Question:

Let the given matrix be $$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}$$ First, we compute the matrix $A+I$: $$A+I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 5 & -1 \\ 0 & 12 & -2 \end{pmatrix}$$ Next, we find the eigenvalues of $A+I$. Since $A+I$ is a block upper triangular matrix, its eigenvalues are the eigenvalues of the diagonal blocks. One eigenvalue is $2$ (from the $1 \times 1$ block). The remaining eigenvalues are found from the $2 \times 2$ block $\begin{pmatrix} 5 & -1 \\ 12 & -2 \end{pmatrix}$. The characteristic equation for this $2 \times 2$ block is given by $\det \begin{pmatrix} 5-\lambda & -1 \\ 12 & -2-\lambda \end{pmatrix} = 0$. $$(5-\lambda)(-2-\lambda) - (-1)(12) = 0$$ $$-10 - 5\lambda + 2\lambda + \lambda^2 + 12 = 0$$ $$\lambda^2 - 3\lambda + 2 = 0$$ Factoring the quadratic equation yields: $$(\lambda-1)(\lambda-2) = 0$$ Thus, the eigenvalues for the $2 \times 2$ block are $\lambda_1 = 1$ and $\lambda_2 = 2$. Combining these with the first eigenvalue, the eigenvalues of $A+I$ are $1, 2, 2$. If $\lambda$ is an eigenvalue of a matrix $M$, then $\lambda^k$ is an eigenvalue of $M^k$. Therefore, the eigenvalues of $(A+I)^{11}$ are $1^{11}, 2^{11}, 2^{11}$. These values are $1, 2048, 2048$. The sum of the diagonal elements of a matrix (its trace) is equal to the sum of its eigenvalues. Therefore, the sum of the diagonal elements of $(A+I)^{11}$ is: $$\text{Trace}((A+I)^{11}) = 1^{11} + 2^{11} + 2^{11} = 1 + 2048 + 2048 = 4097$$