$A = \left[ {\begin{array}{*{20}{c}}
  3&{ - 2}&3 \\ 
  2&1&{ - 1} \\ 
  4&{ - 3}&2 
\end{array}} \right],X = \left[ {\begin{array}{*{20}{l}}
  x \\ 
  y \\ 
  z 
\end{array}} \right]{\text{ and }}B = \left[ {\begin{array}{*{20}{l}}
  8 \\ 
  1 \\ 
  4 
\end{array}} \right]$

We see that

$|A|=3(2-3)+2(4+4)+3(-6-4)=-17 \neq 0$

Hence, $A$ is nonsingular and so its inverse exists. Now

$\begin{gathered}
  {{\text{A}}_{11}} =  - 1,\,\,\,{{\text{A}}_{12}} =  - 8,\,\,\,{{\text{A}}_{13}} =  - 10 \hfill \\
  {{\text{A}}_{21}} =  - 5,\,\,\,{{\text{A}}_{22}} =  - 6,\,\,\,{{\text{A}}_{23}} = 1 \hfill \\
  {{\text{A}}_{31}} =  - 1,\,\,\,{{\text{A}}_{32}} = 9,\,\,\,{{\text{A}}_{33}} = 7 \hfill \\ 
\end{gathered} $

$Therefore\,\,\,\,\,{A^{ - 1}} =  - \frac{1}{{17}}\left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 5}&{ - 1} \\ 
  { - 8}&{ - 6}&9 \\ 
  { - 10}&1&7 
\end{array}} \right]\,$

$So\quad \,X = {A^{ - 1}}B =  - \frac{1}{{17}}\left[ {\begin{array}{*{20}{l}}
  { - 1}&{ - 5}&{ - 1} \\ 
  { - 8}&{ - 6}&9 \\ 
  { - 10}&1&7 
\end{array}} \right]\left[ {\begin{array}{*{20}{l}}
  8 \\ 
  1 \\ 
  4 
\end{array}} \right]$

$i.e.\,\,\,\,\,\left[ {\begin{array}{*{20}{l}}
  x \\ 
  y \\ 
  z 
\end{array}} \right] =  - \frac{1}{{17}}\left[ {\begin{array}{*{20}{l}}
  { - 17} \\ 
  { - 34} \\ 
  { - 51} 
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  1 \\ 
  2 \\ 
  3 
\end{array}} \right]$

$Hence\quad x = 1,y = 2\,and\,z = 3$