If $A = \left[ {\begin{array}{llllllllllllllllllll}1&2\\4&1\end{array}} \right]$, then find ${A^2} + 2A + 7I$.

We have, $A = \left[ {\begin{array}{llllllllllllllllllll}1&2\\4&1\end{array}} \right]$

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

$= \left[ {\begin{array}{llllllllllllllllllll}{1 + 8}&{2 + 2}\\{4 + 4}&{8 + 1}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}9&4\\8&9\end{array}} \right]$

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