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(D) Given $A_\alpha = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.The determinant of matrix $A_\alpha$ is $

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$1$

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(D) Given $A_\alpha = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.The determinant of matrix $A_\alpha$ is $\det(A_\alpha) = \cos^2 \alpha - (-\sin^2 \alpha) = \cos^2 \alpha + \sin^2 \alpha = 1$.We know that for any square matrices $A$ and $B$, $\det(AB) = \det(A) \cdot \det(B)$.Therefore, $\det(A_{\pi / 5} A_{\pi / 4} A_{3 \pi / 10}) = \det(A_{\pi / 5}) \cdot \det(A_{\pi / 4}) \cdot \det(A_{3 \pi / 10})$.Since $\det(A_\alpha) = 1$ for any value of $\alpha$, we have $\det(A_{\pi / 5}) = 1$, $\det(A_{\pi / 4}) = 1$, and $\det(A_{3 \pi / 10}) = 1$.Thus, $\det(A_{\pi / 5} A_{\pi / 4} A_{3 \pi / 10}) = 1 	imes 1 	imes 1 = 1$.

If $A_\alpha = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$, then the determinant of $A_{\pi / 5} A_{\pi / 4} A_{3 \pi / 10}$ is:

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