Solve the following systems of equations ( Matrices) Solve the following systems of equations: 
\begin{Bmatrix}
       x_1 & -x_2 & -x_3 & +0x_4 & = 2 \\
        -x_1 & +2x_2 & +0x_3 & +3x_4 &= 1 \\
        x_1 & +0x_2 & +x_3 & +0x_4 & =5\\
        \end{Bmatrix}
I try to make a zero triangle but always reach a step where I'm in a cycle trying to do the same thing over and over. I once got $x_3=1/3$ 
Working: 
1-
\begin{Bmatrix}
       1 & -1 & -1 & +0 &  2 \\
        -1 & +2 & +0 & +3 & 1 \\
        1 & +0 & +1 & +0 & 5\\
        \end{Bmatrix}
2- Eq2 to Eq2 + Eq1
\begin{Bmatrix}
       1 & -1 & -1 & +0 &  2 \\
        0 & 1 & -1 & 3 & 3 \\
        1 & +0 & +1 & +0 & 5\\
        \end{Bmatrix}
3- Eq3 to Eq3 + Eq1
\begin{Bmatrix}
       1 & -1 & -1 & +0 &  2 \\
        0 & 1 & -1 & 3 & 3 \\
        0 & 1 & 2 & 0 & 3\\
        \end{Bmatrix}
4- Eq3 to Eq3 - Eq2
\begin{Bmatrix}
       1 & -1 & -1 & +0 &  2 \\
        0 & 1 & -1 & 3 & 3 \\
        0 & 0 & 3 & 0 & 1\\
        \end{Bmatrix}
Now I can say $3x_3=1$ 
Thus, $x_3=1/3$
After that i tried but couldn't complete...
Any suggestions would be appreciated
EDIT: The last step is incorrect.
It should be :
\begin{Bmatrix}
       1 & -1 & -1 & +0 &  2 \\
        0 & 1 & -1 & 3 & 3 \\
        0 & 0 & 3 & -3 & 0\\
        \end{Bmatrix}
So what I did next is:
From Third Row: $x_3+x_4=0$
so, $x_4=-x_3$
Then, From the Second row:
$x_2-x_3+3x_4=3$
$x_2-x_3+3(-x_3)=3$
$x_2-x_3=3$
$x_2=3+x_3$
And I stopped

You should now do backwards elimination; first divide the third equation by $3$ and add the third equation to the second and the first, getting
$$
\begin{bmatrix}
1 & -1 & 0 & 0 & 7/3\\
0 & 1 & 0 & 3 & 10/3\\
0 & 0 & 1 & 0 & 1/3
\end{bmatrix}
$$
Now add the second equation to the first one:
$$
\begin{bmatrix}
1 & 0 & 0 & 3 & 17/3\\
0 & 1 & 0 & 3 & 10/3\\
0 & 0 & 1 & 0 & 1/3
\end{bmatrix}
$$
Now you know that you can give $x_4$ any value, say $x_4=h$, and determine consequently the other unknowns:
$$
\begin{cases}
x_1=\frac{17}{3}-3h\\
x_2=\frac{10}{3}-3h\\
x_3=\frac{1}{3}\\
x_4=h
\end{cases}
$$
