Let A = [ cases 1 & 0 & 0 0 & 4 & - 1 0 & 12 & - 3 cases ] . Then...

Given,

A = \left[\begin{cases}1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3\end{cases}\right]

{A}^{2} = \left[\begin{cases}1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3\end{cases}\right]\left[\begin{cases}1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3\end{cases}\right]

⇒ {A}^{2} = \left[\begin{cases}1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3\end{cases}\right] = A

$⇒ {A}^{2} = A$

$⇒ {A}^{3} = A \cdot {A}^{2} = {A}^{2} = A$

$⇒ {A}^{4} = {A}^{2} \cdot {A}^{2} = {A}^{2} = A$

$⇒ {A}^{3} = {A}^{4} = \ldots \ldots = A$

${\left(A + I\right)}^{11} = C0 11 {A}^{11} + C1 11 {A}^{10} + \ldots . C10 11 A + C11 11 I$

$= \left(C0 11 + C1 11 + \ldots . C10 11\right)A + I$

$= \left({2}^{11} - 1\right)A + I = 2047A + I$

$⇒ {\left(A + I\right)}^{11} = 2047A + I$

⇒ {\left(A + I\right)}^{11} = \left[\begin{cases}2047 + 1 & 0 & 0 \\ 0 & 4 \times 2047 + 1 & - 1 \times 2047 \\ 0 & 12 \times 2047 & - 3 \times 2047 + 1\end{cases}\right]

$∴$ Sum of diagonal elements

$= 2047\left(1 + 4 - 3\right) + 3$

$= 4094 + 3 = 4097$