Consider the product $\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$

$=\left[\begin{array}{rrr}
-2-9+12 & 0-2+2 & 1+3-4 \\
0+18-18 & 0+4-3 & 0-6+6 \\
-6-18+24 & 0-4+4 & 3+6-8
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]$

$\text { Hence } \quad\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]^{-1}=\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]$

Now, given system of equations can be written, in matrix form, as follows

$\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right]$

or ${\begin{array}{*{20}{c}}
  x \\ 
  y \\ 
  z 
\end{array} = {{\left[ {\begin{array}{*{20}{r}}
  1&{ – 1}&2 \\ 
  0&2&{ – 3} \\ 
  3&{ – 2}&4 
\end{array}} \right]}^{ – 1}}\left[ {\begin{array}{*{20}{l}}
  1 \\ 
  1 \\ 
  2 
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  2&0&1 \\ 
  9&2&3 \\ 
  6&1&2 
\end{array}} \right]\left[ {\begin{array}{*{20}{l}}
  1 \\ 
  1 \\ 
  2 
\end{array}} \right]}$

$=\left[\begin{array}{r}
-2+0+2 \\
9+2-6 \\
6+1-4
\end{array}\right]=\left[\begin{array}{l}
0 \\
5 \\
3
\end{array}\right]$

Hence $\quad x=0, y=5$ and $z=3$