Answer
The values of x, y, z are $6,-4,2$ respectively.
Work Step by Step
Let us assume a matrix $ A $ be formed from the given system of equations,
$ A=\left[ \begin{matrix}
1 & -2 & 1 \\
2 & -1 & -1 \\
3 & 5 & -4 \\
\end{matrix} \right]$
Then, matrix $ B $ is obtained by writing the right-hand side of the equations as follows:
$ B=\left[ \begin{matrix}
16 \\
14 \\
-10 \\
\end{matrix} \right]$
Write the augmented matrix $\left[ A|B \right]$ as shown below:
$\left[ \begin{matrix}
1 & -2 & 1 & 16 \\
2 & -1 & -1 & 14 \\
3 & 5 & -4 & -10 \\
\end{matrix} \right]$
Then, one can apply ${{R}_{2}}\to 2{{R}_{1}}-{{R}_{2}}$ and one gets:
$\left[ \begin{matrix}
1 & -2 & 1 & 16 \\
0 & -3 & 3 & 18 \\
3 & 5 & -4 & -10 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to 3{{R}_{1}}-{{R}_{3}}$ and one gets:
$\left[ \begin{matrix}
1 & -2 & 1 & 16 \\
0 & -3 & 3 & 18 \\
0 & -11 & 7 & 58 \\
\end{matrix} \right]$
Apply ${{R}_{2}}\to \frac{{{R}_{2}}}{-3}$ and one gets:
$\left[ \begin{matrix}
1 & -2 & 1 & 16 \\
0 & 1 & -1 & -6 \\
0 & -11 & 7 & 58 \\
\end{matrix} \right]$
Apply ${{R}_{1}}\to {{R}_{1}}+2{{R}_{2}}$ and one gets:
$\left[ \begin{matrix}
1 & 0 & -1 & 4 \\
0 & 1 & -1 & -6 \\
0 & -11 & 7 & 58 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to 11{{R}_{2}}+{{R}_{3}}$ and one gets:
$\left[ \begin{matrix}
1 & 0 & -1 & 4 \\
0 & 1 & -1 & -6 \\
0 & 0 & -4 & -8 \\
\end{matrix} \right]$
Apply ${{R}_{1}}\to {{R}_{1}}-\frac{{{R}_{3}}}{4}$ and one gets:
$\left[ \begin{matrix}
1 & 0 & 0 & 6 \\
0 & 1 & -1 & -6 \\
0 & 0 & -4 & -8 \\
\end{matrix} \right]$
Apply ${{R}_{2}}\to {{R}_{2}}-\frac{{{R}_{3}}}{4}$ and one gets:
$\left[ \begin{matrix}
1 & 0 & 0 & 6 \\
0 & 1 & 0 & -4 \\
0 & 0 & -4 & -8 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to \frac{{{R}_{3}}}{-4}$ and one gets:
$\left[ \begin{matrix}
1 & 0 & 0 & 6 \\
0 & 1 & 0 & -4 \\
0 & 0 & 1 & 2 \\
\end{matrix} \right]$
Thus, the values of $ x,y,z $ are $6,-4,2$.
