Row reduce $A$,
\begin{equation*}
A\rref
\begin{bmatrix}
\leading{1} & 0 & 0 & 1 & 1 \\
0 & \leading{1} & 0 & -3 & -1 \\
0 & 0 & \leading{1} & -2 & -2 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{equation*}
So $r=3$ for this matrix. Then
\begin{align*}
\dimension{\nsp{A}}
&=\nullity{A}&&\knowl{./knowls/definition.NOM.knowl}{\text{Definition NOM}}\\
&=\left(\nullity{A}+\rank{A}\right)-\rank{A}\\
&=5-\rank{A}&&\knowl{./knowls/theorem.RPNC.knowl}{\text{Theorem RPNC}}\\
&=5-3&&\knowl{./knowls/theorem.CRN.knowl}{\text{Theorem CRN}}\\
&=2
\end{align*}
We could also use Theorem BNS and create a basis for $\nsp{A}$ with $n-r=5-3=2$ vectors (because the solutions are described with 2 free variables) and arrive at the dimension as the size of this basis.