Solution The system of equations can be written in the form $\mathrm{AX}=\mathrm{B}$, where

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

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} $

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

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

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

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