If `A=[(cos alpha,sin alpha),(-sin alpha,cos alpha)]` and A. adj `A=[(k,0),(0,k)]` , then k is equal to

Options

A. 0
B. 1
C. `sin alpha cos alpha `
D. ` cos 2alpha`

Answer

To solve the problem, we need to find the value of \( k \) given the matrix \( A \) and the equation \( A \cdot \text{adj} A = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \).

### Step-by-step Solution:

1. **Define the matrix \( A \)**:
\[ A = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \]

2. **Find the adjoint of matrix \( A \)**:
The adjoint of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:
\[ \text{adj} A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
For our matrix \( A \):
- \( a = \cos \alpha \)
- \( b = \sin \alpha \)
- \( c = -\sin \alpha \)
- \( d = \cos \alpha \)

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

3. **Multiply \( A \) and \( \text{adj} A \)**:
We need to compute \( A \cdot \text{adj} A \):
\[ A \cdot \text{adj} A = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \cdot \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]

Performing the multiplication:
- First row, first column:
\[ \cos \alpha \cdot \cos \alpha + \sin \alpha \cdot \sin \alpha = \cos^2 \alpha + \sin^2 \alpha = 1 \]
- First row, second column:
\[ \cos \alpha \cdot (-\sin \alpha) + \sin \alpha \cdot \cos \alpha = -\sin \alpha \cos \alpha + \sin \alpha \cos \alpha = 0 \]
- Second row, first column:
\[ -\sin \alpha \cdot \cos \alpha + \cos \alpha \cdot \sin \alpha = -\sin \alpha \cos \alpha + \sin \alpha \cos \alpha = 0 \]
- Second row, second column:
\[ -\sin \alpha \cdot (-\sin \alpha) + \cos \alpha \cdot \cos \alpha = \sin^2 \alpha + \cos^2 \alpha = 1 \]

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

4. **Set the result equal to the given matrix**:
We know from the problem statement:
\[ A \cdot \text{adj} A = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \]
Therefore, we can equate:
\[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \]

5. **Compare the matrices**:
From the comparison, we find:
\[ k = 1 \]

### Final Answer:
Thus, the value of \( k \) is \( 1 \).