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

$A^{2}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right]$ <br/><br/>$$ =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{array}\right]=A $$ <br/><br/>$\Rightarrow \mathrm{A}_{3}=\mathrm{A}_{4}=.......=\mathrm{A}$ <br/><br/>Now, <br/><br/>$$ \begin{aligned} (A+I)^{11} & ={ }^{11} C_{0} A^{11}+{ }^{11} C_{1} A^{10}+\ldots{ }^{11} C_{11} I \\\\ & =A\left({ }^{11} C_{0}+{ }^{11} C_{1} \ldots{ }^{11} C_{10}\right)+I \\\\ & =A\left(2^{11}-1\right)+I \end{aligned} $$ <br/><br/>Trace of <br/><br/>$$ \begin{aligned} (A+I)^{11} & =2^{11}+4\left(2^{11}-1\right)+1-3\left(2^{11}-1\right)+1 \\\\ & =2 \times 2^{11}-4+3+2 \\\\ & =2^{12}+1 \\\\ & =4097 \end{aligned} $$