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

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

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(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \ -2 & 3 & -1 \ 3 & 1 & 2 \end{bmatrix}$.First,calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 1 & 2 & 3 \ -2 & 3 & -1 \ 3 & 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ -2 & 3 & -1 \ 3 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 11 & 7 \ -11 & 4 & -11 \ 7 & 11 & 12 \end{bmatrix}$.Now,calculate $9I = 9 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 9 \end{bmatrix}$.Finally,add $A^2$ and $9I$:$A^2 + 9I = \begin{bmatrix} 6 & 11 & 7 \ -11 & 4 & -11 \ 7 & 11 & 12 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 9 \end{bmatrix} = \begin{bmatrix} 15 & 11 & 7 \ -11 & 13 & -11 \ 7 & 11 & 21 \end{bmatrix}$.Comparing this result with the given options,it does not match $2A$,$4A$,or $6A$. Therefore,the correct option is $(d)$.

If $A = \begin{bmatrix} 1 & 2 & 3 \\ -2 & 3 & -1 \\ 3 & 1 & 2 \end{bmatrix}$ and $I$ is a unit matrix of $3^{rd}$ order,then $(A^2 + 9I)$ equals

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