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(C) Given $A = \begin{bmatrix} 0 & \sin \alpha \ \sin \alpha & 0 \end{bmatrix}$.First,calculate $A^{2}$:$A^{2} = \begin{bmatrix} 0 & \sin \alpha \ \

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$\frac{\pi}{4}$

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(C) Given $A = \begin{bmatrix} 0 & \sin \alpha \ \sin \alpha & 0 \end{bmatrix}$.First,calculate $A^{2}$:$A^{2} = \begin{bmatrix} 0 & \sin \alpha \ \sin \alpha & 0 \end{bmatrix} \begin{bmatrix} 0 & \sin \alpha \ \sin \alpha & 0 \end{bmatrix} = \begin{bmatrix} \sin^{2} \alpha & 0 \ 0 & \sin^{2} \alpha \end{bmatrix} = \sin^{2} \alpha I$.Now,substitute this into the determinant equation:$\det\left(A^{2} - \frac{1}{2} Iight) = \det\left(\sin^{2} \alpha I - \frac{1}{2} Iight) = \det\left(\left(\sin^{2} \alpha - \frac{1}{2}ight) Iight) = 0$.Since $I$ is a $2 	imes 2$ identity matrix,$\det(kI) = k^{2} \det(I) = k^{2}$.Thus,$\left(\sin^{2} \alpha - \frac{1}{2}ight)^{2} = 0$.This implies $\sin^{2} \alpha = \frac{1}{2}$,so $\sin \alpha = \pm \frac{1}{\sqrt{2}}$.For $\sin \alpha = \frac{1}{\sqrt{2}}$,a possible value is $\alpha = \frac{\pi}{4}$.

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

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