Question

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

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

Hint:

Compute \(A^2\); it is a scalar multiple of \(I\), so the determinant becomes a perfect square.

Answer 1

The Correct Option is A

\(A^2 = \begin{pmatrix} 0 & \sin\alpha \\ \sin\alpha & 0 \end{pmatrix}\begin{pmatrix} 0 & \sin\alpha \\ \sin\alpha & 0 \end{pmatrix} = \begin{pmatrix} \sin^2\alpha & 0 \\ 0 & \sin^2\alpha \end{pmatrix} = \sin^2\alpha\, I\).
So \(A^2 - \frac{1}{2}I = \left(\sin^2\alpha - \frac{1}{2}\right)I\).
\(\det = \left(\sin^2\alpha - \frac{1}{2}\right)^2 = 0 \Rightarrow \sin^2\alpha = \frac{1}{2}\).
\(\sin\alpha = \pm\frac{1}{\sqrt{2}}\Rightarrow \alpha = \frac{\pi}{4}\) is a possible value.
Answer: (1) \(\frac{\pi}{4}\)