<s>[INST] Justify that the vectors are linearly dependent Let $V$ be a real vector space and $a,b,c,d,e\in V$. 
We have the vectors \begin{align*}&v_1=a+b+c \\ &v_2=2a+2b+2c-d \\ &v_3=a-b-e \\ &v_4=5a+6b-c+d+e \\ &v_5=a-c+3e \\ &v_6=a+b+d+e\end{align*} 
I want to justify that these vectors are linearly dependent. 
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These six vectors are linearly dependetn iff we can get the zer0-vector by a linear combination of these vectors, \begin{equation*}\lambda_1v_1+\lambda_2v_2+\lambda_3v_3+\lambda_4v_4+\lambda_5v_5+\lambda_6v_6=0\end{equation*} where at least of the coefficients $\lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_4$ or $\lambda_6$ is not equal to zero. 
We have the following: 
\begin{align*}&\lambda_1v_1+\lambda_2v_2+\lambda_3v_3+\lambda_4v_4+\lambda_5v_5+\lambda_6v_6=0  \\ & \Rightarrow \lambda_1\left (a+b+c\right )+\lambda_2\left (2a+2b+2c-d\right )+\lambda_3\left (a-b-e\right )+\lambda_4\left (5a+6b-c+d+e\right )+\lambda_5\left (a-c+3e\right )+\lambda_6\left (a+b+d+e\right )=0 \\ & \Rightarrow \lambda_1a+\lambda_1b+\lambda_1c+2\lambda_2 a+2\lambda_2 b+2\lambda_2 c-\lambda_2d+\lambda_3 a-\lambda_3 b-\lambda_3 e+5\lambda_4 a+6\lambda_4 b-\lambda_4 c+\lambda_4 d+\lambda_4 e+\lambda_5 a-\lambda_5 c+3\lambda_5 e+\lambda_6a+\lambda_6b+\lambda_6d+\lambda_6e=0 \\ & \Rightarrow a\left (\lambda_1+2\lambda_2+\lambda_3+5\lambda_4+\lambda_5+\lambda_6\right )+b\left (\lambda_1+2\lambda_2-\lambda_3+6\lambda_4+\lambda_6\right )+c\left (\lambda_1+2\lambda_2 -\lambda_4-\lambda_5\right )+d\left (-\lambda_2+\lambda_4+\lambda_6\right ) +e\left (-\lambda_3   +\lambda_4  +3\lambda_5 +\lambda_6\right )=0\end{align*}
Is everything correct? How could we continue? 
Do we use the fact that $a,b,c,d,e\in V$ ? But how? 
 [/INST] If you select the set $\{v_k \}_{k=1}^{\color{red}{5}}$ you can write
$$
\left(\begin{array}{c}v_1\\ v_2 \\ v_3 \\v_4 \\ v_5 \end{array}\right) = \left(\begin{array}{ccccc} 
1 &  1 &  1 &  0 &  0 \\ 
2 &  2 &  2 & -1 &  0 \\ 
1 & -1 &  0 &  0 & -1 \\
5 &  6 & -1 &  1 &  1 \\
1 &  0 & -1 &  0 &  3
\end{array}\right) \left(\begin{array}{c} a \\ b \\ c  \\ d \\ e\end{array}\right)
$$
Which can be solved for $\{a, \cdots, e \}$ since the array as determinant different from zero (-44).  This means it is possible to write $a, \cdots, e$ as a linear combination of $v_1, \cdots, v_5$. 
This means that $v_6$ can be written as a linear combination of $\{v_k \}_{k=1}^{\color{red}{5}}$
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