If $A=$ $\begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}$ and $I =$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, find k so that $A^2 = 5A + kI$
$A^2 = A \times A$
$=\begin{pmatrix}
3 & 1 \\
-1 & 2
\end{pmatrix}$
$\begin{pmatrix}
3 & 1 \\
-1 & 2
\end{pmatrix} =$
$\begin{pmatrix}
3 \times 3 + 1 \times \left(-1\right) & 3 \times 1 + 1 \times 2 \\
-1 \times 3 + 2 \times \left(-1\right) & -1 \times 1 + 2 \times 2
\end{pmatrix}$
$=\begin{pmatrix}
8 & 5 \\
-5 & 3
\end{pmatrix}$
$5A = 5$
$\begin{pmatrix}
3 & 1 \\
-1 & 2
\end{pmatrix} = $
$\begin{pmatrix}
15 & 5 \\
-5 & 10
\end{pmatrix}$
$kI = k$
$\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} =$
$\begin{pmatrix}
k & 0 \\
0 & k
\end{pmatrix}$
$\therefore$ $A^2 = 5A + kI \implies$
$\begin{pmatrix}
8 & 5 \\
-5 & 3
\end{pmatrix} =$
$\begin{pmatrix}
15 & 5 \\
-5 & 10
\end{pmatrix} +$
$\begin{pmatrix}
k & 0 \\
0 & k
\end{pmatrix}$
i.e.
$\begin{pmatrix}
8 & 5 \\
-5 & 3
\end{pmatrix} =$
$\begin{pmatrix}
15+k & 5 \\
-5 & 10+k
\end{pmatrix}$
$\implies 8 = 15+k \implies k = -7$