Question

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

• A. $\frac{\pi}{16}$
• B. 0
• C. $\frac{\pi}{32}$
• D. $\frac{\pi}{64}$

Answer 1

Answer: (D)

$A = \begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix} $
$ 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} $
$ = \begin{bmatrix}\cos2\alpha&-\sin2\alpha\\ \sin2\alpha&\cos2\alpha\end{bmatrix} $
$A^{3} = \begin{bmatrix}\cos2\alpha&-\sin2\alpha\\ \sin2\alpha&\cos2\alpha\end{bmatrix} \begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix} $
$= \begin{bmatrix}\cos3\alpha&-\sin3\alpha\\ \sin3\alpha&\cos3\alpha\end{bmatrix} $
Similarly $A^{32} = \begin{bmatrix}\cos32\alpha&-\sin32\alpha\\ \sin32\alpha&\cos32\alpha\end{bmatrix} = \begin{bmatrix}0&-1\\ 1&0\end{bmatrix} $
$\Rightarrow \cos32\alpha = 0 \& \sin 32 \alpha = 1 $
$ \Rightarrow 32\alpha = \left(4n+1\right) \frac{\pi}{2} , n \in I$
$ \alpha = \left(4n+1\right) \frac{\pi}{64}, n \in I $
$ \alpha = \frac{\pi}{64}, \, n = 0 $