Answer
$N(A)=1$
Work Step by Step
We are given: $A=\begin{bmatrix}
1 & 3
-2 & -6
\end{bmatrix}$
Since $A$ is $1 \times 5$ matrix, X is $5 \times 1$
Set $AX=0 \rightarrow \begin{bmatrix}
1 & 3
-2 & -6
\end{bmatrix} \begin{bmatrix}
x \\
y \end{bmatrix}=0$
$\rightarrow \begin{bmatrix}
x+3y\\
-2x-6y
\end{bmatrix}=0$
We have $x+3=0$ and $-2x-6y=0$
$\rightarrow x=-3y$
$v=(-3y,y)=y(-3,1)$ means that $v \in $ span $\{ (-3,1) \}$ and the $N(A)$ is a subset of span $\{-3,1\}$
We can obtain:
$\begin{bmatrix}
1 & 3\\
-2 & -6
\end{bmatrix}\begin{bmatrix}
-3a \\
a
\end{bmatrix}=\begin{bmatrix}
-3a+ 3a\\
6a-6a
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
Hence the null space of $A$ is span $\{(-3,1)\}$ and since the set of vectors is linearly independent we have $\{-3, 1\}$ is a basis for the null space of $A$.
Thus $N(A)=1$
