Context from this video and the lecturer did not proceed to solve for the basis vectors of this null space.:
I need to find the basis vector of the null space and I get that we need to first find the null space followed by finding the basis. I follow through until I'm instructed to apply the gaussian algorithm. But all I get is the following after gausian elimination: $\begin{pmatrix}1 & 0\\0&1\\0&0\\0&0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}=x_1\begin{pmatrix}1\\0\\0\\0\end{pmatrix} + x_2\begin{pmatrix}0\\1\\0\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}$.
Does that mean the basis vectors are... $\begin{pmatrix} x_1 \\ x_2\end{pmatrix}$...?.. how do I find the values?....
The basis vectors are: $$\begin{pmatrix}-2 \\ 1 \\ 0 \\ 0\end{pmatrix} \quad \text{ and } \quad \begin{pmatrix}0 \\ 0 \\ -3 \\ 1\end{pmatrix}.$$ Why? Recall that bases are defined by two properties: linear independence and spanning.
As Gauss-Jordan elimination of whatever homogenous system you started with showed, a general vector in $\operatorname{Null} A$ took the form $$\vec{x} = s\begin{pmatrix}-2 \\ 1 \\ 0 \\ 0\end{pmatrix} + t\begin{pmatrix}0 \\ 0 \\ -3 \\ 1\end{pmatrix},$$ where $s$ and $t$ are arbitrary scalars. That is, every element of $\operatorname{Null} A$ is a linear combination of the above vectors, which means the above vectors span $\operatorname{Null} A$.
To verify linear independence, we simply need to prove that $$s\begin{pmatrix}-2 \\ 1 \\ 0 \\ 0\end{pmatrix} + t\begin{pmatrix}0 \\ 0 \\ -3 \\ 1\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0 \\ 0\end{pmatrix} \implies s = t = 0.$$ Equating the second coordinates, $1s + 0t = 0 \implies s = 0$. Equating the fourth coordinates, $0s + 1t = 0 \implies t = 0$ (that's why the second and fourth rows are highlighted). So, the only possible solution is $s = t = 0$, and the vectors are linearly independent. This completes the proof that they form a basis for $\operatorname{Null} A$.