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(A) We know that for any square matrix $A$,the product of the matrix and its adjoint is given by the property: $A \cdot (adj(A)) = |A| \cdot I$,where

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$\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

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(A) We know that for any square matrix $A$,the product of the matrix and its adjoint is given by the property: $A \cdot (adj(A)) = |A| \cdot I$,where $I$ is the identity matrix of the same order.First,we calculate the determinant of matrix $A$:$|A| = \begin{vmatrix} \cos x & \sin x \ -\sin x & \cos x \end{vmatrix} = (\cos x)(\cos x) - (\sin x)(-\sin x) = \cos^2 x + \sin^2 x = 1$.Since $|A| = 1$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$,we have:$A \cdot (adj(A)) = 1 \cdot \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$,then $A \cdot (adj(A)) = $

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