Chapter 2 Matrices (Q & A)

The order of a matrix is given by the number of its rows multiplied by the number of its columns, expressed in the form (rows $\times$ columns).

The given matrix is $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$.

Let's count the number of rows (horizontal arrangements):

Row 1: [1 2 3]

Row 2: [4 5 6]

So, there are 2 rows.

Now, let's count the number of columns (vertical arrangements):

Column 1: $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$

Column 2: $\begin{bmatrix} 2 \\ 5 \end{bmatrix}$

Column 3: $\begin{bmatrix} 3 \\ 6 \end{bmatrix}$

So, there are 3 columns.

Thus, the order of the matrix A is $2 \times 3$.

Therefore, the correct option is (A).

A square matrix is defined as a matrix that has an equal number of rows and columns.

The order of a general matrix $A = [a_{ij}]$ is given by $m \times n$, where:

• $m$ is the number of rows.
• $n$ is the number of columns.

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

Mathematically, this condition is expressed as:

$m = n$

Let's analyze the options:

• (A) $m = n$: This is the correct definition for a square matrix.
• (B) $m < n$: This describes a rectangular matrix (specifically, a horizontal matrix) where the number of rows is less than the number of columns.
• (C) $m > n$: This describes a rectangular matrix (specifically, a vertical matrix) where the number of rows is greater than the number of columns.
• (D) $a_{ij} = 0$ for all $i, j$: This describes a zero or null matrix. While a zero matrix can be square (e.g., a $2 \times 2$ zero matrix), this condition defines the elements, not the shape (order) of the matrix required for it to be square.

Therefore, the condition for a matrix to be a square matrix is that the number of its rows equals the number of its columns.

The correct option is (A).

A scalar matrix is a special type of diagonal matrix. For a matrix to be a scalar matrix, it must satisfy the following three conditions:

1. It must be a square matrix (number of rows = number of columns).
2. All of its non-diagonal elements must be zero.
3. All of its diagonal elements must be equal.

Let's examine each option based on these conditions:

(A) $\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$

This is a square matrix and its non-diagonal elements are zero. However, its diagonal elements (2 and 3) are not equal. So, it is a diagonal matrix, but not a scalar matrix.

(B) $\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

This is a square matrix. Its non-diagonal elements are zero. Its diagonal elements are both equal to 5. This matrix satisfies all the conditions. Therefore, it is a scalar matrix.

(C) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

This is a square matrix, its non-diagonal elements are zero, and its diagonal elements are equal (both are 1). This is a scalar matrix. It is also known by the more specific name, an identity matrix. Since every identity matrix is a scalar matrix, this is a correct example.

(D) $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

This is a square matrix, its non-diagonal elements are zero, and its diagonal elements are equal (both are 0). This is a scalar matrix. It is also known as a zero matrix or null matrix. A square zero matrix is a type of scalar matrix.

Among the given options, (B), (C), and (D) are all technically scalar matrices. However, option (B) is the most general example of a scalar matrix that is not a more specific type like an identity or zero matrix. In such questions, the most representative example is usually the intended answer.

Therefore, the best choice representing a scalar matrix is (B).

For two matrices, say $A = [a_{ij}]$ and $B = [b_{ij}]$, to be considered equal, two conditions must be satisfied simultaneously:

1. Same Order: The matrices must have the same dimensions, meaning they must have the same number of rows and the same number of columns. If $A$ is an $m \times n$ matrix, then $B$ must also be an $m \times n$ matrix.
2. Corresponding Elements are Equal: Each element in matrix $A$ must be equal to the corresponding element in matrix $B$. That is, $a_{ij} = b_{ij}$ for all values of $i$ and $j$.

Let's analyze the given options:

• (A) They must have the same order. This is a necessary condition, but it is not sufficient on its own. For example, $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$ have the same order ($2 \times 2$) but are not equal.
• (B) Their corresponding elements must be equal. This is also a necessary condition, but it is not sufficient on its own because it implicitly assumes the matrices have the same order so that "corresponding elements" can be compared.
• (C) Both (A) and (B). This option states that both conditions—having the same order and having equal corresponding elements—must be met. This is the complete and correct definition of matrix equality.
• (D) Their determinants must be equal. This is incorrect. First, determinants are only defined for square matrices. Second, even for square matrices, having equal determinants does not guarantee that the matrices are equal. For instance, $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ both have a determinant of 1, but they are not equal matrices.

Since both conditions are required for two matrices to be equal, the correct option is (C).

According to the principle of matrix equality, two matrices are equal if and only if they have the same order and their corresponding elements are equal.

The given equation is:

$\begin{bmatrix} x+y & 2 \\ 5 & x-y \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 4 \end{bmatrix}$

By equating the corresponding elements, we get a system of linear equations:

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

$\begin{array}{rcl} & x + y & = & 6 \\ + & x - y & = & 4 \\ \hline & 2x \phantom{+ y} & = & 10 \\ \hline \end{array}$

$2x = 10$

$x = \frac{10}{2}$

$x = 5$

Now, substitute the value of $x$ into equation (i):

$5 + y = 6$

$y = 6 - 5$

$y = 1$

So, the values are $x=5$ and $y=1$.

This corresponds to option (A). Therefore, the correct option is (A).

The transpose of a matrix is an operation that flips the matrix over its main diagonal. This is achieved by interchanging the rows and columns of the matrix.

A row matrix is a matrix that has only a single row. The order of a row matrix is $1 \times n$, where $n$ is the number of columns.

Let's consider a general row matrix $A$:

$A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \end{bmatrix}$

This matrix has 1 row and $n$ columns, so its order is $1 \times n$.

When we find the transpose of $A$, which we denote as $A^T$, we convert the single row into a single column. The new matrix will have $n$ rows and 1 column.

$A^T = \begin{bmatrix} a_{11} \\ a_{12} \\ \vdots \\ a_{1n} \end{bmatrix}$

The order of $A^T$ is $n \times 1$.

A matrix with only one column is defined as a column matrix.

For example, if we take the row matrix $B = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$, its transpose is:

$B^T = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

The resulting matrix $B^T$ is a column matrix.

Therefore, the transpose of a row matrix is always a column matrix. The correct option is (A).

This question deals with a fundamental property of the transpose of a matrix.

The operation $A'$ (or $A^T$) denotes the transpose of matrix $A$. The transpose is formed by interchanging the rows and columns of the original matrix. If an element in $A$ is at position $(i, j)$ (row $i$, column $j$), its corresponding element in $A'$ will be at position $(j, i)$ (row $j$, column $i$).

The expression $(A')'$ means we are taking the transpose of the matrix $A'$, which is already the transpose of $A$.

Let's consider a general matrix $A$ of order $m \times n$:

$A = [a_{ij}]_{m \times n}$

When we take the first transpose, we get $A'$:

$A' = [a_{ji}]_{n \times m}$

Now, when we take the transpose of $A'$, we interchange its rows and columns again. The element at position $(j, i)$ in $A'$ moves back to position $(i, j)$. The resulting matrix will have the order $m \times n$.

$(A')' = [a_{ij}]_{m \times n}$

This resulting matrix is identical to the original matrix $A$. Therefore, the property is:

$(A')' = A$

Example:

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

First, find the transpose $A'$:

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

Now, find the transpose of $A'$:

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

As we can see, $(A')'$ is the same as the original matrix $A$.

Therefore, the correct option is (A).

A symmetric matrix is a special type of square matrix that is defined by its relationship with its own transpose.

By definition, a square matrix $A$ is said to be symmetric if it is equal to its transpose. The transpose of matrix $A$ is denoted by $A'$ or $A^T$.

So, the condition for a matrix $A$ to be symmetric is:

$A = A'$

This means that for every element $a_{ij}$ (in the $i$-th row and $j$-th column), it must be equal to the element $a_{ji}$ (in the $j$-th row and $i$-th column). For example:

If $A = \begin{bmatrix} 1 & 5 & 7 \\ 5 & 2 & 9 \\ 7 & 9 & 3 \end{bmatrix}$, then its transpose is $A' = \begin{bmatrix} 1 & 5 & 7 \\ 5 & 2 & 9 \\ 7 & 9 & 3 \end{bmatrix}$. Since $A=A'$, this matrix is symmetric.

Let's examine the other options:

• (B) $A = -A'$: This is the definition of a skew-symmetric matrix.
• (C) $A^2 = I$: This is the definition of an involutory matrix.
• (D) $A^2 = A$: This is the definition of an idempotent matrix.

Therefore, the correct condition for a square matrix $A$ to be symmetric is that it equals its transpose.

The correct option is (A).

A skew-symmetric matrix is a square matrix $A$ that satisfies the condition that its transpose is equal to its negative. The transpose of $A$ is denoted by $A'$.

The defining property of a skew-symmetric matrix $A = [a_{ij}]$ is:

$A' = -A$

In terms of the elements of the matrix, this means that for all row indices $i$ and column indices $j$:

$a_{ji} = -a_{ij}$

The question asks about the diagonal elements. For any diagonal element, the row index is equal to the column index. So, for a diagonal element, we have $j = i$.

Let's apply this condition to our definition:

Substitute $j=i$ into the equation $a_{ji} = -a_{ij}$:

$a_{ii} = -a_{ii}$

To solve for $a_{ii}$, we can add $a_{ii}$ to both sides of the equation:

$a_{ii} + a_{ii} = 0$

$2a_{ii} = 0$

$a_{ii} = 0$

This result shows that every diagonal element of a skew-symmetric matrix must be zero.

Example:

The matrix $A = \begin{bmatrix} 0 & 5 & -2 \\ -5 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix}$ is a skew-symmetric matrix. As we can see, all of its diagonal elements are 0.

Therefore, the diagonal elements of a skew-symmetric matrix are always 0. The correct option is (C).

The problem asks to calculate the sum of a matrix $A$ and its transpose $A'$.

The given matrix is:

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

First, we need to find the transpose of matrix $A$, which is denoted by $A'$. The transpose is obtained by interchanging the rows and columns of $A$.

The first row of $A$, [1 2], becomes the first column of $A'$.

The second row of $A$, [3 4], becomes the second column of $A'$.

So, the transpose of $A$ is:

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

Next, we perform the matrix addition $A + A'$. We add the corresponding elements of the two matrices.

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

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

$A + A' = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix}$

Note that for any square matrix $A$, the matrix $A+A'$ is always a symmetric matrix, which is true for our result.

Comparing our result with the given options, we find that it matches option (A).

Therefore, the correct option is (A).

To determine if the matrix is symmetric, skew-symmetric, both, or neither, we need to compare it with its transpose.

The given matrix is:

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

1. Check for Symmetry

A matrix is symmetric if it is equal to its transpose, i.e., $A = A'$.

First, let's find the transpose of $A$, denoted as $A'$. We interchange the rows and columns.

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

Now, we compare $A$ and $A'$:

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

Since $A = A'$, the matrix is symmetric.

2. Check for Skew-Symmetry

A matrix is skew-symmetric if it is equal to the negative of its transpose, i.e., $A = -A'$. A necessary condition for a matrix to be skew-symmetric is that all its diagonal elements must be zero.

The diagonal elements of matrix $A$ are 1 and -1, which are not zero. Therefore, the matrix cannot be skew-symmetric.

Alternatively, let's calculate $-A'$:

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

Comparing $A$ with $-A'$:

$\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \neq \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$

Since $A \neq -A'$, the matrix is not skew-symmetric.

Conclusion

The matrix $A$ is symmetric but not skew-symmetric.

Therefore, the correct option is (A).

This question asks about a fundamental property of the transpose operation when applied to the sum of two matrices.

The property is known as the additivity of the transpose operation. It states that the transpose of the sum of two matrices is equal to the sum of their individual transposes. For this operation to be defined, the matrices $A$ and $B$ must have the same order.

The formula for this property is:

$(A+B)' = A' + B'$

Let's verify this with an example.

Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$.

Step 1: Calculate the left-hand side (LHS), $(A+B)'$.

First, find the sum $A+B$:

$A + B = \begin{bmatrix} 1+5 & 2+6 \\ 3+7 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}$

Now, find the transpose of this sum:

$(A+B)' = \begin{bmatrix} 6 & 10 \\ 8 & 12 \end{bmatrix}$

Step 2: Calculate the right-hand side (RHS), $A' + B'$.

First, find the individual transposes $A'$ and $B'$:

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

$B' = \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix}$

Now, add these transposes:

$A' + B' = \begin{bmatrix} 1+5 & 3+7 \\ 2+6 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 10 \\ 8 & 12 \end{bmatrix}$

Step 3: Compare LHS and RHS.

We see that LHS = RHS, as both are equal to $\begin{bmatrix} 6 & 10 \\ 8 & 12 \end{bmatrix}$.

This confirms the property $(A+B)' = A' + B'$.

Therefore, the correct option is (A).

The problem involves two main matrix operations: scalar multiplication and matrix subtraction.

The given matrices are:

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$

We need to calculate $2A - B$.

Step 1: Scalar Multiplication

First, we multiply the matrix $A$ by the scalar 2. This means every element inside matrix $A$ is multiplied by 2.

$2A = 2 \times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times 3 & 2 \times 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}$

Step 2: Matrix Subtraction

Now, we subtract matrix $B$ from the resulting matrix $2A$. This is done by subtracting the corresponding elements.

$2A - B = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix} - \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$

$2A - B = \begin{bmatrix} 2 - 5 & 4 - 6 \\ 6 - 7 & 8 - 8 \end{bmatrix}$

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

The resulting matrix is $\begin{bmatrix} -3 & -2 \\ -1 & 0 \end{bmatrix}$, which corresponds to option (A).

Therefore, the correct option is (A).

The multiplication of two matrices, say $A$ and $B$, to form the product $AB$, is not always possible. There is a specific condition on the dimensions (or orders) of the matrices that must be met.

Let's define the orders of the two matrices:

• Let matrix $A$ have an order of $m \times n$ (meaning it has $m$ rows and $n$ columns).
• Let matrix $B$ have an order of $p \times q$ (meaning it has $p$ rows and $q$ columns).

The matrix product $AB$ is defined if and only if the number of columns in the first matrix ($A$) is exactly equal to the number of rows in the second matrix ($B$).

In terms of our defined orders, this means the condition is:

$n = p$

If this condition is met, the resulting matrix, $C = AB$, will have an order of $m \times q$.

Let's analyze the given options:

• (A) Number of columns in A equals number of rows in B. This is the correct condition ($n = p$).
• (B) Number of rows in A equals number of columns in B. This condition ($m = q$) is not required for the multiplication to be possible.
• (C) Number of rows in A equals number of rows in B. This condition ($m = p$) is not required.
• (D) A and B are square matrices of the same order. While multiplication is possible in this case (if $A$ is $n \times n$ and $B$ is $n \times n$, the condition is met since columns of A = n and rows of B = n), this is a specific case and not the general rule. The rule in (A) is more general and applies to rectangular matrices as well.

Therefore, the fundamental condition for matrix multiplication $AB$ to be possible is that the number of columns in A must equal the number of rows in B.

The correct option is (A).

The problem asks to calculate $A^2$. For a matrix $A$, the expression $A^2$ represents the matrix multiplication of $A$ with itself.

$A^2 = A \times A$

The given matrix is:

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

Now, we perform the multiplication:

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

The multiplication is done by taking the dot product of rows from the first matrix with columns from the second matrix.

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

Calculating the values for each element:

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

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

The resulting matrix is the $2 \times 2$ identity matrix, denoted by $I$. A matrix for which $A^2 = I$ is called an involutory matrix.

Comparing our result with the given options:

(A) $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

(B) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

(C) $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

(D) $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$

The calculated matrix matches option (B).

Therefore, the correct option is (B).

This question asks for a fundamental property related to the transpose of a product of two matrices. This property is often called the reversal law of transposes.

Let $A$ and $B$ be two matrices such that their product $AB$ is defined. The transpose of their product, denoted by $(AB)'$ or $(AB)^T$, is equal to the product of their individual transposes, but in the reverse order.

The mathematical formula for this property is:

$(AB)' = B'A'$

This is analogous to putting on socks and then shoes; to reverse the process, you must take off the shoes first and then the socks.

Let's analyze the given options:

• (A) $A'B'$: This is incorrect. The order of multiplication is not preserved. The product $A'B'$ may not even be defined, even if $AB$ is.
• (B) $B'A'$: This is the correct statement of the reversal law for transposes.
• (C) $AB$: The transpose of a product is generally not equal to the product itself.
• (D) $BA$: The product $BA$ is generally not equal to $(AB)'$. In fact, matrix multiplication is not commutative, so $AB \neq BA$ in most cases.

Therefore, the correct property for the transpose of a product of two matrices is that the order of the matrices is reversed.

The correct option is (B).

The determinant of a $2 \times 2$ matrix is calculated using a specific formula.

For a general $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, its determinant, denoted as $|A|$ or $\det(A)$, is given by:

$|A| = ad - bc$

This is the product of the main diagonal elements ($a$ and $d$) minus the product of the off-diagonal elements ($b$ and $c$).

The given matrix is:

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

Here, we have:

• $a = 2$
• $b = 3$
• $c = 4$
• $d = 5$

Applying the formula for the determinant:

Determinant = $(2 \times 5) - (3 \times 4)$

Determinant = $10 - 12$

Determinant = $-2$

The calculated determinant is -2, which corresponds to option (D).

Therefore, the correct option is (D).

This question asks about a fundamental property of matrices related to symmetry. We want to determine the nature of the matrix formed by adding a square matrix $A$ to its transpose $A'$.

Let's define a new matrix $B$ such that:

$B = A + A'$

To determine if $B$ is symmetric or skew-symmetric, we need to find its transpose, $B'$, and compare it to $B$.

• If $B' = B$, the matrix is symmetric.
• If $B' = -B$, the matrix is skew-symmetric.

Let's find the transpose of $B$:

$B' = (A + A')'$

Using the property of transposes that $(X+Y)' = X' + Y'$, we get:

$B' = A' + (A')'$

Now, using the property that the transpose of a transpose is the original matrix, i.e., $(X')' = X$, we get:

$B' = A' + A$

Since matrix addition is commutative ($X+Y = Y+X$), we can write:

$B' = A + A'$

By our initial definition, $B = A + A'$. Therefore, we have shown that:

$B' = B$

This is the definition of a symmetric matrix. Thus, for any square matrix $A$, the matrix $A+A'$ is always symmetric.

Therefore, the correct option is (B).

We need to find the value of the determinant of the given $3 \times 3$ matrix.

The matrix is: $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$

The determinant of a $3 \times 3$ matrix $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$ can be calculated by expanding along the first row as:

Determinant = $a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}$

Applying this formula to our matrix, we get:

Value = $1 \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}$

Now, we calculate the determinants of the $2 \times 2$ matrices:

Value = $1((5 \times 9) - (6 \times 8)) - 2((4 \times 9) - (6 \times 7)) + 3((4 \times 8) - (5 \times 7))$

Value = $1(45 - 48) - 2(36 - 42) + 3(32 - 35)$

Value = $1(-3) - 2(-6) + 3(-3)$

Value = $-3 - (-12) - 9$

Value = $-3 + 12 - 9$

Value = $9 - 9 = 0$

Alternate Method (Using Properties of Determinants):

Let the determinant be $\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$

We apply the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$. These operations do not change the value of the determinant.

$\Delta = \begin{vmatrix} 1 & 2-1 & 3-1 \\ 4 & 5-4 & 6-4 \\ 7 & 8-7 & 9-7 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 2 \\ 4 & 1 & 2 \\ 7 & 1 & 2 \end{vmatrix}$

Now, we can take out the common factor 2 from the third column ($C_3$):

$\Delta = 2 \begin{vmatrix} 1 & 1 & 1 \\ 4 & 1 & 1 \\ 7 & 1 & 1 \end{vmatrix}$

According to the properties of determinants, if two rows or two columns of a matrix are identical, the value of its determinant is zero. In this matrix, the second and third columns are identical.

Therefore, $\Delta = 2 \times 0 = 0$.

Both methods yield the same result.

The value of the determinant is 0, which corresponds to option (A).

Therefore, the correct option is (A).

This question asks for a standard property relating the determinant of a matrix to the determinant of its adjugate.

Let $A$ be a square matrix of order $n$. The fundamental relationship between a matrix, its adjugate (or adjoint), and its determinant is given by the formula:

To find the determinant of the adjugate, $|\text{adj}(A)|$, we can take the determinant of both sides of this equation:

$|A(\text{adj}(A))| = ||A|I_n|$

Now, we simplify both sides using properties of determinants:

Left-Hand Side (LHS):

Using the property that the determinant of a product of matrices is the product of their determinants, i.e., $|XY| = |X||Y|$:

$|A(\text{adj}(A))| = |A| \cdot |\text{adj}(A)|$

Right-Hand Side (RHS):

The term $|A|$ is a scalar. Using the property that for a scalar $k$ and a matrix $B$ of order $n$, $|kB| = k^n|B|$:

$||A|I_n| = |A|^n |I_n|$

Since the determinant of the identity matrix $|I_n|$ is always 1, this simplifies to:

$|A|^n \cdot 1 = |A|^n$

Equating LHS and RHS:

Now we equate the simplified expressions for the LHS and RHS:

$|A| \cdot |\text{adj}(A)| = |A|^n$

If $A$ is a non-singular matrix ($|A| \neq 0$), we can divide both sides by $|A|$:

$|\text{adj}(A)| = \frac{|A|^n}{|A|}$

$|\text{adj}(A)| = |A|^{n-1}$

This formula also holds true if $A$ is a singular matrix ($|A|=0$), because in that case, both sides of the equation become 0.

Therefore, for a square matrix $A$ of order $n$, the determinant of its adjugate is $|A|^{n-1}$.

The correct option is (C).

A square matrix $A$ is said to be invertible (or non-singular) if its inverse, denoted as $A^{-1}$, exists. The existence of the inverse is directly linked to the value of the matrix's determinant.

The formula to calculate the inverse of a matrix $A$ is:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

where:

• $|A|$ is the determinant of matrix $A$.
• $\text{adj}(A)$ is the adjugate (or adjoint) of matrix $A$.

For the inverse $A^{-1}$ to be defined, the expression $\frac{1}{|A|}$ must be a well-defined number. This is only possible if the denominator, $|A|$, is not equal to zero.

• If $|A| \neq 0$, the inverse exists, and the matrix is invertible.
• If $|A| = 0$, we would have to divide by zero, which is an undefined operation. Therefore, the inverse does not exist, and the matrix is called non-invertible or singular.

Let's analyze the options:

• (A) $|A| = 0$: This is the condition for a matrix to be singular or non-invertible.
• (B) $|A| \neq 0$: This is the condition for a matrix to be non-singular or invertible. This is the correct condition.
• (C) $A$ is symmetric: A symmetric matrix can be either invertible or non-invertible. For example, the identity matrix is symmetric and invertible, but the matrix $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ is symmetric but not invertible (its determinant is 0).
• (D) $A$ is skew-symmetric: A skew-symmetric matrix can also be invertible or not. For example, $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ is skew-symmetric and invertible (determinant is 1), but $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ is skew-symmetric and not invertible.

Therefore, a matrix $A$ is invertible if and only if its determinant is non-zero.

The correct option is (B).

To find the inverse of a matrix $A$, denoted as $A^{-1}$, we use the formula:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

where $|A|$ is the determinant of $A$ and $\text{adj}(A)$ is the adjugate of $A$.

The given matrix is:

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

Step 1: Calculate the determinant of A, $|A|$.

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is $ad-bc$.

$|A| = (2)(5) - (3)(4)$

$|A| = 10 - 12 = -2$

Since the determinant is non-zero, the inverse exists.

Step 2: Find the adjugate of A, adj(A).

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjugate is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

So, we swap the main diagonal elements (2 and 5) and negate the off-diagonal elements (3 and 4).

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

Step 3: Calculate the inverse, $A^{-1}$.

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

Now, we multiply each element of the adjugate matrix by $-\frac{1}{2}$:

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

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

Comparing this result with the given options, it matches option (A).

Therefore, the correct option is (A).

A system of linear equations represented in the matrix form is given by:

$AX = B$

where $A$ is the coefficient matrix, $X$ is the column matrix of variables, and $B$ is the column matrix of constants.

To solve for $X$, we can pre-multiply both sides by the inverse of $A$, provided the inverse exists:

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

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

$IX = A^{-1}B$

$X = A^{-1}B$

This solution for $X$ is unique. However, this entire process is contingent on the existence of the inverse matrix, $A^{-1}$.

A square matrix $A$ is invertible (meaning $A^{-1}$ exists) if and only if its determinant is non-zero.

Condition for invertibility: $|A| \neq 0$.

Therefore, for the system of linear equations $AX=B$ to have a unique solution, the determinant of the coefficient matrix $A$ must not be equal to zero.

Let's analyze the other cases for completeness:

• If $|A| = 0$, the matrix is singular, and a unique solution does not exist. In this situation, we check the value of $(\text{adj } A)B$:
  • If $(\text{adj } A)B \neq 0$, the system is inconsistent and has no solution.
  • If $(\text{adj } A)B = 0$, the system is consistent and has infinitely many solutions.

Based on this, the condition for a unique solution is $|A| \neq 0$.

The correct option is (B).

Explanation:

For a system of two linear equations in two variables:

Cramer's rule provides a method to solve for the variables $x$ and $y$ using determinants.

First, we define the determinant of the coefficient matrix, denoted as $D$ (or $\Delta$):

$D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}$

Next, we find the determinant for $x$, denoted as $D_x$, by replacing the first column (the coefficients of $x$) with the constant terms from the right side of the equations:

$D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}$

Similarly, we find the determinant for $y$, denoted as $D_y$, by replacing the second column (the coefficients of $y$) with the constant terms:

$D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}$

According to Cramer's rule, the solution for $x$ is the ratio of $D_x$ to $D$, provided that $D \neq 0$.

$x = \frac{D_x}{D} = \frac{\begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}}$

And the solution for $y$ is:

$y = \frac{D_y}{D} = \frac{\begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}}$

By comparing the derived formula for $x$ with the given options, we can see that option (A) matches our result.

The correct answer is (A)

Explanation:

The rule for matrix multiplication states that if a matrix $A$ has an order of $m \times n$ and a matrix $B$ has an order of $p \times q$, their product $AB$ is only defined if the number of columns in matrix $A$ ($n$) is equal to the number of rows in matrix $B$ ($p$). The resulting matrix $AB$ will have an order of $m \times q$.

Finding the order of AB:

Given:

Order of matrix A = $3 \times 4$

Order of matrix B = $4 \times 3$

Here, the number of columns of A (4) is equal to the number of rows of B (4). So, the product $AB$ is defined.

The order of the resulting matrix $AB$ will be (rows of A) $\times$ (columns of B), which is $3 \times 3$.

Finding the order of BA:

Given:

Order of matrix B = $4 \times 3$

Order of matrix A = $3 \times 4$

Here, the number of columns of B (3) is equal to the number of rows of A (3). So, the product $BA$ is defined.

The order of the resulting matrix $BA$ will be (rows of B) $\times$ (columns of A), which is $4 \times 4$.

Thus, the order of $AB$ is $3 \times 3$ and the order of $BA$ is $4 \times 4$, respectively.

The correct answer is (A)

Explanation:

Let's examine each of the options provided to determine which one is not a standard type of matrix in linear algebra.

(A) Column matrix: A column matrix (or column vector) is a matrix that has only one column. For example, $\begin{bmatrix} a \\ b \\ c \end{bmatrix}$ is a $3 \times 1$ column matrix. This is a valid and commonly used type of matrix.

(B) Diagonal matrix: A diagonal matrix is a square matrix in which all the entries outside the main diagonal are zero. For example, $\begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$ is a diagonal matrix. This is a valid type of matrix.

(C) Identity matrix: An identity matrix is a special type of diagonal matrix where all the elements on the main diagonal are 1. It is denoted by $I$. For example, the $3 \times 3$ identity matrix is $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. This is a fundamental type of matrix.

(D) Cubic matrix: The term "cubic matrix" is not a standard or recognized term in the study of matrices. Matrices are defined as two-dimensional rectangular arrays of numbers. The term "cubic" implies a three-dimensional structure, which would be described by a tensor of rank 3, not a matrix. Therefore, "cubic matrix" is not a type of matrix.

The correct answer is (D)

Explanation:

We are given the matrix A:

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

First, we need to find the transpose of matrix A, denoted as $A'$. The transpose of a matrix is obtained by interchanging its rows and columns.

The first row of A, $\begin{bmatrix} 1 & 2 \end{bmatrix}$, becomes the first column of $A'$.

The second row of A, $\begin{bmatrix} 0 & 3 \end{bmatrix}$, becomes the second column of $A'$.

So, the transpose of A is:

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

Now, we need to find the sum of A and A', which is $A + A'$. We add the corresponding elements of the two matrices.

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

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

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

Comparing this result with the given options, it matches option (A).

The correct answer is (A)

Explanation:

To determine the type of the matrix, let's denote the given matrix by $P$.

$P = \frac{1}{2}(A - A')$

Now, let's find the transpose of $P$, which is $P'$.

$P' = \left( \frac{1}{2}(A - A') \right)'$

Using the properties of the transpose of a matrix:

1. $(kM)' = kM'$ (where $k$ is a scalar)

2. $(M - N)' = M' - N'$

3. $(M')' = M$

Applying these properties to $P'$:

$P' = \frac{1}{2} (A - A')'$

$P' = \frac{1}{2} (A' - (A')')$

$P' = \frac{1}{2} (A' - A)$

Now, let's compare $P'$ with $P$. We can factor out a $-1$ from the expression for $P'$:

$P' = -\frac{1}{2} (A - A')$

We know that $P = \frac{1}{2}(A - A')$. Substituting this back into the equation for $P'$:

$P' = -P$

A matrix $P$ is called a skew-symmetric matrix if its transpose is equal to its negative, i.e., $P' = -P$.

Since we have shown that $\left( \frac{1}{2}(A - A') \right)' = - \left( \frac{1}{2}(A - A') \right)$, the matrix is always skew-symmetric.

This is a fundamental result in matrix algebra, where any square matrix $A$ can be expressed as the sum of a symmetric matrix ($\underbrace{\frac{1}{2}(A+A')}_{\text{Symmetric}}$) and a skew-symmetric matrix ($\underbrace{\frac{1}{2}(A-A')}_{\text{Skew-Symmetric}}$).

The correct answer is (B)

Explanation:

We are given the matrix A:

$A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$

First, we find the transpose of matrix A, which is denoted by $A'$. The transpose is obtained by interchanging the rows and columns of A.

$A' = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$

Now, we need to compute the product $AA'$.

$AA' = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$

We perform the matrix multiplication:

$AA' = \begin{bmatrix} (\cos\theta)(\cos\theta) + (-\sin\theta)(-\sin\theta) & (\cos\theta)(\sin\theta) + (-\sin\theta)(\cos\theta) \\ (\sin\theta)(\cos\theta) + (\cos\theta)(-\sin\theta) & (\sin\theta)(\sin\theta) + (\cos\theta)(\cos\theta) \end{bmatrix}$

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

Using the fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$, we can simplify the matrix:

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

The resulting matrix is the $2 \times 2$ identity matrix, often denoted as $I$.

Therefore, $AA'$ is the identity matrix.

The correct answer is (B)

Explanation:

By definition, a square matrix $A$ is called a singular matrix if its determinant is equal to zero.

That is, for a matrix $A$ to be singular, the condition is:

$|A| = 0$

This property is crucial because it indicates that the matrix does not have a multiplicative inverse. The formula for the inverse of a matrix $A$ is $A^{-1} = \frac{1}{|A|} \text{adj}(A)$. If $|A|=0$, the expression for the inverse involves division by zero, which is undefined.

Conversely, a matrix for which the determinant is non-zero ($|A| \neq 0$) is called a non-singular or invertible matrix.

Since the question explicitly states that $A$ is a singular matrix, its determinant $|A|$ must be 0 by definition.

The correct answer is (C)

Explanation:

We are given two matrices:

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

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

To find the product $A \times B$, we multiply the rows of matrix A by the columns of matrix B.

Let the resulting matrix be $C = A \times B = \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix}$.

Step 1: Calculate the element $c_{11}$ (Row 1 of A $\times$ Column 1 of B)

$c_{11} = (3)(1) + (1)(3) = 3 + 3 = 6$

Step 2: Calculate the element $c_{12}$ (Row 1 of A $\times$ Column 2 of B)

$c_{12} = (3)(2) + (1)(4) = 6 + 4 = 10$

Step 3: Calculate the element $c_{21}$ (Row 2 of A $\times$ Column 1 of B)

$c_{21} = (2)(1) + (5)(3) = 2 + 15 = 17$

Step 4: Calculate the element $c_{22}$ (Row 2 of A $\times$ Column 2 of B)

$c_{22} = (2)(2) + (5)(4) = 4 + 20 = 24$

Combining these elements, we get the product matrix:

$A \times B = \begin{bmatrix} 6 & 10 \\ 17 & 24 \end{bmatrix}$

This result matches option (A).

The correct answer is (A)

Explanation:

We are given that $A$ is an invertible square matrix of order $n$. The fundamental relationship between the inverse of a matrix, its adjugate (or adjoint), and its determinant is given by the formula:

Our goal is to find the value of $k$ in the given equation:

To do this, we can rearrange the standard formula for the inverse to express $\text{adj}(A)$ in terms of $A^{-1}$ and $|A|$. We can multiply both sides of the inverse formula by $|A|$:

$|A| \times A^{-1} = |A| \times \left( \frac{1}{|A|} \text{adj}(A) \right)$

$|A| A^{-1} = \frac{|A|}{|A|} \text{adj}(A)$

$|A| A^{-1} = \text{adj}(A)$

Rewriting this to match the format of the given equation, we have:

Now, by comparing equation (i) and equation (ii):

$k A^{-1} = |A| A^{-1}$

We can clearly see that the value of the scalar $k$ must be $|A|$.

The correct answer is (A)

Explanation:

A system of linear equations can be represented in the form of a matrix equation, which is typically written as $AX = B$.

The given system of equations is:

To convert this system into the matrix form $AX = B$, we identify the following components:

1. The coefficient matrix (A), which contains the coefficients of the variables $x$ and $y$. The coefficients of the first equation form the first row, and the coefficients of the second equation form the second row.

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

2. The variable matrix (X), which is a column matrix containing the variables.

$X = \begin{bmatrix} x \\ y \end{bmatrix}$

3. The constant matrix (B), which is a column matrix containing the constants from the right-hand side of the equations.

$B = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$

Now, we assemble these matrices into the equation $AX = B$:

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

If we were to multiply the matrices on the left side, we would get $\begin{bmatrix} 1(x) + (-2)(y) \\ 2(x) + 3(y) \end{bmatrix} = \begin{bmatrix} x-2y \\ 2x+3y \end{bmatrix}$, which when set equal to the right side gives back the original system of equations.

This resulting matrix equation matches option (A).

The correct answer is (A)

Explanation:

The key property to consider when dealing with matrix multiplication is that it is not commutative in general. This means that for two matrices $A$ and $B$, the product $AB$ is not always equal to the product $BA$. We must use the distributive property carefully.

Let's evaluate each option:

(A) $(A+B)^2 = A^2 + 2AB + B^2$

Let's expand the left side:

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

This expression is equal to $A^2 + 2AB + B^2$ only if $AB+BA = 2AB$, which implies $BA=AB$. Since this is not always true, this option is generally false.

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

Let's expand the left side:

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

This expression is equal to $A^2 - 2AB + B^2$ only if $-AB-BA = -2AB$, which implies $BA=AB$. Since this is not always true, this option is generally false.

(C) $(A+B)(A-B) = A^2 - B^2$

Let's expand the left side:

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

This expression is equal to $A^2 - B^2$ only if $-AB+BA=0$, which implies $BA=AB$. Since this is not always true, this option is generally false.

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

As shown in the expansion for option (C), the correct expansion of $(A+B)(A-B)$ using the distributive property is indeed $A^2 - AB + BA - B^2$. This identity holds true for any two square matrices of the same order, regardless of whether they commute or not.

The correct answer is (D)

Explanation:

We are asked to find the value of $A^2$, which is the product of the matrix $A$ with itself.

Given the matrix A:

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

Now, we compute $A^2 = A \times A$:

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

We perform the matrix multiplication by multiplying the rows of the first matrix with the columns of the second matrix.

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

Now, we simplify each element:

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

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

This result matches the matrix given in option (A).

The correct answer is (A)

Explanation:

The order of a matrix, denoted as $m \times n$, specifies its dimensions. Here:

• $m$ represents the number of rows in the matrix.
• $n$ represents the number of columns in the matrix.

A matrix is essentially a rectangular array of numbers. To find the total number of elements (or entries) in this array, we need to find the total number of positions within the grid formed by its rows and columns.

The total number of elements is calculated by multiplying the number of rows by the number of columns.

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

Total number of elements = $m \times n$

For example, a $2 \times 3$ matrix has 2 rows and 3 columns, and it contains $2 \times 3 = 6$ elements.

The correct answer is (C)

Explanation:

Let's review the definitions of the terms given in the options:

(A) Singular matrix: A square matrix is defined as singular if its determinant is exactly zero. This property implies that the matrix does not have an inverse.

(B) Non-singular matrix: A square matrix is called non-singular (or invertible) if its determinant is not equal to zero. These matrices have a unique inverse.

(C) Identity matrix: An identity matrix is a square matrix with 1s on the main diagonal and 0s everywhere else. The determinant of any identity matrix is always 1, which is not zero.

(D) Scalar matrix: A scalar matrix is a diagonal matrix where all the elements on the main diagonal are equal to the same scalar, say $k$. Its determinant is $k^n$ (where $n$ is the order of the matrix). Its determinant is zero only if $k=0$, in which case it's a zero matrix. While the zero matrix is singular, the general and precise term for any matrix with a determinant of zero is "singular matrix".

Based on the standard definition, a matrix whose determinant is zero is called a singular matrix.

The correct answer is (A)

Explanation:

A square matrix is defined as symmetric if it is equal to its transpose. For a matrix $A$, this condition is written as $A = A'$.

In terms of the elements of the matrix, if $A = [a_{ij}]$, it is symmetric if $a_{ij} = a_{ji}$ for all values of $i$ and $j$. This means the element in the $i$-th row and $j$-th column must be equal to the element in the $j$-th row and $i$-th column.

We are given the matrix:

$A = \begin{bmatrix} 1 & x \\ y & 1 \end{bmatrix}$

Let's identify the elements:

• $a_{11} = 1$
• $a_{12} = x$
• $a_{21} = y$
• $a_{22} = 1$

For the matrix $A$ to be symmetric, the condition $a_{12} = a_{21}$ must be satisfied. The diagonal elements ($a_{11}$ and $a_{22}$) don't impose any conditions on $x$ and $y$ as they are their own transpose counterparts.

Applying the condition:

$a_{12} = a_{21}$

$x = y$

Therefore, for the given matrix to be symmetric, the value of $x$ must be equal to the value of $y$.

The correct answer is (A)

Explanation:

We are given the condition for a matrix A as $A = -A'$, where $A'$ represents the transpose of matrix A.

Let's review the definitions of the matrix types provided in the options:

(A) Symmetric matrix: A matrix A is called symmetric if its transpose is equal to the matrix itself. The condition is $A' = A$.

(B) Skew-symmetric matrix: A matrix A is called skew-symmetric (or anti-symmetric) if its transpose is equal to the negative of the matrix. The condition is $A' = -A$.

(C) Diagonal matrix: A diagonal matrix is a square matrix where all the elements outside the main diagonal are zero. Its transpose is the matrix itself ($A' = A$), so it is a type of symmetric matrix.

(D) Row matrix: A row matrix is a matrix with only one row. This describes the shape of the matrix, not a property related to its transpose.

The given condition is $A = -A'$. If we multiply both sides of this equation by -1, we get:

$-1 \times A = -1 \times (-A')$

$-A = A'$

This can be written as $A' = -A$, which is the precise definition of a skew-symmetric matrix.

The correct answer is (B)

Explanation:

Let's match each matrix type in the first column with its corresponding property in the second column based on their standard definitions in linear algebra.

• (i) Identity matrix: An identity matrix is a square matrix in which all the elements on the main diagonal are 1 and all other elements are 0. This directly corresponds to property (d) Diagonal elements are 1, others are 0.

• (ii) Zero matrix: A zero matrix (or null matrix) is a matrix in which every element is zero. This matches property (b) All elements are zero.

• (iii) Symmetric matrix: A square matrix $A$ is said to be symmetric if it is equal to its transpose, $A'$. The condition is $A = A'$. This matches property (c) $A = A'$.

• (iv) Skew-symmetric matrix: A square matrix $A$ is said to be skew-symmetric if it is equal to the negative of its transpose. The condition is $A = -A'$. This matches property (a) $A = -A'$.

So, the correct matching is:

(i) → (d)

(ii) → (b)

(iii) → (c)

(iv) → (a)

This set of pairings corresponds to option (A).

The correct answer is (A)

Explanation:

This question deals with a fundamental property of matrix inverses known as the reversal law of inverses. This law states how to find the inverse of a product of two or more invertible matrices.

For any two invertible square matrices $A$ and $B$ of the same order, the inverse of their product $(AB)$ is the product of their inverses in the reverse order.

The formula is:

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

To verify this, we can show that multiplying $(AB)$ by $(B^{-1}A^{-1})$ results in the identity matrix $I$.

$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1}$      (by associativity of matrix multiplication)

$= A(I)A^{-1}$      (since $BB^{-1} = I$)

$= AA^{-1}$      (since $AI = A$)

$= I$      (since $AA^{-1} = I$)

Similarly, multiplying in the other order also yields the identity matrix:

$(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}(I)B = B^{-1}B = I$

Since multiplying $(AB)$ by $(B^{-1}A^{-1})$ gives the identity matrix, $(B^{-1}A^{-1})$ is the inverse of $(AB)$ by definition.

Comparing this result with the given options:

• (A) $A^{-1}B^{-1}$ is incorrect because matrix multiplication is not commutative in general ($AB \neq BA$).
• (B) $B^{-1}A^{-1}$ is the correct formula according to the reversal law.
• (C) $(BA)^{-1}$ is the inverse of the product $BA$, which is generally not equal to $(AB)^{-1}$.
• (D) $AB$ is the original product, not its inverse.

The correct answer is (B)

Explanation:

We are given the following system of linear equations:

The first step is to write down the coefficient matrix. This matrix is formed by the coefficients of the variables $x$ and $y$ from each equation.

The coefficients from the first equation ($2x + 3y = 7$) are 2 and 3.

The coefficients from the second equation ($1x - 1y = 1$) are 1 and -1.

The coefficient matrix, let's call it $A$, is:

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

Next, we need to find the determinant of this matrix, denoted as $|A|$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is calculated as $ad - bc$.

In our case, $a=2$, $b=3$, $c=1$, and $d=-1$.

So, the determinant is:

$|A| = (2)(-1) - (3)(1)$

$|A| = -2 - 3$

$|A| = -5$

The determinant of the coefficient matrix is -5, which corresponds to option (B).

The correct answer is (B)

Explanation:

Let's analyze the given matrix $A = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ against the definitions provided in the options.

(A) Diagonal matrix: A matrix is a diagonal matrix if it is a square matrix and all its non-diagonal elements are zero. In matrix $A$, the non-diagonal elements are $a_{12}=0$ and $a_{21}=0$. So, $A$ is a diagonal matrix.

(B) Scalar matrix: A matrix is a scalar matrix if it is a diagonal matrix and all its main diagonal elements are equal to the same non-zero scalar. In matrix $A$, it is a diagonal matrix, and its main diagonal elements are both equal to 2 ($a_{11}=2$ and $a_{22}=2$). So, $A$ is also a scalar matrix.

(C) Identity matrix: A matrix is an identity matrix if it is a scalar matrix where the diagonal elements are equal to 1. Since the diagonal elements of $A$ are 2, it is not an identity matrix.

Since matrix $A$ satisfies the conditions for both a diagonal matrix and a scalar matrix, the most accurate description is that it is both.

The correct answer is (D)

Explanation:

We need to use a key property of determinants. For any square matrix $A$ of order $n$ and any scalar $k$, the determinant of the matrix $kA$ is given by the formula:

$|kA| = k^n |A|$

In this problem, we are given:

• The matrix $A$ is of order $3 \times 3$, so $n=3$.
• The determinant of $A$ is $|A| = 5$.
• We are asked to find the determinant of $3A$, so the scalar is $k=3$.

Now, we substitute these values into the formula:

$|3A| = 3^n \times |A|$

$|3A| = 3^3 \times 5$

First, calculate $3^3$:

$3^3 = 3 \times 3 \times 3 = 27$

Now, multiply this by $|A|$:

$|3A| = 27 \times 5$

$|3A| = 135$

Therefore, the value of $|3A|$ is 135, which corresponds to option (D).

The correct answer is (D)

Explanation:

A square matrix $A$ is defined as symmetric if it is equal to its transpose, $A'$. This means that for every element $a_{ij}$ in the matrix, it must be equal to the element $a_{ji}$. In simpler terms, the elements across the main diagonal must be mirror images of each other.

Let's check each of the given matrices:

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

Here, the element at row 1, column 2 is $a_{12} = 2$. The element at row 2, column 1 is $a_{21} = 2$. Since $a_{12} = a_{21}$, this matrix is symmetric.

(B) $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$

Here, the element at row 1, column 2 is $a_{12} = 1$. The element at row 2, column 1 is $a_{21} = -1$. Since $a_{12} \neq a_{21}$, this matrix is NOT symmetric. (It is, in fact, a skew-symmetric matrix because $a_{12} = -a_{21}$).

(C) $\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

Here, the element at row 1, column 2 is $a_{12} = 0$. The element at row 2, column 1 is $a_{21} = 0$. Since $a_{12} = a_{21}$, this matrix is symmetric.

(D) $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix}$

We check the corresponding elements: $a_{12}=2, a_{21}=2$; $a_{13}=3, a_{31}=3$; $a_{23}=5, a_{32}=5$. Since all corresponding off-diagonal elements are equal, this matrix is symmetric.

Therefore, the matrix in option (B) is the only one that is not symmetric.

The correct answer is (B)

Explanation:

We are given the matrix $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ and the identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. We need to evaluate the expression $A^2 - 4A + I$.

Step 1: Calculate $A^2$

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

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

$A^2 = \begin{bmatrix} 4+3 & 6+6 \\ 2+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}$

Step 2: Calculate $4A$

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

Step 3: Evaluate the expression $A^2 - 4A + I$

Now, substitute the calculated matrices back into the expression:

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

Perform the subtraction and addition on the corresponding elements:

$= \begin{bmatrix} 7-8+1 & 12-12+0 \\ 4-4+0 & 7-8+1 \end{bmatrix}$

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

The resulting matrix is the zero matrix.

The correct answer is (A)

Explanation:

A matrix is defined as singular if its determinant is equal to zero. We are given the matrix:

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

To find the value of $k$ for which this matrix is singular, we must first calculate its determinant and then set it equal to zero.

The determinant of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by the formula $ad-bc$.

For our matrix, $a = k-1$, $b = 2$, $c = 3$, and $d = k-2$.

So, the determinant $|A|$ is:

$|A| = (k-1)(k-2) - (2)(3)$

Now, we expand the expression:

$|A| = (k^2 - 2k - k + 2) - 6$

$|A| = k^2 - 3k + 2 - 6$

$|A| = k^2 - 3k - 4$

For the matrix to be singular, we set the determinant to zero:

$k^2 - 3k - 4 = 0$

This is a quadratic equation which we can solve by factoring. We look for two numbers that multiply to -4 and add to -3. These numbers are -4 and 1.

$(k-4)(k+1) = 0$

This gives two possible solutions for $k$:

$k - 4 = 0 \implies k = 4$

$k + 1 = 0 \implies k = -1$

Therefore, the matrix is singular when $k=4$ or $k=-1$. This corresponds to option (A).

The correct answer is (A)

Explanation:

The adjoint (or adjugate) of a square matrix is the transpose of its cofactor matrix. For a $2 \times 2$ matrix, there is a simple shortcut to find its adjoint.

Given the matrix:

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

To find the adjoint of $A$, denoted as $\text{adj}(A)$, we perform two steps:

1. Swap the elements on the main diagonal (the top-left to bottom-right diagonal). So, $a$ and $d$ are interchanged.

2. Change the sign (negate) of the elements on the off-diagonal (the top-right to bottom-left diagonal). So, $b$ becomes $-b$ and $c$ becomes $-c$.

Applying these steps to matrix $A$:

The main diagonal elements $a$ and $d$ are swapped, giving $\begin{bmatrix} d & \dots \\ \dots & a \end{bmatrix}$.

The off-diagonal elements $b$ and $c$ are negated, giving $\begin{bmatrix} \dots & -b \\ -c & \dots \end{bmatrix}$.

Combining these, we get the adjoint matrix:

$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

This result matches the matrix given in option (A).

The correct answer is (A)

Explanation:

Let's analyze the Assertion (A) and the Reason (R) separately.

Assertion (A): If $A$ and $B$ are two matrices such that $AB = BA$, then $(A+B)^2 = A^2 + 2AB + B^2$.

To check this, let's expand the left-hand side, $(A+B)^2$, using the distributive property of matrix multiplication:

$(A+B)^2 = (A+B)(A+B)$

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

$= A \cdot A + A \cdot B + B \cdot A + B \cdot B$

$= A^2 + AB + BA + B^2$

The assertion states that this is equal to $A^2 + 2AB + B^2$. The two expressions are equal if and only if:

$A^2 + AB + BA + B^2 = A^2 + 2AB + B^2$

Subtracting $A^2$ and $B^2$ from both sides gives:

$AB + BA = 2AB$

Subtracting $AB$ from both sides gives:

$BA = AB$

The assertion assumes that $AB = BA$ is true. Therefore, under this condition, the statement $(A+B)^2 = A^2 + 2AB + B^2$ holds. Thus, Assertion (A) is true.

Reason (R): Matrix multiplication is always commutative.

This statement claims that for any two matrices $A$ and $B$, it is always true that $AB = BA$. This is a well-known misconception. In general, matrix multiplication is not commutative. For most pairs of matrices, $AB \neq BA$. Therefore, Reason (R) is false.

Since Assertion (A) is true and Reason (R) is false, the correct option is (C).

The correct answer is (C)

Explanation:

To find the total daily cost of production, we need to multiply the number of units of each product by their respective costs and then sum the results.

This can be calculated using matrix multiplication. The total cost is the product of the cost matrix $C$ and the production matrix $P$.

Given matrices:

Production matrix (units of X and Y): $P = \begin{bmatrix} 100 \\ 150 \end{bmatrix}$

Cost matrix (cost per unit of X and Y): $C = \begin{bmatrix} \textsf{₹}200 & \textsf{₹}300 \end{bmatrix}$

The total daily cost is given by the product $CP$:

Total Cost = $C \times P = \begin{bmatrix} 200 & 300 \end{bmatrix} \begin{bmatrix} 100 \\ 150 \end{bmatrix}$

Performing the matrix multiplication:

Total Cost = $\begin{bmatrix} (200 \times 100) + (300 \times 150) \end{bmatrix}$

Total Cost = $\begin{bmatrix} 20000 + 45000 \end{bmatrix}$

Total Cost = $\begin{bmatrix} 65000 \end{bmatrix}$

So, the total daily cost of production is $\textsf{₹}65,000$. This corresponds to option (A).

The correct answer is (A)

Explanation:

We are given the matrix M and asked to compute the expression $2M - M'$.

The given matrix is:

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

Step 1: Calculate the matrix $2M$

We multiply each element of the matrix M by the scalar 2.

$2M = 2 \times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times 3 & 2 \times 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}$

Step 2: Find the transpose of M, denoted as $M'$

The transpose of a matrix is found by interchanging its rows and columns.

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

Step 3: Calculate the final expression $2M - M'$

Now we subtract the transpose $M'$ from the matrix $2M$ by subtracting the corresponding elements.

$2M - M' = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$

$2M - M' = \begin{bmatrix} 2-1 & 4-3 \\ 6-2 & 8-4 \end{bmatrix}$

$2M - M' = \begin{bmatrix} 1 & 1 \\ 4 & 4 \end{bmatrix}$

Upon reviewing the options, there appears to be a typo in the provided choices as none of them match the correctly calculated result. However, option (A) has the correct first row and is repeated in options (C) and (D), suggesting it might be the intended answer with a typo in its second row.

The correct answer is (A)

Explanation:

Let's examine the definitions of the different types of matrices listed in the options to see which one matches the given condition, $A^2 = I$.

• (A) Idempotent matrix: A square matrix $A$ is called idempotent if $A^2 = A$. This does not match the given condition.

• (B) Involuntary matrix (or Involutory matrix): A square matrix $A$ is called involuntary if its square is the identity matrix, i.e., $A^2 = I$. This is the exact definition provided in the question. Such a matrix is its own inverse ($A^{-1} = A$).

• (C) Nilpotent matrix: A square matrix $A$ is called nilpotent if there exists a positive integer $k$ such that $A^k = 0$ (the zero matrix). This does not match the given condition.

• (D) Orthogonal matrix: A square matrix $A$ is called orthogonal if its transpose is equal to its inverse, i.e., $A' = A^{-1}$. This is equivalent to the condition $AA' = I$. While some orthogonal matrices are also involuntary, the condition $A^2 = I$ is the specific definition of an involuntary matrix.

Based on these definitions, a matrix $A$ for which $A^2 = I$ is called an involuntary matrix.

The correct answer is (B)

Explanation:

A matrix is defined as singular if its determinant is equal to zero. We are given the matrix:

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

To find the value of $x$ that makes this matrix singular, we first need to calculate its determinant and then set it equal to 0.

The determinant of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is calculated as $ad - bc$.

For the given matrix A, we have $a=2$, $b=3$, $c=4$, and $d=x$.

The determinant, $|A|$, is:

$|A| = (2)(x) - (3)(4)$

$|A| = 2x - 12$

Now, we set the determinant to zero for the matrix to be singular:

$2x - 12 = 0$

Solving for $x$:

$2x = 12$

$x = \frac{12}{2}$

$x = 6$

Thus, the value of $x$ for which the matrix is singular is 6. This corresponds to option (A).

The correct answer is (A)

Explanation:

We are given the following system of linear equations:

To find the determinant of the coefficient matrix, we first need to construct the matrix. The coefficient matrix is formed by the coefficients of the variables $x$ and $y$.

The coefficients from the first equation ($1x + 1y = 5$) form the first row: [1, 1].

The coefficients from the second equation ($2x + 2y = 10$) form the second row: [2, 2].

So, the coefficient matrix, let's call it $A$, is:

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

Now, we calculate the determinant of this $2 \times 2$ matrix. The formula for the determinant of a matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $ad - bc$.

For our matrix $A$, we have $a=1$, $b=1$, $c=2$, and $d=2$.

The determinant $|A|$ is:

$|A| = (1)(2) - (1)(2)$

$|A| = 2 - 2$

$|A| = 0$

The determinant of the coefficient matrix is 0. This indicates that the system of equations has either no solution or infinitely many solutions (in this case, infinitely many, as the two equations represent the same line).

The correct answer is (A)

Explanation:

This question refers to a fundamental theorem in matrix algebra which states that any square matrix $A$ can be uniquely expressed as the sum of a symmetric matrix ($P$) and a skew-symmetric matrix ($Q$).

The expression is given by:

$A = P + Q$

Where:

• $P$ is a symmetric matrix, meaning $P' = P$.
• $Q$ is a skew-symmetric matrix, meaning $Q' = -Q$.

As stated in the question, the formula for the symmetric part is:

$P = \frac{1}{2}(A+A')$

To find the skew-symmetric part $Q$, we can use the following derivation:

We start with the two equations:

Taking the transpose of the first equation:

$A' = (P+Q)' = P' + Q' = P - Q$

Now, subtract equation (ii) from equation (i):

$A - A' = (P+Q) - (P-Q)$

$A - A' = P + Q - P + Q$

$A - A' = 2Q$

Solving for $Q$, we get:

$Q = \frac{1}{2}(A-A')$

Thus, the skew-symmetric part of the matrix $A$ is $\frac{1}{2}(A-A')$. This corresponds to option (C).

The correct answer is (C)

Explanation:

We are given a matrix with complex entries and need to determine its type. The matrix is:

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

For matrices with complex entries, we often check for Hermitian or Skew-Hermitian properties. Let's review the definitions:

• Symmetric: $A' = A$ (Transpose equals the original)
• Skew-Symmetric: $A' = -A$ (Transpose equals the negative)
• Hermitian: $A^\theta = A$ (Conjugate transpose equals the original)
• Skew-Hermitian: $A^\theta = -A$ (Conjugate transpose equals the negative)

Here, $A^\theta$ denotes the conjugate transpose, which is found by taking the transpose of the matrix of complex conjugates, i.e., $A^\theta = (\overline{A})'$.

Step 1: Find the conjugate of A, denoted as $\overline{A}$

We replace each element with its complex conjugate. The conjugate of $i$ is $-i$, and the conjugate of $-i$ is $i$.

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

Step 2: Find the conjugate transpose of A, denoted as $A^\theta$

We take the transpose of $\overline{A}$.

$A^\theta = (\overline{A})' = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix}' = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix}$

Step 3: Compare $A^\theta$ with $A$ and $-A$

Is $A^\theta = A$? No, because $\begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \neq \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$. So, the matrix is not Hermitian.

Now let's find $-A$:

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

Is $A^\theta = -A$? Yes, because $\begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix}$.

Since the conjugate transpose of A is equal to the negative of A ($A^\theta = -A$), the matrix is Skew-Hermitian.

The correct answer is (D)

Explanation:

The question asks to identify the type of a square matrix $A$ based on the property $A^T = A$, where $A^T$ represents the transpose of the matrix $A$.

Let's examine the definitions for each option:

• (A) Orthogonal matrix: A square matrix $A$ is orthogonal if its transpose is equal to its inverse, i.e., $A^T = A^{-1}$. This is different from the given condition.

• (B) Idempotent matrix: A square matrix $A$ is idempotent if its square is equal to itself, i.e., $A^2 = A$. This is different from the given condition.

• (C) Symmetric matrix: A square matrix $A$ is defined as symmetric if its transpose is equal to the original matrix. The condition for a symmetric matrix is $A^T = A$. This perfectly matches the condition given in the question.

• (D) Skew-symmetric matrix: A square matrix $A$ is skew-symmetric if its transpose is equal to the negative of the original matrix, i.e., $A^T = -A$. This is different from the given condition.

Therefore, based on the standard definitions, if $A^T = A$, the matrix $A$ is a symmetric matrix.

The correct answer is (C)

Explanation:

Let's evaluate each of the given statements about the properties of determinants for square matrices $A$ and $B$ of the same order $n$.

(A) $|A+B| = |A| + |B|$

This statement is generally false. The determinant of the sum of two matrices is not equal to the sum of their determinants. For example, let $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$. Then $|A|=0$ and $|B|=0$, so $|A|+|B|=0$. However, $A+B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, and $|A+B| = 1$. Since $1 \neq 0$, the statement is false.

(B) $|AB| = |A||B|$

This statement is always true. It is a fundamental property of determinants, often called the product rule for determinants. It states that the determinant of the product of two square matrices of the same order is equal to the product of their individual determinants.

(C) $|kA| = k|A|$ (for scalar $k$)

This statement is generally false. The correct property is $|kA| = k^n|A|$, where $n$ is the order of the matrix $A$. The scalar $k$ is raised to the power of the order of the matrix. The given statement would only be true if $n=1$.

(D) $|A'| = -|A|$

This statement is generally false. The determinant of the transpose of a matrix is always equal to the determinant of the original matrix. The correct property is $|A'| = |A|$. The given statement would only be true if $|A|=0$.

Therefore, the only statement that is always true is the one in option (B).

The correct answer is (B)

Explanation:

We are asked to find the inverse of the matrix $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

This matrix is the $2 \times 2$ identity matrix, denoted by $I$. By definition, the inverse of a matrix $A$ is a matrix $A^{-1}$ such that $AA^{-1} = I$. For the identity matrix itself, we have $I \times I = I$. This means that the identity matrix is its own inverse. Therefore, the inverse of $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Alternate Method (Using the formula):

The formula for the inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by:

$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

For our matrix $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, we have $a=1, b=0, c=0, d=1$.

Step 1: Calculate the determinant $|A|$

$|A| = (1)(1) - (0)(0) = 1 - 0 = 1$

Since the determinant is not zero, the inverse exists.

Step 2: Find the adjoint of A, adj(A)

We swap the main diagonal elements (1 and 1) and change the sign of the off-diagonal elements (0 and 0).

$\text{adj}(A) = \begin{bmatrix} 1 & -0 \\ -0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Step 3: Calculate the inverse $A^{-1}$

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

Both methods confirm that the inverse of the given matrix is the matrix itself, which corresponds to option (B).

The correct answer is (B)

Explanation:

The given system of linear equations is:

To analyze the nature of the solution using the matrix method ($AX=B$), we first identify the matrices involved.

The coefficient matrix is $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.

The constant matrix is $B = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$.

Step 1: Calculate the determinant of the coefficient matrix, $|A|$.

$|A| = (1)(1) - (1)(1) = 1 - 1 = 0$

Since the determinant $|A|$ is 0, the system does not have a unique solution. It can either have no solution (inconsistent) or infinitely many solutions (consistent and dependent). This rules out options (A) and (D).

Step 2: Calculate the product of the adjoint of A and the matrix B, i.e., $\text{adj}(A)B$.

First, we find the adjoint of A. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$\text{adj}(A) = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$

Now, we compute the product $\text{adj}(A)B$:

$\text{adj}(A)B = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

$\text{adj}(A)B = \begin{bmatrix} (1)(1) + (-1)(2) \\ (-1)(1) + (1)(2) \end{bmatrix} = \begin{bmatrix} 1 - 2 \\ -1 + 2 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$

The condition for the nature of solutions when $|A|=0$ is:

• If $\text{adj}(A)B = 0$ (zero matrix), the system has infinitely many solutions.
• If $\text{adj}(A)B \neq 0$ (non-zero matrix), the system has no solution.

In our case, $\text{adj}(A)B = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$, which is not the zero matrix.

Therefore, the system has no solution. This corresponds to the statement in option (C).

The correct answer is (C)

Explanation:

Let's examine each of the given properties to determine which one is stated incorrectly. We assume $A$, $B$, and $C$ are matrices of the same order and $k$ is a scalar.

(A) Associative: $(A+B)+C = A+(B+C)$

This is the associative law of matrix addition. Since matrix addition is performed element-wise and the addition of numbers is associative, this property is correct. For any element, $(a_{ij}+b_{ij})+c_{ij} = a_{ij}+(b_{ij}+c_{ij})$.

(B) Commutative: $A+B = B+A$

This is the commutative law of matrix addition. Since matrix addition is element-wise and the addition of numbers is commutative, this property is correct. For any element, $a_{ij}+b_{ij} = b_{ij}+a_{ij}$.

(C) Existence of Identity: $A+0 = A$ (where 0 is zero matrix)

This describes the existence of an additive identity. The zero matrix (a matrix of the same order as A with all elements being zero) serves as the additive identity for matrix addition. This property is correct.

(D) Distributive over scalar multiplication: $k(A+B) = kA + B$

This statement describes the distributive property of a scalar over matrix addition. However, it is stated incorrectly. The correct distributive law is:

$k(A+B) = kA + kB$

The scalar $k$ must be multiplied by both matrices inside the parentheses. The statement $k(A+B) = kA + B$ is therefore incorrect.

The correct answer is (D)

Explanation:

This question asks about a fundamental property of determinants concerning the transpose of a matrix. The notation $\text{det}(A)$ or $|A|$ represents the determinant of a square matrix $A$, and $A^T$ represents its transpose.

One of the basic and essential properties of determinants is that the determinant of a square matrix is always equal to the determinant of its transpose.

The property is stated as:

$\text{det}(A^T) = \text{det}(A)$

This means that interchanging the rows and columns of a matrix does not change the value of its determinant.

Let's examine the given options:

• (A) $-\text{det}(A)$: This is incorrect.
• (B) $\text{det}(A)$: This correctly states the property.
• (C) $1/\text{det}(A)$: This is the formula for the determinant of the inverse matrix, $\text{det}(A^{-1})$.
• (D) $(\text{det}(A))^2$: This is incorrect.

Therefore, the determinant of the transpose of a matrix is equal to the determinant of the matrix itself.

The correct answer is (B)

Explanation:

We are asked to evaluate the matrix expression $A^2 - 5A + 2I$ where $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $I$ is the $2 \times 2$ identity matrix.

Step 1: Calculate $A^2$

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

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

$A^2 = \begin{bmatrix} 1+6 & 2+8 \\ 3+12 & 6+16 \end{bmatrix} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$

Step 2: Calculate $5A$ and $2I$

$5A = 5 \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix}$

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

Step 3: Evaluate the expression $A^2 - 5A + 2I$

$A^2 - 5A + 2I = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} - \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

$= \begin{bmatrix} 7-5+2 & 10-10+0 \\ 15-15+0 & 22-20+2 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$

The calculated result is $\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$, which is not listed in the options.

Note on the likely intended question:

This type of question often relates to the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation, $A^2 - \text{tr}(A)A + \text{det}(A)I = 0$.

For matrix A:

• The trace is $\text{tr}(A) = 1+4=5$.
• The determinant is $\text{det}(A) = (1)(4) - (2)(3) = 4-6 = -2$.

Therefore, the characteristic equation for A is $A^2 - 5A - 2I = 0$.

It is highly probable that the question intended to ask for the value of $A^2 - 5A - 2I$, which, according to the theorem, is the zero matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$. This corresponds to option (A).

The correct answer is (A)

Explanation:

This question asks about the relationship between the inverse and the transpose of a non-singular matrix $A$. We need to find an expression equivalent to $(A^{-1})'$, which is the transpose of the inverse of $A$.

We start with the fundamental definition of an inverse matrix:

Now, let's take the transpose of both sides of this equation:

We use two key properties of the transpose:

1. The transpose of a product of matrices is the product of their transposes in reverse order: $(XY)' = Y'X'$.
2. The transpose of the identity matrix is the identity matrix itself: $I' = I$.

Applying the first property to the left side of our equation:

Applying the second property to the right side:

This equation, $(A^{-1})' A' = I$, shows that the matrix $(A^{-1})'$ is the left inverse of the matrix $A'$. Since $A$ is a square matrix, its left inverse is also its right inverse. By the definition of an inverse, if a matrix $M$ multiplies $N$ to give the identity matrix ($MN=I$), then $M$ is the inverse of $N$ ($M=N^{-1}$).

Therefore, we can conclude that:

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

This means that taking the transpose of the inverse is the same as taking the inverse of the transpose. This corresponds to option (B).

The correct answer is (B)

Explanation:

We are given the system of linear equations:

According to Cramer's rule, the solution for a system of equations can be found using determinants. The determinant $D_x$ is found by replacing the column of coefficients of $x$ in the main coefficient matrix with the column of constants from the right-hand side of the equations.

First, let's identify the coefficient matrix and the constant matrix:

The coefficient matrix is $A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$.

The constant matrix is $B = \begin{bmatrix} 1 \\ 5 \end{bmatrix}$.

To find the determinant $D_x$, we replace the first column of the coefficient matrix (which is $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$) with the constant matrix $B$.

So, the matrix for $D_x$ is formed by putting the constant column $\begin{bmatrix} 1 \\ 5 \end{bmatrix}$ in the first position and the y-coefficient column $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ in the second position:

Matrix for $D_x$ = $\begin{bmatrix} 1 & -1 \\ 5 & 1 \end{bmatrix}$

Therefore, the determinant $D_x$ is:

$D_x = \begin{vmatrix} 1 & -1 \\ 5 & 1 \end{vmatrix}$

This expression matches the one given in option (A).

The correct answer is (A)

Explanation:

The question provides the definition of a type of square matrix $A = [a_{ij}]_{n \times n}$ based on its elements. The condition given is:

$a_{ij} = 0$ for all $i > j$

This condition means that any element where the row index ($i$) is greater than the column index ($j$) must be zero. These are precisely the elements that lie below the main diagonal of the matrix.

For example, in a $3 \times 3$ matrix, the elements where $i > j$ are $a_{21}$, $a_{31}$, and $a_{32}$. Setting these to zero results in a matrix of the form:

$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & a_{33} \end{bmatrix}$

Let's analyze the options based on this structure:

• (A) Diagonal matrix: In a diagonal matrix, all non-diagonal elements are zero ($a_{ij} = 0$ for $i \neq j$). This is a more restrictive condition than the one given.

• (B) Upper triangular matrix: This is the exact definition for an upper triangular matrix. All elements below the main diagonal are zero, and the non-zero elements form a "triangle" in the upper part of the matrix.

• (C) Lower triangular matrix: In a lower triangular matrix, all elements above the main diagonal are zero ($a_{ij} = 0$ for $i < j$). This is the opposite of the given condition.

• (D) Scalar matrix: This is a special type of diagonal matrix where all diagonal elements are equal. It is therefore also a special case of an upper triangular matrix, but the definition provided is for the general class of upper triangular matrices.

Therefore, the condition $a_{ij} = 0$ for $i > j$ defines an upper triangular matrix.

The correct answer is (B)

Explanation:

We are given the determinant:

$ \begin{vmatrix} \text{cosec} \theta & 1 \\ 1 & \sin \theta \end{vmatrix} = \text{cosec} \theta \cdot \sin \theta - 1 \cdot 1 $

We know that:

Substituting this in the expression:

$\frac{1}{\sin \theta} \cdot \sin \theta - 1 = 1 - 1 = 0$

Hence, the value of the determinant is $0$.

Correct Option: (C)

Explanation:

A matrix $A$ is said to be skew-symmetric if $A^T = -A$.

This means the transpose of the matrix is equal to the negative of the original matrix.

Let us apply this condition to matrix $A = \begin{bmatrix} 5 & x \\ y & -2 \end{bmatrix}$

Transpose of $A$, i.e. $A^T = \begin{bmatrix} 5 & y \\ x & -2 \end{bmatrix}$

According to skew-symmetric condition:

Comparing both sides:

• $5 = -5$ ⟹ Contradiction
• Diagonal elements in a skew-symmetric matrix must be $0$

So, for $A$ to be skew-symmetric:

• $5 = 0$ and $-2 = 0$ is not true ⟹ diagonal elements must be 0, hence $5$ and $-2$ are not valid
• Thus, $A$ is not skew-symmetric unless the diagonal elements are zero

But if we assume the diagonal entries are $0$, and the matrix is $A = \begin{bmatrix} 0 & x \\ y & 0 \end{bmatrix}$, then:

• $A^T = \begin{bmatrix} 0 & y \\ x & 0 \end{bmatrix}$
• For $A^T = -A$: $x = -y$

Hence, $x = -y$ and diagonal elements must be 0.

Correct Option: (B)

Explanation:

We are given that $|A| = 4$ and $A$ is a square matrix of order $3 \times 3$.

We know the following property of determinants:

Substituting the given value:

$|A^{-1}| = \frac{1}{4}$

Hence, the determinant of the inverse matrix $A^{-1}$ is $\frac{1}{4}$.

Correct Option: (A)

Explanation:

We are given a system of linear equations $AX = B$ with the following conditions:

• $|A| = 0$
• $\text{adj}(A)B = 0$

Let’s analyze what this means:

If $|A| = 0$, then matrix $A$ is singular and does not have an inverse. So, the system cannot have a unique solution.

Now, if $\text{adj}(A)B = 0$, this implies that the system is consistent and the solution exists.

But since $A$ is singular, and the system is consistent, it must have infinitely many solutions.

This is a standard result in linear algebra: if $|A| = 0$ and $\text{adj}(A)B = 0$, the system has infinite solutions.

Correct Option: (C)

Explanation:

A null matrix is a matrix in which all elements are zero.

Now, let’s examine the given options:

• Option (A): $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ – contains 1’s ⟹ Not a null matrix
• Option (B): $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ – all entries are 1 ⟹ Not a null matrix
• Option (C): $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ – all entries are 0 ⟹ ✅ Null matrix
• Option (D): $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ – contains a 1 ⟹ Not a null matrix

Therefore, the correct answer is:

Correct Option: (C)

Explanation:

We are given a matrix $A = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}$.

We are required to find $A'$, which denotes the transpose of matrix $A$.

The transpose of a matrix is obtained by interchanging its rows and columns.

So,

Correct Option: (B)

Explanation:

Matrix multiplication follows several properties:

Associative Property: $(AB)C = A(BC)$ is always true for compatible matrices.

Distributive Property: $A(B+C) = AB + AC$ also holds for matrices of appropriate sizes.

Identity Property: There exists an identity matrix $I$ such that $IA = AI = A$.

Commutative Property: However, in general, matrix multiplication is not commutative. That is,

Hence, the commutative property does not hold for matrix multiplication.

Answer: (B)

Given:

Determinant of a $2 \times 2$ matrix $= 10$

Scalar $k = 3$

Concept:

If a matrix of order $n \times n$ is multiplied by a scalar $k$, the determinant of the resulting matrix becomes $k^n \cdot \text{det}(A)$.

Here, the matrix is of order $2 \times 2$, so $n = 2$.

$\text{New determinant} = 9 \cdot 10 = 90$

Answer: (B)

Given:

Two matrices are equal:

$\begin{bmatrix} x & 2 \\ 3 & z \end{bmatrix} = \begin{bmatrix} 1 & y \\ 3 & 5 \end{bmatrix}$

To Find: Values of $x$, $y$, and $z$ such that the above matrices are equal.

Solution:

Two matrices are equal if and only if their corresponding elements are equal.

Therefore, the required values are:

$x = 1$, $y = 2$, $z = 5$

Answer: (A)

Concept:

A diagonal matrix is a square matrix in which all the non-diagonal elements are zero.

A scalar matrix is a diagonal matrix in which all the diagonal elements are equal (i.e., a scalar multiple of the identity matrix).

Check each option:

(A) $\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$: Diagonal matrix and also scalar (all diagonal elements are equal)

(B) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$: Diagonal and scalar matrix

(C) $\begin{bmatrix} 5 & 0 \\ 0 & 0 \end{bmatrix}$: Diagonal matrix, but diagonal elements are not equal ⇒ Not scalar

(D) $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$: Diagonal matrix and also considered scalar since all diagonal elements are equal (zero)

Required matrix: Diagonal matrix but NOT scalar ⇒ Option (C)

Answer: (C)

Concept:

A matrix $A$ is called symmetric if $A' = A$, where $A'$ is the transpose of $A$.

If $A$ is symmetric, then for any scalar $k$, the matrix $kA$ also satisfies:

Hence, $kA$ is also symmetric.

Elimination of Other Options:

(A) Skew-symmetric matrix ⇒ requires $A' = -A$ which is not true here.

(C) Identity matrix ⇒ Only specific matrices are identity matrices; scalar multiplication of $A$ doesn't ensure this.

(D) Null matrix ⇒ Only possible if $k = 0$ and $A$ is any matrix, but not always.

Correct option: (B) Symmetric matrix

Answer: (B)

Concept:

A matrix $A$ is called skew-symmetric if $A' = -A$.

We are asked to find the nature of $A^2$.

Let us find the transpose of $A^2$:

So, $(A^2)' = A^2$ ⇒ $A^2$ is symmetric.

Elimination of Other Options:

(B) Skew-symmetric ⇒ $A^2$ is not skew-symmetric as its transpose equals itself.

(C) Identity ⇒ Only true if $A$ specifically satisfies $A^2 = I$; not always.

(D) None of these ⇒ Incorrect as (A) is valid.

Correct option: (A) Symmetric

Answer: (A)

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

To Find: $\text{adj}(A)$ (adjugate of matrix $A$)

Step 1: Cofactor matrix

Find cofactor of each element:

• Cofactor of element at $(1,1)$ = $\begin{vmatrix} 4 \end{vmatrix} = 4$
• Cofactor of element at $(1,2)$ = $-\begin{vmatrix} 3 \end{vmatrix} = -3$
• Cofactor of element at $(2,1)$ = $-\begin{vmatrix} 1 \end{vmatrix} = -1$
• Cofactor of element at $(2,2)$ = $\begin{vmatrix} 2 \end{vmatrix} = 2$

So the cofactor matrix is:

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

Step 2: Take transpose of cofactor matrix

This gives the adjugate matrix:

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

Correct option: (A) $\begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}$

Answer: (A)

Given: Matrix equation $AX = B$, where $A$ is a square matrix and $|A| \neq 0$ (i.e., $A$ is invertible).

To Find: The value of $X$ using the matrix method.

Solution:

Given: $AX = B$

To isolate $X$, we multiply both sides of the equation on the left by $A^{-1}$ (the inverse of $A$):

Correct option: (C) $X = A^{-1}B$

Answer: (C)

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

To Find: $|A'|$, i.e., determinant of the transpose of $A$

Step 1: Find transpose of $A$

$A' = A^T = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$

Step 2: Find determinant of $A'$

$|A'| = \begin{vmatrix} 2 & 1 \\ 3 & 4 \end{vmatrix}$

$= (2)(4) - (1)(3) = 8 - 3 = 5$

Important Note: The determinant of a matrix and its transpose are always equal, i.e., $|A'| = |A|$

Correct option: (A) $5$

Answer: (A)

Concept: For a system of linear equations represented by $AX = B$, where $A$ is the coefficient matrix, the system has a unique solution only when the matrix $A$ is non-singular, i.e., $|A| \neq 0$.

If $|A| = 0$, then the matrix is singular, and the system either has no solution or infinitely many solutions — in either case, it does not have a unique solution.

Analysis of options:

(A) $|A| = 5$ ⇒ Unique solution exists

(B) $|A| = -3$ ⇒ Unique solution exists

(C) $|A| = 0$ ⇒ No unique solution

(D) $|A| = 1$ ⇒ Unique solution exists

Correct option: (C) $|A| = 0$

Answer: (C)

Given:

Sales matrix $S = \begin{bmatrix} 120 & 150 \\ 130 & 160 \end{bmatrix}$

Cost percentage matrix $C = \begin{bmatrix} 0.6 & 0.5 \end{bmatrix}$

To Find: Total cost incurred for Product P1 over Quarter 1 and Quarter 2.

Step 1: Extract Product P1 sales from matrix $S$

Product P1 sales: $120$ (Q1), $150$ (Q2)

Step 2: Use cost percentage for P1 from matrix $C$

Cost percentage for P1 = $0.6$

Step 3: Calculate total cost incurred for P1

Correct option: (A) $\textsf{₹}162$ Lakhs

Answer: (A)

Concept:

A matrix $A$ is called an idempotent matrix if it satisfies the condition:

Explanation of other options:

(B) Involuntary matrix: A matrix $A$ is involuntary if $A^2 = I$, where $I$ is the identity matrix.

(C) Nilpotent matrix: A matrix $A$ is nilpotent if $A^k = 0$ for some positive integer $k$.

(D) Orthogonal matrix: A matrix $A$ is orthogonal if $A^T A = I$.

Correct option: (A) Idempotent matrix

Answer: (A)

Given:

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

To Find: Element $a_{21}$ of the matrix $A^2$

Step 1: Find $A^2 = A \cdot A$

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

Step 2: Multiply the matrices

$\begin{bmatrix} (1 \cdot 1 + 2 \cdot 3) & (1 \cdot 2 + 2 \cdot 4) \\ (3 \cdot 1 + 4 \cdot 3) & (3 \cdot 2 + 4 \cdot 4) \end{bmatrix} = \begin{bmatrix} 1 + 6 & 2 + 8 \\ 3 + 12 & 6 + 16 \end{bmatrix} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$

Step 3: Extract the element $a_{21}$

$a_{21}$ refers to the element in the 2nd row and 1st column of $A^2$

So, $a_{21} = 15$

Correct option: (D) $15$

Answer: (D)

Given:

$|A| = 3$ and $|B| = -2$, where $A$ and $B$ are square matrices of the same order.

To Find: $|AB|$

Concept Used:

If $A$ and $B$ are square matrices of the same order, then the determinant of their product satisfies:

Step-by-step Calculation:

$|AB| = -6$

Correct option: (C) $-6$

Answer: (C)

Given:

Equations: $2x - y = 4$ and $4x - 2y = 8$

To Find: Determinant of the coefficient matrix using Cramer's Rule

Step 1: Write the coefficient matrix

The coefficient matrix is:

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

Step 2: Find the determinant

$\text{Det} = (2)(-2) - (4)(-1)$

$\text{Det} = -4 + 4 = 0$

Therefore, the determinant of the coefficient matrix is $0$.

Correct option: (A) $0$

Answer: (A)

Given:

$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $ad - bc = 0$

To Find: The status of $A^{-1}$

Concept Used:

The inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ exists if and only if the determinant of $A$, i.e., $|A| = ad - bc \ne 0$.

Step-by-step Analysis:

Here, it is given that $ad - bc = 0$, so:

$|A| = 0$

Since the determinant is zero, the matrix is singular.

Therefore, the inverse of $A$ does not exist.

Correct option: (C) Does not exist

Answer: (C)

Given:

The matrix multiplication $\begin{bmatrix} x & y \end{bmatrix} A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$

To Find: The order of matrix $A$

Step 1: Analyze the order of matrices involved

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

Let the order of $A$ be $2 \times n$ so that the multiplication is defined.

Matrix multiplication rule: $(1 \times 2) \cdot (2 \times n) = 1 \times n$

Step 2: Compare with RHS

The right-hand side matrix is $\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$, which is of order $1 \times 3$

So, $1 \times n = 1 \times 3$ implies $n = 3$

Therefore, the order of matrix $A$ is $2 \times 3$

Correct option: (A) $2 \times 3$

Answer: (A)

Given:

$A$ is a symmetric matrix

To Find: Nature of $A^2$

Concept Used:

A matrix $A$ is symmetric if $A' = A$.

We examine whether $A^2$ is symmetric:

Consider $(A^2)' = (AA)' = A'A' = AA$ (since $A$ is symmetric, $A' = A$)

Thus, $(A^2)' = A^2$

Therefore, $A^2$ is also symmetric.

Correct option: (B) Symmetric

Answer: (B)

Given:

$A$ is any square matrix

To Find: The nature of the matrix $A - A'$

Concept Used:

Let us examine the transpose of $A - A'$:

This implies $(A - A')' = - (A - A')$

Therefore, $A - A'$ is a skew-symmetric matrix.

Correct option: (B) Skew-symmetric

Answer: (B)

Given:

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

To Find: Product matrix $AB$

Solution:

We perform matrix multiplication $AB$:

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

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

$= \begin{bmatrix} 2 + 2 & 4 + 4 \\ 4 + 4 & 8 + 8 \end{bmatrix}$

$= \begin{bmatrix} 4 & 8 \\ 8 & 16 \end{bmatrix}$

Correct option: (A) $\begin{bmatrix} 4 & 8 \\ 8 & 16 \end{bmatrix}$

Answer: (A)

Given:

Determinant: $\begin{vmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{vmatrix}$

To Find: Value of the determinant

Formula Used:

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is given by:

$\text{det} = ad - bc$

Solution:

Here, $a = \sin \theta$, $b = -\cos \theta$, $c = \cos \theta$, $d = \sin \theta$

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

Final Answer: $1$

Correct option: (B)

Answer: (B)

Given:

$A (\text{adj} A) = (\text{adj} A) A = |A| I_n$, where $A$ is a square matrix of order $n$ and $I_n$ is the identity matrix of order $n$.

To Find: What this identity is used to find

Solution:

This is a fundamental identity in matrix theory and is especially useful in deriving the formula for the inverse of a square matrix.

If $|A| \ne 0$, then we can divide both sides of the identity by $|A|$ to get:

This formula shows that the property $A (\text{adj} A) = |A| I_n$ is used to find the inverse of a matrix when its determinant is non-zero.

Correct option: (C) Inverse of $A$

Answer: (C)

Given: System of equations

To Find: Nature of solution using matrix method

Matrix Form: $AX = B$, where

$A = \begin{bmatrix} 1 & 2 \\ 3 & 6 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 5 \\ 15 \end{bmatrix}$

Step 1: Find the determinant of matrix $A$

Since $|A| = 0$, the matrix is singular. Hence, inverse of $A$ does not exist, and we cannot apply the usual inverse method directly.

Step 2: Check for consistency using adjoint matrix:

Note: Equation (ii) is just 3 times equation (i), so both equations represent the same line. This implies the system is dependent and consistent.

Therefore, the system has infinitely many solutions.

Also, for such a system, $\text{adj}(A)B = 0$

Correct option: (C) Infinitely many solutions exist because $|A| = 0$ and $\text{adj}(A)B = 0$.

Answer: (C)

Given: The determinant of the coefficient matrix is non-zero.

To Find: The nature of the solution to the system using Cramer's rule

Concept:

Cramer's Rule is applicable only when the determinant of the coefficient matrix (say $|A|$) is non-zero.

If $|A| \ne 0$, then the system of linear equations has a unique solution.

This is because in Cramer's rule, the solution for each variable is given by the ratio of the determinant of a modified matrix to $|A|$, i.e.,

Since $|A| \ne 0$, all the denominators are non-zero, so all variables are uniquely determined.

Correct option: (B) A unique solution

Answer: (B)

Given: $A' = A$

To Find: The nature of the matrix $A$

Concept:

The transpose of a matrix $A$ is denoted by $A'$ (or $A^T$), and it is defined as: if $A = [a_{ij}]$ of order $m \times n$, then $A' = [a_{ji}]$ of order $n \times m$.

Now, if $A' = A$, it implies that the matrix is equal to its transpose. This condition is called symmetry.

But this can only happen if $A$ is a square matrix, i.e., $m = n$.

Therefore, $A$ must be a square matrix and symmetric.

Correct option: (A) It must be a square matrix and symmetric.

Answer: (A)

Given: Matrices

To Find: $A \circ B$ (Hadamard product)

Concept:

The Hadamard product of two matrices $A$ and $B$, denoted as $A \circ B$, is the element-wise multiplication:

$A \circ B = \begin{bmatrix} a_{11} \cdot b_{11} & a_{12} \cdot b_{12} \\ a_{21} \cdot b_{21} & a_{22} \cdot b_{22} \end{bmatrix}$

Substitute the values:

$A \circ B = \begin{bmatrix} 1 \cdot 0 & 2 \cdot 1 \\ 3 \cdot 1 & 4 \cdot 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 3 & 0 \end{bmatrix}$

Correct option: (A) $\begin{bmatrix} 0 & 2 \\ 3 & 0 \end{bmatrix}$

Answer: (A)

Given: Determinant

$\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$

To Find: Cofactor of the element $3$

Concept:

The cofactor of an element $a_{ij}$ in a determinant is given by:

$C_{ij} = (-1)^{i+j} \cdot M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$ obtained by deleting the $i^{\text{th}}$ row and $j^{\text{th}}$ column.

The element '3' lies in the first row and third column, so $i = 1$, $j = 3$.

Now, eliminate first row and third column from the determinant to find the minor:

$M_{13} = \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}$

Therefore, the cofactor of 3 is:

$C_{13} = 1 \cdot \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}$

Correct option: (A) $\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}$

Answer: (A)

Given: $AB = 0$, where $A$ and $B$ are matrices.

To Find: The statement which is NOT necessarily true.

Concept:

If $AB = 0$ (the zero matrix), it does not imply that $A = 0$ or $B = 0$ individually. This is different from multiplication of real numbers.

There can be non-zero matrices $A$ and $B$ such that their product is the zero matrix.

For example, consider:

Then,

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

So $AB = 0$, but neither $A$ nor $B$ is the null matrix.

Option analysis:

(A) $A$ is a null matrix → Not necessarily true

(B) $B$ is a null matrix → Not necessarily true

(C) Both $A$ and $B$ are null matrices → Not necessarily true

(D) The product $AB$ is the null matrix → Always true (Given)

So, (C) is the strongest incorrect assumption: if $AB=0$, both $A$ and $B$ need not be null matrices.

Correct option: (C) Both $A$ and $B$ are null matrices.

Answer: (C)

Given: $A$ is a $3 \times 3$ matrix and each element of the first row is multiplied by $k$.

To Find: The new determinant value in terms of $|A|$.

Concept:

If one row of a determinant of order $n$ is multiplied by a scalar $k$, then the determinant is multiplied by $k$.

In this case, since only the first row is multiplied by $k$, and the matrix is of order $3$, the new determinant becomes:

Correct option: (B) $k|A|$

Answer: (B)

Given: $A$ is an invertible matrix and $A^{-1}$ is its inverse.

To Find: The value of $AA^{-1}$.

Concept:

By the definition of an inverse matrix:

This property always holds for any invertible (non-singular) matrix $A$.

Correct option: (C) Identity matrix

Answer: (C)

Given: Different types of systems of equations.

To Find: Which system Cramer's rule cannot solve.

Concept:

Cramer's Rule is used to solve a system of $n$ linear equations in $n$ variables, provided the coefficient matrix has a non-zero determinant:

If $|A| = 0$, the system either has:

• No solution (inconsistent system), or
• Infinitely many solutions (dependent system).

In both these cases, Cramer’s rule fails because it involves dividing by $|A|$.

Correct option: (C) A system with infinitely many solutions

Answer: (C)

Definition:

A matrix is a rectangular array or arrangement of numbers, symbols, or expressions, arranged in rows and columns. The numbers or elements in a matrix are called its entries.

General Representation:

A matrix with $m$ rows and $n$ columns is called an $m \times n$ matrix.

It is usually represented as:

$A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix}$

Example of a $3 \times 2$ Matrix:

A $3 \times 2$ matrix has 3 rows and 2 columns. An example is:

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

Here, matrix $B$ has 3 rows and 2 columns, so it is a $3 \times 2$ matrix.

Given:

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

Order of the Matrix:

Matrix $A$ has 2 rows and 3 columns.

Therefore, the order of matrix $A$ is $2 \times 3$.

Total Number of Elements:

To find the total number of elements in a matrix, we multiply the number of rows by the number of columns:

∴ Matrix $A$ has 6 elements.

Definition of Square Matrix:

A square matrix is a matrix in which the number of rows is equal to the number of columns. That is, for a square matrix of order $n$, it has $n$ rows and $n$ columns.

Definition of Diagonal Matrix:

A diagonal matrix is a square matrix in which all the elements outside the main diagonal are zero. The elements on the main diagonal may or may not be zero.

Example of a $3 \times 3$ Diagonal Matrix:

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

Here, matrix $D$ is a square matrix of order $3 \times 3$, and all elements outside the main diagonal are zero. Hence, it is a diagonal matrix.

Definition of Scalar Matrix:

A scalar matrix is a special type of diagonal matrix in which all the elements on the main diagonal are equal, and all other elements are zero.

That is, a scalar matrix has the form:

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

where $k$ is a constant.

Justification:

Yes, every scalar matrix is a diagonal matrix because it satisfies the condition that all elements outside the main diagonal are zero.

However, the converse is not always true. That is, not every diagonal matrix is a scalar matrix, because in a diagonal matrix, the diagonal elements can be unequal.

Example to Justify:

Scalar Matrix:

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

Here, all diagonal elements are equal to 4, and all other elements are zero. So, it is a scalar matrix.

Diagonal Matrix (not scalar):

$D = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}$

Here, all off-diagonal elements are zero, but diagonal elements are not equal. So, it is a diagonal matrix but not a scalar matrix.

Given:

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

Step-by-step comparison of corresponding elements:

From the first row, first column:

From the second row, second column:

Substitute equation (ii) into equation (i):

Final Answer:

$x = 7$, $y = -1$

Given:

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

Step-by-step comparison of corresponding elements:

From first row, first column:

From first row, second column:

From second row, first column:

From second row, second column:

Solving equations (i) and (iii):

From (i): $a = b - 1$

Substitute into (iii):

Substitute (v) into (i):

Now solve (ii) and (iv) using (vi):

Substitute $a = 1$ into (ii):

Substitute $c = 3$ into (iv):

Final Answer:

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

Given:

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

To Find:

Transpose of matrix $A$, i.e., $A^T$ or $A'$

Concept:

The transpose of a matrix is obtained by interchanging its rows and columns. If $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$, then:

$A^T = \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \\ a_{13} & a_{23} \end{bmatrix}$

Solution:

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

Interchanging rows and columns:

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

Final Answer:

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

Given:

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

To Find:

Transpose of matrix $B$, i.e., $B^T$

Concept:

The transpose of a matrix is obtained by interchanging its rows and columns. That is, the element in the $i^{th}$ row and $j^{th}$ column becomes the element in the $j^{th}$ row and $i^{th}$ column.

Solution:

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

Now, interchange rows and columns:

$B^T = \begin{bmatrix} -1 & 2 & 1 \\ 0 & 4 & 1 \\ 5 & -3 & 6 \end{bmatrix}$

Final Answer:

$B^T = \begin{bmatrix} -1 & 2 & 1 \\ 0 & 4 & 1 \\ 5 & -3 & 6 \end{bmatrix}$

Definition:

A symmetric matrix is a square matrix that is equal to its transpose. That is, a matrix $A$ is said to be symmetric if:

Example:

Let $A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}$

Find the transpose of $A$:

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

Since $A = A^T$, matrix $A$ is a symmetric matrix.

Final Answer:

$\begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}$ is a $2 \times 2$ symmetric matrix.

Definition:

A skew-symmetric matrix is a square matrix whose transpose is equal to its negative. That is, a matrix $A$ is skew-symmetric if:

Also, all the diagonal elements of a skew-symmetric matrix are always zero.

Example:

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

Now, find the transpose of $A$:

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

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

Since $A^T = -A$, matrix $A$ is skew-symmetric.

Final Answer:

$\begin{bmatrix} 0 & 2 & -1 \\ -2 & 0 & -4 \\ 1 & 4 & 0 \end{bmatrix}$ is a $3 \times 3$ skew-symmetric matrix.

Given:

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

Step 1: Find the transpose of matrix A

$A^T = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix}$

Step 2: Compare A with $A^T$ and $-A$

$A \ne A^T$ $\Rightarrow$ $A$ is not symmetric

Also, $-A = \begin{bmatrix} -2 & -3 \\ -4 & -5 \end{bmatrix}$ and $A^T \ne -A$

$\Rightarrow$ $A$ is not skew-symmetric

Final Answer:

Matrix $A$ is neither symmetric nor skew-symmetric.

Given:

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

To Find:

$A + B$

Solution:

Add corresponding elements of matrices $A$ and $B$:

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

$\Rightarrow A + B = \begin{bmatrix} 0 & 1 \\ 7 & -1 \end{bmatrix}$

Final Answer:

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

Given:

$A = \begin{bmatrix} 5 & 0 \\ -2 & 3 \end{bmatrix}$, $\quad B = \begin{bmatrix} 1 & 1 \\ 6 & -4 \end{bmatrix}$

To Find:

$A - B$

Solution:

Subtract corresponding elements of matrices $A$ and $B$:

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

$\Rightarrow A - B = \begin{bmatrix} 4 & -1 \\ -8 & 7 \end{bmatrix}$

Final Answer:

$A - B = \begin{bmatrix} 4 & -1 \\ -8 & 7 \end{bmatrix}$

Given:

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

To Find:

$3A$ and $-2A$

Solution:

1. To find $3A$:

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

$\Rightarrow 3A = \begin{bmatrix} 3 & 6 \\ -3 & 0 \end{bmatrix}$

2. To find $-2A$:

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

$\Rightarrow -2A = \begin{bmatrix} -2 & -4 \\ 2 & 0 \end{bmatrix}$

Final Answers:

$3A = \begin{bmatrix} 3 & 6 \\ -3 & 0 \end{bmatrix}$, $\quad -2A = \begin{bmatrix} -2 & -4 \\ 2 & 0 \end{bmatrix}$

Given:

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

To Find:

$AB$

Solution:

Matrix multiplication is done by taking the dot product of rows of the first matrix with columns of the second matrix:

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

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

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

$= \begin{bmatrix} 1 & 4 \\ 1 & 8 \end{bmatrix}$

Final Answer:

$AB = \begin{bmatrix} 1 & 4 \\ 1 & 8 \end{bmatrix}$

Given:

$A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$, which is a $3 \times 1$ matrix

$B = \begin{bmatrix} 4 & 5 & 6 \end{bmatrix}$, which is a $1 \times 3$ matrix

To Find:

$AB$ and whether $BA$ is possible. If yes, what is its order?

Solution:

1. To find $AB$:

$A$ is $3 \times 1$, $B$ is $1 \times 3$ $\Rightarrow$ product $AB$ is possible and will be a $3 \times 3$ matrix.

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

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

$= \begin{bmatrix} 4 & 5 & 6 \\ 8 & 10 & 12 \\ 12 & 15 & 18 \end{bmatrix}$

2. Is $BA$ possible?

$B$ is $1 \times 3$, $A$ is $3 \times 1$ $\Rightarrow$ product $BA$ is possible and will be a $1 \times 1$ matrix.

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

$= \begin{bmatrix} 4 \times 1 + 5 \times 2 + 6 \times 3 \end{bmatrix}$

$= \begin{bmatrix} 4 + 10 + 18 \end{bmatrix} = \begin{bmatrix} 32 \end{bmatrix}$

Final Answers:

$AB = \begin{bmatrix} 4 & 5 & 6 \\ 8 & 10 & 12 \\ 12 & 15 & 18 \end{bmatrix}$ (order $3 \times 3$)

$BA = \begin{bmatrix} 32 \end{bmatrix}$ is possible and its order is $1 \times 1$

Given:

$A = \begin{bmatrix} 3 & -1 \\ 2 & 4 \end{bmatrix}$, which is a $2 \times 2$ matrix

To Find:

$\det(A)$

Solution:

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is given by:

$\det(A) = ad - bc$

Here, $a = 3$, $b = -1$, $c = 2$, $d = 4$

So, $\det(A) = (3)(4) - (-1)(2) = 12 + 2 = 14$

Final Answer:

$\det(A) = 14$

Given:

$\begin{vmatrix} x & 2 \\ 3 & 5 \end{vmatrix} = 4$

To Find:

Value of $x$

Solution:

For a $2 \times 2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$

Comparing, $a = x$, $b = 2$, $c = 3$, $d = 5$

So, $\begin{vmatrix} x & 2 \\ 3 & 5 \end{vmatrix} = x \cdot 5 - 2 \cdot 3$

$\Rightarrow 5x - 6 = 4$

$\Rightarrow 5x = 10$

$\Rightarrow x = \frac{10}{5} = 2$

Final Answer:

$x = 2$

Given:

$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$, a $3 \times 3$ diagonal matrix

To Find:

$\det(A)$

Solution:

The determinant of a diagonal matrix is the product of its diagonal elements.

So, $\det(A) = 1 \cdot 2 \cdot 3 = 6$

Final Answer:

$\det(A) = 6$

Given:

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

To Find:

$\text{adj}(A)$

Solution:

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is given by:

$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

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

So, $\text{adj}(A) = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$

Final Answer:

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

Given:

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

To Find:

$A^{-1}$ (Inverse of matrix $A$)

Formula:

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is given by:

$A^{-1} = \frac{1}{\det(A)} \cdot \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$, where $\det(A) \ne 0$

Solution:

Here, $a = 3$, $b = 1$, $c = 5$, $d = 2$

First, find the determinant:

$\det(A) = (3)(2) - (5)(1) = 6 - 5 = 1$

Since $\det(A) \ne 0$, the inverse exists.

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

Final Answer:

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

Given:

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

To Find:

$A^{-1}$ (Inverse of matrix $A$)

Solution:

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is given by:

$A^{-1} = \frac{1}{\det(A)} \cdot \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$, provided $\det(A) \ne 0$

Here, $a = 1$, $b = 2$, $c = 2$, $d = 4$

$\det(A) = (1)(4) - (2)(2) = 4 - 4 = 0$

Since $\det(A) = 0$, the inverse of $A$ does not exist.

Observation:

The matrix $A$ is singular, as its determinant is zero.

Also, the second row is a scalar multiple of the first row ($R_2 = 2 \times R_1$), hence the matrix is not of full rank.

Therefore, $A$ is non-invertible.

Given:

$2x + 3y = 7$ …(i)            $x - y = 1$ …(ii)

To Write:

The system of equations in the form $AX = B$

Matrix Form:

Comparing the equations:

Coefficients matrix: $A = \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}$

Variable matrix: $X = \begin{bmatrix} x \\ y \end{bmatrix}$

Constant matrix: $B = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$

Therefore, the system of equations in matrix form is:

$AX = B \Rightarrow \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$

Given:

$x + y = 3$ …(i)            $x - y = 1$ …(ii)

To Find:

Values of $x$ and $y$ using Cramer's Rule

Step 1: Coefficient Matrix

From equations (i) and (ii), the coefficient matrix is: $A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$

The determinant of $A$, denoted as $|A|$:

$|A| = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = (1)(-1) - (1)(1) = -1 - 1 = -2$

Step 2: Replace columns to find $|A_x|$ and $|A_y|$

Replace first column with constants to get $A_x$: $A_x = \begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix}$

$|A_x| = \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (1)(1) = -3 - 1 = -4$

Replace second column with constants to get $A_y$: $A_y = \begin{bmatrix} 1 & 3 \\ 1 & 1 \end{bmatrix}$

$|A_y| = \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} = (1)(1) - (3)(1) = 1 - 3 = -2$

Step 3: Apply Cramer's Rule

$x = \frac{|A_x|}{|A|} = \frac{-4}{-2} = 2$

$y = \frac{|A_y|}{|A|} = \frac{-2}{-2} = 1$

Final Answer:

$x = 2,\quad y = 1$

Definition:

A matrix is called a singular matrix if its determinant is equal to zero.

In other words, a matrix $A$ is singular if $|A| = 0$.

Example:

Consider the matrix: $A = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$

Calculate its determinant:

$|A| = \begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix} = (2)(2) - (4)(1) = 4 - 4 = 0$

Since the determinant is zero, the matrix $A$ is a singular matrix.

Given:

Matrix $A$ is both symmetric and skew-symmetric.

To Find:

The type of matrix $A$.

Concept:

A matrix $A$ is:

• Symmetric if $A^T = A$
• Skew-symmetric if $A^T = -A$

Solution:

Since $A$ is both symmetric and skew-symmetric, we have:

From these two, we equate:

Adding $A$ to both sides of equation (i):

$\Rightarrow 2A = 0 \Rightarrow A = 0$

Final Answer:

The matrix $A$ must be a zero matrix.

To determine if $AB = BA$, we need to calculate both $AB$ and $BA$ and then compare the results.

First, let's calculate $AB$:

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

Next, let's calculate $BA$:

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

Now, let's compare $AB$ and $BA$:

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

Since $AB \neq BA$, the answer is no.

Therefore, $AB$ is not equal to $BA$.

To find the determinant of the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}$, we can use the formula for the determinant of a $3 \times 3$ matrix.

Since the given matrix $A$ is an upper triangular matrix, the determinant is simply the product of the diagonal elements.

The determinant of $A$, denoted as $|A|$ or $\text{det}(A)$, is calculated as follows:

$|A| = 1 \cdot 4 \cdot 6$

Calculating the product:

$|A| = 1 \times 4 \times 6 = 24$

Therefore, the determinant of matrix $A$ is 24.

To show that $(A^T)^T = A$, we need to first find the transpose of matrix $A$, denoted as $A^T$, and then find the transpose of $A^T$, denoted as $(A^T)^T$.

Given the matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, the transpose $A^T$ is obtained by interchanging the rows and columns of $A$.

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

Now, we find the transpose of $A^T$, which is $(A^T)^T$. This is obtained by interchanging the rows and columns of $A^T$.

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

Comparing $(A^T)^T$ with the original matrix $A$:

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

Since $(A^T)^T = A$, we have shown that the transpose of the transpose of matrix $A$ is equal to the original matrix $A$.

Therefore, $(A^T)^T = A$ is proven.

To find $(AB)^T$, we first need to calculate the product of matrices $A$ and $B$, and then find the transpose of the resulting matrix.

Given matrices $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, let's calculate $AB$:

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

Now, we find the transpose of $AB$, which is $(AB)^T$. This is obtained by interchanging the rows and columns of $AB$.

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

Therefore, $(AB)^T = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}$.

To show that $A^2 = 2A$, we first need to calculate $A^2$, which means multiplying matrix $A$ by itself, and then calculate $2A$, which means multiplying each element of matrix $A$ by 2. Finally, we compare the results.

Given the matrix $A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$, let's calculate $A^2$:

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

Now, let's calculate $2A$:

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

Comparing $A^2$ and $2A$:

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

Since $A^2 = 2A$, we have shown that the square of matrix $A$ is equal to 2 times matrix $A$.

Therefore, $A^2 = 2A$ is proven.

If $A$ is an invertible matrix and $AB = AC$, we can conclude something specific about the relationship between matrices $B$ and $C$.

Since $A$ is invertible, there exists a matrix $A^{-1}$ such that $A^{-1}A = I$, where $I$ is the identity matrix.

Given $AB = AC$, we can multiply both sides of the equation by $A^{-1}$ on the left:

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

Using the associative property of matrix multiplication, we have:

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

Since $A^{-1}A = I$, we get:

$IB = IC$

Since the identity matrix $I$ multiplied by any matrix results in the same matrix (i.e., $IB = B$ and $IC = C$), we have:

$B = C$

Therefore, if $A$ is an invertible matrix and $AB = AC$, we can conclude that $B = C$.

The value of the determinant $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{vmatrix}$ can be stated without explicitly calculating it, based on a property of determinants.

The determinant of a matrix is zero if any two rows (or columns) are identical.

In the given matrix, we observe that the first row and the third row are identical:

Row 1: $\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ Row 3: $\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$

Since the first and third rows are identical, the determinant of the matrix is zero.

Therefore, the value of the determinant $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{vmatrix}$ is 0.

To find the determinant of the matrix $\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$, we use the formula for the determinant of a $2 \times 2$ matrix.

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is given by $ad - bc$.

In our case, $a = \cos \theta$, $b = -\sin \theta$, $c = \sin \theta$, and $d = \cos \theta$. So, the determinant is:

$\text{det} = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta)$

Simplifying the expression:

$\text{det} = \cos^2 \theta + \sin^2 \theta$

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

$\text{det} = 1$

Therefore, the determinant of the matrix $\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ is 1.

A matrix is singular if its determinant is equal to zero. So, to find the value of $x$ for which the matrix $A = \begin{bmatrix} 2 & x \\ 1 & 3 \end{bmatrix}$ is singular, we need to set its determinant to zero and solve for $x$.

The determinant of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $ad - bc$.

In our case, $a = 2$, $b = x$, $c = 1$, and $d = 3$. So, the determinant of $A$ is:

$\text{det}(A) = (2)(3) - (x)(1) = 6 - x$

Since $A$ is a singular matrix, $\text{det}(A) = 0$. Therefore, we have:

$6 - x = 0$

Solving for $x$:

$x = 6$

Therefore, the value of $x$ for which the matrix $A$ is singular is 6.

To find $A^2 - 5A + 2I$, we first need to calculate $A^2$, then $5A$, then $2I$, and finally combine the results.

Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, let's calculate $A^2$:

$A^2 = A \times A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} (1 \times 1) + (2 \times 3) & (1 \times 2) + (2 \times 4) \\ (3 \times 1) + (4 \times 3) & (3 \times 2) + (4 \times 4) \end{bmatrix} = \begin{bmatrix} 1 + 6 & 2 + 8 \\ 3 + 12 & 6 + 16 \end{bmatrix} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$

Now, let's calculate $5A$:

$5A = 5 \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 5 \times 1 & 5 \times 2 \\ 5 \times 3 & 5 \times 4 \end{bmatrix} = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix}$

Now, let's calculate $2I$:

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

Now, let's calculate $A^2 - 5A + 2I$:

$A^2 - 5A + 2I = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} - \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 7 - 5 + 2 & 10 - 10 + 0 \\ 15 - 15 + 0 & 22 - 20 + 2 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$

Therefore, $A^2 - 5A + 2I = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$.

The cofactor of an element $a_{ij}$ in a matrix is defined as follows:

Definition: The cofactor of an element $a_{ij}$ in a square matrix is given by $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$. The minor $M_{ij}$ is the determinant of the submatrix formed by deleting the $i$-th row and the $j$-th column of the original matrix.

Now, let's find the cofactor of the element '5' in the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$.

The element '5' is located at $a_{22}$, which means $i = 2$ and $j = 2$.

First, we find the minor $M_{22}$. To do this, we delete the 2nd row and the 2nd column of the matrix:

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \xrightarrow{\text{Delete row 2 and column 2}} \begin{bmatrix} 1 & 3 \\ 7 & 9 \end{bmatrix}$

So, $M_{22} = \begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} = (1 \times 9) - (3 \times 7) = 9 - 21 = -12$.

Next, we find the cofactor $C_{22}$ using the formula $C_{ij} = (-1)^{i+j}M_{ij}$:

$C_{22} = (-1)^{2+2}M_{22} = (-1)^4 \times (-12) = 1 \times (-12) = -12$

Therefore, the cofactor of the element '5' in the given matrix is -12.

To find $|\text{adj}(A)|$, where $A$ is a $3 \times 3$ matrix with $|A| = 5$, we can use the property relating the determinant of the adjoint of a matrix to the determinant of the original matrix.

The property is: For an $n \times n$ matrix $A$, $|\text{adj}(A)| = |A|^{n-1}$, where $\text{adj}(A)$ is the adjugate (or adjoint) of $A$ and $|A|$ is the determinant of $A$.

In this case, $A$ is a $3 \times 3$ matrix, so $n = 3$. We are given that $|A| = 5$.

Using the property, we have:

$|\text{adj}(A)| = |A|^{3-1} = |A|^2$

Substituting the given value of $|A|$:

$|\text{adj}(A)| = 5^2 = 25$

Therefore, $|\text{adj}(A)| = 25$.

For matrix multiplication to be defined, the number of columns in the first matrix must be equal to the number of rows in the second matrix. We need to determine the conditions for both $AB$ and $BA$ to be defined.

Condition for $AB$ to be defined:

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

$n = p$

Condition for $BA$ to be defined:

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

$q = m$

Conditions for both $AB$ and $BA$ to be defined:

For both $AB$ and $BA$ to be defined, we need both conditions to be satisfied:

1. $n = p$
2. $q = m$

Therefore, the conditions for $AB$ and $BA$ both to be defined are $n = p$ and $q = m$.

To show that $A$ is a skew-symmetric matrix, we need to show that $A^T = -A$, where $A^T$ is the transpose of matrix $A$.

Given the matrix $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$, let's find the transpose $A^T$ by interchanging the rows and columns:

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

Now, let's find $-A$:

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

Comparing $A^T$ and $-A$:

$A^T = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ and $-A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Since $A^T = -A$, the matrix $A$ is a skew-symmetric matrix.

Therefore, $A$ is a skew-symmetric matrix is proven.

Given the matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ such that $ad - bc \neq 0$, the inverse of $A$, denoted as $A^{-1}$, can be found using a specific formula.

The determinant of $A$ is given by $|A| = ad - bc$. Since $ad - bc \neq 0$, the matrix $A$ is invertible.

The formula for the inverse of a $2 \times 2$ matrix $A$ is:

$A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Substituting $|A| = ad - bc$, we get:

$A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Therefore, the formula for $A^{-1}$ is $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

To solve the matrix equation $\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} X = \begin{bmatrix} 5 \\ 5 \end{bmatrix}$ for matrix $X$, we first need to find the inverse of the matrix $A = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}$.

Let $A = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}$. The determinant of $A$ is $|A| = (2 \times 3) - (1 \times 1) = 6 - 1 = 5$.

The inverse of $A$ is given by:

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

Now, we can solve for $X$ by multiplying both sides of the equation by $A^{-1}$ on the left:

$A^{-1}AX = A^{-1} \begin{bmatrix} 5 \\ 5 \end{bmatrix}$

$IX = A^{-1} \begin{bmatrix} 5 \\ 5 \end{bmatrix}$

$X = A^{-1} \begin{bmatrix} 5 \\ 5 \end{bmatrix}$

Substituting $A^{-1}$ and performing the multiplication:

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

Therefore, the solution for matrix $X$ is $X = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$.

To find the minors of the elements of the first row of the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$, we need to calculate the minors $M_{11}$, $M_{12}$, and $M_{13}$.

Finding $M_{11}$:

To find $M_{11}$, we delete the first row and the first column of the matrix:

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \xrightarrow{\text{Delete row 1 and column 1}} \begin{bmatrix} 5 & 6 \\ 8 & 9 \end{bmatrix}$

So, $M_{11} = \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = (5 \times 9) - (6 \times 8) = 45 - 48 = -3$.

Finding $M_{12}$:

To find $M_{12}$, we delete the first row and the second column of the matrix:

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \xrightarrow{\text{Delete row 1 and column 2}} \begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix}$

So, $M_{12} = \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} = (4 \times 9) - (6 \times 7) = 36 - 42 = -6$.

Finding $M_{13}$:

To find $M_{13}$, we delete the first row and the third column of the matrix:

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \xrightarrow{\text{Delete row 1 and column 3}} \begin{bmatrix} 4 & 5 \\ 7 & 8 \end{bmatrix}$

So, $M_{13} = \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = (4 \times 8) - (5 \times 7) = 32 - 35 = -3$.

Therefore, the minors of the elements of the first row are: $M_{11} = -3$, $M_{12} = -6$, and $M_{13} = -3$.

Let's define an identity matrix and discuss its order.

Definition: An identity matrix, denoted by $I$ (or $I_n$ to specify its size), is a square matrix with ones on the main diagonal and zeros everywhere else. The main diagonal consists of the elements from the upper left corner to the lower right corner.

For example, the $2 \times 2$ identity matrix is $I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, and the $3 \times 3$ identity matrix is $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

Order of an Identity Matrix:

The order of an identity matrix is $n \times n$, where $n$ is the number of rows and columns. Since it is a square matrix, the number of rows is equal to the number of columns.

Therefore, an identity matrix is a square matrix with ones on the main diagonal and zeros elsewhere, and its order is $n \times n$.

To find $A^2$, we need to multiply matrix $A$ by itself.

Given the matrix $A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$, let's calculate $A^2$:

$A^2 = A \times A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} (2 \times 2) + (0 \times 0) & (2 \times 0) + (0 \times 3) \\ (0 \times 2) + (3 \times 0) & (0 \times 0) + (3 \times 3) \end{bmatrix} = \begin{bmatrix} 4 + 0 & 0 + 0 \\ 0 + 0 & 0 + 9 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 9 \end{bmatrix}$

Therefore, $A^2 = \begin{bmatrix} 4 & 0 \\ 0 & 9 \end{bmatrix}$.

To find the determinant of the coefficient matrix for the given system of equations, we first need to identify the coefficient matrix.

The system of equations is:

$x - 2y = 5$ $3x + y = 1$

The coefficient matrix $A$ is formed by the coefficients of $x$ and $y$ in the equations:

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

Now, we find the determinant of $A$, denoted as $|A|$:

$|A| = (1 \times 1) - (-2 \times 3) = 1 - (-6) = 1 + 6 = 7$

Therefore, the determinant of the coefficient matrix is 7.

To show that $A + A^T$ is a symmetric matrix, we need to prove that $(A + A^T)^T = A + A^T$.

We know that for any matrices $B$ and $C$, $(B + C)^T = B^T + C^T$. Also, for any matrix $B$, $(B^T)^T = B$.

Let's find the transpose of $A + A^T$:

$(A + A^T)^T = A^T + (A^T)^T$

Using the property $(A^T)^T = A$, we have:

$(A + A^T)^T = A^T + A$

Since matrix addition is commutative, $A^T + A = A + A^T$. Therefore:

$(A + A^T)^T = A + A^T$

Since $(A + A^T)^T = A + A^T$, the matrix $A + A^T$ is a symmetric matrix.

Therefore, $A + A^T$ is a symmetric matrix is proven.

To show that $A - A^T$ is a skew-symmetric matrix, we need to prove that $(A - A^T)^T = -(A - A^T)$.

We know that for any matrices $B$ and $C$, $(B - C)^T = B^T - C^T$. Also, for any matrix $B$, $(B^T)^T = B$.

Let's find the transpose of $A - A^T$:

$(A - A^T)^T = A^T - (A^T)^T$

Using the property $(A^T)^T = A$, we have:

$(A - A^T)^T = A^T - A$

Now, let's factor out a -1:

$A^T - A = -(A - A^T)$

Therefore:

$(A - A^T)^T = -(A - A^T)$

Since $(A - A^T)^T = -(A - A^T)$, the matrix $A - A^T$ is a skew-symmetric matrix.

Therefore, $A - A^T$ is a skew-symmetric matrix is proven.

To find the determinant of $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 4 \\ 5 & 0 & 6 \end{bmatrix}$ by expanding along the second row, we use the formula for determinant expansion.

The determinant of a matrix $A$ can be found by expanding along any row or column. Expanding along the second row, the formula is:

$|A| = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}$

where $a_{ij}$ are the elements of the matrix and $C_{ij}$ are their corresponding cofactors.

In this case, the elements of the second row are $a_{21} = 0$, $a_{22} = 0$, and $a_{23} = 4$.

Now, we find the cofactors:

• $C_{21} = (-1)^{2+1} \begin{vmatrix} 2 & 3 \\ 0 & 6 \end{vmatrix} = (-1)(2 \times 6 - 3 \times 0) = (-1)(12 - 0) = -12$
• $C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 3 \\ 5 & 6 \end{vmatrix} = (1)(1 \times 6 - 3 \times 5) = (1)(6 - 15) = -9$
• $C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 5 & 0 \end{vmatrix} = (-1)(1 \times 0 - 2 \times 5) = (-1)(0 - 10) = 10$

Substituting these values into the formula:

$|A| = 0 \times (-12) + 0 \times (-9) + 4 \times (10) = 0 + 0 + 40 = 40$

Therefore, the determinant of the matrix $A$ is 40.

Let's discuss the conditions for a unique solution using determinants (Cramer's rule) and for non-trivial solutions in a homogeneous system.

Condition for a Unique Solution (Cramer's Rule):

Consider a system of $n$ linear equations with $n$ variables. Let $A$ be the coefficient matrix of the system. The system has a unique solution if and only if the determinant of the coefficient matrix is non-zero, i.e., $|A| \neq 0$.

Cramer's rule states that if $|A| \neq 0$, then the solution for each variable $x_i$ can be found as $x_i = \frac{|A_i|}{|A|}$, where $A_i$ is the matrix formed by replacing the $i$-th column of $A$ with the column vector of constants from the right-hand side of the equations.

Condition for Non-Trivial Solutions in a Homogeneous System:

A homogeneous system of linear equations is a system where all the constant terms are zero. For example, $Ax = 0$, where $A$ is the coefficient matrix and $x$ is the column vector of variables.

A homogeneous system always has a trivial solution, which is $x = 0$ (all variables are zero). For a homogeneous system to have non-trivial solutions (i.e., solutions other than the trivial solution), the determinant of the coefficient matrix must be zero, i.e., $|A| = 0$.

Therefore:

• A system of linear equations has a unique solution (using Cramer's rule) if the determinant of the coefficient matrix is non-zero ($|A| \neq 0$).
• A homogeneous system has non-trivial solutions when the determinant of the coefficient matrix is zero ($|A| = 0$).

To prove that $|A^{-1}| = \frac{1}{|A|}$, where $A$ is an invertible matrix of order $n$, we can use the property that the product of a matrix and its inverse is the identity matrix.

Since $A$ is an invertible matrix, there exists a matrix $A^{-1}$ such that:

$AA^{-1} = I$

where $I$ is the identity matrix of order $n$.

Now, we take the determinant of both sides of the equation:

$|AA^{-1}| = |I|$

Using the property that the determinant of the product of two matrices is the product of their determinants, we have:

$|A| \cdot |A^{-1}| = |I|$

The determinant of the identity matrix $I$ is always 1, i.e., $|I| = 1$. So,

$|A| \cdot |A^{-1}| = 1$

Now, we solve for $|A^{-1}|$:

$|A^{-1}| = \frac{1}{|A|}$

Therefore, if $A$ is an invertible matrix of order $n$, then $|A^{-1}| = \frac{1}{|A|}$ is proven.

To determine if $AB = BA$, we need to calculate both $AB$ and $BA$ and then compare the results.

First, let's calculate $AB$:

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

$AB = \begin{bmatrix} 2-1-2 & 1+0+4 & 0-3+2 \\ 0+3-1 & 0+0+2 & 0+9+1 \\ -4+1+0 & -2+0+0 & 0+3+0 \end{bmatrix} = \begin{bmatrix} -1 & 5 & -1 \\ 2 & 2 & 10 \\ -3 & -2 & 3 \end{bmatrix}$

Next, let's calculate $BA$:

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

$BA = \begin{bmatrix} 2+0+0 & -2+3+0 & 4+1+0 \\ 1+0-6 & -1+0+3 & 2+0+0 \\ -1+0-2 & 1+6+1 & -2+2+0 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 5 \\ -5 & 2 & 2 \\ -3 & 8 & 0 \end{bmatrix}$

Now, let's compare $AB$ and $BA$:

$AB = \begin{bmatrix} -1 & 5 & -1 \\ 2 & 2 & 10 \\ -3 & -2 & 3 \end{bmatrix}$ and $BA = \begin{bmatrix} 2 & 1 & 5 \\ -5 & 2 & 2 \\ -3 & 8 & 0 \end{bmatrix}$

Since $AB \neq BA$, the answer is no.

Therefore, $AB$ is not equal to $BA$.

To verify that $(AB)^T = B^T A^T$, we need to first calculate $AB$, then find its transpose $(AB)^T$. Next, we calculate $B^T$ and $A^T$, and then find the product $B^T A^T$. Finally, we compare the results.

First, let's calculate $AB$:

$AB = \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} (2 \times 1) + (3 \times 0) & (2 \times -2) + (3 \times 3) \\ (1 \times 1) + (-1 \times 0) & (1 \times -2) + (-1 \times 3) \end{bmatrix} = \begin{bmatrix} 2 + 0 & -4 + 9 \\ 1 + 0 & -2 - 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & -5 \end{bmatrix}$

Now, let's find $(AB)^T$:

$(AB)^T = \begin{bmatrix} 2 & 5 \\ 1 & -5 \end{bmatrix}^T = \begin{bmatrix} 2 & 1 \\ 5 & -5 \end{bmatrix}$

Next, let's calculate $A^T$ and $B^T$:

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

$B^T = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}^T = \begin{bmatrix} 1 & 0 \\ -2 & 3 \end{bmatrix}$

Now, let's calculate $B^T A^T$:

$B^T A^T = \begin{bmatrix} 1 & 0 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} (1 \times 2) + (0 \times 3) & (1 \times 1) + (0 \times -1) \\ (-2 \times 2) + (3 \times 3) & (-2 \times 1) + (3 \times -1) \end{bmatrix} = \begin{bmatrix} 2 + 0 & 1 + 0 \\ -4 + 9 & -2 - 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 5 & -5 \end{bmatrix}$

Now, let's compare $(AB)^T$ and $B^T A^T$:

$(AB)^T = \begin{bmatrix} 2 & 1 \\ 5 & -5 \end{bmatrix}$ and $B^T A^T = \begin{bmatrix} 2 & 1 \\ 5 & -5 \end{bmatrix}$

Since $(AB)^T = B^T A^T$, the verification is complete.

Therefore, $(AB)^T = B^T A^T$ is verified.

To express a matrix $A$ as the sum of a symmetric and a skew-symmetric matrix, we use the following decomposition:

$A = \underbrace{\frac{1}{2}(A+A')}_{\text{Symmetric}} + \underbrace{\frac{1}{2}(A-A')}_{\text{Skew-Symmetric}}$

First, find the transpose of $A$:

$A = \begin{bmatrix} 3 & -1 & 2 \\ 1 & 4 & -2 \\ 5 & 0 & 6 \end{bmatrix} \implies A' = A^T = \begin{bmatrix} 3 & 1 & 5 \\ -1 & 4 & 0 \\ 2 & -2 & 6 \end{bmatrix}$

Now, calculate the symmetric part $\frac{1}{2}(A+A')$:

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

$\frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 6 & 0 & 7 \\ 0 & 8 & -2 \\ 7 & -2 & 12 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 3.5 \\ 0 & 4 & -1 \\ 3.5 & -1 & 6 \end{bmatrix}$

Now, calculate the skew-symmetric part $\frac{1}{2}(A-A')$:

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

$\frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & -2 & -3 \\ 2 & 0 & -2 \\ 3 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 & -1.5 \\ 1 & 0 & -1 \\ 1.5 & 1 & 0 \end{bmatrix}$

Thus, $A$ can be written as the sum of a symmetric and a skew-symmetric matrix as:

$A = \begin{bmatrix} 3 & 0 & 3.5 \\ 0 & 4 & -1 \\ 3.5 & -1 & 6 \end{bmatrix} + \begin{bmatrix} 0 & -1 & -1.5 \\ 1 & 0 & -1 \\ 1.5 & 1 & 0 \end{bmatrix}$

Therefore, we have expressed A as the sum of a symmetric and a skew-symmetric matrix.

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

To find:

(i) Determinant of A i.e., $|A|$

(ii) Adjoint of A i.e., $adj(A)$

Solution:

(i) Determinant of $A$:

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

$|A| = 1 \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix}$

$|A| = 1(3 \cdot 2 - 1 \cdot 1) - 2(2 \cdot 2 - 1 \cdot 3) + 3(2 \cdot 1 - 3 \cdot 3)$

$|A| = 1(6 - 1) - 2(4 - 3) + 3(2 - 9)$

$|A| = 1(5) - 2(1) + 3(-7)$

$|A| = 5 - 2 - 21$

$|A| = -18$

(ii) Adjoint of $A$:

First, we find the matrix of cofactors:

$C_{11} = \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = 6 - 1 = 5$

$C_{12} = - \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = -(4 - 3) = -1$

$C_{13} = \begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix} = 2 - 9 = -7$

$C_{21} = - \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = -(4 - 3) = -1$

$C_{22} = \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} = 2 - 9 = -7$

$C_{23} = - \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = -(1 - 6) = 5$

$C_{31} = \begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix} = 2 - 9 = -7$

$C_{32} = - \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = -(1 - 6) = 5$

$C_{33} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = 3 - 4 = -1$

The matrix of cofactors is:

$C = \begin{bmatrix} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{bmatrix}$

The adjoint of $A$ is the transpose of the cofactor matrix:

$adj(A) = C^T = \begin{bmatrix} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{bmatrix}^T = \begin{bmatrix} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{bmatrix}$

Results:

(i) $|A| = -18$

(ii) $adj(A) = \begin{bmatrix} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{bmatrix}$

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

To find: Inverse of matrix A i.e., $A^{-1}$

Solution:

The inverse of a matrix $A$ is given by $A^{-1} = \frac{1}{|A|} adj(A)$, where $|A|$ is the determinant of $A$ and $adj(A)$ is the adjoint of $A$.

First, we find the determinant of $A$:

$|A| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{vmatrix}$

$|A| = 1 \begin{vmatrix} 5 & 7 \\ -4 & -5 \end{vmatrix} - 2 \begin{vmatrix} 2 & 7 \\ -2 & -5 \end{vmatrix} + 3 \begin{vmatrix} 2 & 5 \\ -2 & -4 \end{vmatrix}$

$|A| = 1(5 \cdot (-5) - 7 \cdot (-4)) - 2(2 \cdot (-5) - 7 \cdot (-2)) + 3(2 \cdot (-4) - 5 \cdot (-2))$

$|A| = 1(-25 + 28) - 2(-10 + 14) + 3(-8 + 10)$

$|A| = 1(3) - 2(4) + 3(2)$

$|A| = 3 - 8 + 6$

$|A| = 1$

Next, we find the adjoint of $A$. First, we find the matrix of cofactors:

$C_{11} = \begin{vmatrix} 5 & 7 \\ -4 & -5 \end{vmatrix} = -25 - (-28) = 3$

$C_{12} = - \begin{vmatrix} 2 & 7 \\ -2 & -5 \end{vmatrix} = -(-10 - (-14)) = -4$

$C_{13} = \begin{vmatrix} 2 & 5 \\ -2 & -4 \end{vmatrix} = -8 - (-10) = 2$

$C_{21} = - \begin{vmatrix} 2 & 3 \\ -4 & -5 \end{vmatrix} = -(-10 - (-12)) = -2$

$C_{22} = \begin{vmatrix} 1 & 3 \\ -2 & -5 \end{vmatrix} = -5 - (-6) = 1$

$C_{23} = - \begin{vmatrix} 1 & 2 \\ -2 & -4 \end{vmatrix} = -(-4 - (-4)) = 0$

$C_{31} = \begin{vmatrix} 2 & 3 \\ 5 & 7 \end{vmatrix} = 14 - 15 = -1$

$C_{32} = - \begin{vmatrix} 1 & 3 \\ 2 & 7 \end{vmatrix} = -(7 - 6) = -1$

$C_{33} = \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = 5 - 4 = 1$

The matrix of cofactors is:

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

The adjoint of $A$ is the transpose of the cofactor matrix:

$adj(A) = C^T = \begin{bmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{bmatrix}$

Now, we find the inverse of $A$:

$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{1} \begin{bmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{bmatrix}$

Result:

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

Given system of equations:

$x + y + z = 6$

$x - y + z = 2$

$2x + y - z = 1$

To find: Values of $x, y, z$ using the matrix inverse method.

Solution:

We can write the system of equations in matrix form as $AX = B$, where

$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 6 \\ 2 \\ 1 \end{bmatrix}$

To solve for $X$, we can use the formula $X = A^{-1}B$.

First, we find the determinant of $A$:

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

$|A| = 1 \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + 1 \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix}$

$|A| = 1((-1)(-1) - (1)(1)) - 1((1)(-1) - (1)(2)) + 1((1)(1) - (-1)(2))$

$|A| = 1(1 - 1) - 1(-1 - 2) + 1(1 + 2)$

$|A| = 1(0) - 1(-3) + 1(3)$

$|A| = 0 + 3 + 3 = 6$

Next, we find the adjoint of $A$. First, we find the matrix of cofactors:

$C_{11} = \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} = 1 - 1 = 0$

$C_{12} = - \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -(-1 - 2) = 3$

$C_{13} = \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1 - (-2) = 3$

$C_{21} = - \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -(-1 - 1) = 2$

$C_{22} = \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -1 - 2 = -3$

$C_{23} = - \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = -(1 - 2) = 1$

$C_{31} = \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = 1 - (-1) = 2$

$C_{32} = - \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = -(1 - 1) = 0$

$C_{33} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -1 - 1 = -2$

The matrix of cofactors is:

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

The adjoint of $A$ is the transpose of the cofactor matrix:

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

Now, we find the inverse of $A$:

$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{6} \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix}$

So, $X = A^{-1}B = \frac{1}{6} \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix} \begin{bmatrix} 6 \\ 2 \\ 1 \end{bmatrix}$

$X = \frac{1}{6} \begin{bmatrix} 0(6) + 2(2) + 2(1) \\ 3(6) + (-3)(2) + 0(1) \\ 3(6) + 1(2) + (-2)(1) \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 0 + 4 + 2 \\ 18 - 6 + 0 \\ 18 + 2 - 2 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 6 \\ 12 \\ 18 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Therefore, $x = 1, y = 2, z = 3$

Result:

$x = 1, y = 2, z = 3$

Given system of equations:

$2x - y + 3z = 9$

$x + y + z = 6$

$x - y + z = 2$

To find: Values of $x, y, z$ using Cramer's rule.

Solution:

We can write the system of equations in matrix form as $AX = B$, where

$A = \begin{bmatrix} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 9 \\ 6 \\ 2 \end{bmatrix}$

According to Cramer's rule:

$x = \frac{|A_x|}{|A|}, \$\$ y = \frac{|A_y|}{|A|}, \$\$ z = \frac{|A_z|}{|A|}$

where $A_x, A_y, A_z$ are the matrices obtained by replacing the corresponding columns of $A$ with the column matrix $B$.

First, we find the determinant of $A$:

$|A| = \begin{vmatrix} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$

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

$|A| = 2((1)(1) - (1)(-1)) + 1((1)(1) - (1)(1)) + 3((1)(-1) - (1)(1))$

$|A| = 2(1 + 1) + 1(1 - 1) + 3(-1 - 1)$

$|A| = 2(2) + 1(0) + 3(-2)$

$|A| = 4 + 0 - 6 = -2$

Now, we find $|A_x|$:

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

$|A_x| = \begin{vmatrix} 9 & -1 & 3 \\ 6 & 1 & 1 \\ 2 & -1 & 1 \end{vmatrix}$

$|A_x| = 9 \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - (-1) \begin{vmatrix} 6 & 1 \\ 2 & 1 \end{vmatrix} + 3 \begin{vmatrix} 6 & 1 \\ 2 & -1 \end{vmatrix}$

$|A_x| = 9(1 - (-1)) + 1(6 - 2) + 3(-6 - 2)$

$|A_x| = 9(2) + 1(4) + 3(-8)$

$|A_x| = 18 + 4 - 24 = -2$

$x = \frac{|A_x|}{|A|} = \frac{-2}{-2} = 1$

Now, we find $|A_y|$:

$A_y = \begin{bmatrix} 2 & 9 & 3 \\ 1 & 6 & 1 \\ 1 & 2 & 1 \end{bmatrix}$

$|A_y| = \begin{vmatrix} 2 & 9 & 3 \\ 1 & 6 & 1 \\ 1 & 2 & 1 \end{vmatrix}$

$|A_y| = 2 \begin{vmatrix} 6 & 1 \\ 2 & 1 \end{vmatrix} - 9 \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + 3 \begin{vmatrix} 1 & 6 \\ 1 & 2 \end{vmatrix}$

$|A_y| = 2(6 - 2) - 9(1 - 1) + 3(2 - 6)$

$|A_y| = 2(4) - 9(0) + 3(-4)$

$|A_y| = 8 - 0 - 12 = -4$

$y = \frac{|A_y|}{|A|} = \frac{-4}{-2} = 2$

Now, we find $|A_z|$:

$A_z = \begin{bmatrix} 2 & -1 & 9 \\ 1 & 1 & 6 \\ 1 & -1 & 2 \end{bmatrix}$

$|A_z| = \begin{vmatrix} 2 & -1 & 9 \\ 1 & 1 & 6 \\ 1 & -1 & 2 \end{vmatrix}$

$|A_z| = 2 \begin{vmatrix} 1 & 6 \\ -1 & 2 \end{vmatrix} - (-1) \begin{vmatrix} 1 & 6 \\ 1 & 2 \end{vmatrix} + 9 \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix}$

$|A_z| = 2(2 - (-6)) + 1(2 - 6) + 9(-1 - 1)$

$|A_z| = 2(8) + 1(-4) + 9(-2)$

$|A_z| = 16 - 4 - 18 = -6$

$z = \frac{|A_z|}{|A|} = \frac{-6}{-2} = 3$

Result:

$x = 1, y = 2, z = 3$

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

To prove: $A^2 - 4A + 5I = 0$ and find $A^{-1}$.

Solution:

First, we find $A^2$:

$A^2 = A \cdot A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} (1)(1) + (-1)(2) & (1)(-1) + (-1)(3) \\ (2)(1) + (3)(2) & (2)(-1) + (3)(3) \end{bmatrix} = \begin{bmatrix} 1 - 2 & -1 - 3 \\ 2 + 6 & -2 + 9 \end{bmatrix} = \begin{bmatrix} -1 & -4 \\ 8 & 7 \end{bmatrix}$

Next, we find $4A$:

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

And, $5I = 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

Now, we calculate $A^2 - 4A + 5I$:

$A^2 - 4A + 5I = \begin{bmatrix} -1 & -4 \\ 8 & 7 \end{bmatrix} - \begin{bmatrix} 4 & -4 \\ 8 & 12 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} -1 - 4 + 5 & -4 - (-4) + 0 \\ 8 - 8 + 0 & 7 - 12 + 5 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$

Thus, $A^2 - 4A + 5I = 0$

To find $A^{-1}$, we start with the equation $A^2 - 4A + 5I = 0$

Multiply both sides by $A^{-1}$:

$A^{-1}(A^2 - 4A + 5I) = A^{-1} \cdot 0$

$A^{-1}A^2 - 4A^{-1}A + 5A^{-1}I = 0$

$A - 4I + 5A^{-1} = 0$

$5A^{-1} = 4I - A$

$A^{-1} = \frac{1}{5}(4I - A) = \frac{1}{5} \left( 4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \right) = \frac{1}{5} \left( \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \right)$

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

Results:

$A^2 - 4A + 5I = 0$

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

Given: $A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$

To verify: $\text{adj}(AB) = \text{adj}(B) \text{adj}(A)$

Solution:

First, we find $AB$:

$AB = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} (3)(1) + (1)(-1) & (3)(2) + (1)(3) \\ (7)(1) + (5)(-1) & (7)(2) + (5)(3) \end{bmatrix} = \begin{bmatrix} 3 - 1 & 6 + 3 \\ 7 - 5 & 14 + 15 \end{bmatrix} = \begin{bmatrix} 2 & 9 \\ 2 & 29 \end{bmatrix}$

Now, we find $\text{adj}(AB)$:

For a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

So, $\text{adj}(AB) = \begin{bmatrix} 29 & -9 \\ -2 & 2 \end{bmatrix}$

Now, we find $\text{adj}(A)$ and $\text{adj}(B)$:

$\text{adj}(A) = \begin{bmatrix} 5 & -1 \\ -7 & 3 \end{bmatrix}$

$\text{adj}(B) = \begin{bmatrix} 3 & -2 \\ 1 & 1 \end{bmatrix}$

Now, we find $\text{adj}(B) \text{adj}(A)$:

$\text{adj}(B) \text{adj}(A) = \begin{bmatrix} 3 & -2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ -7 & 3 \end{bmatrix} = \begin{bmatrix} (3)(5) + (-2)(-7) & (3)(-1) + (-2)(3) \\ (1)(5) + (1)(-7) & (1)(-1) + (1)(3) \end{bmatrix} = \begin{bmatrix} 15 + 14 & -3 - 6 \\ 5 - 7 & -1 + 3 \end{bmatrix} = \begin{bmatrix} 29 & -9 \\ -2 & 2 \end{bmatrix}$

Thus, $\text{adj}(AB) = \text{adj}(B) \text{adj}(A) = \begin{bmatrix} 29 & -9 \\ -2 & 2 \end{bmatrix}$

Result:

Verified: $\text{adj}(AB) = \text{adj}(B) \text{adj}(A)$

Given: Determinant $\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix}$

To prove: $\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3$

Solution:

Let $D = \begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix}$

Apply $C_1 \rightarrow C_1 + C_2 + C_3$

$D = \begin{vmatrix} a+b+2c+a+b & a & b \\ c+b+c+2a+b & b+c+2a & b \\ c+a+c+a+2b & a & c+a+2b \end{vmatrix}$

$D = \begin{vmatrix} 2a+2b+2c & a & b \\ 2a+2b+2c & b+c+2a & b \\ 2a+2b+2c & a & c+a+2b \end{vmatrix}$

Take $2(a+b+c)$ common from $C_1$

$D = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix}$

Apply $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$

$D = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 0 & b+c+2a-a & b-b \\ 0 & a-a & c+a+2b-b \end{vmatrix}$

$D = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 0 & a+b+c & 0 \\ 0 & 0 & a+b+c \end{vmatrix}$

Expanding along $C_1$, we get:

$D = 2(a+b+c) \left[ 1 \begin{vmatrix} a+b+c & 0 \\ 0 & a+b+c \end{vmatrix} \right]$

$D = 2(a+b+c) [(a+b+c)(a+b+c) - 0]$

$D = 2(a+b+c)(a+b+c)^2$

$D = 2(a+b+c)^3$

Hence, $\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3$

Result:

Proved: $\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3$

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

To find: Inverse of matrix A i.e., $A^{-1}$ using elementary row operations.

Solution:

We write $A = IA$, where $I$ is the identity matrix. Then we perform elementary row operations on $A$ and $I$ simultaneously until $A$ becomes $I$. The matrix that $I$ becomes is $A^{-1}$.

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

Apply $R_1 \rightarrow \frac{1}{2} R_1$

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

Apply $R_2 \rightarrow R_2 - 4R_1$ and $R_3 \rightarrow R_3 + R_1$

$\begin{bmatrix} 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & -1 & -6 \\ 0 & \frac{3}{2} & \frac{7}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ -2 & 1 & 0 \\ \frac{1}{2} & 0 & 1 \end{bmatrix} A$

Apply $R_2 \rightarrow -R_2$

$\begin{bmatrix} 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 6 \\ 0 & \frac{3}{2} & \frac{7}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 2 & -1 & 0 \\ \frac{1}{2} & 0 & 1 \end{bmatrix} A$

Apply $R_1 \rightarrow R_1 - \frac{1}{2} R_2$ and $R_3 \rightarrow R_3 - \frac{3}{2} R_2$

$\begin{bmatrix} 1 & 0 & -\frac{3}{2} \\ 0 & 1 & 6 \\ 0 & 0 & - \frac{11}{2} \end{bmatrix} = \begin{bmatrix} - \frac{1}{2} & \frac{1}{2} & 0 \\ 2 & -1 & 0 \\ -\frac{5}{2} & \frac{3}{2} & 1 \end{bmatrix} A$

Apply $R_3 \rightarrow -\frac{2}{11} R_3$

$\begin{bmatrix} 1 & 0 & -\frac{3}{2} \\ 0 & 1 & 6 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} - \frac{1}{2} & \frac{1}{2} & 0 \\ 2 & -1 & 0 \\ \frac{5}{11} & -\frac{3}{11} & -\frac{2}{11} \end{bmatrix} A$

Apply $R_1 \rightarrow R_1 + \frac{3}{2} R_3$ and $R_2 \rightarrow R_2 - 6 R_3$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} - \frac{1}{2} + \frac{3}{2} \cdot \frac{5}{11} & \frac{1}{2} + \frac{3}{2} \cdot (-\frac{3}{11}) & 0 + \frac{3}{2} \cdot (-\frac{2}{11}) \\ 2 - 6 \cdot \frac{5}{11} & -1 - 6 \cdot (-\frac{3}{11}) & 0 - 6 \cdot (-\frac{2}{11}) \\ \frac{5}{11} & -\frac{3}{11} & -\frac{2}{11} \end{bmatrix} A$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} - \frac{1}{2} + \frac{15}{22} & \frac{1}{2} - \frac{9}{22} & -\frac{3}{11} \\ 2 - \frac{30}{11} & -1 + \frac{18}{11} & \frac{12}{11} \\ \frac{5}{11} & -\frac{3}{11} & -\frac{2}{11} \end{bmatrix} A$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{-11 + 15}{22} & \frac{11 - 9}{22} & -\frac{3}{11} \\ \frac{22 - 30}{11} & \frac{-11 + 18}{11} & \frac{12}{11} \\ \frac{5}{11} & -\frac{3}{11} & -\frac{2}{11} \end{bmatrix} A$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{4}{22} & \frac{2}{22} & -\frac{3}{11} \\ -\frac{8}{11} & \frac{7}{11} & \frac{12}{11} \\ \frac{5}{11} & -\frac{3}{11} & -\frac{2}{11} \end{bmatrix} A$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{11} & \frac{1}{11} & -\frac{3}{11} \\ -\frac{8}{11} & \frac{7}{11} & \frac{12}{11} \\ \frac{5}{11} & -\frac{3}{11} & -\frac{2}{11} \end{bmatrix} A$

Thus, $A^{-1} = \begin{bmatrix} \frac{2}{11} & \frac{1}{11} & -\frac{3}{11} \\ -\frac{8}{11} & \frac{7}{11} & \frac{12}{11} \\ \frac{5}{11} & -\frac{3}{11} & -\frac{2}{11} \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 2 & 1 & -3 \\ -8 & 7 & 12 \\ 5 & -3 & -2 \end{bmatrix}$

Result:

$A^{-1} = \frac{1}{11} \begin{bmatrix} 2 & 1 & -3 \\ -8 & 7 & 12 \\ 5 & -3 & -2 \end{bmatrix}$

Given system of equations:

$x - y + z = 4$

$2x + y - 3z = 0$

$x + y + z = 2$

To find: Values of $x, y, z$ using the matrix inverse method.

Solution:

We can write the system of equations in matrix form as $AX = B$, where

$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$

To solve for $X$, we can use the formula $X = A^{-1}B$.

First, we find the determinant of $A$:

$|A| = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{vmatrix}$

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

$|A| = 1((1)(1) - (-3)(1)) + 1((2)(1) - (-3)(1)) + 1((2)(1) - (1)(1))$

$|A| = 1(1 + 3) + 1(2 + 3) + 1(2 - 1)$

$|A| = 1(4) + 1(5) + 1(1)$

$|A| = 4 + 5 + 1 = 10$

Next, we find the adjoint of $A$. First, we find the matrix of cofactors:

$C_{11} = \begin{vmatrix} 1 & -3 \\ 1 & 1 \end{vmatrix} = 1 - (-3) = 4$

$C_{12} = - \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = -(2 - (-3)) = -5$

$C_{13} = \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = 2 - 1 = 1$

$C_{21} = - \begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix} = -(-1 - 1) = 2$

$C_{22} = \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 1 - 1 = 0$

$C_{23} = - \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = -(1 - (-1)) = -2$

$C_{31} = \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} = 3 - 1 = 2$

$C_{32} = - \begin{vmatrix} 1 & 1 \\ 2 & -3 \end{vmatrix} = -(-3 - 2) = 5$

$C_{33} = \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1 - (-2) = 3$

The matrix of cofactors is:

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

The adjoint of $A$ is the transpose of the cofactor matrix:

$adj(A) = C^T = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$

Now, we find the inverse of $A$:

$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$

So, $X = A^{-1}B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$

$X = \frac{1}{10} \begin{bmatrix} 4(4) + 2(0) + 2(2) \\ -5(4) + 0(0) + 5(2) \\ 1(4) + (-2)(0) + 3(2) \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$

Therefore, $x = 2, y = -1, z = 1$

Result:

$x = 2, y = -1, z = 1$

Given system of equations:

$x + 2y + 3z = 6$

$2x + 4y + z = 7$

$3x + 2y + 9z = 14$

To find: Values of $x, y, z$ using Cramer's rule.

Solution:

We can write the system of equations in matrix form as $AX = B$, where

$A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 1 \\ 3 & 2 & 9 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 6 \\ 7 \\ 14 \end{bmatrix}$

According to Cramer's rule:

$x = \frac{|A_x|}{|A|}, \$\$ y = \frac{|A_y|}{|A|}, \$\$ z = \frac{|A_z|}{|A|}$

where $A_x, A_y, A_z$ are the matrices obtained by replacing the corresponding columns of $A$ with the column matrix $B$.

First, we find the determinant of $A$:

$|A| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 1 \\ 3 & 2 & 9 \end{vmatrix}$

$|A| = 1 \begin{vmatrix} 4 & 1 \\ 2 & 9 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ 3 & 9 \end{vmatrix} + 3 \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix}$

$|A| = 1((4)(9) - (1)(2)) - 2((2)(9) - (1)(3)) + 3((2)(2) - (4)(3))$

$|A| = 1(36 - 2) - 2(18 - 3) + 3(4 - 12)$

$|A| = 1(34) - 2(15) + 3(-8)$

$|A| = 34 - 30 - 24 = -20$

Now, we find $|A_x|$:

$A_x = \begin{bmatrix} 6 & 2 & 3 \\ 7 & 4 & 1 \\ 14 & 2 & 9 \end{bmatrix}$

$|A_x| = \begin{vmatrix} 6 & 2 & 3 \\ 7 & 4 & 1 \\ 14 & 2 & 9 \end{vmatrix}$

$|A_x| = 6 \begin{vmatrix} 4 & 1 \\ 2 & 9 \end{vmatrix} - 2 \begin{vmatrix} 7 & 1 \\ 14 & 9 \end{vmatrix} + 3 \begin{vmatrix} 7 & 4 \\ 14 & 2 \end{vmatrix}$

$|A_x| = 6(36 - 2) - 2(63 - 14) + 3(14 - 56)$

$|A_x| = 6(34) - 2(49) + 3(-42)$

$|A_x| = 204 - 98 - 126 = -20$

$x = \frac{|A_x|}{|A|} = \frac{-20}{-20} = 1$

Now, we find $|A_y|$:

$A_y = \begin{bmatrix} 1 & 6 & 3 \\ 2 & 7 & 1 \\ 3 & 14 & 9 \end{bmatrix}$

$|A_y| = \begin{vmatrix} 1 & 6 & 3 \\ 2 & 7 & 1 \\ 3 & 14 & 9 \end{vmatrix}$

$|A_y| = 1 \begin{vmatrix} 7 & 1 \\ 14 & 9 \end{vmatrix} - 6 \begin{vmatrix} 2 & 1 \\ 3 & 9 \end{vmatrix} + 3 \begin{vmatrix} 2 & 7 \\ 3 & 14 \end{vmatrix}$

$|A_y| = 1(63 - 14) - 6(18 - 3) + 3(28 - 21)$

$|A_y| = 1(49) - 6(15) + 3(7)$

$|A_y| = 49 - 90 + 21 = -20$

$y = \frac{|A_y|}{|A|} = \frac{-20}{-20} = 1$

Now, we find $|A_z|$:

$A_z = \begin{bmatrix} 1 & 2 & 6 \\ 2 & 4 & 7 \\ 3 & 2 & 14 \end{bmatrix}$

$|A_z| = \begin{vmatrix} 1 & 2 & 6 \\ 2 & 4 & 7 \\ 3 & 2 & 14 \end{vmatrix}$

$|A_z| = 1 \begin{vmatrix} 4 & 7 \\ 2 & 14 \end{vmatrix} - 2 \begin{vmatrix} 2 & 7 \\ 3 & 14 \end{vmatrix} + 6 \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix}$

$|A_z| = 1(56 - 14) - 2(28 - 21) + 6(4 - 12)$

$|A_z| = 1(42) - 2(7) + 6(-8)$

$|A_z| = 42 - 14 - 48 = -20$

$z = \frac{|A_z|}{|A|} = \frac{-20}{-20} = 1$

Result:

$x = 1, y = 1, z = 1$

Given: $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$ and the system of equations: $2x - 3y + 5z = 11$, $3x + 2y - 4z = -5$, $x + y - 2z = -3$.

To find: $A^{-1}$ and the solution of the system of equations using $A^{-1}$.

Solution:

First, we find the determinant of $A$:

$|A| = \begin{vmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{vmatrix}$

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

$|A| = 2((2)(-2) - (-4)(1)) + 3((3)(-2) - (-4)(1)) + 5((3)(1) - (2)(1))$

$|A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2)$

$|A| = 2(0) + 3(-2) + 5(1)$

$|A| = 0 - 6 + 5 = -1$

Next, we find the adjoint of $A$. First, we find the matrix of cofactors:

$C_{11} = \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} = -4 - (-4) = 0$

$C_{12} = - \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} = -(-6 - (-4)) = -(-2) = 2$

$C_{13} = \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} = 3 - 2 = 1$

$C_{21} = - \begin{vmatrix} -3 & 5 \\ 1 & -2 \end{vmatrix} = -(6 - 5) = -1$

$C_{22} = \begin{vmatrix} 2 & 5 \\ 1 & -2 \end{vmatrix} = -4 - 5 = -9$

$C_{23} = - \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = -(2 - (-3)) = -5$

$C_{31} = \begin{vmatrix} -3 & 5 \\ 2 & -4 \end{vmatrix} = 12 - 10 = 2$

$C_{32} = - \begin{vmatrix} 2 & 5 \\ 3 & -4 \end{vmatrix} = -(-8 - 15) = 23$

$C_{33} = \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix} = 4 - (-9) = 13$

The matrix of cofactors is:

$C = \begin{bmatrix} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{bmatrix}$

The adjoint of $A$ is the transpose of the cofactor matrix:

$adj(A) = C^T = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}$

Now, we find the inverse of $A$:

$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$

To solve the system of equations, we can write it in matrix form as $AX = B$, where

$A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

To solve for $X$, we can use the formula $X = A^{-1}B$.

$X = A^{-1}B = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

$X = \begin{bmatrix} 0(11) + 1(-5) + (-2)(-3) \\ -2(11) + 9(-5) + (-23)(-3) \\ -1(11) + 5(-5) + (-13)(-3) \end{bmatrix} = \begin{bmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Therefore, $x = 1, y = 2, z = 3$

Results:

$A^{-1} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$

$x = 1, y = 2, z = 3$

Given: Determinant $\begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix}$

To prove: $\begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x)$

Solution:

Let $D = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix}$

Take $x$, $y$, and $z$ common from $C_1$, $C_2$, and $C_3$ respectively:

$D = xyz \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix}$

Apply $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$

$D = xyz \begin{vmatrix} 1 & 0 & 0 \\ x & y-x & z-x \\ x^2 & y^2-x^2 & z^2-x^2 \end{vmatrix}$

$D = xyz \begin{vmatrix} 1 & 0 & 0 \\ x & y-x & z-x \\ x^2 & (y-x)(y+x) & (z-x)(z+x) \end{vmatrix}$

Take $(y-x)$ and $(z-x)$ common from $C_2$ and $C_3$ respectively:

$D = xyz (y-x)(z-x) \begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ x^2 & y+x & z+x \end{vmatrix}$

Expanding along $R_1$:

$D = xyz (y-x)(z-x) \left[ 1 \begin{vmatrix} 1 & 1 \\ y+x & z+x \end{vmatrix} \right]$

$D = xyz (y-x)(z-x) [ (1)(z+x) - (1)(y+x) ]$

$D = xyz (y-x)(z-x) [ z+x - y - x ]$

$D = xyz (y-x)(z-x) (z-y)$

$D = xyz (x-y)(y-z)(z-x)$

Hence, $\begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x)$

Result:

Proved: $\begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x)$

Given: $A$ is a square matrix such that $A^2 = A$.

To prove: $(I+A)^3 = I + 7A$

Solution:

We have $(I+A)^3 = (I+A)(I+A)(I+A)$

First, let's expand $(I+A)(I+A)$:

$(I+A)(I+A) = I^2 + IA + AI + A^2 = I + A + A + A^2 = I + 2A + A^2$

Since $A^2 = A$, we can substitute $A^2$ with $A$:

$(I+A)(I+A) = I + 2A + A = I + 3A$

Now, let's multiply $(I+3A)(I+A)$:

$(I+3A)(I+A) = I^2 + IA + 3AI + 3A^2 = I + A + 3A + 3A^2 = I + 4A + 3A^2$

Since $A^2 = A$, we can substitute $A^2$ with $A$:

$(I+3A)(I+A) = I + 4A + 3A = I + 7A$

Therefore, $(I+A)^3 = I + 7A$

Result:

Proved: $(I+A)^3 = I + 7A$

Given: $A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{bmatrix}$

To find: Inverse of matrix A i.e., $A^{-1}$ using elementary column operations.

Solution:

We write $A = AI$, where $I$ is the identity matrix. Then we perform elementary column operations on $A$ and $I$ simultaneously until $A$ becomes $I$. The matrix that $I$ becomes is $A^{-1}$.

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

Apply $C_2 \rightarrow C_2 - 2C_1$

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

Apply $C_3 \rightarrow C_3 - C_2$

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

Apply $C_3 \rightarrow \frac{1}{5}C_3$

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

Apply $C_1 \rightarrow C_1 - 2C_3$ and $C_2 \rightarrow C_2 + 4C_3$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = A \begin{bmatrix} 1-\frac{4}{5} & -2+\frac{8}{5} & \frac{2}{5} \\ 0+\frac{2}{5} & 1-\frac{4}{5} & -\frac{1}{5} \\ 0-\frac{2}{5} & 0+\frac{4}{5} & \frac{1}{5} \end{bmatrix}$

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

Thus, $A^{-1} = \begin{bmatrix} \frac{1}{5} & -\frac{2}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{1}{5} & -\frac{1}{5} \\ -\frac{2}{5} & \frac{4}{5} & \frac{1}{5} \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 1 & -2 & 2 \\ 2 & 1 & -1 \\ -2 & 4 & 1 \end{bmatrix}$

Result:

$A^{-1} = \frac{1}{5} \begin{bmatrix} 1 & -2 & 2 \\ 2 & 1 & -1 \\ -2 & 4 & 1 \end{bmatrix}$

Let $u = \frac{1}{x}$, $v = \frac{1}{y}$, and $w = \frac{1}{z}$. The system of equations becomes:

$2u - 3v + 5w = -2$

$u + 2v + w = 5$

$3u + v - 4w = 8$

We can represent this system as a matrix equation $AX = B$, where:

$A = \begin{bmatrix} 2 & -3 & 5 \\ 1 & 2 & 1 \\ 3 & 1 & -4 \end{bmatrix}$, $X = \begin{bmatrix} u \\ v \\ w \end{bmatrix}$, and $B = \begin{bmatrix} -2 \\ 5 \\ 8 \end{bmatrix}$

To solve for $X$, we need to find the inverse of matrix $A$, denoted as $A^{-1}$, and then calculate $X = A^{-1}B$.

First, we find the determinant of $A$:

$|A| = 2(2(-4) - 1(1)) - (-3)(1(-4) - 1(3)) + 5(1(1) - 2(3)) = 2(-8-1) + 3(-4-3) + 5(1-6) = 2(-9) + 3(-7) + 5(-5) = -18 - 21 - 25 = -64$

Now, we find the adjugate of $A$, which is the transpose of the cofactor matrix:

$C_{11} = (2)(-4) - (1)(1) = -8 - 1 = -9$

$C_{12} = -(1(-4) - 1(3)) = -(-4-3) = 7$

$C_{13} = (1)(1) - (2)(3) = 1 - 6 = -5$

$C_{21} = -((-3)(-4) - (5)(1)) = -(12 - 5) = -7$

$C_{22} = (2)(-4) - (5)(3) = -8 - 15 = -23$

$C_{23} = -(2(1) - (-3)(3)) = -(2 + 9) = -11$

$C_{31} = (-3)(1) - (2)(5) = -3 - 10 = -13$

$C_{32} = -(2(1) - (1)(5)) = -(2 - 5) = 3$

$C_{33} = (2)(2) - (-3)(1) = 4 + 3 = 7$

The cofactor matrix is:

$\begin{bmatrix} -9 & 7 & -5 \\ -7 & -23 & -11 \\ -13 & 3 & 7 \end{bmatrix}$

The adjugate matrix is the transpose of the cofactor matrix:

$adj(A) = \begin{bmatrix} -9 & -7 & -13 \\ 7 & -23 & 3 \\ -5 & -11 & 7 \end{bmatrix}$

Now we can find $A^{-1} = \frac{1}{|A|}adj(A)$:

$A^{-1} = \frac{1}{-64}\begin{bmatrix} -9 & -7 & -13 \\ 7 & -23 & 3 \\ -5 & -11 & 7 \end{bmatrix} = \begin{bmatrix} \frac{9}{64} & \frac{7}{64} & \frac{13}{64} \\ -\frac{7}{64} & \frac{23}{64} & -\frac{3}{64} \\ \frac{5}{64} & \frac{11}{64} & -\frac{7}{64} \end{bmatrix}$

Now we can find $X = A^{-1}B$:

$X = \begin{bmatrix} \frac{9}{64} & \frac{7}{64} & \frac{13}{64} \\ -\frac{7}{64} & \frac{23}{64} & -\frac{3}{64} \\ \frac{5}{64} & \frac{11}{64} & -\frac{7}{64} \end{bmatrix} \begin{bmatrix} -2 \\ 5 \\ 8 \end{bmatrix} = \begin{bmatrix} \frac{-18 + 35 + 104}{64} \\ \frac{14 + 115 - 24}{64} \\ \frac{-10 + 55 - 56}{64} \end{bmatrix} = \begin{bmatrix} \frac{121}{64} \\ \frac{105}{64} \\ \frac{-11}{64} \end{bmatrix}$

So, $u = \frac{121}{64}$, $v = \frac{105}{64}$, and $w = -\frac{11}{64}$.

Therefore, $x = \frac{64}{121}$, $y = \frac{64}{105}$, and $z = -\frac{64}{11}$.

Solution: $x = \frac{64}{121}$, $y = \frac{64}{105}$, $z = -\frac{64}{11}$

Let $\Delta = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix}$

Taking $a$ common from $R_1$, $b$ from $R_2$ and $c$ from $R_3$, we get:

$\Delta = abc \begin{vmatrix} \frac{1+a}{a} & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1+b}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1+c}{c} \end{vmatrix}$

$\Delta = abc \begin{vmatrix} 1 + \frac{1}{a} & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & 1 + \frac{1}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & 1 + \frac{1}{c} \end{vmatrix}$

Applying $C_1 \rightarrow C_1 - C_2$ and $C_2 \rightarrow C_2 - C_3$

$\Delta = abc \begin{vmatrix} 1 & 0 & \frac{1}{a} \\ -\frac{1}{b} & 1 & \frac{1}{b} \\ 0 & -1 & 1 + \frac{1}{c} \end{vmatrix}$

Expanding along $R_1$, we get:

$\Delta = abc \left[ 1 \begin{vmatrix} 1 & \frac{1}{b} \\ -1 & 1 + \frac{1}{c} \end{vmatrix} - 0 + \frac{1}{a} \begin{vmatrix} -\frac{1}{b} & 1 \\ 0 & -1 \end{vmatrix} \right]$

$\Delta = abc \left[ 1(1 + \frac{1}{c} + \frac{1}{b}) + \frac{1}{a} (\frac{1}{b}) \right]$

$\Delta = abc \left[ 1 + \frac{1}{b} + \frac{1}{c} + \frac{1}{ab} \right]$

Another method is as follows

Applying $C_1 \rightarrow C_1 + C_2 + C_3$

$\Delta = abc \begin{vmatrix} 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & \frac{1}{a} & \frac{1}{a} \\ 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{b} & \frac{1}{b} \\ 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & \frac{1}{c} & 1+\frac{1}{c} \end{vmatrix}$

Taking $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ common from $C_1$

$\Delta = abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\begin{vmatrix} 1 & \frac{1}{a} & \frac{1}{a} \\ 1 & 1+\frac{1}{b} & \frac{1}{b} \\ 1 & \frac{1}{c} & 1+\frac{1}{c} \end{vmatrix}$

Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$

$\Delta = abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\begin{vmatrix} 1 & \frac{1}{a} & \frac{1}{a} \\ 0 & 1+\frac{1}{b}-\frac{1}{a} & \frac{1}{b}-\frac{1}{a} \\ 0 & \frac{1}{c}-\frac{1}{a} & 1+\frac{1}{c}-\frac{1}{a} \end{vmatrix}$

Expanding along $C_1$, we get

$\Delta = abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})[(1+\frac{1}{b}-\frac{1}{a})(1+\frac{1}{c}-\frac{1}{a})-(\frac{1}{c}-\frac{1}{a})(\frac{1}{b}-\frac{1}{a})]$

$\Delta = abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})[1+\frac{1}{c}-\frac{1}{a}+\frac{1}{b}+\frac{1}{bc}-\frac{1}{ab}-\frac{1}{a}-\frac{1}{ac}+\frac{1}{a^2}-(\frac{1}{bc}-\frac{1}{ac}-\frac{1}{ab}+\frac{1}{a^2})]$

$\Delta = abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})[1+\frac{1}{c}-\frac{1}{a}+\frac{1}{b}+\frac{1}{bc}-\frac{1}{ab}-\frac{1}{a}-\frac{1}{ac}+\frac{1}{a^2}-\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}-\frac{1}{a^2}]$

$\Delta = abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})[1+\frac{1}{b}+\frac{1}{c}-\frac{2}{a}]$

Final solution:

$\Delta = abc(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

Hence Proved

The given system of equations is:

$x + y + 2z = 4$

$2x - y + 3z = 9$

$3x + y - z = 2$

We can write the system of equations in matrix form as $AX = B$, where:

$A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & -1 & 3 \\ 3 & 1 & -1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} 4 \\ 9 \\ 2 \end{bmatrix}$

Let's solve this using the matrix inverse method.

First, we find the determinant of matrix A:

$|A| = 1 \begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} - 1 \begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} + 2 \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} = 1(1 - 3) - 1(-2 - 9) + 2(2 + 3) = -2 + 11 + 10 = 19$

Next, we find the adjugate (transpose of the cofactor matrix) of matrix A:

$C_{11} = \begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} = 1 - 3 = -2$

$C_{12} = -\begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} = -(-2 - 9) = 11$

$C_{13} = \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} = 2 + 3 = 5$

$C_{21} = -\begin{vmatrix} 1 & 2 \\ 1 & -1 \end{vmatrix} = -(-1 - 2) = 3$

$C_{22} = \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = -1 - 6 = -7$

$C_{23} = -\begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} = -(1 - 3) = 2$

$C_{31} = \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} = 3 + 2 = 5$

$C_{32} = -\begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = -(3 - 4) = 1$

$C_{33} = \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -1 - 2 = -3$

The cofactor matrix is:

$\begin{bmatrix} -2 & 11 & 5 \\ 3 & -7 & 2 \\ 5 & 1 & -3 \end{bmatrix}$

The adjugate matrix is the transpose of the cofactor matrix:

$adj(A) = \begin{bmatrix} -2 & 3 & 5 \\ 11 & -7 & 1 \\ 5 & 2 & -3 \end{bmatrix}$

Now we find $A^{-1} = \frac{1}{|A|} adj(A)$:

$A^{-1} = \frac{1}{19} \begin{bmatrix} -2 & 3 & 5 \\ 11 & -7 & 1 \\ 5 & 2 & -3 \end{bmatrix} = \begin{bmatrix} -\frac{2}{19} & \frac{3}{19} & \frac{5}{19} \\ \frac{11}{19} & -\frac{7}{19} & \frac{1}{19} \\ \frac{5}{19} & \frac{2}{19} & -\frac{3}{19} \end{bmatrix}$

Now we can find $X = A^{-1}B$:

$X = \begin{bmatrix} -\frac{2}{19} & \frac{3}{19} & \frac{5}{19} \\ \frac{11}{19} & -\frac{7}{19} & \frac{1}{19} \\ \frac{5}{19} & \frac{2}{19} & -\frac{3}{19} \end{bmatrix} \begin{bmatrix} 4 \\ 9 \\ 2 \end{bmatrix} = \begin{bmatrix} -\frac{8}{19} + \frac{27}{19} + \frac{10}{19} \\ \frac{44}{19} - \frac{63}{19} + \frac{2}{19} \\ \frac{20}{19} + \frac{18}{19} - \frac{6}{19} \end{bmatrix} = \begin{bmatrix} \frac{29}{19} \\ -\frac{17}{19} \\ \frac{32}{19} \end{bmatrix}$

So, $x = \frac{29}{19}$, $y = -\frac{17}{19}$, and $z = \frac{32}{19}$.

Solution: $x = \frac{29}{19}$, $y = -\frac{17}{19}$, $z = \frac{32}{19}$

Given $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$, we need to show that $A^3 - 6A^2 + 7A + 2I = 0$.

First, let's find $A^2$:

$A^2 = A \cdot A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}$

Now, let's find $A^3$:

$A^3 = A^2 \cdot A = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39 \end{bmatrix} = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}$

Now, we can substitute these values into the given equation:

$A^3 - 6A^2 + 7A + 2I = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - 6 \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

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

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

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

Therefore, $A^3 - 6A^2 + 7A + 2I = 0$.

Hence Proved

We are given the matrix equation $XA = B$, where $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 3 \\ 1 & 4 \end{bmatrix}$. We want to solve for the matrix $X$.

To find $X$, we can multiply both sides of the equation by the inverse of $A$, denoted as $A^{-1}$, on the right:

$XA A^{-1} = B A^{-1}$

$XI = B A^{-1}$

$X = B A^{-1}$

First, we need to find the inverse of matrix A:

$|A| = (2)(2) - (1)(3) = 4 - 3 = 1$

$A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$

Now, we can find $X = B A^{-1}$:

$X = \begin{bmatrix} 5 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} (5)(2) + (3)(-3) & (5)(-1) + (3)(2) \\ (1)(2) + (4)(-3) & (1)(-1) + (4)(2) \end{bmatrix} = \begin{bmatrix} 10 - 9 & -5 + 6 \\ 2 - 12 & -1 + 8 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -10 & 7 \end{bmatrix}$

Therefore, the solution for $X$ is:

$X = \begin{bmatrix} 1 & 1 \\ -10 & 7 \end{bmatrix}$

Given matrix $A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$, we need to verify that $A(\text{adj} A) = (\text{adj} A)A = |A|I$.

First, let's find the determinant of A:

$|A| = (2)(3) - (-1)(1) = 6 + 1 = 7$

Now, let's find the adjugate of A:

The cofactor matrix is:

$C_{11} = 3$, $C_{12} = -1$, $C_{21} = -(-1) = 1$, $C_{22} = 2$

The adjugate matrix is the transpose of the cofactor matrix:

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

Now, let's find $A(\text{adj} A)$:

$A(\text{adj} A) = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} (2)(3) + (-1)(-1) & (2)(1) + (-1)(2) \\ (1)(3) + (3)(-1) & (1)(1) + (3)(2) \end{bmatrix} = \begin{bmatrix} 6 + 1 & 2 - 2 \\ 3 - 3 & 1 + 6 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Next, let's find $(\text{adj} A)A$:

$(\text{adj} A)A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} (3)(2) + (1)(1) & (3)(-1) + (1)(3) \\ (-1)(2) + (2)(1) & (-1)(-1) + (2)(3) \end{bmatrix} = \begin{bmatrix} 6 + 1 & -3 + 3 \\ -2 + 2 & 1 + 6 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Finally, let's find $|A|I$:

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

Since $A(\text{adj} A) = (\text{adj} A)A = |A|I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$, the given condition is verified.

Hence Verified

Let $\Delta = \begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

Applying $C_1 \rightarrow C_1 - C_2$ and $C_2 \rightarrow C_2 - C_3$, we have

$\Delta = \begin{vmatrix} a-3b-c & 0 & 2a \\ 2b - b +c + a & b -3c -a & 2b \\ 0 & 2c-c+a+b & c-a-b \end{vmatrix} = \begin{vmatrix} a-3b-c & 0 & 2a \\ a + b + c & b -3c -a & 2b \\ 0 & a+b+c & c-a-b \end{vmatrix}$

Applying $R_1 \rightarrow R_1 + R_2 + R_3$

$\Delta = \begin{vmatrix} a-b-c+2b+2c & 2a+b-c-a+2c & 2a+2b+c-a-b \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

$\Delta = \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

Taking $(a+b+c)$ common from $R_1$

$\Delta = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$

$\Delta = (a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & b-c-a-2b & 0 \\ 2c & 0 & c-a-b-2c \end{vmatrix}$

$\Delta = (a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & -a-b-c & 0 \\ 2c & 0 & -a-b-c \end{vmatrix}$

Expanding along $R_1$

$\Delta = (a+b+c)[1\{(-a-b-c)(-a-b-c) - 0\}]$

$\Delta = (a+b+c)(a+b+c)^2$

$\Delta = (a+b+c)^3$

Hence Proved

Given system of equations is:

$x - y + z = 3$

$2x + y - z = 2$

$-x - 2y + 2z = 1$

We can write this in matrix form as $AX = B$, where

$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$

First, let's find the determinant of A:

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

Since $|A| = 0$, we need to find the adjoint of A and check if $(\text{adj } A)B = 0$.

The cofactor matrix is:

$C_{11} = \begin{vmatrix} 1 & -1 \\ -2 & 2 \end{vmatrix} = 2 - 2 = 0$

$C_{12} = -\begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} = -(4 - 1) = -3$

$C_{13} = \begin{vmatrix} 2 & 1 \\ -1 & -2 \end{vmatrix} = -4 + 1 = -3$

$C_{21} = -\begin{vmatrix} -1 & 1 \\ -2 & 2 \end{vmatrix} = -(-2 + 2) = 0$

$C_{22} = \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} = 2 + 1 = 3$

$C_{23} = -\begin{vmatrix} 1 & -1 \\ -1 & -2 \end{vmatrix} = -(-2 - 1) = 3$

$C_{31} = \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} = 1 - 1 = 0$

$C_{32} = -\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -(-1 - 2) = 3$

$C_{33} = \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1 + 2 = 3$

The cofactor matrix is:

$\begin{bmatrix} 0 & -3 & -3 \\ 0 & 3 & 3 \\ 0 & 3 & 3 \end{bmatrix}$

The adjugate matrix is the transpose of the cofactor matrix:

$\text{adj } A = \begin{bmatrix} 0 & 0 & 0 \\ -3 & 3 & 3 \\ -3 & 3 & 3 \end{bmatrix}$

Now, we find $(\text{adj } A)B$:

$(\text{adj } A)B = \begin{bmatrix} 0 & 0 & 0 \\ -3 & 3 & 3 \\ -3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0(3) + 0(2) + 0(1) \\ -3(3) + 3(2) + 3(1) \\ -3(3) + 3(2) + 3(1) \end{bmatrix} = \begin{bmatrix} 0 \\ -9 + 6 + 3 \\ -9 + 6 + 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Since $|A| = 0$ and $(\text{adj } A)B = 0$, the system is consistent and has infinitely many solutions.

To solve the system, we can rewrite the first two equations:

$x - y + z = 3 \implies x = 3 + y - z$

$2x + y - z = 2$

Substituting $x = 3 + y - z$ into the second equation:

$2(3 + y - z) + y - z = 2$

$6 + 2y - 2z + y - z = 2$

$3y - 3z = -4$

$y = z - \frac{4}{3}$

Now substituting $y$ back into the expression for $x$

$x = 3 + z - \frac{4}{3} - z = \frac{5}{3}$

Let $z = k$, where $k$ is any real number. Then, $x = \frac{5}{3}$ and $y = k - \frac{4}{3}$.

The general solution is $x = \frac{5}{3}$, $y = k - \frac{4}{3}$, $z = k$, where $k \in \mathbb{R}$.

Given matrix $A = \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix}$, we want to find its inverse using the adjoint method.

First, let's find the determinant of A:

$|A| = 1 \begin{vmatrix} 0 & -5 \\ 5 & 0 \end{vmatrix} - 3 \begin{vmatrix} -3 & -5 \\ 2 & 0 \end{vmatrix} + (-2) \begin{vmatrix} -3 & 0 \\ 2 & 5 \end{vmatrix} = 1(0 + 25) - 3(0 + 10) - 2(-15 - 0) = 25 - 30 + 30 = 25$

Next, we find the adjugate (transpose of the cofactor matrix) of matrix A:

$C_{11} = \begin{vmatrix} 0 & -5 \\ 5 & 0 \end{vmatrix} = 0 - (-25) = 25$

$C_{12} = -\begin{vmatrix} -3 & -5 \\ 2 & 0 \end{vmatrix} = -(0 - (-10)) = -10$

$C_{13} = \begin{vmatrix} -3 & 0 \\ 2 & 5 \end{vmatrix} = -15 - 0 = -15$

$C_{21} = -\begin{vmatrix} 3 & -2 \\ 5 & 0 \end{vmatrix} = -(0 - (-10)) = -10$

$C_{22} = \begin{vmatrix} 1 & -2 \\ 2 & 0 \end{vmatrix} = 0 - (-4) = 4$

$C_{23} = -\begin{vmatrix} 1 & 3 \\ 2 & 5 \end{vmatrix} = -(5 - 6) = 1$

$C_{31} = \begin{vmatrix} 3 & -2 \\ 0 & -5 \end{vmatrix} = -15 - 0 = -15$

$C_{32} = -\begin{vmatrix} 1 & -2 \\ -3 & -5 \end{vmatrix} = -(-5 - 6) = 11$

$C_{33} = \begin{vmatrix} 1 & 3 \\ -3 & 0 \end{vmatrix} = 0 - (-9) = 9$

The cofactor matrix is:

$\begin{bmatrix} 25 & -10 & -15 \\ -10 & 4 & 1 \\ -15 & 11 & 9 \end{bmatrix}$

The adjugate matrix is the transpose of the cofactor matrix:

$\text{adj } A = \begin{bmatrix} 25 & -10 & -15 \\ -10 & 4 & 11 \\ -15 & 1 & 9 \end{bmatrix}$

Now we find $A^{-1} = \frac{1}{|A|} adj(A)$:

$A^{-1} = \frac{1}{25} \begin{bmatrix} 25 & -10 & -15 \\ -10 & 4 & 11 \\ -15 & 1 & 9 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{bmatrix}$

Therefore, the inverse of A is:

$A^{-1} = \begin{bmatrix} 1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{bmatrix}$

Given equations:

$x + y = 5 \quad$ …(i)

$y + z = 3 \quad$ …(ii)

$x + z = 4 \quad$ …(iii)

Step 1: Represent the system in matrix form

We rewrite the system in the form $AX = B$.

The equations can be written as:

$\begin{aligned} x + y + 0z &= 5 \\ 0x + y + z &= 3 \\ x + 0y + z &= 4 \end{aligned}$

So,

$A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$, \quad $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, \quad $B = \begin{bmatrix} 5 \\ 3 \\ 4 \end{bmatrix}$

Step 2: Use inverse method to solve

We solve using $X = A^{-1}B$

First, compute $|A|$:

$|A| = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix}$ $= 1 \cdot (1 \cdot 1 - 1 \cdot 0) - 1 \cdot (0 \cdot 1 - 1 \cdot 1) + 0$ $= 1(1) - 1(-1) + 0 = 2$

Step 3: Find adjoint of A

We find cofactors and then transpose to get adjoint.

$\text{adj}(A) = \begin{bmatrix} \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} & -\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} & \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} \\ -\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} & \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} & -\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \\ \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} & -\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} & \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} \end{bmatrix}$

$= \begin{bmatrix} 1 & -(-1) & -1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & -1 & 1 \end{bmatrix}$

Step 4: Find inverse of A

$A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)$

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

Step 5: Multiply $A^{-1}$ with $B$

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

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

Final Answer:

$x = 3$, $y = 2$, $z = -2$

Theorem: For any two square matrices $A$ and $B$ of the same order, the determinant of the product is equal to the product of the determinants.

That is, $|AB| = |A||B|$

Proof:

The determinant of a product of two matrices equals the product of their determinants. This is a fundamental property of determinants. It holds true for all square matrices of the same order.

This means that if $A$ and $B$ are both $n \times n$ matrices, then:

Illustration using $2 \times 2$ matrices:

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

Step 1: Calculate $|A|$ and $|B|$

$|A| = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2$

$|B| = \begin{vmatrix} 2 & 0 \\ 1 & 2 \end{vmatrix} = 2 \cdot 2 - 0 \cdot 1 = 4$

Step 2: Find $AB$

$AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 \cdot 2 + 2 \cdot 1 & 1 \cdot 0 + 2 \cdot 2 \\ 3 \cdot 2 + 4 \cdot 1 & 3 \cdot 0 + 4 \cdot 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 10 & 8 \end{bmatrix}$

Step 3: Calculate $|AB|$

$|AB| = \begin{vmatrix} 4 & 4 \\ 10 & 8 \end{vmatrix} = 4 \cdot 8 - 4 \cdot 10 = 32 - 40 = -8$

Conclusion:

From (i) and (ii), we get: