MCQ
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
  • A
    $\frac{\pi}{2}$
  • B
    $\frac{\pi}{3}$
  • $\frac{\pi}{4}$
  • D
    $\frac{\pi}{6}$

Answer

Correct option: C.
$\frac{\pi}{4}$
c
$A^{2}=\sin ^{2} \alpha I$

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

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

$\alpha=$ $\frac{\pi}{4}$

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