Question

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

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

Answer 1

The correct option is D

$\frac{\mathrm{\pi }}{4}$

Step 1: Determine the value of $\left(A^2-\frac{1}{2}I\right)$

The given matrix $A=\left[\begin{array}{cc}0& \sin \left(\alpha \right)\\ \sin \left(\alpha \right)& 0\end{array}\right]$

$\Rightarrow A=\sin \left(\alpha \right)\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$.

Thus, $A^2-\frac{1}{2}I=\left[\sin \left(\alpha \right)\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\times \sin \left(\alpha \right)\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\right]-\frac{1}{2}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$$\begin{gathered} \Rightarrow A^2-\frac{1}{2}I={\sin }^2\left(\alpha \right)\left[\begin{array}{cc}0+1& 0+0\\ 0+0& 1+0\end{array}\right]-\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right] \\ \Rightarrow A^2-\frac{1}{2}I=\left[\begin{array}{cc}{\sin }^2\left(\alpha \right)& 0\\ 0& {\sin }^2\left(\alpha \right)\end{array}\right]-\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right] \\ \Rightarrow A^2-\frac{1}{2}I=\left[\begin{array}{cc}{\sin }^2\left(\alpha \right)-\frac{1}{2}& 0\\ 0& {\sin }^2\left(\alpha \right)-\frac{1}{2}\end{array}\right] \end{gathered}$$

Step 2: Solve the given equation

It is given that $\det \left(A^2-\frac{1}{2}I\right)=0$

$$\begin{gathered} \Rightarrow \left|\begin{array}{cc}{\sin }^2\left(\alpha \right)-\frac{1}{2}& 0\\ 0& {\sin }^2\left(\alpha \right)-\frac{1}{2}\end{array}\right|=0 \\ \Rightarrow {\left({\sin }^2\left(\alpha \right)-\frac{1}{2}\right)}^2-0=0 \\ \Rightarrow {\sin }^2\left(\alpha \right)-\frac{1}{2}=0 \\ \Rightarrow {\sin }^2\left(\alpha \right)=\frac{1}{2} \\ \Rightarrow \sin \left(\alpha \right)=\frac{1}{\sqrt{2}} \\ \Rightarrow \sin \left(\alpha \right)=\sin \left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \alpha =n\mathrm{\pi }+{\left(-1\right)}^{\mathrm{n}}\left(\frac{\mathrm{\pi }}{4}\right) \end{gathered}$$

Step 3: Determine the possible solution of $\alpha$

Put $n=0$.

Thus, $\alpha =\left(0\right)\mathrm{\pi }+{\left(-1\right)}^0\left(\frac{\mathrm{\pi }}{4}\right)=\frac{\mathrm{\pi }}{4}$.

Therefore, a possible value of $\alpha$ is $\frac{\mathrm{\pi }}{4}$.

Hence, (D) is the correct option.