$A^{2}=A .A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$

$\left[ {\begin{array}{*{20}{l}}
  {3(3) + ( – 2)(4)}&{3( – 2) + ( – 2)( – 2)} \\ 
  {4(3) + ( – 2)(4)}&{4( – 2) + ( – 2)( – 2)} 
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{l}}
  1&{ – 2} \\ 
  4&{ – 4} 
\end{array}} \right]$

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

$ \Rightarrow \left[ {\begin{array}{*{20}{l}}
  1&{ – 2} \\ 
  4&{ – 4} 
\end{array}} \right] = $ $k\left[ {\begin{array}{*{20}{l}}
  3&{ – 2} \\ 
  4&{ – 2} 
\end{array}} \right] – 2\left[ {\begin{array}{*{20}{l}}
  1&0 \\ 
  0&1 
\end{array}} \right]$

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

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

Comparing the corresponding elements, we have:

$3 k-2=1$

$\Rightarrow 3 k=3$

$\Rightarrow $  $k=1$

Thus, the value of $k$ is $1$.