How to find linear dependency relationship 
The Question I am attempting to answer gives vectors of,
    (2,1,4,3) (2,3,2,3) (4,11,-1,6)   and asks for an explicit dependency relationship.

I have found them to be linearly dependent. (found the determinant = 0)
How do i then find a dependency relationship?
Looking at this explanation, I understand up until the "which implies..." in the final line.
If someone could explain it that would be much appreciated.
 Thanks, JD
 A: Look for the relationship between the three vectors:
$$
xu+yv+zw=
x
\left[\begin{matrix}
2\\
1\\
4\\
3
\end{matrix}\right]
+
y
\left[\begin{matrix}
2\\
3\\
2\\
3
\end{matrix}\right]
+
z
\left[\begin{matrix}
4\\
11\\
-1\\
6
\end{matrix}\right]
=
\left[\begin{matrix}
2 & 2 & 4\\
1 & 3 & 11\\
4 & 2 & -1\\
3 & 3 & 6\\
\end{matrix}\right]
\left[\begin{matrix}
x\\
y\\
z\\
\end{matrix}\right]
=
\left[\begin{matrix}
0\\
0\\
0\\
0
\end{matrix}\right].
$$
Writing the matrix in augmented form, and then performing Gaussian elimination gives
$$
\left[\begin{array}{ccc|c}
1 & 0 & \frac{-5}{2} & 0\\
0 & 1 & \frac{9}{2} & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{array}\right],
$$
where we see that there is one free variable, $z$. Two relationships are obtained from this matrix, namely
$$x=\frac{5z}{2},y=-\frac{9z}{2}.$$
Return to the first equation mentioned, and plug in these two relationships.
$$xu+yv+zw=\left(\frac{5z}{2}\right)u-\left(\frac{9z}{2}\right)v+zw=z\left(\frac{5}{2}u-\frac{9}{2}v+w\right)=0.$$
Notice that we can choose $z$ as we please to obtain a relationship between these three vectors. Divide $z$ out. Then we have
$$\frac{5}{2}u-\frac{9}{2}v+w=0,$$
and we can solve for $w$, obtaining a relationship between the vectors. For completeness, it is
$$w=-\frac{5}{2}u+\frac{9}{2}v.$$
A: Please check the following link
Linear Dependence/Independence of Vectors and Rank of a Matrix
and use the following tool
Linear Dependency / NULL Space / Rank / Solution to Homogeneous System of Linear Equations Calculator
