3/5&1/25&9/25 \end{bmatrix} A$

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix}1&-10/25&-15/25\\-2/5&4/25&11/25\\-3/5&1/25&9/25 \end{bmatrix} A$

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix}1&-2/5&-3/5\\-2/5&4/25&11/25\\-3/5&1/25&9/25 \end{bmatrix} A$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}1&-2/5&-3/5\\-2/5&4/25&11/25\\-3/5&1/25&9/25 \end{bmatrix} = \frac{1}{25} \begin{bmatrix}25&-10&-15\\-10&4&11\\-15&1&9 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Apply $C_2 \to C_2 - 3C_1$ and $C_3 \to C_3 + 2C_1$:

$\begin{bmatrix}1&0&0\\−3&9&−11\\2&-1&4 \end{bmatrix} = A \begin{bmatrix}1&-3&2\\0&1&0\\0&0&1 \end{bmatrix}$

Apply $C_2 \to \frac{1}{9}C_2$:

$\begin{bmatrix}1&0&0\\−3&1&−11\\2&-1/9&4 \end{bmatrix} = A \begin{bmatrix}1&-1/3&2\\0&1/9&0\\0&0&1 \end{bmatrix}$

Apply $C_1 \to C_1 + 3C_2$ and $C_3 \to C_3 + 11C_2$:

$\begin{bmatrix}1&0&0\\0&1&0\\5/3&-1/9&25/9 \end{bmatrix} = A \begin{bmatrix}0&-1/3&-5/3\\1/3&1/9&11/9\\0&0&1 \end{bmatrix}$

Apply $C_3 \to \frac{9}{25}C_3$:

$\begin{bmatrix}1&0&0\\0&1&0\\5/3&-1/9&1 \end{bmatrix} = A \begin{bmatrix}0&-1/3&-3/5\\1/3&1/9&11/25\\0&0&9/25 \end{bmatrix}$

Apply $C_1 \to C_1 - \frac{5}{3}C_3$ and $C_2 \to C_2 + \frac{1}{9}C_3$:

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = A \begin{bmatrix}0 - \frac{5}{3}(-3/5)&-1/3 + \frac{1}{9}(-3/5)&-3/5\\1/3 - \frac{5}{3}(11/25)&1/9 + \frac{1}{9}(11/25)&11/25\\0 - \frac{5}{3}(9/25)&0 + \frac{1}{9}(9/25)&9/25 \end{bmatrix}$

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = A \begin{bmatrix}1&-2/5&-3/5\\-2/5&4/25&11/25\\-3/5&1/25&9/25 \end{bmatrix}$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}1&-2/5&-3/5\\-2/5&4/25&11/25\\-3/5&1/25&9/25 \end{bmatrix} = \frac{1}{25} \begin{bmatrix}25&-10&-15\\-10&4&11\\-15&1&9 \end{bmatrix}$.

Both methods yield the same result.

Given:

Let the given matrix be $A = \begin{bmatrix}2&0&−1\\5&1&0\\0&1&3 \end{bmatrix}$.

To Find:

The inverse of matrix $A$, denoted as $A^{-1}$, using elementary transformations.

Solution (Using Elementary Row Operations):

We write $A = IA$, where $I$ is the identity matrix of order 3.

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

Apply $R_1 \to \frac{1}{2}R_1$:

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

Apply $R_2 \to R_2 - 5R_1$:

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

Apply $R_3 \to R_3 - R_2$:

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

Apply $R_3 \to 2R_3$:

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

Apply $R_1 \to R_1 + \frac{1}{2}R_3$ and $R_2 \to R_2 - \frac{5}{2}R_3$:

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix}1/2 + 5/2&0 - 1&0 + 1\\-5/2 - 25/2&1 + 5&0 - 5\\5&-2&2 \end{bmatrix} A$

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2 \end{bmatrix} A$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2 \end{bmatrix}$.

Alternate Solution (Using Elementary Column Operations):

We write $A = AI$.

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

Apply $C_1 \to \frac{1}{2}C_1$:

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

Apply $C_3 \to C_3 + C_1$:

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

Apply $C_1 \to C_1 - \frac{5}{2}C_2$ and $C_3 \to C_3 - \frac{5}{2}C_2$:

$\begin{bmatrix}1&0&0\\0&1&0\\-5/2&1&1/2 \end{bmatrix} = A \begin{bmatrix}1/2&0&1/2\\-5/2&1&-5/2\\0&0&1 \end{bmatrix}$

Apply $C_3 \to 2C_3$:

$\begin{bmatrix}1&0&0\\0&1&0\\-5/2&1&1 \end{bmatrix} = A \begin{bmatrix}1/2&0&1\\-5/2&1&-5\\0&0&2 \end{bmatrix}$

Apply $C_1 \to C_1 + \frac{5}{2}C_3$ and $C_2 \to C_2 - C_3$:

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = A \begin{bmatrix}1/2 + 5/2&0 - 1&1\\-5/2 - 25/2&1 - (-5)&-5\\0 + 5&0 - 2&2 \end{bmatrix}$

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix} = A \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2 \end{bmatrix}$

Since we have obtained the identity matrix on the LHS, the matrix on the RHS is the inverse of $A$.

Thus, $A^{-1} = \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2 \end{bmatrix}$.

Both methods yield the same result.

Explanation:

By definition, two square matrices $A$ and $B$ of the same order are said to be inverses of each other if their product is the identity matrix $I$ of the same order.

This means that $A$ is the inverse of $B$ (denoted as $A = B^{-1}$) and $B$ is the inverse of $A$ (denoted as $B = A^{-1}$) if and only if:

$AB = I$ and $BA = I$.

Combining these two conditions, we get $AB = BA = I$.

Let's look at the given options:

(A) $AB = BA$: This condition defines matrices that commute, but they are not necessarily inverses.

(B) $AB = BA = 0$: This means the product is the zero matrix, not the identity matrix. This describes pairs of matrices where at least one might be singular, or they are zero divisors in the ring of matrices.

(C) $AB = 0, BA = I$: If $BA = I$, then both $A$ and $B$ must be invertible. If $A$ is invertible, then $AB = 0$ implies $A^{-1}(AB) = A^{-1}0$, which simplifies to $IB = 0$, meaning $B = 0$. But if $B=0$, then $BA=0$, which contradicts $BA=I$. Therefore, this case is impossible for square matrices.

(D) $AB = BA = I$: This is the definition of matrices being inverses of each other.

Therefore, the correct condition is $AB = BA = I$.

Final Answer: The final answer is $\boxed{(D)}$

Given:

The matrix $A = \begin{bmatrix}\cosθ&\sinθ\\−\sinθ&\cosθ \end{bmatrix}$.

To Prove:

$A^n = \begin{bmatrix}\cos nθ&\sin nθ\\−\sin nθ&\cos nθ \end{bmatrix}$ for all $n \in \mathbb{N}$ (natural numbers).

Proof (Using Mathematical Induction):

Let $P(n)$ be the statement: $A^n = \begin{bmatrix}\cos nθ&\sin nθ\\−\sin nθ&\cos nθ \end{bmatrix}$.

Base Case (n=1):

We need to show that $P(1)$ is true.

For $n=1$, $A^1 = A = \begin{bmatrix}\cosθ&\sinθ\\−\sinθ&\cosθ \end{bmatrix}$.

The right-hand side (RHS) for $n=1$ is $\begin{bmatrix}\cos(1 \cdot θ)&\sin(1 \cdot θ)\\−\sin(1 \cdot θ)&\cos(1 \cdot θ) \end{bmatrix} = \begin{bmatrix}\cosθ&\sinθ\\−\sinθ&\cosθ \end{bmatrix}$.

Since LHS = RHS, $P(1)$ is true.

Inductive Hypothesis:

Assume that $P(k)$ is true for some positive integer $k \in \mathbb{N}$.

That is, assume $A^k = \begin{bmatrix}\cos kθ&\sin kθ\\−\sin kθ&\cos kθ \end{bmatrix}$.

Inductive Step:

We need to prove that $P(k+1)$ is true, i.e., $A^{k+1} = \begin{bmatrix}\cos (k+1)θ&\sin (k+1)θ\\−\sin (k+1)θ&\cos (k+1)θ \end{bmatrix}$.

Consider $A^{k+1} = A^k \cdot A$.

Using the inductive hypothesis for $A^k$:

$A^{k+1} = \begin{bmatrix}\cos kθ&\sin kθ\\−\sin kθ&\cos kθ \end{bmatrix} \begin{bmatrix}\cosθ&\sinθ\\−\sinθ&\cosθ \end{bmatrix}$

Performing matrix multiplication:

$A^{k+1} = \begin{bmatrix} (\cos kθ)(\cosθ) + (\sin kθ)(−\sinθ) & (\cos kθ)(\sinθ) + (\sin kθ)(\cosθ) \\ (−\sin kθ)(\cosθ) + (\cos kθ)(−\sinθ) & (−\sin kθ)(\sinθ) + (\cos kθ)(\cosθ) \end{bmatrix}$

$A^{k+1} = \begin{bmatrix} \cos kθ \cosθ - \sin kθ \sinθ & \cos kθ \sinθ + \sin kθ \cosθ \\ −(\sin kθ \cosθ + \cos kθ \sinθ) & \cos kθ \cosθ - \sin kθ \sinθ \end{bmatrix}$

Using the trigonometric sum identities:

$\cos(A+B) = \cos A \cos B - \sin A \sin B$

$\sin(A+B) = \sin A \cos B + \cos A \sin B$

Applying these identities with $A = kθ$ and $B = θ$:

$\cos kθ \cosθ - \sin kθ \sinθ = \cos(kθ + θ) = \cos((k+1)θ)$

$\cos kθ \sinθ + \sin kθ \cosθ = \sin(kθ + θ) = \sin((k+1)θ)$

Therefore,

$A^{k+1} = \begin{bmatrix} \cos((k+1)θ) & \sin((k+1)θ) \\ −\sin((k+1)θ) & \cos((k+1)θ) \end{bmatrix}$

This shows that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

By the principle of mathematical induction, the statement $P(n)$ is true for all $n \in \mathbb{N}$.

Hence, $A^n = \begin{bmatrix}\cos nθ&\sin nθ\\−\sin nθ&\cos nθ \end{bmatrix}$ for all $n \in \mathbb{N}$.

Given:

A and B are symmetric matrices of the same order.

This means $A' = A$ and $B' = B$, where $A'$ and $B'$ denote the transpose of matrices A and B respectively.

To Prove:

AB is symmetric if and only if $AB = BA$.

This requires proving two parts:

Part 1: If $AB = BA$, then AB is symmetric.

Part 2: If AB is symmetric, then $AB = BA$.

Proof:

Part 1: Assume $AB = BA$. Show that AB is symmetric.

To show that AB is symmetric, we need to prove that $(AB)' = AB$.

Consider the transpose of the product AB:

Since A and B are symmetric matrices, we have $A' = A$ and $B' = B$. Substituting these into the equation:

Now, using the assumption that A and B commute ($AB = BA$):

Thus, if $AB = BA$, then $(AB)' = AB$, which means AB is symmetric.

Part 2: Assume AB is symmetric. Show that $AB = BA$.

Assuming AB is symmetric, by definition, we have:

We also know the property of the transpose of a product:

Since A and B are symmetric matrices, $A' = A$ and $B' = B$. Substituting these:

Now, equating the two expressions for $(AB)'$ from the assumption and the property:

Thus, if AB is symmetric, then $AB = BA$.

Conclusion:

From Part 1 and Part 2, we have shown that AB is symmetric if and only if $AB = BA$.

Hence proved.

Given:

Matrices A, B, and C are given as:

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

$B = \begin{bmatrix}5&2\\7&4 \end{bmatrix}$

$C = \begin{bmatrix}2&5\\3&8 \end{bmatrix}$

The equation $CD – AB = O$, where O is the null/zero matrix of the same order.

To Find:

The matrix D.

Solution:

The given equation is $CD - AB = O$.

This can be rewritten as $CD = AB$.

First, we calculate the product AB:

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

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

$AB = \begin{bmatrix} 10 - 7 & 4 - 4 \\ 15 + 28 & 6 + 16 \end{bmatrix}$

$AB = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix}$

Now the equation is $CD = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix}$.

$\begin{bmatrix}2&5\\3&8 \end{bmatrix} D = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix}$

To find D, we need to pre-multiply both sides by the inverse of C, $C^{-1}$. First, let's find $C^{-1}$.

The determinant of C is $det(C) = (2)(8) - (5)(3) = 16 - 15 = 1$.

Since $det(C) \neq 0$, the inverse exists.

The inverse of a 2x2 matrix $\begin{bmatrix}a&b\\c&d \end{bmatrix}$ is $\frac{1}{ad-bc} \begin{bmatrix}d&-b\\-c&a \end{bmatrix}$.

$C^{-1} = \frac{1}{1} \begin{bmatrix}8&-5\\-3&2 \end{bmatrix} = \begin{bmatrix}8&-5\\-3&2 \end{bmatrix}$.

Now, pre-multiply the equation $CD = AB$ by $C^{-1}$:

$C^{-1}(CD) = C^{-1}(AB)$

$(C^{-1}C)D = C^{-1}(AB)$

$ID = C^{-1}(AB)$, where I is the identity matrix.

$D = C^{-1}(AB)$

$D = \begin{bmatrix}8&-5\\-3&2 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix}$

$D = \begin{bmatrix} (8)(3) + (-5)(43) & (8)(0) + (-5)(22) \\ (-3)(3) + (2)(43) & (-3)(0) + (2)(22) \end{bmatrix}$

$D = \begin{bmatrix} 24 - 215 & 0 - 110 \\ -9 + 86 & 0 + 44 \end{bmatrix}$

$D = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}$

Therefore, the required matrix D is $\begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}$.

Given:

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

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

$a, b$ are scalars.

$n \in \mathbb{N}$ (set of natural numbers).

To Prove:

$(aI + bA)^n = a^n I + na^{n-1} bA$ for all $n \in \mathbb{N}$.

Proof (Using Mathematical Induction):

Let $P(n)$ be the statement: $(aI + bA)^n = a^n I + na^{n-1} bA$.

First, let's calculate $A^2$:

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

Where O is the zero matrix of order 2. This implies that $A^k = O$ for all $k \ge 2$.

Base Case (n=1):

We need to show that $P(1)$ is true.

LHS = $(aI + bA)^1 = aI + bA$.

RHS = $a^1 I + (1)a^{1-1} bA = aI + a^0 bA = aI + (1)bA = aI + bA$.

Since LHS = RHS, $P(1)$ is true.

Inductive Hypothesis:

Assume that $P(k)$ is true for some positive integer $k \in \mathbb{N}$.

That is, assume $(aI + bA)^k = a^k I + ka^{k-1} bA$.

Inductive Step:

We need to prove that $P(k+1)$ is true, i.e., $(aI + bA)^{k+1} = a^{k+1} I + (k+1)a^k bA$.

Consider the LHS:

$(aI + bA)^{k+1} = (aI + bA)^k (aI + bA)$

Using the inductive hypothesis:

$(aI + bA)^{k+1} = (a^k I + ka^{k-1} bA) (aI + bA)$

Expand the product using distributive property:

$(aI + bA)^{k+1} = (a^k I)(aI) + (a^k I)(bA) + (ka^{k-1} bA)(aI) + (ka^{k-1} bA)(bA)$

$(aI + bA)^{k+1} = a^{k+1} (I \cdot I) + a^k b (I \cdot A) + ka^k b (A \cdot I) + k a^{k-1} b^2 (A \cdot A)$

Using properties $I \cdot I = I$, $I \cdot A = A$, $A \cdot I = A$, and $A \cdot A = A^2 = O$:

$(aI + bA)^{k+1} = a^{k+1} I + a^k b A + ka^k b A + k a^{k-1} b^2 O$

$(aI + bA)^{k+1} = a^{k+1} I + (a^k b + ka^k b) A + O$

$(aI + bA)^{k+1} = a^{k+1} I + (1 + k) a^k b A$

$(aI + bA)^{k+1} = a^{k+1} I + (k+1) a^k b A$

This is exactly the RHS required for $P(k+1)$.

Thus, $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

By the principle of mathematical induction, the statement $P(n)$ is true for all $n \in \mathbb{N}$.

Hence, it is proved that $(aI + bA)^n = a^n I + na^{n-1} bA$ for all $n \in \mathbb{N}$.

Given:

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

To Prove:

$A^n = \begin{bmatrix}3^{n−1}&3^{n−1}&3^{n−1}\\3^{n−1}&3^{n−1}&3^{n−1}\\3^{n−1}&3^{n−1}&3^{n−1} \end{bmatrix}$ for all $n \in \mathbb{N}$ (natural numbers).

Proof (Using Mathematical Induction):

Let $P(n)$ be the statement: $A^n = \begin{bmatrix}3^{n−1}&3^{n−1}&3^{n−1}\\3^{n−1}&3^{n−1}&3^{n−1}\\3^{n−1}&3^{n−1}&3^{n−1} \end{bmatrix}$.

Base Case (n=1):

We need to show that $P(1)$ is true.

For $n=1$, the Left Hand Side (LHS) is $A^1 = A = \begin{bmatrix}1&1&1\\1&1&1\\1&1&1 \end{bmatrix}$.

The Right Hand Side (RHS) for $n=1$ is $\begin{bmatrix}3^{1−1}&3^{1−1}&3^{1−1}\\3^{1−1}&3^{1−1}&3^{1−1}\\3^{1−1}&3^{1−1}&3^{1−1} \end{bmatrix} = \begin{bmatrix}3^{0}&3^{0}&3^{0}\\3^{0}&3^{0}&3^{0}\\3^{0}&3^{0}&3^{0} \end{bmatrix} = \begin{bmatrix}1&1&1\\1&1&1\\1&1&1 \end{bmatrix}$.

Since LHS = RHS, $P(1)$ is true.

Inductive Hypothesis:

Assume that $P(k)$ is true for some positive integer $k \in \mathbb{N}$.

That is, assume $A^k = \begin{bmatrix}3^{k−1}&3^{k−1}&3^{k−1}\\3^{k−1}&3^{k−1}&3^{k−1}\\3^{k−1}&3^{k−1}&3^{k−1} \end{bmatrix}$.

Inductive Step:

We need to prove that $P(k+1)$ is true, i.e., $A^{k+1} = \begin{bmatrix}3^{(k+1)−1}&3^{(k+1)−1}&3^{(k+1)−1}\\3^{(k+1)−1}&3^{(k+1)−1}&3^{(k+1)−1}\\3^{(k+1)−1}&3^{(k+1)−1}&3^{(k+1)−1} \end{bmatrix} = \begin{bmatrix}3^{k}&3^{k}&3^{k}\\3^{k}&3^{k}&3^{k}\\3^{k}&3^{k}&3^{k} \end{bmatrix}$.

Consider $A^{k+1} = A^k \cdot A$.

Using the inductive hypothesis for $A^k$ and the given matrix A:

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

Performing matrix multiplication, let's find the element in the i-th row and j-th column of the product. For example, the element in the first row, first column is:

$(3^{k-1})(1) + (3^{k-1})(1) + (3^{k-1})(1) = 3^{k-1} + 3^{k-1} + 3^{k-1} = 3 \cdot (3^{k-1}) = 3^1 \cdot 3^{k-1} = 3^{1+k-1} = 3^k$.

Since all elements in A and $A^k$ (by hypothesis) are the same, every element in the product matrix $A^{k+1}$ will be calculated similarly and will result in $3^k$.

Therefore,

$A^{k+1} = \begin{bmatrix}3^k&3^k&3^k\\3^k&3^k&3^k\\3^k&3^k&3^k \end{bmatrix}$

This shows that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

By the principle of mathematical induction, the statement $P(n)$ is true for all $n \in \mathbb{N}$.

Hence, $A^n = \begin{bmatrix}3^{n−1}&3^{n−1}&3^{n−1}\\3^{n−1}&3^{n−1}&3^{n−1}\\3^{n−1}&3^{n−1}&3^{n−1} \end{bmatrix}$ for all $n \in \mathbb{N}$.

Given:

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

To Prove:

$A^n = \begin{bmatrix}1+2n&−4n\\n&1−2n \end{bmatrix}$ for all positive integers $n$ (i.e., $n \in \mathbb{N}$).

Proof (Using Mathematical Induction):

Let $P(n)$ be the statement: $A^n = \begin{bmatrix}1+2n&−4n\\n&1−2n \end{bmatrix}$.

Base Case (n=1):

We need to show that $P(1)$ is true.

For $n=1$, the Left Hand Side (LHS) is $A^1 = A = \begin{bmatrix}3&−4\\1&−1 \end{bmatrix}$.

The Right Hand Side (RHS) for $n=1$ is $\begin{bmatrix}1+2(1)&−4(1)\\1&1−2(1) \end{bmatrix} = \begin{bmatrix}1+2&−4\\1&1−2 \end{bmatrix} = \begin{bmatrix}3&−4\\1&−1 \end{bmatrix}$.

Since LHS = RHS, $P(1)$ is true.

Inductive Hypothesis:

Assume that $P(k)$ is true for some positive integer $k \in \mathbb{N}$.

That is, assume $A^k = \begin{bmatrix}1+2k&−4k\\k&1−2k \end{bmatrix}$.

Inductive Step:

We need to prove that $P(k+1)$ is true, i.e., $A^{k+1} = \begin{bmatrix}1+2(k+1)&−4(k+1)\\k+1&1−2(k+1) \end{bmatrix} = \begin{bmatrix}1+2k+2&−4k-4\\k+1&1−2k-2 \end{bmatrix} = \begin{bmatrix}3+2k&−4k-4\\k+1&−1−2k \end{bmatrix}$.

Consider $A^{k+1} = A^k \cdot A$.

Using the inductive hypothesis for $A^k$ and the given matrix A:

$A^{k+1} = \begin{bmatrix}1+2k&−4k\\k&1−2k \end{bmatrix} \begin{bmatrix}3&−4\\1&−1 \end{bmatrix}$

Performing matrix multiplication:

$A^{k+1} = \begin{bmatrix} (1+2k)(3) + (−4k)(1) & (1+2k)(-4) + (−4k)(-1) \\ (k)(3) + (1−2k)(1) & (k)(-4) + (1−2k)(-1) \end{bmatrix}$

$A^{k+1} = \begin{bmatrix} 3 + 6k - 4k & -4 - 8k + 4k \\ 3k + 1 - 2k & -4k - 1 + 2k \end{bmatrix}$

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

This resulting matrix is exactly the matrix we expected for $P(k+1)$: $\begin{bmatrix}3+2k&−4k-4\\k+1&−1−2k \end{bmatrix}$.

This shows that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

By the principle of mathematical induction, the statement $P(n)$ is true for all positive integers $n$.

Hence, $A^n = \begin{bmatrix}1+2n&−4n\\n&1−2n \end{bmatrix}$ for all $n \in \mathbb{N}$.

Given:

A and B are symmetric matrices of the same order.

By definition of symmetric matrices, we have:

where $X'$ denotes the transpose of matrix X.

To Prove:

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

By definition of a skew-symmetric matrix, we need to show that $(AB - BA)' = -(AB - BA)$.

Proof:

Consider the transpose of $(AB - BA)$:

Now, apply the property of the transpose of a product of matrices, $(XY)' = Y'X'$:

Since A and B are symmetric matrices, we can substitute $A' = A$ and $B' = B$ using equations (i) and (ii):

Factor out $-1$ from the right side:

This shows that the transpose of $(AB - BA)$ is equal to the negative of $(AB - BA)$.

Therefore, by the definition of a skew-symmetric matrix, $(AB - BA)$ is a skew-symmetric matrix.

Hence proved.

Given:

Matrices A and B such that the product $B'AB$ is defined.

$B'$ denotes the transpose of matrix B.

To Prove:

The matrix $B'AB$ is symmetric if A is symmetric.

The matrix $B'AB$ is skew-symmetric if A is skew-symmetric.

Proof:

Let $C = B'AB$. We need to examine the transpose of C, denoted as $C'$.

$C' = (B'AB)'$

Using the reversal law for the transpose of a product $(XYZ)' = Z'Y'X'$:

Using the property that the transpose of a transpose is the original matrix, $(X')' = X$:

Now, we consider the two cases based on the properties of matrix A.

Case 1: A is symmetric

If A is symmetric, then by definition, $A' = A$.

Substitute $A' = A$ into the expression for $C'$:

$C' = B'AB$

Comparing this with the original matrix $C = B'AB$, we see that $C' = C$.

Therefore, if A is symmetric, the matrix $B'AB$ is symmetric.

Case 2: A is skew-symmetric

If A is skew-symmetric, then by definition, $A' = -A$.

Substitute $A' = -A$ into the expression for $C'$:

$C' = -(B'AB)$

Comparing this with the original matrix $C = B'AB$, we see that $C' = -C$.

Therefore, if A is skew-symmetric, the matrix $B'AB$ is skew-symmetric.

Conclusion:

We have shown that the matrix $B'AB$ is symmetric or skew-symmetric depending on whether A is symmetric or skew-symmetric, respectively.

Hence proved.

Given:

The matrix $A = \begin{bmatrix}0&2y&z\\x&y&−z\\x&−y&z \end{bmatrix}$.

The condition $A'A = I$, where $A'$ is the transpose of A and $I$ is the identity matrix of order 3.

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

To Find:

The values of x, y, and z.

Solution:

First, find the transpose of matrix A:

$A' = \begin{bmatrix}0&x&x\\2y&y&-y\\z&-z&z \end{bmatrix}$

Now, 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:

$A'A = \begin{bmatrix} (0)(0)+(x)(x)+(x)(x) & (0)(2y)+(x)(y)+(x)(-y) & (0)(z)+(x)(-z)+(x)(z) \\ (2y)(0)+(y)(x)+(-y)(x) & (2y)(2y)+(y)(y)+(-y)(-y) & (2y)(z)+(y)(-z)+(-y)(z) \\ (z)(0)+(-z)(x)+(z)(x) & (z)(2y)+(-z)(y)+(z)(-y) & (z)(z)+(-z)(-z)+(z)(z) \end{bmatrix}$

$A'A = \begin{bmatrix} 0+x^2+x^2 & 0+xy-xy & 0-xz+xz \\ 0+yx-yx & 4y^2+y^2+y^2 & 2yz-yz-yz \\ 0-zx+zx & 2zy-zy-zy & z^2+z^2+z^2 \end{bmatrix}$

$A'A = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix}$

Now, we are given the condition $A'A = I$. Substitute the calculated $A'A$ and the identity matrix $I$:

$\begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Equating the corresponding elements of the matrices:

Solving these equations for x, y, and z:

From (i): $x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$

From (ii): $y^2 = \frac{1}{6} \implies y = \pm \frac{1}{\sqrt{6}}$

From (iii): $z^2 = \frac{1}{3} \implies z = \pm \frac{1}{\sqrt{3}}$

Therefore, the values are:

$x = \pm \frac{1}{\sqrt{2}}$

$y = \pm \frac{1}{\sqrt{6}}$

$z = \pm \frac{1}{\sqrt{3}}$

Given:

The matrix equation: $\begin{bmatrix}1&2&1 \end{bmatrix} \begin{bmatrix}1&2&0\\2&0&1\\1&0&2 \end{bmatrix} \begin{bmatrix}0\\2\\x \end{bmatrix} = O$, where O is the null matrix.

Since the product results in a 1x1 matrix, O is the 1x1 null matrix, $O = [0]$.

To Find:

The value of $x$.

Solution:

Let the given matrices be $A = \begin{bmatrix}1&2&1 \end{bmatrix}$, $B = \begin{bmatrix}1&2&0\\2&0&1\\1&0&2 \end{bmatrix}$, and $C = \begin{bmatrix}0\\2\\x \end{bmatrix}$.

The equation is $ABC = O$.

First, calculate the product $AB$:

$AB = \begin{bmatrix}1&2&1 \end{bmatrix} \begin{bmatrix}1&2&0\\2&0&1\\1&0&2 \end{bmatrix}$

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

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

$AB = \begin{bmatrix} 6 & 2 & 4 \end{bmatrix}$

Now, multiply the result $AB$ by $C$:

$(AB)C = \begin{bmatrix} 6 & 2 & 4 \end{bmatrix} \begin{bmatrix}0\\2\\x \end{bmatrix}$

$(AB)C = [(6)(0) + (2)(2) + (4)(x)]$

$(AB)C = [0 + 4 + 4x]$

$(AB)C = [4 + 4x]$

We are given that $ABC = O$. Therefore,

$[4 + 4x] = [0]$

Equating the elements:

$4 + 4x = 0$

$4x = -4$

$x = \frac{-4}{4}$

$x = -1$

Thus, the value of $x$ for which the given equation holds true is -1.

Given:

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

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

$O$ is the zero matrix of order 2, $O = \begin{bmatrix}0&0\\0&0 \end{bmatrix}$.

To Show:

$A^2 - 5A + 7I = O$.

Solution:

First, calculate $A^2$:

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

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

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

$A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

Next, calculate $5A$:

$5A = 5 \begin{bmatrix}3&1\\−1&2 \end{bmatrix} = \begin{bmatrix} 5(3) & 5(1) \\ 5(-1) & 5(2) \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$

Then, calculate $7I$:

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

Now, substitute these results into the expression $A^2 - 5A + 7I$:

$A^2 - 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$A^2 - 5A + 7I = \begin{bmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 - (-5) + 0 & 3 - 10 + 7 \end{bmatrix}$

$A^2 - 5A + 7I = \begin{bmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 + 5 + 0 & 3 - 10 + 7 \end{bmatrix}$

$A^2 - 5A + 7I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Since the result is the zero matrix $O$, we have shown that:

$A^2 - 5A + 7I = O$

Hence proved.

Given:

The matrix equation: $\begin{bmatrix}x&−5&−1 \end{bmatrix} \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix} \begin{bmatrix}x\\4\\1 \end{bmatrix} = O$.

Here, O represents the 1x1 null matrix, $O = [0]$.

To Find:

The value(s) of $x$.

Solution:

Let's perform the matrix multiplication step by step.

First, multiply the first two matrices:

$\begin{bmatrix}x&−5&−1 \end{bmatrix} \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

$= \begin{bmatrix} (x)(1) + (-5)(0) + (-1)(2) & (x)(0) + (-5)(2) + (-1)(0) & (x)(2) + (-5)(1) + (-1)(3) \end{bmatrix}$

$= \begin{bmatrix} x - 0 - 2 & 0 - 10 - 0 & 2x - 5 - 3 \end{bmatrix}$

$= \begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix}$

Now, multiply this resulting matrix by the third matrix:

$\begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix} \begin{bmatrix}x\\4\\1 \end{bmatrix}$

$= [(x - 2)(x) + (-10)(4) + (2x - 8)(1)]$

$= [x^2 - 2x - 40 + 2x - 8]$

$= [x^2 - 48]$

We are given that this product is equal to the null matrix O = [0]:

$[x^2 - 48] = [0]$

Equating the elements, we get the equation:

$x^2 - 48 = 0$

$x^2 = 48$

$x = \pm \sqrt{48}$

$x = \pm \sqrt{16 \times 3}$

$x = \pm 4\sqrt{3}$

Therefore, the values of x that satisfy the given equation are $x = 4\sqrt{3}$ and $x = -4\sqrt{3}$.

Given:

The annual sales for products x, y, z in two markets I and II can be represented by the sales matrix S:

$S = \begin{bmatrix}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}$

The first row represents Market I and the second row represents Market II. The columns represent the units sold for products x, y, and z respectively.

(a) Total Revenue in Each Market:

The unit sale prices for x, y, and z are given as $\textsf{₹} 2.50$, $\textsf{₹} 1.50$, and $\textsf{₹} 1.00$. We can represent these prices as a column matrix P:

$P = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$

The total revenue in each market is obtained by multiplying the sales matrix S by the price matrix P. Let R be the revenue matrix:

$R = S \times P = \begin{bmatrix}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$

Performing the matrix multiplication:

$R = \begin{bmatrix} (10000)(2.50) + (2000)(1.50) + (18000)(1.00) \\ (6000)(2.50) + (20000)(1.50) + (8000)(1.00) \end{bmatrix}$

$R = \begin{bmatrix} 25000 + 3000 + 18000 \\ 15000 + 30000 + 8000 \end{bmatrix}$

$R = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}$

The resulting matrix R shows the total revenue for each market:

Total revenue in Market I = $\textsf{₹} 46,000$.

Total revenue in Market II = $\textsf{₹} 53,000$.

(b) Gross Profit:

The unit costs for x, y, and z are given as $\textsf{₹} 2.00$, $\textsf{₹} 1.00$, and 50 paise ($\textsf{₹} 0.50$). We represent these costs as a column matrix C:

$C = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$

The total cost for the products sold in each market is obtained by multiplying the sales matrix S by the cost matrix C. Let TC be the total cost matrix:

$TC = S \times C = \begin{bmatrix}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$

Performing the matrix multiplication:

$TC = \begin{bmatrix} (10000)(2.00) + (2000)(1.00) + (18000)(0.50) \\ (6000)(2.00) + (20000)(1.00) + (8000)(0.50) \end{bmatrix}$

$TC = \begin{bmatrix} 20000 + 2000 + 9000 \\ 12000 + 20000 + 4000 \end{bmatrix}$

$TC = \begin{bmatrix} 31000 \\ 36000 \end{bmatrix}$

Gross profit is calculated as Total Revenue - Total Cost.

Let GP be the gross profit matrix:

$GP = R - TC = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix} - \begin{bmatrix} 31000 \\ 36000 \end{bmatrix}$

$GP = \begin{bmatrix} 46000 - 31000 \\ 53000 - 36000 \end{bmatrix}$

$GP = \begin{bmatrix} 15000 \\ 17000 \end{bmatrix}$

The gross profit from Market I is $\textsf{₹} 15,000$.

The gross profit from Market II is $\textsf{₹} 17,000$.

The total gross profit for the manufacturer is the sum of the profits from both markets:

Total Gross Profit = $\textsf{₹} 15,000 + \textsf{₹} 17,000 = \textsf{₹} 32,000$.

Given:

The matrix equation $X A = B$, where:

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

$B = \begin{bmatrix}−7&−8&−9\\2&4&6 \end{bmatrix}$

To Find:

The matrix X.

Solution:

Let the matrix A be of order $2 \times 3$ and matrix B be of order $2 \times 3$.

Let the matrix X be of order $m \times n$.

For the product XA to be defined, the number of columns of X must be equal to the number of rows of A. Thus, $n = 2$.

The order of the product matrix XA is $m \times 3$.

Since $XA = B$, the order of XA must be the same as the order of B, which is $2 \times 3$.

Therefore, $m \times 3 = 2 \times 3$, which implies $m = 2$.

So, the matrix X must be of order $2 \times 2$.

Let $X = \begin{bmatrix}a&b\\c&d \end{bmatrix}$.

Substitute X into the given equation:

$\begin{bmatrix}a&b\\c&d \end{bmatrix} \begin{bmatrix}1&2&3\\4&5&6 \end{bmatrix} = \begin{bmatrix}−7&−8&−9\\2&4&6 \end{bmatrix}$

Perform matrix multiplication on the left side:

$\begin{bmatrix} a(1)+b(4) & a(2)+b(5) & a(3)+b(6) \\ c(1)+d(4) & c(2)+d(5) & c(3)+d(6) \end{bmatrix} = \begin{bmatrix}−7&−8&−9\\2&4&6 \end{bmatrix}$

$\begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix} = \begin{bmatrix}−7&−8&−9\\2&4&6 \end{bmatrix}$

Equating the corresponding elements of the two matrices, we get the following systems of linear equations:

From the first row:

Multiply equation (i) by 2:

$2a+8b = -14$

Subtract equation (ii) from this:

$(2a+8b) - (2a+5b) = -14 - (-8)$

$3b = -6$

$b = -2$

Substitute $b = -2$ into equation (i):

$a + 4(-2) = -7$

$a - 8 = -7$

$a = 1$

Checking with equation (iii): $1 + 2(-2) = 1 - 4 = -3$. It holds true.

From the second row:

Multiply equation (iv) by 2:

$2c+8d = 4$

Subtract equation (v) from this:

$(2c+8d) - (2c+5d) = 4 - 4$

$3d = 0$

$d = 0$

Substitute $d = 0$ into equation (iv):

$c + 4(0) = 2$

$c = 2$

Checking with equation (vi): $2 + 2(0) = 2$. It holds true.

So, we have $a=1, b=-2, c=2, d=0$.

The matrix X is:

$X = \begin{bmatrix}a&b\\c&d \end{bmatrix} = \begin{bmatrix}1&-2\\2&0 \end{bmatrix}$.

Given:

A and B are square matrices of the same order.

$AB = BA$.

To Prove:

1. $AB^n = B^n A$ for all $n \in \mathbb{N}$.

2. $(AB)^n = A^n B^n$ for all $n \in \mathbb{N}$.

Proof of $AB^n = B^n A$ by Induction:

Let $P(n)$ be the statement: $AB^n = B^n A$.

Base Case (n=1):

We need to show that $P(1)$ is true, i.e., $AB^1 = B^1 A$.

LHS = $AB^1 = AB$.

RHS = $B^1 A = BA$.

Since $AB = BA$ is given, LHS = RHS. Thus, $P(1)$ is true.

Inductive Hypothesis:

Assume that $P(k)$ is true for some positive integer $k \in \mathbb{N}$.

That is, assume $AB^k = B^k A$.

Inductive Step:

We need to prove that $P(k+1)$ is true, i.e., $AB^{k+1} = B^{k+1} A$.

Consider the LHS:

$AB^{k+1} = A(B^k B)$

Using the inductive hypothesis ($AB^k = B^k A$):

Using the given condition $AB = BA$:

$ = B^{k+1} A$

This is the RHS for $P(k+1)$. Thus, $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

By the principle of mathematical induction, the statement $P(n)$ is true for all $n \in \mathbb{N}$.

Hence, $AB^n = B^n A$ for all $n \in \mathbb{N}$.

Proof of $(AB)^n = A^n B^n$ by Induction:

Let $Q(n)$ be the statement: $(AB)^n = A^n B^n$.

Base Case (n=1):

We need to show that $Q(1)$ is true, i.e., $(AB)^1 = A^1 B^1$.

LHS = $(AB)^1 = AB$.

RHS = $A^1 B^1 = AB$.

Since LHS = RHS, $Q(1)$ is true.

Inductive Hypothesis:

Assume that $Q(k)$ is true for some positive integer $k \in \mathbb{N}$.

That is, assume $(AB)^k = A^k B^k$.

Inductive Step:

We need to prove that $Q(k+1)$ is true, i.e., $(AB)^{k+1} = A^{k+1} B^{k+1}$.

Consider the LHS:

$(AB)^{k+1} = (AB)^k (AB)$

Using the inductive hypothesis ($(AB)^k = A^k B^k$):

From the first part of the proof, we know $B^k A = AB^k$ (since A and B commute):

$ = A^{k+1} B^{k+1}$

This is the RHS for $Q(k+1)$. Thus, $Q(k+1)$ is true whenever $Q(k)$ is true.

Conclusion:

By the principle of mathematical induction, the statement $Q(n)$ is true for all $n \in \mathbb{N}$.

Hence, $(AB)^n = A^n B^n$ for all $n \in \mathbb{N}$ when $AB=BA$.

Given:

The matrix $A = \begin{bmatrix}α&β\\γ&−α \end{bmatrix}$.

The condition $A^2 = I$, where $I$ is the identity matrix of order 2, $I = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$.

To Find:

The correct relationship between $\alpha$, $\beta$, and $\gamma$.

Solution:

First, calculate $A^2$:

$A^2 = A \cdot A = \begin{bmatrix}α&β\\γ&−α \end{bmatrix} \begin{bmatrix}α&β\\γ&−α \end{bmatrix}$

$A^2 = \begin{bmatrix} (α)(α) + (β)(γ) & (α)(β) + (β)(-α) \\ (γ)(α) + (-α)(γ) & (γ)(β) + (-α)(-α) \end{bmatrix}$

$A^2 = \begin{bmatrix} α^2 + βγ & αβ - αβ \\ γα - αγ & γβ + α^2 \end{bmatrix}$

$A^2 = \begin{bmatrix} α^2 + βγ & 0 \\ 0 & α^2 + βγ \end{bmatrix}$

Now, set $A^2 = I$:

$\begin{bmatrix} α^2 + βγ & 0 \\ 0 & α^2 + βγ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Equating the corresponding elements of the matrices:

$α^2 + βγ = 1$

Rearranging the equation to match the options:

$1 - (α^2 + βγ) = 0$

$1 - α^2 - βγ = 0$

Comparing this result with the given options:

(A) $1 + α² + βγ = 0$

(B) $1 \ – \ α² + βγ = 0$

(C) $1 \ – \ α² \ – \ βγ = 0$

(D) $1 + α² \ – \ βγ = 0$

The equation $1 - α^2 - βγ = 0$ matches option (C).

Final Answer: The final answer is $\boxed{(C)}$

Given:

A matrix A is both symmetric and skew-symmetric.

Definitions:

1. A matrix A is symmetric if its transpose is equal to itself, i.e., $A' = A$.

2. A matrix A is skew-symmetric if its transpose is equal to its negative, i.e., $A' = -A$.

Solution:

Since matrix A is both symmetric and skew-symmetric, it must satisfy both conditions simultaneously:

From equations (i) and (ii), we can equate the expressions for A':

$A = -A$

Add A to both sides:

$A + A = -A + A$

$2A = O$

Where O is the zero matrix of the same order as A.

Dividing by 2 (or multiplying by the scalar 1/2):

$A = \frac{1}{2} O$

$A = O$

This means that every element of matrix A must be zero. Therefore, A is the zero matrix.

Let's check the options:

(A) A is a diagonal matrix: While the zero matrix is technically a diagonal matrix (all off-diagonal elements are zero), this option is not specific enough. Other diagonal matrices (like the identity matrix) are symmetric but not skew-symmetric.

(B) A is a zero matrix: This is the direct conclusion from our derivation.

(C) A is a square matrix: Symmetric and skew-symmetric matrices are inherently square matrices. While true, it's not the most defining characteristic based on the given conditions.

(D) None of these: Option (B) is correct.

Final Answer: The final answer is $\boxed{(B)}$

Given:

A is a square matrix such that $A^2 = A$.

$I$ is the identity matrix of the same order as A.

To Find:

The value of the expression $(I + A)^3 – 7 A$.

Solution:

We need to expand $(I + A)^3$. Since $I$ commutes with $A$ (i.e., $IA = AI = A$), we can use algebraic identities similar to binomial expansion.

First, let's find $(I+A)^2$:

$(I + A)^2 = (I + A)(I + A)$

$ = I(I + A) + A(I + A)$

$ = I^2 + IA + AI + A^2$

Since $I^2 = I$, $IA = A$, $AI = A$, and $A^2 = A$ (given):

$ = I + A + A + A$

$ = I + 3A$

Now, let's find $(I+A)^3$:

$(I + A)^3 = (I + A)^2 (I + A)$

$ = (I + 3A)(I + A)$

$ = I(I + A) + 3A(I + A)$

$ = I^2 + IA + 3AI + 3A^2$

Substitute the properties $I^2 = I$, $IA = A$, $AI = A$, and $A^2 = A$:

$ = I + A + 3A + 3A$

$ = I + 7A$

Now substitute this result back into the expression we need to evaluate:

$(I + A)^3 – 7 A = (I + 7A) - 7A$

$ = I + 7A - 7A$

$ = I$

The result is the identity matrix $I$.

Comparing this with the given options:

(A) A

(B) I – A

(C) I

(D) 3A

Final Answer: The final answer is $\boxed{(C)}$