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

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

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$-A$

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(C) Given $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$.First,calculate $A^2 = A \cdot A$:$A^2 = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix}$.Next,calculate $A^3 = A^2 \cdot A$:$A^3 = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 41 & 42 & 42 \ 42 & 41 & 42 \ 42 & 42 & 41 \end{bmatrix}$.Now,compute the expression $A^3 - 4A^2 - 6A$:$A^3 - 4A^2 - 6A = \begin{bmatrix} 41 & 42 & 42 \ 42 & 41 & 42 \ 42 & 42 & 41 \end{bmatrix} - 4 \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix} - 6 \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$$= \begin{bmatrix} 41-36-6 & 42-32-12 & 42-32-12 \ 42-32-12 & 41-36-6 & 42-32-12 \ 42-32-12 & 42-32-12 & 41-36-6 \end{bmatrix}$$= \begin{bmatrix} -1 & -2 & -2 \ -2 & -1 & -2 \ -2 & -2 & -1 \end{bmatrix} = -\begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} = -A$.

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$,then $A^3 - 4A^2 - 6A$ is equal to:

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