Answer
$N(A)=2$
Work Step by Step
We are given: $A=\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0\\
0 & 1 & 0
\end{bmatrix}$
Since $A$ is $1 \times 5$ matrix, X is $5 \times 1$
Set $AX=0 \rightarrow \begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0\\
0 & 1 & 0
\end{bmatrix}\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}$
$\rightarrow y=0$
$v=(x,0,z)=(x,0,0)+(0,0,z)=x(1,0,0)+z(0,0,1)$ means that $v \in $ span $\{ (1,0,0);(0,0,1) \}$ and the $N(A)$ is a subset of span $\{ (1,0,0);(0,0,1) \}$
We can set: $v=a(1,0,0)+b(0,0,1) \in$ span $\{ (1,0,0);(0,0,1) \}$ then $v=(a,0,b)$ is in the nullspace of $A$.
Hence the null space of $A$ is span $\{ (1,0,0);(0,0,1) \}$ and since the set of vectors is linearly independent in $R^3$ we have $\{-3, 1\}$ is a basis for the null space of $A$.
Thus $N(A)=2$
