If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then find $k$ such that $A^2 = kA - 2 I$
$A = \begin{bmatrix}
3 & -2 \\
4 & -2
\end{bmatrix}$, $\;$ $I = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} = 2 \times 2$ $\;$ identity matrix
$\begin{aligned}
A^2 = A \times A & = \begin{bmatrix}
3 & -2 \\
4 & -2
\end{bmatrix} \times \begin{bmatrix}
3 & -2 \\
4 & -2
\end{bmatrix} \\\\
& = \begin{bmatrix}
3 \times 3 - 2 \times 4 & 3 \times \left(-2\right) - 2 \times \left(-2\right) \\
4 \times 3 - 2 \times 4 & 4 \times \left(-2\right) - 2 \times \left(-2\right)
\end{bmatrix} \\\\
& = \begin{bmatrix}
1 & -2 \\
4 & -4
\end{bmatrix}
\end{aligned}$
$\begin{aligned}
kA & = k \begin{bmatrix}
3 & -2 \\
4 & -2
\end{bmatrix} \\\\
& = \begin{bmatrix}
3k & - 2k \\
4k & -2k
\end{bmatrix}
\end{aligned}$
$\begin{aligned}
2I & = 2 \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\\\
& = \begin{bmatrix}
2 & 0 \\
0 & 2
\end{bmatrix}
\end{aligned}$
Given: $\;$ $A^2 = kA - 2I$
i.e. $\;$ $\begin{bmatrix}
1 & -2 \\
4 & -4
\end{bmatrix} = \begin{bmatrix}
3k & -2k \\
4k & -2k
\end{bmatrix} - \begin{bmatrix}
2 & 0 \\
0 & 2
\end{bmatrix}$
i.e. $\;$ $\begin{bmatrix}
1 & -2 \\
4 & -4
\end{bmatrix} = \begin{bmatrix}
3k - 2 & -2k \\
4k & -2k - 2
\end{bmatrix}$
When two matrices are equal, their corresponding elements are equal.
$\;$ We have, $\;$ $1 = 3k - 2$; $\;$ $-2 = -2 k$; $\;$ $4 = 4k$; $\;$ $-4 = -2k - 2$
These equations give $\;$ $k = 1$