Question

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

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

Answer 1

The correct option is B $\frac{\pi }{64}$

$A^2=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc}\cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{array}\right]A^4=\left[\begin{array}{cc}\cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{array}\right]\left[\begin{array}{cc}\cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{array}\right]=\left[\begin{array}{cc}\cos 4\alpha & -\sin 4\alpha \\ \sin 4\alpha & \cos 4\alpha \end{array}\right]$

$\Rightarrow A^{32}=\left[\begin{array}{cc}\cos 32\alpha & -\sin 32\alpha \\ \sin 32\alpha & \cos 32\alpha \end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\Rightarrow \cos 32\alpha =0\text{and}\sin 32\alpha =1\Rightarrow 32\alpha =2n\pi +\frac{\pi }{2}\Rightarrow \alpha =\frac{n\pi }{16}+\frac{\pi }{64}$
For $n=0,\alpha =\frac{\pi }{64}$