If `A=[(cos theta, sin theta),(-sin theta, cos theta)]` and `A(adjA)=lamdaI` then `lamda` is equal to

Options

A. 1
B. 2
C. 3
D. `sin theta cos theta`

Answer

To solve the problem, we need to find the value of \(\lambda\) given that \(A \cdot \text{adj}(A) = \lambda I\), where \(A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\).

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

The adjoint of a matrix \(A\) is defined as the transpose of its cofactor matrix. For a \(2 \times 2\) matrix, the cofactor matrix can be computed as follows:

1. The cofactor \(C_{11}\) is obtained by removing the first row and first column from \(A\):
\[ C_{11} = \cos \theta \]

2. The cofactor \(C_{12}\) is obtained by removing the first row and second column from \(A\):
\[ C_{12} = -(-\sin \theta) = \sin \theta \]

3. The cofactor \(C_{21}\) is obtained by removing the second row and first column from \(A\):
\[ C_{21} = -\sin \theta \]

4. The cofactor \(C_{22}\) is obtained by removing the second row and second column from \(A\):
\[ C_{22} = \cos \theta \]

Thus, the cofactor matrix is:
\[ C = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]

Now, the adjoint of \(A\) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = C^T = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

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

Now we need to multiply \(A\) and \(\text{adj}(A)\):
\[ A \cdot \text{adj}(A) = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \cdot \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

Calculating the product:
1. First row, first column:
\[ \cos \theta \cdot \cos \theta + \sin \theta \cdot \sin \theta = \cos^2 \theta + \sin^2 \theta = 1 \]

2. First row, second column:
\[ \cos \theta \cdot (-\sin \theta) + \sin \theta \cdot \cos \theta = -\cos \theta \sin \theta + \sin \theta \cos \theta = 0 \]

3. Second row, first column:
\[ -\sin \theta \cdot \cos \theta + \cos \theta \cdot \sin \theta = -\sin \theta \cos \theta + \cos \theta \sin \theta = 0 \]

4. Second row, second column:
\[ -\sin \theta \cdot (-\sin \theta) + \cos \theta \cdot \cos \theta = \sin^2 \theta + \cos^2 \theta = 1 \]

Putting it all together, we get:
\[ A \cdot \text{adj}(A) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]

### Step 3: Relate to \(\lambda I\)

From the equation \(A \cdot \text{adj}(A) = \lambda I\), we have:
\[ I = \lambda I \]

This implies:
\[ \lambda = 1 \]

### Final Answer
Thus, the value of \(\lambda\) is:
\[ \lambda = 1 \]