Let $A=\left[\begin{array}{l}23\\ -12\end{array}\right]$ and $f\left(x\right)=x^2-4x+7$  Show that $f\left(A\right)=O$ Use this result to find $A^5$

we have,

$f\left(x\right)=x^2-4x+7$.Therefore, $f\left(A\right)=A^2-4A+7I_2$

$A^2=\left[\begin{array}{l}23\\ -12\end{array}\right]\left[\begin{array}{l}23\\ -12\end{array}\right]$$=\left[\begin{array}{l}4-36+6\\ -2-2-3+4\end{array}\right]$$=\left[\begin{array}{l}\text{ 1 12}\\ -4\text{ 1 }\end{array}\right]$

$-4A=\left[\begin{array}{l}\text{-8 -12}\\ 4\text{ -8 }\end{array}\right]$and$7I_2=\left[\begin{array}{l}\text{7 0}\\ 0\text{ 7}\end{array}\right]$

$\therefore f\left(A\right)=A^2-4A+7I_2$

$=\left[\begin{array}{l}\text{ 1 12}\\ -4\text{ 1 }\end{array}\right]$$+\left[\begin{array}{l}\text{-8 -12}\\ 4\text{ -8 }\end{array}\right]$$+\left[\begin{array}{l}7\text{ 0}\\ 0\text{ 7}\end{array}\right]$

$=\left[\begin{array}{l}\text{ 1-8+7 12-12+0}\\ -4+4+0\text{ 1-8+7 }\end{array}\right]$$=\left[\begin{array}{l}\text{0 0}\\ 0\text{ 0}\end{array}\right]=O$

$f\left(A\right)=O$

$\Rightarrow A^2-4A+7I_2=O$

$\Rightarrow A^2=4A-7I_2$

$\Rightarrow A^3=A^2.A=(4A-7I_2)A=4A^2-7I_{2}A$

$\begin{array}{ll}\Rightarrow & A^3=4\left(4A-7I_2\right)-7A\\ \Rightarrow & A^3=9A-28I_2\\ \Rightarrow & A^4=A^{3}A=\left(9A-28I_2\right)A\\ \Rightarrow & A^4=9A^2-28A=9\left(4A-7I_2\right)-28A\\ \Rightarrow & A^4=36A-63I_2-28A=8A-63I_2\\ \Rightarrow & A^5=A^4·A=\left(8A-63I_2\right)A=8A^2-63I_{2}A\\ \Rightarrow & A^5=8\left(4A-7I_2\right)-63A=-31A-56I_2\\ \Rightarrow & A^5=-31\left[\begin{array}{rr}2& 3\\ -1& 2\end{array}\right]-56\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ \Rightarrow & A^5=\left[\begin{array}{cc}-62& -93\\ 31& -62\end{array}\right]-\left[\begin{array}{lr}56& 0\\ 0& 56\end{array}\right]\\ \Rightarrow & A^5=\left[\begin{array}{lr}-118& -93\\ 31& -118\end{array}\right]\end{array}$