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(B) Given $A = nI$,where $I$ is the identity matrix of order $3 	imes 3$.Then $A^2 = (nI)^2 = n^2 I^2 = n^2 I = \begin{bmatrix} n^2 & 0 & 0 \ 0 & n^

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$n(2nI + B)$

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(B) Given $A = nI$,where $I$ is the identity matrix of order $3 	imes 3$.Then $A^2 = (nI)^2 = n^2 I^2 = n^2 I = \begin{bmatrix} n^2 & 0 & 0 \ 0 & n^2 & 0 \ 0 & 0 & n^2 \end{bmatrix}$.Next,$B^2 = \begin{bmatrix} 0 & 0 & n \ 0 & n & 0 \ n & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & n \ 0 & n & 0 \ n & 0 & 0 \end{bmatrix} = \begin{bmatrix} n^2 & 0 & 0 \ 0 & n^2 & 0 \ 0 & 0 & n^2 \end{bmatrix} = n^2 I$.Also,$AB = (nI)B = n(IB) = nB = \begin{bmatrix} 0 & 0 & n^2 \ 0 & n^2 & 0 \ n^2 & 0 & 0 \end{bmatrix}$.Now,$A^2 + B^2 + AB = n^2 I + n^2 I + nB = 2n^2 I + nB$.Factoring out $n$,we get $n(2nI + B)$.

Let $A = \begin{bmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 & n \\ 0 & n & 0 \\ n & 0 & 0 \end{bmatrix}$. Then,$A^2 + B^2 + AB =$

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