$A = {({A^{ - 1}})^{ - 1}} = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}
  {\frac{{ - 1}}{2}} \\ 
  { - 1} 
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
  1 \\ 
  4 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\frac{1}{2}} \\ 
  1 
\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  { - 4} 
\end{array}} \right]$
$3{A^{ - 1}}{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  2 \\ 
  0 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  0 \\ 
  2 
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  { - 2} \\ 
  1 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 4} \\ 
  { - 1} 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  4 \\ 
  { - 1} 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  4 \\ 
  3 
\end{array}} \right]$
$ \to 3{B^{ - 1}} = A\left[ {\begin{array}{*{20}{c}}
  4 \\ 
  { - 1} 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  4 \\ 
  3 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\frac{1}{2}} \\ 
  1 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  { - 4} 
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  4 \\ 
  { - 1} 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  4 \\ 
  3 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  3 \\ 
  8 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  { - 8} 
\end{array}} \right]$
$2A - 3{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  1 \\ 
  2 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 2} \\ 
  { - 8} 
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  3 \\ 
  8 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  { - 8} 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 2} \\ 
  { - 6} 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  0 
\end{array}} \right]$