We will find the left null space of \begin{equation*} A= \begin{bmatrix} 1 & -3 & 1 \\ -2 & 1 & 1 \\ 1 & 5 & 1 \\ 9 & -4 & 0 \end{bmatrix} \end{equation*}
We transpose $A$ and row-reduce, \begin{equation*} \transpose{A}= \begin{bmatrix} 1 & -2 & 1 & 9 \\ -3 & 1 & 5 & -4 \\ 1 & 1 & 1 & 0 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 2 \\ 0 & \leading{1} & 0 & -3 \\ 0 & 0 & \leading{1} & 1 \end{bmatrix} \end{equation*}
Applying Definition LNS and Theorem BNS we have \begin{equation*} \lns{A}=\nsp{\transpose{A}}= \spn{\set{ \colvector{-2\\3\\-1\\1} }} \end{equation*}
If you row-reduce $A$ you will discover one zero row in the reduced row-echelon form. This zero row is created by a sequence of row operations, which in total amounts to a linear combination, with scalars $a_1=-2$, $a_2=3$, $a_3=-1$ and $a_4=1$, on the rows of $A$ and which results in the zero vector (check this!). So the components of the vector describing the left null space of $A$ provide a relation of linear dependence on the rows of $A$.