Gaussian-Jordan Elimination question? I have the linear system 
$$
\begin{align*}
2x-y-z+v&=0 \\
x-2y-z+5u-v&=1 \\
2x-z+v&=1
\end{align*}$$
Very well. I form the matrix   
$$
\left[
\begin{array}{@{}ccccc|c@{}}
2&-1&-1 & 0 & 1 &0 \\
1&-2&-1 & 5 & -1 &1 \\
2&0&-1 & 0&1&1 \\
\end{array}
\right]
$$
So I thought about exchanging the first row with the second one,and the first column with the last column. The first row is $a1$,the second $a$2 and the third $a3$. So what I do is find $a3-a2$ ,then multiply $a1$ by $-2$, then do $a2+a1$ , then $a3-a1$, this way I will have a $scaled$  $matrix$ and it will look like 
$$
\left[
\begin{array}{@{}ccccc|c@{}}
-2&2&2 & -10 & -4 &-2 \\
0&-3&1 &-10 & -5 &-2 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
When I proceed with the Gaussian method, I don't know what to multiply or divide to get some of the numbers in the matrix zero. Help me ? Please.
 A: Hints:
Updated
Variables arranged using $(x, y, z, u, v)$.
From your system, we have:
$$
\left[
\begin{array}{@{}ccccc|c@{}}
2&-1&-1&0&1  &0 \\
1&-2&-1 &5&-1& 1 \\
2&0&-1 &0&1& 1 \\
\end{array}
\right]
$$
The row-reduced-echelon-form should be:
$$
\left[
\begin{array}{@{}ccccc|c@{}}
1&0& 0 & -5 & 2 & -2\\
0&1&0 & 0 & 0 &1 \\
0&0&1 & -10&3&-5 \\
\end{array}
\right]
$$
Regards
A: Assuming that you've done everything right so far
$$
\left[
\begin{array}{@{}ccccc|c@{}}
-2&2&2 & -10 & -4 &-2 \\
0&-3&1 &-10 & -5 &-2 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
Now, let the first row be L1, the second L2, the third L3.
L1$\times$(-0.5) and then, L1-2L2.
You got
$$
\left[
\begin{array}{@{}ccccc|c@{}}
1&-1&-1 & 5 & 0&-1 \\
0&-3&1 &-10 & -5 &-2 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
next, you do L2$\times$(-1/3)-(5/3)L3. You got
$$
\left[
\begin{array}{@{}ccccc|c@{}}
1&-1&-1 & 5 & 0&-1 \\
0&1&-1/3 &10/3 & 0 &-2/3 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
Finally, L1+L2, you have:
$$
\left[
\begin{array}{@{}ccccc|c@{}}
1& 0& -4/3 & 25/3 & 0&-2 \\
0&1&-1/3 &10/3 & 0 &-1 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
