Question

If $A=\left[\begin{array}{cc}0& \sin \alpha \\ \sin \alpha & 0\end{array}\right]$ and $det(A^2-\frac{1}{2}I)=0,$ then a possible value of $a$ is:

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

Answer 1

The correct option is D $\frac{\pi }{4}$

Given : $A=\left[\begin{array}{cc}0& \sin \alpha \\ \sin \alpha & 0\end{array}\right]$
Now,
$A^2=\left[\begin{array}{cc}0& \sin \alpha \\ \sin \alpha & 0\end{array}\right]\left[\begin{array}{cc}0& \sin \alpha \\ \sin \alpha & 0\end{array}\right]\Rightarrow A^2=\left[\begin{array}{cc}{\sin }^2\alpha & 0\\ 0& {\sin }^2\alpha \end{array}\right]\Rightarrow A^2-\frac{1}{2}I=\left[\begin{array}{cc}{\sin }^2\alpha -\frac{1}{2}& 0\\ 0& {\sin }^2\alpha -\frac{1}{2}\end{array}\right]\Rightarrow |A^2-\frac{1}{2}I|=0\Rightarrow {({\sin }^2\alpha -\frac{1}{2})}^2=0\Rightarrow {\sin }^2\alpha =\frac{1}{2}\Rightarrow \sin \alpha =\pm \frac{1}{\sqrt{2}}$
Hence, from the given options,
$\alpha =\frac{\pi }{4}$