If and, find k so that

$\begin{aligned} A^{2}=A \cdot A &=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right] \\ &=\left[\begin{array}{ll}3(3)+(-2)(4) & 3(-2)+(-2)(-2) \\ 4(3)+(-2)(4) & 4(-2)+(-2)(-2)\end{array}\right]=\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right] \end{aligned}$

Now $A^{2}=k A-2 I$

$\Rightarrow\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]=k\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]-2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]=\left[\begin{array}{ll}3 k & -2 k \\ 4 k & -2 k\end{array}\right]-\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]=\left[\begin{array}{lc}3 k-2 & -2 k \\ 4 k & -2 k-2\end{array}\right]$

Comparing the corresponding elements, we have:

$3 k-2=1$

$\Rightarrow 3 k=3$

$\Rightarrow k=1$

Thus, the value of $k$ is 1 .