Let A=pmatrix&-&pmatrix, R, such that A^32=pmatrix0&-11&0pmatrix.

Question

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

Accepted Answer

$A^n=\begin{pmatrix}\cos n\alpha&-\sin n\alpha\\sin n\alpha&\cos n\alpha\end{pmatrix}$, so $A^{32}=\begin{pmatrix}0&-1\1&0\end{pmatrix}$ requires $\cos32\alpha=0,\ \sin32\alpha=1$, i.e. $32\alpha=\dfrac\pi2+2k\pi$.

Taking $k=0$: $\alpha=\dfrac\pi{64}$.

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