Given that: \[ A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ - \sin \alpha & \cos \alpha \end{bmatrix} \] We are tasked with finding \( A^2 \). To compute this, we multiply the matrix \( A \) by itself: \[ A^2 = \begin{bmatrix} \cos \alpha & \sin \alpha \\ - \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ - \sin \alpha & \cos \alpha \end{bmatrix} \] Carrying out the multiplication: \[ A^2 = \begin{bmatrix} \cos^2 \alpha - \sin^2 \alpha & 2 \sin \alpha \cos \alpha \\ - 2 \sin \alpha \cos \alpha & \cos^2 \alpha - \sin^2 \alpha \end{bmatrix} \] Using the double-angle identities: \[ \cos^2 \alpha - \sin^2 \alpha = \cos 2\alpha, \quad 2 \sin \alpha \cos \alpha = \sin 2\alpha \] We get: \[ A^2 = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ - \sin 2\alpha & \cos 2\alpha \end{bmatrix} \] Similarly, it can be proved that for any \( n \): \[ A^n = \begin{bmatrix} \cos n\alpha & \sin n\alpha \\ - \sin n\alpha & \cos n\alpha \end{bmatrix} \]