Linear dependency in $\mathbb{R}^4$ Show that the three vectors are linearly dependent in $\mathbb{R}^4$.
$u=(0,3,1,-1)\ v=(6,0,5,1)\ w=(4,-7,1,3)$
They are linearly independent if $c_{1}u+c_{2}v+c_{3}w=0\Rightarrow c_{1}=c_{2}=c_{3}=0$
$$
c_{1}u+c_{2}v+c_{3}w\Rightarrow \begin{pmatrix}
0 & 6 &4 \\ 
 3&0  &-7 \\ 
1 &5  &1 \\ 
-1 &1  &3 
\end{pmatrix}
\Rightarrow
\begin{pmatrix}
1 &0  &-\frac{7}{3} \\ 
0 &1  &\frac{2}{3} \\ 
0 &0  &0 \\0 
 &0  &0 
\end{pmatrix}
$$

*

*How does that show that the vectors are linearly dependent?


*Write each vector as a linear combination of the other two.
EDIT.
$$\begin{pmatrix}
0\\3 
\\1 
\\-1 
\end{pmatrix}=a\begin{pmatrix}
6\\0 
\\5 
\\1 
\end{pmatrix}\cdot b\begin{pmatrix}
4\\-7 
\\1 
\\3 
\end{pmatrix}\Rightarrow a=\frac{2}{7},\ b=-\frac{3}{7}$$
Have i understood 2. correct?

We have found, by row reduced echelon form (RREF), that the system $Ac=0$ has a not trivial solution (i.e. a solution with $c\neq 0$) that is precisely the definition for linear dependency for the columns of $A$.
