If $A = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}$, $B = \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}$ and $A^2 - 5B^2 = 5C$, find $C$.
Given: $\;$ $A^2 - 5B^2 = 5 C$
i.e. $\;$ $C = \dfrac{1}{5} \times A^2 - B^2$ $\;\;\; \cdots \; (1)$
$\begin{aligned}
A^2 & = \begin{bmatrix}
1 & 3 \\
3 & 4
\end{bmatrix} \times \begin{bmatrix}
1 & 3 \\
3 & 4
\end{bmatrix} \\\\
& = \begin{bmatrix}
1 \times 1 + 3 \times 3 & 1 \times 3 + 3 \times 4 \\
3 \times 1 + 4 \times 3 & 3 \times 3 + 4 \times 4
\end{bmatrix} \\\\
& = \begin{bmatrix}
10 & 15 \\
15 & 25
\end{bmatrix}
\end{aligned}$
$\begin{aligned}
\therefore \; \dfrac{1}{5} \times A^2 & = \dfrac{1}{5} \times \begin{bmatrix}
10 & 15 \\
15 & 25
\end{bmatrix} \\\\
& = \begin{bmatrix}
2 & 3 \\
3 & 5
\end{bmatrix} \;\;\; \cdots \; (2)
\end{aligned}$
$\begin{aligned}
B^2 & = \begin{bmatrix}
-2 & 1 \\
-3 & 2
\end{bmatrix} \times \begin{bmatrix}
-2 & 1 \\
-3 & 2
\end{bmatrix} \\\\
& = \begin{bmatrix}
-2 \times \left(-2\right) + 1 \times \left(-3\right) & -2 \times 1 + 1 \times 2 \\
-3 \times \left(-2\right) + 2 \times \left(-3\right) & -3 \times 1 + 2 \times 2
\end{bmatrix} \\\\
& = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \;\;\; \cdots \; (3)
\end{aligned}$
In view of equations $(2)$ and $(3)$, equation $(1)$ becomes
$\begin{aligned}
C & = \begin{bmatrix}
2 & 3 \\
3 & 5
\end{bmatrix} - \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\\\
& = \begin{bmatrix}
2 -1 & 3 - 0 \\
3 - 0 & 5 - 1
\end{bmatrix} \\\\
& = \begin{bmatrix}
1 & 3 \\
3 & 4
\end{bmatrix}
\end{aligned}$