Finding all solutions to Ax = b Edit: I've rewritten this whole question because it was unclear.  I'm not sure if this is any better.  This is a multi-step problem using MATLAB, so I've tried to whittle it down a bit without losing anything.

Let $A$ be a random 6x6 matrix, and $b$ be a random 6x1 matrix.  Since $A$ was generated randomly, we would expect it to be nonsingular.  The system $Ax = b$ should have a unique solution.
Now let's change $A$ so that it is singular.  Replace the third column of $A$ with a linear combination of its first two columns, namely $a_3 = 4*a_1 + 3*a_2$, where $a_1, a_2,$ and $a_3$ are first, second and third columns of $A$.
Now let $y$ be a random 6x1 matrix and let $c = Ay$.
Compute the reduced row echelon form $U$ of $[A$ $c]$.
The free variable determined by the echelon form should be $x_3$.  By examining the system corresponding to the matrix $U$, you should be able to determine the solution corresponding to $x_3 = 0$.  Let the column vector $w$ be this solution.
Now set the last column of $U$ to zeros.  So $U$ should now correspond to the reduced row echelon form of $(A | 0)$.  Use $U$ to determine the solution to the homogeneous system when the free variable $x_3 = 1$ and enter your result as vector $z$.
Set $v = w + 3 * z$.  The vector $v$ should be a solution to the system $Ax = c$.  Why?  Explain. . . . What is the value of the free variable $x_3$ for this solution?  

Here is the part I'm stuck on:

How could we determine all possible solutions to the system in terms of the vectors $w$ and $z$?  Explain.

 A: To fill out my comment above - the construction tells us that we are looking for a 1-dimensional subspace. 
If $Aw=c$ and $Ay=c$ then, by linearity $A(w-y)=0$ 
and since we have a solution: $Az=0$, and we are dealing with a 1-dimensional subspace, $(w-y)$ must be a multiple of that solution, i.e.
$w-y=\lambda z$
As joriki has stated.
A: Your reference to "the free variable" is unclear. Generally, if $A$ is a singular matrix, the solution space can have any number of dimensions. Even if it has one dimension, there is generally not one specific free variable; in the general case any variable can be regarded as "free" and the others expressed in terms of it. Thus, "with the free variable set to zero" is not a well-defined description of $w$.
For general singular $A$, it's not possible to express all solutions of the system in terms of only two vectors. However, you seem to be implicitly assuming that the solution space is one-dimensional. In that case, all solutions are of the form $w+\lambda z$ with arbitrary $\lambda$.
By the way, the fact that $v=w+3z$ is also a solution to $Ax=b$ follows from the other facts given and isn't required for the solution.
