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