Answer
$\{(1,1,1)\}$.
Work Step by Step
The given system of equations is
$x+y+z=3$
$-y+2z=1$
$-x+z=0$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 3\\
0 &-1 & 2& 1 \\
-1&0&1&0
\end{array}\right]$
Perform $R_3\rightarrow R_3+R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 3\\
0 &-1 & 2& 1 \\
-1+1&0+1&1+1&0+3
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 3\\
0 &-1 & 2& 1 \\
0&1&2&3
\end{array}\right]$
Perform $R_2\Rightarrow R_2/(-1)$
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 3\\
0/(-1) &-1/(-1) & 2/(-1)& 1 /(-1)\\
0&1&2&3
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 3\\
0 &1 & -2& -1 \\
0&1&2&3
\end{array}\right]$
Perform $R_1\rightarrow R_1- R_2$ and $R_3\rightarrow R_3- R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-0 & 1-1 & 1-(-2)& 3-(-1)\\
0 &1 & -2& -1 \\
0-0&1-1&2-(-2)&3-(-1)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 3& 4\\
0 &1 & -2& -1 \\
0&0&4&4
\end{array}\right]$
Perform $R_3\Rightarrow R_3/4$
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 3& 4\\
0 &1 & -2& -1 \\
0/4&0/4&4/4&4/4
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 3& 4\\
0 &1 & -2& -1 \\
0&0&1&1
\end{array}\right]$
Perform $R_1\rightarrow R_1-3R_3$ and $R_2\rightarrow R_2+2 R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-3(0) & 0-3(0) & 3-3(1)& 4-3(1)\\
0+2(0) &1+2(0) & -2+2(1)& -1+2(1) \\
0&0&1&1
\end{array}\right]$
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 $\{(x,y,z)\}=\{(1,1,1)\}$.
