Answer
$\{(1,-1,1)\}$.
Work Step by Step
The given system of equations is
$x-2y-z=2$
$2x-y+z=4$
$-x+y-2z=-4$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -2 & -1& 2\\
2 & -1 & 1& 4 \\
-1&1&-2&-4
\end{array}\right]$
Perform $R_2\rightarrow R_2-2\times R_1$ and $R_3\rightarrow R_3+ R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -2 & -1& 2\\
2-2\times (1) & -1-2\times (-2) & 1-2\times (-1)& 4-2\times (2) \\
-1+1&1-2&-2-1&-4+2
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -2 & -1& 2\\
0 & 3 & 3& 0 \\
0&-1&-3&-2
\end{array}\right]$
Perform $R_2\leftrightarrow R_3$
Change the sign of the second row and swap the 3rd and 2nd row.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -2 & -1& 2\\
0&1&3&2 \\
0 & 3 & 3& 0
\end{array}\right]$
Perform $R_1\rightarrow R_1+2\times R_2$ and $R_3\rightarrow R_3-3 R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1+2\times (0) & -2+2\times (1) & -1+2\times (3)& 2+2\times (2)\\
0&1&3&2 \\
0 -3\times (0)& 3-3\times (1) & 3-3\times (3)& 0-3\times (2)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 5& 6\\
0&1&3&2 \\
0 & 0 & -6& -6
\end{array}\right]$
Perform $R_3\rightarrow \frac{R_3}{-6}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 5& 6\\
0&1&3&2 \\
0/(-6) & 0 /(-6)& -6/(-6)& -6/(-6)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 5& 6\\
0&1&3&2 \\
0 & 0& 1& 1
\end{array}\right]$
Perform $R_1\rightarrow R_1-5\times R_3$ and $R_2\rightarrow R_2-3 R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-5(0) & 0-5(0) & 5-5(1)& 6-5(1)\\
0-3(0)&1-3(0)&3-3(1)&2-3(1) \\
0 & 0& 1& 1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& 1\\
0&1&0&-1 \\
0 & 0& 1& 1
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=1$
and
$\Rightarrow y=-1$.
and
$\Rightarrow z=1$.
The solution set is $\{(x,y,z)\}=\{(1,-1,1)\}$.
