JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
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 \(\frac{\pi}{3}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(A^{2}=\sin ^{2} \alpha I\) 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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