If A = begin{bmatrix} 0 & 1 & -1 2 & 1 & 3 | vedclass

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(B) We know that $A \cdot (	ext{adj } A) = |A| I$,where $I$ is the identity matrix. Given expression is $(A \cdot (	ext{adj } A) \cdot A^{-1}) A$. U

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$\begin{bmatrix} 6 & 0 & 0 \ 0 & 6 & 0 \ 0 & 0 & 6 \end{bmatrix}$

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(B) We know that $A \cdot (	ext{adj } A) = |A| I$,where $I$ is the identity matrix. Given expression is $(A \cdot (	ext{adj } A) \cdot A^{-1}) A$. Using the property of matrix multiplication,this simplifies to $(|A| I \cdot A^{-1}) A = |A| (I \cdot A^{-1} \cdot A) = |A| (I \cdot I) = |A| I$. First,calculate the determinant $|A|$: $|A| = 0(1-6) - 1(2-9) - 1(4-3) = 0 - 1(-7) - 1(1) = 7 - 1 = 6$. Thus,the expression becomes $6I = 6 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \ 0 & 6 & 0 \ 0 & 0 & 6 \end{bmatrix}$. This matches option $B$.

If $A = \begin{bmatrix} 0 & 1 & -1 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{bmatrix}$,then evaluate $(A \cdot (\text{adj } A) \cdot A^{-1}) A$.

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