If $A = \begin{bmatrix}2 & 1 \\ 1 & 3\end{bmatrix}$, what is $A^2 - 5A + 6I$?

Explanation

1. 01

   $A^2 = \begin{bmatrix}5 & 5 \\ 5 & 10\end{bmatrix}$ and $5A = \begin{bmatrix}10 & 5 \\ 5 & 15\end{bmatrix}$, so $A^2 - 5A = \begin{bmatrix}-5 & 0 \\ 0 & -5\end{bmatrix}$.

2. 02

   The constant term is $6I$, which adds 6 to the diagonal only: the result is $\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = I$.