Let A = begin{bmatrix} 1 & 0 & 0 0 & 4 & -1 | vedclass

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(C) First,calculate $A^2$: $A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 &

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$4097$

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(C) First,calculate $A^2$: $A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 16-12 & -4+3 \ 0 & 48-36 & -12+9 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix} = A$. Since $A^2 = A$,it follows that $A^n = A$ for all $n \geq 1$. Using the binomial expansion for $(A + I)^{11}$: $(A + I)^{11} = \sum_{k=0}^{11} \binom{11}{k} A^k I^{11-k} = \binom{11}{0} I + \sum_{k=1}^{11} \binom{11}{k} A^k$. Since $A^k = A$ for $k \geq 1$,we have: $(A + I)^{11} = I + A \left( \sum_{k=1}^{11} \binom{11}{k} ight) = I + A (2^{11} - 1) = I + 2047A$. $(A + I)^{11} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} + 2047 \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix} = \begin{bmatrix} 2048 & 0 & 0 \ 0 & 8189 & -2047 \ 0 & 24564 & -6140 \end{bmatrix}$. The sum of the diagonal elements (trace) is $2048 + 8189 - 6140 = 4097$.

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix}$. Then the sum of the diagonal elements of the matrix $(A + I)^{11}$ is equal to:

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