Question

Let $A = \begin{bmatrix}cos \,\alpha&-sin\,\alpha\\ sin \,\alpha&cos\, \alpha\end{bmatrix} , \left(\alpha\in R\right)$ such that $A^{32} = \begin{bmatrix}0&-1\\ 1&0\end{bmatrix}.$ If the value of $\alpha$ is $\frac{\pi}{2^k}$, then value of $k$ is

Answer 1

$A = \begin{bmatrix}cos \,\alpha&-sin\,\alpha\\ sin\, \alpha&cos \,\alpha\end{bmatrix} $
$A = \begin{bmatrix}cos \,\alpha&-sin\,\alpha\\ sin \,\alpha&cos\, \alpha\end{bmatrix} \begin{bmatrix}cos\, \alpha&-sin\,\alpha\\ sin\, \alpha&cos\, \alpha\end{bmatrix}$
$= \begin{bmatrix}cos \,2\alpha&-sin\,2\alpha\\ sin \,2\alpha&cos \,2\alpha\end{bmatrix} $
Similarly, $A^4 = A^2 . A^2 = \begin{bmatrix}cos \,4\alpha&-sin\,4\alpha\\ sin \,4\alpha&cos \,4\alpha\end{bmatrix} $
and so on $A^{32} = \begin{bmatrix}cos \,32\alpha&-sin\,32\alpha\\ sin \,32\alpha&cos \,32\alpha\end{bmatrix} = \begin{bmatrix}0&-1\\ 1&0\end{bmatrix}$
Then $sin \,32\alpha = 1$ and $cos \,32\alpha = 0$
$ 32\,\alpha = n\pi + (-1)^n \frac{\pi}{2}$ and $32 \alpha = 2n\pi + \frac{\pi}{2}$
$\alpha = \frac{n\pi}{32} + (-1)^n \frac{\pi}{64}$ and $\alpha = \frac{n\pi}{16} + \frac{\pi}{64}$ where $ n \in Z$
Put $n = 0, \alpha = \frac{n\pi}{16} + \frac{\pi}{64}$ where $n \in Z$
Put $ n = 0, \alpha = \frac{\pi}{64} = \frac{\pi}{2^6}$
Then, the value of $\alpha $ is $\frac{\pi}{2^6} = \frac{\pi}{2^k} $
$\Rightarrow k = 6$