If A = begin{bmatrix} cos alpha & sin alpha | vedclass

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(B) Given $A = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.We know that for any square matrix $A$,$A \cdot

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$1$

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(B) Given $A = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.We know that for any square matrix $A$,$A \cdot 	ext{adj}(A) = |A| I$,where $I$ is the identity matrix.First,calculate the determinant of $A$:$|A| = (\cos \alpha)(\cos \alpha) - (\sin \alpha)(-\sin \alpha) = \cos^2 \alpha + \sin^2 \alpha = 1$.Thus,$A \cdot 	ext{adj}(A) = 1 \cdot \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.Comparing this with the given matrix $\begin{bmatrix} k & 0 \ 0 & k \end{bmatrix}$,we get $k = 1$.

If $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$ and $A \cdot \text{adj}(A) = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$,then $k$ is equal to

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