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 $A^{32}=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right].$ If the value of $\alpha$ is $\frac{\pi }{2^k}$, then value of $k$ is

Answer 1

Answer: 6

$A=\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]$
$A=\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]$
$=\left[\begin{array}{cc}cos2\alpha & -sin2\alpha \\ sin2\alpha & cos2\alpha \end{array}\right]$
Similarly, $A^4=A^2.A^2=\left[\begin{array}{cc}cos4\alpha & -sin4\alpha \\ sin4\alpha & cos4\alpha \end{array}\right]$
and so on $A^{32}=\left[\begin{array}{cc}cos32\alpha & -sin32\alpha \\ sin32\alpha & cos32\alpha \end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$
Then $sin32\alpha =1$ and $cos32\alpha =0$
$32\alpha =n\pi +(-1{)}^n\frac{\pi }{2}$ 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 $\frac{\pi }{2^6}=\frac{\pi }{2^k}$
$\Rightarrow k=6$