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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
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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 $A=\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)$
$\Rightarrow \quad A=\left(\begin{array}{cc}k & k \\ -k & k\end{array}\right)$
$\Rightarrow A^{2}=\left(\begin{array}{cc}k & k \\ -k & k\end{array}\right)\left(\begin{array}{cc}k & k \\ -k & k\end{array}\right)=\left(\begin{array}{cc}0 & 2 k^{2} \\ -2 k^{2} & 0\end{array}\right)$
$A^{4}=\left(\begin{array}{cc}0 & 2 k^{2} \\ -2 k^{2} & 0\end{array}\right)\left(\begin{array}{cc}0 & 2 k^{2} \\ -2 k^{2} & 0\end{array}\right)$
$A^{4}=\left(\begin{array}{cc}-4 k^{4} & 0 \\ 0 & -4 k^{4}\end{array}\right)=\left(\begin{array}{cc}-4 \times \frac{1}{4} & 0 \\ 0 & -4 \times \frac{1}{4}\end{array}\right)$
$\Rightarrow A^{4}=\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)$
$\Rightarrow A^{B}=\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
Let $A=\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)$
$\Rightarrow \quad A=\left(\begin{array}{cc}k & k \\ -k & k\end{array}\right)$
$\Rightarrow A^{2}=\left(\begin{array}{cc}k & k \\ -k & k\end{array}\right)\left(\begin{array}{cc}k & k \\ -k & k\end{array}\right)=\left(\begin{array}{cc}0 & 2 k^{2} \\ -2 k^{2} & 0\end{array}\right)$
$A^{4}=\left(\begin{array}{cc}0 & 2 k^{2} \\ -2 k^{2} & 0\end{array}\right)\left(\begin{array}{cc}0 & 2 k^{2} \\ -2 k^{2} & 0\end{array}\right)$
$A^{4}=\left(\begin{array}{cc}-4 k^{4} & 0 \\ 0 & -4 k^{4}\end{array}\right)=\left(\begin{array}{cc}-4 \times \frac{1}{4} & 0 \\ 0 & -4 \times \frac{1}{4}\end{array}\right)$
$\Rightarrow A^{4}=\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)$
$\Rightarrow A^{B}=\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
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