Question:
If $\mathrm{A}=\left(\begin{array}{cc}0 & \sin \alpha \\ \sin \alpha & 0\end{array}\right)$ and $\operatorname{det}\left(\mathrm{A}^{2}-\frac{1}{2} \mathrm{I}\right)=0$, then a possible value of $\alpha$ is
Correct Option: , 3
Solution:
$A^{2}=\sin ^{2} \alpha I$
So, $\left|A^{2}-\frac{1}{2}\right|=\left(\sin ^{2} \alpha-\frac{1}{2}\right)^{2}=0$
$\Rightarrow \sin \alpha=\pm \frac{1}{\sqrt{2}}$
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