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(A) Given $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3 \end{bmatrix}$.First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 1 & 1 & 1 \

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$\frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \ 9 & -1 & -4 \ 5 & -3 & -1 \end{bmatrix}$

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(A) Given $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3 \end{bmatrix}$.First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 1 \ -3 & 8 & -14 \ 7 & -3 & 14 \end{bmatrix}$.Next,calculate $A^{3} = A^{2} \cdot A = \begin{bmatrix} 4 & 2 & 1 \ -3 & 8 & -14 \ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 7 & 1 \ -23 & 27 & -69 \ 32 & -13 & 58 \end{bmatrix}$.Now,substitute into $A^{3} - 6A^{2} + 5A + 11I$:$= \begin{bmatrix} 8 & 7 & 1 \ -23 & 27 & -69 \ 32 & -13 & 58 \end{bmatrix} - 6 \begin{bmatrix} 4 & 2 & 1 \ -3 & 8 & -14 \ 7 & -3 & 14 \end{bmatrix} + 5 \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3 \end{bmatrix} + 11 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} = 0$.To find $A^{-1}$,multiply $A^{3} - 6A^{2} + 5A + 11I = 0$ by $A^{-1}$:$A^{2} - 6A + 5I + 11A^{-1} = 0 \Rightarrow 11A^{-1} = -(A^{2} - 6A + 5I)$.$A^{2} - 6A + 5I = \begin{bmatrix} 4 & 2 & 1 \ -3 & 8 & -14 \ 7 & -3 & 14 \end{bmatrix} - \begin{bmatrix} 6 & 6 & 6 \ 6 & 12 & -18 \ 12 & -6 & 18 \end{bmatrix} + \begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -4 & -5 \ -9 & 1 & 4 \ -5 & 3 & 1 \end{bmatrix}$.Thus,$A^{-1} = -\frac{1}{11} \begin{bmatrix} 3 & -4 & -5 \ -9 & 1 & 4 \ -5 & 3 & 1 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \ 9 & -1 & -4 \ 5 & -3 & -1 \end{bmatrix}$.

For the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$,show that $A^{3} - 6A^{2} + 5A + 11I = 0$. Hence,find $A^{-1}$.

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