Question

The least positive integer $n$ such that ${\left(\begin{array}{cc}\cos \frac{\pi }{4}& \sin \frac{\pi }{4}\\ -\sin \frac{\pi }{4}& \cos \frac{\pi }{4}\end{array}\right)}^n$ is an identity matrix of order $2$ is

• A. 4
• B. 8
• C. 12
• D. 16

Answer 1

Answer: (B)

We have, ${\left(\begin{array}{cc}\cos \pi /4& \sin \pi /4\\ -\sin \frac{\pi }{4}& \cos \frac{\pi }{4}\end{array}\right)}^n$
Let $A=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)$
$\Rightarrow A=\left(\begin{array}{cc}k& k\\ -k& k\end{array}\right)($ where, $k=\frac{1}{\sqrt{2}})$
$\Rightarrow A^2=\left(\begin{array}{cc}k& k\\ -k& k\end{array}\right)\left(\begin{array}{cc}k& k\\ -k& k\end{array}\right)=\left(\begin{array}{cc}0& 2k^2\\ -2k^2& 0\end{array}\right)$
$A^4=\left(\begin{array}{cc}0& 2k^2\\ -2k^2& 0\end{array}\right)\left(\begin{array}{cc}0& 2k^2\\ -2k^2& 0\end{array}\right)$
$A^4=\left(\begin{array}{cc}-4k^4& 0\\ 0& -4k^4\end{array}\right)=\left(\begin{array}{cc}-4\times \frac{1}{4}& 0\\ 0& -4\times \frac{1}{4}\end{array}\right)$
$\Rightarrow A^4=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$
$\Rightarrow A^8=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$