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(B) We know that for any square matrix $A$ of order $n$,the property $A 	ext{ adj } A = |A| I$ holds,where $I$ is the identity matrix of order $n$. G

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$1$

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(B) We know that for any square matrix $A$ of order $n$,the property $A 	ext{ adj } A = |A| I$ holds,where $I$ is the identity matrix of order $n$. Given $A = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$. The determinant of $A$ is $|A| = (\cos \alpha)(\cos \alpha) - (\sin \alpha)(-\sin \alpha) = \cos^2 \alpha + \sin^2 \alpha = 1$. Thus,$A 	ext{ adj } A = |A| I = 1 	imes \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$. 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 $A \text{ adj } A = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$,then $k$ is equal to :-

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