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