Let A = begin{bmatrix} cos theta & -sin theta | vedclass

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(A) Given $A = \begin{bmatrix} \cos 	heta & -\sin 	heta \ -\sin 	heta & -\cos 	heta \end{bmatrix}$.First,we find the determinant of $A$:$|A| = (\

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

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

Let $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$,then the inverse of $A$ is

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$\begin{aligned} & A(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^\beta\end{array}\right] \\ & \Rightarrow[A(\alpha, \beta)]^{-1}=\end{aligned}$

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