Solution
= Solution
The <reduced row echelon form> gives $x_1+x_4=0$, $x_2-x_4=0$, and $x_3+x_4=0$. With free parameter $x_4=t$,
$$
\mathbf x=t\begin{pmatrix}-1\\1\\-1\\1\end{pmatrix}.
$$
The displayed nonzero <vector> therefore forms a <basis> for the <null space>:
$$
\boxed{\mathcal B_{\rm null}=\{(-1,1,-1,1)^T\},\qquad\dim\ker A=1.}
$$
This agrees with the <rank-nullity theorem>: there are four columns and <matrix rank> three.