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(D) Let $A = \begin{bmatrix} \cos 2	heta & -\sin 2	heta \ \sin 2	heta & \cos 2	heta \end{bmatrix}$.First,calculate the determinant of $A$:$|A| =

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$\begin{bmatrix} \cos 2	heta & \sin 2	heta \ -\sin 2	heta & \cos 2	heta \end{bmatrix}$

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(D) Let $A = \begin{bmatrix} \cos 2	heta & -\sin 2	heta \ \sin 2	heta & \cos 2	heta \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (\cos 2	heta)(\cos 2	heta) - (-\sin 2	heta)(\sin 2	heta) = \cos^2 2	heta + \sin^2 2	heta = 1$.Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:$adj(A) = \begin{bmatrix} \cos 2	heta & \sin 2	heta \ -\sin 2	heta & \cos 2	heta \end{bmatrix}$.The inverse of the matrix is given by $A^{-1} = \frac{1}{|A|} adj(A)$:$A^{-1} = \frac{1}{1} \begin{bmatrix} \cos 2	heta & \sin 2	heta \ -\sin 2	heta & \cos 2	heta \end{bmatrix} = \begin{bmatrix} \cos 2	heta & \sin 2	heta \ -\sin 2	heta & \cos 2	heta \end{bmatrix}$.

Inverse of the matrix $\begin{bmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{bmatrix}$ is

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