Mathematics - Matrices Question with Solution | TestHub

Given, 

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

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

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

$\Rightarrow A^2=A$

$\Rightarrow A^3=A·A^2=A^2=A$

$\Rightarrow A^4=A^2·A^2=A^2=A$

$\Rightarrow A^3=A^4=\dots \dots =A$

${\left(A+I\right)}^{11}={}^{11}C_0 A^{11}+{}^{11}C_1 A^{10}+\dots .{}^{11}C_{10} A+{}^{11}C_{11}I$

$=\left({}^{11}C_0+{}^{11}C_1+\dots .{}^{11}C_{10}\right)A+I$

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

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

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

$\therefore$  Sum of diagonal elements

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

$=4094+3=4097$