TITLE: How to solve this to find the Null Space
QUESTION [13 upvotes]: What I did: 
I put this into reduced row echelon form: 
$$\begin{bmatrix} 1 & -2 & 2 & 4 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
It is clear that the $r(M)=2$, because there are two independent rows.

Now for the null space, I wrote down the equations from the reduced row echelon form:
$$x-2y+2z+4t=0$$
$$z+t=0$$
I can't seem to write $x$ and $y$ separately in terms of $z$ and $t$. Any hints?

REPLY [3 votes]: Just to make the answer a bit more algorithmic: a "pivot entry" is an entry which is the first non-zero entry in its row. A "pivot column" is a column containing a pivot entry. In your matrix, columns 1 and 3 are pivot columns. 
Name the variables after the columns as you did (so, $x, y, z, w$). Then the ``free variables'' are the ones that don't occur in pivot columns. In your case, these are $y$ and $w$. The remaining rows of the matrix express the bound variables in terms of the free variables.