Gauss Jordan elimination method. Solve by Gauss-Jordan elimination method.
$$x + y  + z = 2 \\
x + 3y + 3z = 0 \\
x + 3y + 5z = 2$$
Putting this in reduced row-echelon form:
$$\left(\begin{array}{ccc|r}
1 & 0 & 0 &  3 \\
0 & 1 & 0 &  -2 \\
0 & 0 & 1 &  1
\end{array}\right)
$$
So then
$\begin{pmatrix} 
x \\ y \\ z
\end{pmatrix} = 
\begin{pmatrix} 
3 \\ -2 \\ 1
\end{pmatrix}$ and is 
therefore consistent system.
Is my final answer correct?

$$\begin{cases} x+y+z=2 \\ x+3y+3z=0 \\ x+3y+5z=2 \end{cases}\Rightarrow \begin{cases} x+y+z=2 \\ x+3y+3z=0 \\ 2z=2 \end{cases}\Rightarrow \begin{cases} x+y=1 \\ x+3y=-3 \\ z=1 \end{cases}\Rightarrow \begin{cases} x+y=1 \\ 2y=-4 \\ z=1 \end{cases}\begin{cases} x=3 \\ y=-2 \\ z=1 \end{cases}\\ \quad \quad \quad \quad \quad \quad \\ \quad \quad \\ \\ $$
