TITLE: How do I get the parametric form solution of a linear system from reduced row-echelon form?
QUESTION [10 upvotes]: 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?

REPLY [14 votes]: 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:
$$\begin{cases} 
x_1 = -6s - 11t\\
x_2 = s\\
x_3 = 8t\\
x_4 = t
 \end{cases}$$