Answer
$nullity (A)=1$
Work Step by Step
We are given $A=\begin{bmatrix}
2 & -3\\
0 & 0 \\
-4 & 6\\
22 & -33
\end{bmatrix}$
$nullspace (A)\\
=\{5t+s,-t-s,t,s: s,t \in R\}\\
=span \{(5,-1,1,0),(1,-1,0,1)\}$
In this problem, A is a $2 \times 4$ matrix, and so, in the Rank-Nullity Theorem, $n = 2$. Further, from the foregoing row-echelon form of the augmented matrix of the system
Ax = 0, we see that $rank(A) = 1$.
Hence:
$$rank (A)+nullity(A)=n \\
nullity (A)=n-rank (A)=2-1=1$$
