Answer
a basis for the nullspace of $A$ consists of the vectors
$$\left[\begin{aligned}-1\\ 1\\1 \\0 \end{aligned}\right], \left[\begin{aligned} 2\\ -2\\0\\1 \end{aligned}\right].$$
Work Step by Step
Given the matrix
$$
A= \left[ \begin {array}{cccc} 1&3&-2&4\\ 0&1&-1&2
\\ -2&-6&4&-8\end {array} \right]
.
$$
The reduced row echelon form is given by
$$
\left[ \begin {array}{cccc} 1&0&1&-2\\ 0&1&-1&2
\\ 0&0&0&0\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned} x_{1} +x_3-2x_4&=0
\\
x_{2} -x_3+2x_4&=0
\end{aligned}.
$$
The solution of the above system is $x_1=-s+2t$,$x_2=s-2t$, $x_3=s$ and $x_4=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2} \\x_3 \\x_4 \end{aligned}\right]= \left[\begin{aligned} -s+2t\\ s-2t\\s\\t \end{aligned}\right]=s\left[\begin{aligned}-1\\ 1\\1 \\0 \end{aligned}\right]+t\left[\begin{aligned} 2\\ -2\\0\\1 \end{aligned}\right] .$$
So, a basis for the nullspace of $A$ consists of the vectors
$$\left[\begin{aligned}-1\\ 1\\1 \\0 \end{aligned}\right], \left[\begin{aligned} 2\\ -2\\0\\1 \end{aligned}\right].$$
