Solve the Following Questions
Question:

If $A=\left(\begin{array}{cc}0 & \sin \alpha \\ \sin \alpha & 0\end{array}\right)$ and $\operatorname{det}\left(A^{2}-\frac{1}{2} I\right)=0$, then

a possible value of $\alpha$ is

1. $\frac{\pi}{2}$

2. $\frac{\pi}{3}$

3. $\frac{\pi}{4}$

4. $\frac{\pi}{6}$

Correct Option: , 3

Solution:

$\mathrm{A}^{2}=\sin ^{2} \alpha \mathrm{I}$

So, $\left|A^{2}-\frac{I}{2}\right|=\left(\sin ^{2} \alpha-\frac{1}{2}\right)^{2}=0$

$\Rightarrow \sin \alpha=\pm \frac{1}{\sqrt{2}}$