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(B) Given the expression $(A^2 - 5A)A^{-1}$.Distributing $A^{-1}$ inside the parentheses,we get:$(A^2 \cdot A^{-1}) - (5A \cdot A^{-1})$Since $A^2 \cd

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$\begin{bmatrix} -4 & 2 & 3 \ -1 & -4 & 2 \ 1 & 2 & -1 \end{bmatrix}$

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(B) Given the expression $(A^2 - 5A)A^{-1}$.Distributing $A^{-1}$ inside the parentheses,we get:$(A^2 \cdot A^{-1}) - (5A \cdot A^{-1})$Since $A^2 \cdot A^{-1} = A$ and $A \cdot A^{-1} = I$ (where $I$ is the identity matrix),the expression simplifies to:$A - 5I$Now,substitute the matrix $A$ and the identity matrix $I$:$\begin{bmatrix} 1 & 2 & 3 \ -1 & 1 & 2 \ 1 & 2 & 4 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$$= \begin{bmatrix} 1 & 2 & 3 \ -1 & 1 & 2 \ 1 & 2 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix}$$= \begin{bmatrix} 1-5 & 2-0 & 3-0 \ -1-0 & 1-5 & 2-0 \ 1-0 & 2-0 & 4-5 \end{bmatrix}$$= \begin{bmatrix} -4 & 2 & 3 \ -1 & -4 & 2 \ 1 & 2 & -1 \end{bmatrix}$

If $A = \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$,then $(A^2 - 5A)A^{-1} = $

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