Question

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

• A. 4
• B. 8
• C. 12
• D. 16

Answer 1

The Correct Option is B

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