The null space of $A$ is the set of solutions to $A{\bf x}={\bf 0}$. To find this, you may take the augmented matrix $[A|0]$ and row reduce to an echelon form. Note that every entry in the rightmost column of this matrix will always be 0 in the row reduction steps. So, we may as well just row reduce $A$, and when finding solutions to $A{\bf x}={\bf 0}$, just keep in mind that the missing column is all 0's.
Suppose after doing this, you obtain
$$
\left[\matrix{1&0&0&0&-1 \cr 0&0&1&1&0 \cr 0&0&0&0&0 \cr 0&0&0&0&0 \cr }\right]
$$
Now, look at the columns that do not contain any of the leading row entries. These columns correspond to the free variables of the solution set to $A{\bf x}={\bf 0}$ Note that at this point, we know the dimension of the null space is 3, since there are three free variables. That the null space has dimension 3 (and thus the solution set to $A{\bf x}={\bf 0}$ has three free variables) could have also been obtained by knowing that the dimension of the column space is 2 from the rank-nullity theorem.
The "free columns" in question are 2,4, and 5. We may assign any value to their corresponding variable.
So, we set $x_2=a$, $x_4=b$, and $x_5=c$, where $a$, $b$, and $c$ are arbitrary.
Now solve for $x_1$ and $x_3$:
The second row tells us $x_3=-x_4=-b$ and the first row tells us $x_1=x_5=c$.
So, the general solution to $A{\bf x}={\bf 0}$ is
$$
{\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right]
$$
Let's pause for a second. We know:
1) The null space of $A$ consists of all vectors of the form $\bf x $ above.
2) The dimension of the null space is 3.
3) We need three independent vectors for our basis for the null space.
So what we can do is take $\bf x$ and split it up as follows:
$$\eqalign{
{\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right]
&=\left[ \matrix{0\cr a\cr 0\cr 0\cr 0}\right]+
\left[\matrix{c\cr 0\cr 0\cr 0\cr c}\right]+
\left[\matrix{0\cr 0\cr -b\cr b\cr 0}\right]\cr
&=
a\left[ \matrix{0\cr1\cr0\cr 0\cr 0}\right]+
c\left[ \matrix{1\cr 0\cr 0\cr 0\cr 1}\right]+
b\left[ \matrix{0\cr 0\cr -1\cr 1\cr 0}\right]\cr
}
$$
Each of the column vectors above are in the null space of $A$. Moreover, they are independent. Thus, they form a basis.
I'm not sure that this answers your question. I did a bit of "hand waving" here. What I glossed over were the facts:
1)The columns of the echelon form of $A$ that do not contain leading row entries correspond to the "free variables" to $A{\bf x}={\bf 0}$. If the number of these columns is $r$, then the dimension of the null space is $r$ (again, if you know the dimension of the column space, you can see that the dimension of the null space must be the number of these columns from the rank-nullity theorem).
2) If you split up the general solution to $A{\bf x}={\bf 0}$ as done above, then these vectors will be independent (and span of course since you'll have $r$ of them).
Write the system
$$\left\{\begin{array}{rcl} x-2y+2z+4t& =& 0 \\ z+t & = & 0\end{array}\right.$$
as
$$\left\{\begin{array}{rcl} x+2z& =& 2y-4t \\ z & = & -t\end{array}\right.$$ and solve it. You get,
$$\left\{\begin{array}{rcl} x& =& 2y-2t \\ z & = & -t\end{array}\right.$$ That is, $$(2y-2t,y,-t,t)$$ is an element of the null space for any $y,t.$ Now, look for two linearly independent vectors.
(Note that the kernel has dimension $2.$ So the system has infinitely many solutions that have to depend on two parameters.)
Best Answer
Great that you have found a row echelon form.
From the second row, we can conclude that $x_4=0$.
Also, from there, and the first row of the row echelon form, we have
$$x_1+2x_2-3x_3=0$$
Now you have a relationship between the variables. Hopefully you can take it from here.