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(D) Given $A = \begin{bmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 2 & 0 & 2 \end{bmatrix}$.We can write $A = 2I + B$, where $I = \begin{bmatrix} 1 & 0 & 0 \ 0

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$a = 2^n, b = n 2^n$

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(D) Given $A = \begin{bmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 2 & 0 & 2 \end{bmatrix}$.We can write $A = 2I + B$, where $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 2 & 0 & 0 \end{bmatrix}$.Note that $B^2 = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 2 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 2 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} = O$.Since $2I$ and $B$ commute, we use the Binomial Theorem:$A^n = (2I + B)^n = \sum_{k=0}^n \binom{n}{k} (2I)^{n-k} B^k = \binom{n}{0} (2I)^n + \binom{n}{1} (2I)^{n-1} B + 0 + ...$$A^n = 2^n I + n(2^{n-1}) B = 2^n \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} + n 2^{n-1} \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 2 & 0 & 0 \end{bmatrix}$$A^n = \begin{bmatrix} 2^n & 0 & 0 \ 0 & 2^n & 0 \ 0 & 0 & 2^n \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ n 2^n & 0 & 0 \end{bmatrix} = \begin{bmatrix} 2^n & 0 & 0 \ 0 & 2^n & 0 \ n 2^n & 0 & 2^n \end{bmatrix}$.Comparing this with the given form, we get $a = 2^n$ and $b = n 2^n$.

If the matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{bmatrix}$, then $A^n = \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a \end{bmatrix}$, for $n \in N$, where:

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