Question

$A=\left[\begin{array}{cc}\cos \left(\alpha \right)& -\sin \left(\alpha \right)\\ \sin \left(\alpha \right)& \cos \left(\alpha \right)\end{array}\right]$ and $A^{32}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$, then $\alpha$ may be

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

Answer 1

The correct option is C

$\frac{\pi }{64}$

Explanation for the correct option:

Step 1: Express the given data

We have been given that,

$A=\left[\begin{array}{cc}\cos \left(\alpha \right)& -\sin \left(\alpha \right)\\ \sin \left(\alpha \right)& \cos \left(\alpha \right)\end{array}\right]\&A^{32}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

We need to find the value of $\alpha$.

We know, $A^n=\left[\begin{array}{cc}\cos \left(n\alpha \right)& -\sin \left(n\alpha \right)\\ \sin \left(n\alpha \right)& \cos \left(n\alpha \right)\end{array}\right]$

For, $n=32,$

$\Rightarrow$ $A^{32}=\left[\begin{array}{cc}\cos \left(32\alpha \right)& -\sin \left(32\alpha \right)\\ \sin \left(32\alpha \right)& \cos \left(32\alpha \right)\end{array}\right]$

Step 2: Find the value of $\alpha$

$\begin{array}{rcl}{A}^{32}& =& \left[\begin{array}{cc}\cos \left(32\alpha \right)& -\sin \left(32\alpha \right)\\ \sin \left(32\alpha \right)& \cos \left(32\alpha \right)\end{array}\right]\end{array}$

$\Rightarrow$$\begin{array}{rcl}\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]& =& \left[\begin{array}{cc}\cos \left(32\alpha \right)& -\sin \left(32\alpha \right)\\ \sin \left(32\alpha \right)& \cos \left(32\alpha \right)\end{array}\right]\end{array}$

$\Rightarrow$$\begin{array}{rcl}\cos \left(32\alpha \right)& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}32\alpha & =& {\cos }^{-1}\left(0\right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}32\alpha & =& \frac{\pi }{2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\alpha & =& \frac{\pi }{32\times 2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\mathit{\alpha }& \mathbf{=}& \frac{\mathbf{\pi }}{\mathbf{64}}\end{array}$

Therefore, option (C) is the correct answer.