$\left| A \right| = \left| {\begin{array}{*{20}{c}}
  1 \\ 
  4 \\ 
  2 
\end{array}\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  1 \\ 
  1 
\end{array}\,\,\,\begin{array}{*{20}{c}}
  { - 3} \\ 
  2 \\ 
  3 
\end{array}} \right| = \left( {3 - 4 - 12} \right) - \left( { - 6 - 12 + 2} \right) =  - 13 + 16 = 3$

$\left[ {\begin{array}{*{20}{c}}
  6 \\ 
  1 
\end{array}\,\,\,\begin{array}{*{20}{c}}
  3 \\ 
  {\frac{2}{3}} 
\end{array}} \right]X = \left[ {\begin{array}{*{20}{c}}
  3 \\ 
  { - 2} 
\end{array}\,\,\begin{array}{*{20}{c}}
  0 \\ 
  1 
\end{array}} \right] \to X = \frac{1}{{4 - 3}}\left[ {\begin{array}{*{20}{c}}
  {\frac{2}{3}} \\ 
  { - 1} 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 3} \\ 
  6 
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  3 \\ 
  { - 2} 
\end{array}\,\,\,\begin{array}{*{20}{c}}
  0 \\ 
  1 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  8 \\ 
  { - 15} 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 3} \\ 
  6 
\end{array}} \right]$