The basis of a kernel So I am having trouble constructing the basis of the kernel of a matrix, after I doing rref to the matrix. 
For example, I want to find the basis of the kernel of  {$\begin{bmatrix}2&2&0&-2&4\\ 1&1&1&2&1\\2&2&1&1&3\end{bmatrix}$
and rreft(A) = {$\begin{bmatrix}1&1&0&-1&2\\ 0&0&1&3&-1\\0&0&0&0&0\end{bmatrix}$
I try to set the x1,x2,x3 as free variable, the I get 
x1=x1
x2=x2
x3=x3
x4=x1+x2+2x5
x5=x3+3x4
so the basis are [1,0,0,1,0]$^T$, [0,1,0,1,0]$^T$,[0,0,1,0,1]$^T$,[0,0,0,1,3]$^T$,[0,0,0,2,1]$^T$ but I really think I messed them up. 
What the solution of the question says: 
Letting s, t, u be the free variables corresponding to the columns of rref(A).
Then the general solution of Av = 0 is
s$\begin{bmatrix}-1\\ 1\\0\\0\\0\end{bmatrix}$+t$\begin{bmatrix}1\\ 0\\-3\\1\\0\end{bmatrix}$+u$\begin{bmatrix}-2\\ 0\\1\\0\\1\end{bmatrix}$
But how do they get there, I mean I see why they are setting the first,second and the third variables as the free variables, but how do they get to the next step? 

Let's start from you rref (which I'll assume you found correctly).
$$
M = \pmatrix{
1&1&0&-1&2\\ 
0&0&1&3&-1\\
0&0&0&0&0
}
$$
The pivots are in columns $1$ and $3$, so every other column (columns 2,4, and 5) corresponds to a free variable.  With the free variables $x_2,x_4,x_5$, the system of equations that defines the kernel of the matrix (the system $Mx = 0$) looks like this:
$$
x_1 + x_2 - x_4 + 2x_5 = 0\\
x_3 + 3x_4 - x_5 = 0
$$
we can solve these equations for the non-free variables.  In particular, we have
$$
x_1 = -x_2 + x_4 - 2x_5\\
x_3 = -3x_4 + x_5
$$
So, the solution to the system of equations $Mx = 0$ can be written in the form
$$
x = \pmatrix{x_1\\x_2\\x_3\\x_4\\x_5} = \pmatrix{-x_2 + x_4 - 2x_5\\x_2\\-3x_4 + x_5\\x_4\\x_5} = x_2 \pmatrix{-1\\1\\0\\0\\0} + x_4\pmatrix{1\\0\\-3\\1\\0} + x_5\pmatrix{-2\\0\\1\\0\\1}
$$
Importantly, the kernel of the rref matrix $M$ is exactly the same as the kernel of the matrix $A$ that we started with, so we've now answered the question.
