JEE Main 2021 (Online) 17th March Morning ShiftQuestion 1
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 :
Matrices
Solution
$A^2=\sin^2\alpha\, I$, so $A^2-\tfrac12I=(\sin^2\alpha-\tfrac12)I$. $\det=(\sin^2\alpha-\tfrac12)^2=0\Rightarrow\sin^2\alpha=\tfrac12$, so $\alpha=\pi/4$. Option (A).