If A = begin{bmatrix} 1 & 4 & 4 4 & 1 & 4 4 | vedclass

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(A) Given $A = \begin{bmatrix} 1 & 4 & 4 \ 4 & 1 & 4 \ 4 & 4 & 1 \end{bmatrix}$.First,calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 1 & 4 & 4

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$27$

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(A) Given $A = \begin{bmatrix} 1 & 4 & 4 \ 4 & 1 & 4 \ 4 & 4 & 1 \end{bmatrix}$.First,calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 1 & 4 & 4 \ 4 & 1 & 4 \ 4 & 4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 4 & 4 \ 4 & 1 & 4 \ 4 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 1+16+16 & 4+4+16 & 4+16+4 \ 4+4+16 & 16+1+16 & 16+4+4 \ 4+16+4 & 16+4+4 & 16+16+1 \end{bmatrix} = \begin{bmatrix} 33 & 24 & 24 \ 24 & 33 & 24 \ 24 & 24 & 33 \end{bmatrix}$.Now,calculate $6A$:$6A = 6 \begin{bmatrix} 1 & 4 & 4 \ 4 & 1 & 4 \ 4 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 24 & 24 \ 24 & 6 & 24 \ 24 & 24 & 6 \end{bmatrix}$.Finally,calculate $A^2 - 6A$:$A^2 - 6A = \begin{bmatrix} 33 & 24 & 24 \ 24 & 33 & 24 \ 24 & 24 & 33 \end{bmatrix} - \begin{bmatrix} 6 & 24 & 24 \ 24 & 6 & 24 \ 24 & 24 & 6 \end{bmatrix} = \begin{bmatrix} 27 & 0 & 0 \ 0 & 27 & 0 \ 0 & 0 & 27 \end{bmatrix} = 27 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = 27 I_3$.Thus,the correct option is $A$.

If $A = \begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$,then $A^2 - 6A =$ . . . . . . (in $I_3$)

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