query about finding null space of matrices I was watching gilbert strangs linear algebra lecture series and got a bit stuck when trying to reason where his use of free variables came from. 
At 8:17 he introduces the idea of free variables as the variables which multiply non pivot columns. What I am confused about is why we are only allowed to freely set those variables which we choose to label as free. 
Example used in lecture:
Consider the matrix $A$:
\begin{pmatrix}
1 & 2 & 2 & 2 \\
0 & 0 & 2 & 4 \\
0 & 0 & 0 & 0
\end{pmatrix}
To find the null space we are looking for solutions to the equations:
$x_1+2x_2+2x_3+2x_4=0 $,
$2x_3+4x_4=0$ 
I see that we have 4 variables and 2 independant equations. This would mean that we should be free to choose the value of ANY 2 variables and solve the resulting equations.
What I am confused about is how we know that we are only allowed to choose the variables $x_2$ and $x_4$, simply because they are the ones which multiply against the non pivot columns, in the equation $Ax=0$. Why cant we freely set the pivot variables to any value? 
 A: In fact, there is nothing wrong (for this particular problem) with setting $x_1$ and $x_3$ (or any two variables) to an arbitrary value, and then solving the resulting equations.  In fact, note that if we wrote the equation in the unorthodox form
$$
2x_2 + x_1 + 2x_4 + 2x_3 = 0\\
0x_2 + 0x_1 + 4x_4 + 2x_3 = 0,
$$
then the standard approach would dictate that $x_2,x_4$ are the "pivot variables", so that $x_1,x_3$ are now the "free variables" that could be set to arbitrary values.
The point of this method, however, is that while it is always possible (and easy, once the system has been reduced) to solve an equation for a certain choice of values for the free variables, the same cannot be said for the pivot variables.  For instance, consider the nullspace of the matrix
$$
A = \pmatrix{1&1&0&0\\0&0&2&4}.
$$
In this case, the free variables are $x_2,x_4$.  If we choose the wrong pair of variables, $x_1$ and $x_2$ for instance, it is not true that the equation can be solved for any choice of values for these two variables. If we take $x_1 = 1$ and $x_2 = 2$, then there is no choice of $x_3,x_4$ for which the system
$$
x_1 + x_2 = 0\\
2x_3 + 4x_4 = 0
$$
has a solution.
