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(B) First,calculate $A^2 = A 	imes A$: $A^2 = \begin{bmatrix} 1 & 1 & 0 \ 1 & 2 & 1 \ 2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \ 1 & 2 &

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$A^3 - 3A^2 - I = O$

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(B) First,calculate $A^2 = A 	imes A$: $A^2 = \begin{bmatrix} 1 & 1 & 0 \ 1 & 2 & 1 \ 2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \ 1 & 2 & 1 \ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 1 \ 5 & 6 & 2 \ 3 & 4 & 1 \end{bmatrix}$ Next,calculate $A^3 = A^2 	imes A$: $A^3 = \begin{bmatrix} 2 & 3 & 1 \ 5 & 6 & 2 \ 3 & 4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \ 1 & 2 & 1 \ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 7 & 9 & 3 \ 15 & 19 & 6 \ 9 & 12 & 4 \end{bmatrix}$ Now,calculate $A^3 - 3A^2$: $A^3 - 3A^2 = \begin{bmatrix} 7 & 9 & 3 \ 15 & 19 & 6 \ 9 & 12 & 4 \end{bmatrix} - 3 \begin{bmatrix} 2 & 3 & 1 \ 5 & 6 & 2 \ 3 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 9 & 3 \ 15 & 19 & 6 \ 9 & 12 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 9 & 3 \ 15 & 18 & 6 \ 9 & 12 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$ Therefore,$A^3 - 3A^2 = I$,which implies $A^3 - 3A^2 - I = O$.

For the matrix $A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix}$,which of the following is correct?

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