Identify the matrix form

The given matrix $A=\begin{pmatrix}\cos\alpha & -\sin\alpha\\[4pt]\sin\alpha & \cos\alpha\end{pmatrix}$ is the standard rotation matrix through angle $\alpha$.

Hence, $A^n=\begin{pmatrix}\cos(n\alpha) & -\sin(n\alpha)\\[4pt]\sin(n\alpha) & \cos(n\alpha)\end{pmatrix}$ for any positive integer $n$.

Use the condition on $A^{32}$

We are given $A^{32}=\begin{pmatrix}0 & -1\\[4pt]1 & 0\end{pmatrix}.$

But this is also a rotation matrix. Compare it with $\begin{pmatrix}\cos\theta & -\sin\theta\\[4pt]\sin\theta & \cos\theta\end{pmatrix}.$

So we must have $\cos(32\alpha)=0, \qquad \sin(32\alpha)=1.$

Therefore, $32\alpha=\frac{\pi}{2}+2k\pi, \quad k\in\mathbb{Z}.$

Solve for $\alpha$

Check the options

• A: $\alpha=0$ $32\alpha=0 \Rightarrow A^{32}=I \neq \begin{pmatrix}0&-1\\1&0\end{pmatrix}$ So A is incorrect.

• B: $\alpha=\frac{\pi}{16}$ $32\alpha=2\pi \Rightarrow A^{32}=I$ Incorrect.

• C: $\alpha=\frac{\pi}{32}$ $32\alpha=\pi \Rightarrow A^{32}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ Incorrect.

• D: $\alpha=\frac{\pi}{64}$ $32\alpha=\frac{\pi}{2}$ Hence $A^{32}=\begin{pmatrix}0&-1\\1&0\end{pmatrix},$ which matches.

Final answer

A value of $\alpha$ is $\boxed{\frac{\pi}{64}}.$