If |A| = -3 and A^{-1} = begin{bmatrix} 1 & 0 | vedclass

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(A) We know that the inverse of a matrix $A$ is given by the formula $A^{-1} = \frac{\operatorname{adj}(A)}{|A|}$.From this,we can express the adjoint

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$\begin{bmatrix} -3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3 \end{bmatrix}$

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(A) We know that the inverse of a matrix $A$ is given by the formula $A^{-1} = \frac{\operatorname{adj}(A)}{|A|}$.From this,we can express the adjoint of $A$ as $\operatorname{adj}(A) = |A| \cdot A^{-1}$.Given $|A| = -3$ and $A^{-1} = \begin{bmatrix} 1 & 0 & 0 \ -1 & \frac{1}{3} & 0 \ 3 & \frac{2}{3} & -1 \end{bmatrix}$.Substituting these values,we get:$\operatorname{adj}(A) = -3 	imes \begin{bmatrix} 1 & 0 & 0 \ -1 & \frac{1}{3} & 0 \ 3 & \frac{2}{3} & -1 \end{bmatrix} = \begin{bmatrix} -3 	imes 1 & -3 	imes 0 & -3 	imes 0 \ -3 	imes (-1) & -3 	imes \frac{1}{3} & -3 	imes 0 \ -3 	imes 3 & -3 	imes \frac{2}{3} & -3 	imes (-1) \end{bmatrix} = \begin{bmatrix} -3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3 \end{bmatrix}$.Thus,the correct option is $A$.

If $|A| = -3$ and $A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{bmatrix}$,then $(\operatorname{adj} A)$ is

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