# How do I get the parametric form solution of a linear system from reduced row-echelon form?

I have the following system of equations:

x1 + 6x2 + 2x3 - 5x4 = 0
-x1 - 6x2 - x3 - 3x4 = 0
2x1 + 612x2 + 5x3 - 18x4 = 0

and I understand that it translated into the following matrix:

1       6     2     -5     0
-1     -6    -1     -3     0
2      12     5     -18    0

Finally, I understand how to use Gauss-Jordan elimination to change this to reduced row-echelon form:

1    6    0   11    0
0    0    1   -8    0
0    0    0    0    0

However, in an example solution that my instructor has prepared, this is then used to find the general solution in parametric form:

x1 = -6s - 11t
x2 = s
x3 = 8t
x4 = t

No intermediate steps are given. I can see that a similarity in the numbers, but I'm not sure exactly what to do.

It looks like arbitrary letter variables have been assigned to those columns which don't start any row with a one and then these variables are used to complete equations for the columns which do start rows. Is that all this is? Or is there something that I'm missing?

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(By the way, it looks like you have a typo: Your row-reduced matrix has a $1$ in the $(1,4)$-entry, but your general solution indicates you should have a $-11$ there.) – Neal Dec 2 '12 at 6:32
It was a couple of typos. The general solution is supposed to have a negative 11 and there's actually supposed to be a positive 11 where the one was. Sorry. – mimicocotopus Dec 2 '12 at 6:36
Okay, hah, and it looks like I made a typo too. All fine :) – Neal Dec 2 '12 at 6:37

Remember that augmented matrices correspond to systems of linear equations. Once you've finished row-reducing, turn the row-reduced matrix back into a system of equations and solve for the variables in the pivot columns:

$$\begin{pmatrix}1 & 6 & 0 & 11 & | & 0 \\ 0 & 0 & 1 & -8 & | & 0 \\ 0 & 0 & 0 & 0 & | & 0\end{pmatrix} \longrightarrow \begin{cases}x_1 + 6x_2 + 11x_4 = 0 \\ x_3 -8x_4 = 0\end{cases}\longrightarrow \begin{cases}x_1 = -6x_2 - 11x_4 \\ x_3 = 8x_4.\end{cases}$$ The free variables $x_2,x_4$ are now parameters. Once you specify them, you specify a single solution to the equation. So subsitute $x_2 = s,x_4 = t$ and arrive at the parametrized form: $$\begin{cases} x_1 = -6s - 11t\\ x_2 = s\\ x_3 = 8t\\ x_4 = t \end{cases}$$

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