Answer
$N(A)=\{ (X_1,X_2,X_3,X_4,X_5) \in R^5: X_1=\frac{1}{8}(9X_2-3X_3-3X_4+5X_5)\}$
Work Step by Step
We are given: $A=\begin{bmatrix}
8 & -9 & 3 & 3 & 5
\end{bmatrix}$
Since $A$ is $1 \times 5$ matrix, X is $5 \times 1$
Set $AX=0 \rightarrow \begin{bmatrix}
8 & -9 & 3 & 3 & -5
\end{bmatrix}\begin{bmatrix}
X_1 \\
X_2 \\
X_3 \\
X_4 \\
X_5
\end{bmatrix}=0$
$\rightarrow 8X_1-9X_2+3X_3+3X_4-5X_5=0$
$X_1=\frac{1}{8}(9X_2-3X_3-3X_4+5X_5)$
Hence null space of the given matrix $A$ is:
$N(A)=\{ (X_1,X_2,X_3,X_4,X_5) \in R^5: X_1=\frac{1}{8}(9X_2-3X_3-3X_4+5X_5)\}$
