Let $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$, $a \in \mathbb{R}$ such that $A^{32} = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$. Then, a value of $\alpha$ is:

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Aicharya · Accepted Answer

Here's the corrected text with proper LaTeX formatting: $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$ $A^{2} = \begin{bmatrix} \cos^{2} \alpha & -\sin^{2} \alpha \ \sin^{2} \alpha & \cos^{2} \alpha \end{bmatrix}$ $A^{3} = \begin{bmatrix} \cos^{3} \alpha & -\sin^{3} \alpha \ \sin^{3} \alpha & \cos^{3} \alpha \end{bmatrix}$ Similarly, $A^{32} = \begin{bmatrix} \cos^{32} \alpha & -\sin^{32} \alpha \ \sin^{32} \alpha & \cos^{32} \alpha \end{bmatrix}$ Given, $A^{32} = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$ $\begin{bmatrix} \cos^{32} \alpha & -\sin^{32} \alpha \ \sin^{32} \alpha & \cos^{32} \alpha \end{bmatrix} = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$ On comparing $\cos^{32} \alpha = 0$ and $\sin^{32} \alpha = 1$, we get, $\alpha = \frac{\pi}{64}$

Let $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$, $a \in…

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