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(C) Given $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -2 \end{

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$2I+P$

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(C) Given $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -2 \end{bmatrix}$.We calculate $P^2$ and $P^3$:$P^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -2 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 4 \end{bmatrix}$.$P^3 = P^2 \cdot P = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -8 \end{bmatrix}$.Now, calculate $P^3 + 2P^2$:$P^3 + 2P^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -8 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 1+2 & 0 & 0 \ 0 & -1+2 & 0 \ 0 & 0 & -8+8 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 \end{bmatrix}$.Comparing this with the options, we check $2I + P$:$2I + P = 2 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -2 \end{bmatrix} = \begin{bmatrix} 2+1 & 0 & 0 \ 0 & 2-1 & 0 \ 0 & 0 & 2-2 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 \end{bmatrix}$.Thus, $P^3 + 2P^2 = 2I + P$.

If $I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $P=\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{bmatrix}$, then the matrix $P^{3}+2P^{2}$ is equal to

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