If A=left[begin{array}{ccc}1 & -2 & 2 0 & 2 | vedclass

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(C) We know that for any square matrix $A$,$A \cdot \operatorname{adj}(A) = |A| I$,where $|A|$ is the determinant of $A$ and $I$ is the identity matri

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$\left[\begin{array}{lll}8 & 0 & 0 \ 0 & 8 & 0 \ 0 & 0 & 8\end{array}ight]$

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(C) We know that for any square matrix $A$,$A \cdot \operatorname{adj}(A) = |A| I$,where $|A|$ is the determinant of $A$ and $I$ is the identity matrix of the same order.First,calculate the determinant of $A$:$|A| = 1(2 	imes 4 - (-3) 	imes (-2)) - (-2)(0 	imes 4 - (-3) 	imes 3) + 2(0 	imes (-2) - 2 	imes 3)$$|A| = 1(8 - 6) + 2(0 + 9) + 2(0 - 6)$$|A| = 1(2) + 2(9) + 2(-6)$$|A| = 2 + 18 - 12 = 8$Since $A$ is a $3 	imes 3$ matrix,$I = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight]$.Therefore,$A \cdot \operatorname{adj}(A) = |A| I = 8 \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight] = \left[\begin{array}{ccc}8 & 0 & 0 \ 0 & 8 & 0 \ 0 & 0 & 8\end{array}ight]$.

If $A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$,then $A \cdot \operatorname{adj}(A)$ is equal to

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