Question

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

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

Answer 1

Answer: (D)

$A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
$A^2=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
$=\left[\begin{array}{cc}\cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{array}\right]$
$A^3=\left[\begin{array}{cc}\cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
$=\left[\begin{array}{cc}\cos 3\alpha & -\sin 3\alpha \\ \sin 3\alpha & \cos 3\alpha \end{array}\right]$
Similarly $A^{32}=\left[\begin{array}{cc}\cos 32\alpha & -\sin 32\alpha \\ \sin 32\alpha & \cos 32\alpha \end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$
$\Rightarrow \cos 32\alpha =0\&\sin 32\alpha =1$
$\Rightarrow 32\alpha =\left(4n+1\right)\frac{\pi }{2},n\in I$
$\alpha =\left(4n+1\right)\frac{\pi }{64},n\in I$
$\alpha =\frac{\pi }{64},n=0$