Prove the following

Given,

$A=\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right]$

$A^{2}=A \cdot A=\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right]=\left[\begin{array}{ll}1+8 & 2+2 \\ 4+4 & 8+1\end{array}\right]=\left[\begin{array}{ll}9 & 4 \\ 8 & 9\end{array}\right]$

$A^{2}+2 A+7 I=\left[\begin{array}{ll}9 & 4 \\ 8 & 9\end{array}\right]+\left[\begin{array}{ll}2 & 4 \\ 8 & 2\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]=\left[\begin{array}{cc}18 & 8 \\ 16 & 18\end{array}\right]$