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