Question

If \[ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{bmatrix} \] , find \( \text{adj}(A) \) and \( A^{-1} \).

Hint:

When finding the inverse of a matrix, use the formula \( A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \). Make sure to calculate the adjugate matrix carefully and pay attention to signs.

Answer 1

Step 1: First, compute the determinant of the matrix \( A \): \[ \text{det}(A) = 1 \cdot \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} = 1 \cdot (-\cos^2 \alpha - \sin^2 \alpha). \] Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \), we get: \[ \text{det}(A) = -1. \] Step 2: Find the adjugate matrix \( \text{adj}(A) \), which is the transpose of the cofactor matrix: \[ \text{adj}(A) = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & \sin \alpha & \cos \alpha \end{bmatrix}. \] Step 3: Compute the inverse using the formula: \[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A). \] Since \( \text{det}(A) = -1 \), we obtain: \[ A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{bmatrix}. \] Thus, the adjugate and inverse of \( A \) are: \[ \text{adj}(A) = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & \sin \alpha & \cos \alpha \end{bmatrix}, \] \[ A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{bmatrix}. \]