If `A=[(cosx,sinx),(-sinx,cosx)] and A.(adjA)=k[(1,0),(0,1)]` then the value of `k` is

Options

A. `sinx cos x`
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the value of \( k \) given the matrix \( A \) and the relationship \( A \cdot (\text{adj} A) = k \cdot I \), where \( I \) is the identity matrix.

Given:
\[ A = \begin{pmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{pmatrix} \]

### Step 1: Find the adjoint of matrix \( A \)

The adjoint of a matrix is the transpose of its cofactor matrix. For a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), the adjoint is given by:
\[ \text{adj} A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

For our matrix \( A \):
- \( a = \cos x \)
- \( b = \sin x \)
- \( c = -\sin x \)
- \( d = \cos x \)

Thus, the adjoint of \( A \) is:
\[ \text{adj} A = \begin{pmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{pmatrix} \]

### Step 2: Compute \( A \cdot (\text{adj} A) \)

Now we compute the product \( A \cdot (\text{adj} A) \):
\[ A \cdot (\text{adj} A) = \begin{pmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{pmatrix} \cdot \begin{pmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{pmatrix} \]

Calculating the elements of the product:
1. First row, first column:
\[ \cos x \cdot \cos x + \sin x \cdot \sin x = \cos^2 x + \sin^2 x = 1 \]
2. First row, second column:
\[ \cos x \cdot (-\sin x) + \sin x \cdot \cos x = -\cos x \sin x + \sin x \cos x = 0 \]
3. Second row, first column:
\[ -\sin x \cdot \cos x + \cos x \cdot \sin x = -\sin x \cos x + \cos x \sin x = 0 \]
4. Second row, second column:
\[ -\sin x \cdot (-\sin x) + \cos x \cdot \cos x = \sin^2 x + \cos^2 x = 1 \]

Thus, we have:
\[ A \cdot (\text{adj} A) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]

### Step 3: Relate to \( k \cdot I \)

From the problem statement, we have:
\[ A \cdot (\text{adj} A) = k \cdot I \]
Substituting our result:
\[ I = k \cdot I \]

### Step 4: Solve for \( k \)

Since \( I \) is the identity matrix, we can equate:
\[ 1 = k \]

Thus, the value of \( k \) is:
\[ \boxed{1} \]