Contradictory result when testing Linear independence using Gaussian elimination Consider a set of vectors - (2,3,1) , (1,-1,2) and (7,3,8).
I want to find if its linearly dependent or independent.
Putting it as:
\begin{equation}
2a + b  + 7c = 0 \\
3a - b  + 3c = 0 \\
 a + 2b + 8c = 0
\end{equation}
If I use Gaussian elimination of equations to calculate row echelon form of the matrix:
\begin{bmatrix}
 2&1&7&0 \\ 
 3&-1&3&0 \\ 
 1&2&8&0 \\
\end{bmatrix}
I get:
\begin{bmatrix}
 1&0&0&0 \\ 
 0&1&0&0 \\ 
 0&0&1&0 \\
\end{bmatrix}
which means a = b = c = 0


*

*So, as per
Determine if vectors are linearly independent
the vectors are linearly independent.

*But I know that they are linearly dependent, with a = 2, b = 3, c =
-1
Why is there such a contradiction? did i do something procedurally wrong?
Note: I just want to understand the mistake in above procedure, dont want an alternate solution like using determinant etc (unless this procedure itself is totally wrong!)
 A: You don't need all that linear equations system and etc. mess. This just messes you, annoys you and distracts you away from more important things.
Just take your vectors, put them in matrix form (as all rows or all columns: it does not matter) and do Gauss. You can even choose to put in the upper row a vector which has $\;1\;$ as first entry, in case there is one...and there is in your case!:
$$\begin{pmatrix}1&-1&2\\2&3&1\\7&3&8\end{pmatrix}\stackrel{R_2-2R_1,\,R_3-7R_1}\longrightarrow\begin{pmatrix}1&-1&2\\0&5&-3\\0&10&-6\end{pmatrix}$$
and we can see at once that the third row equals twice the second one and we're done: the vectors are lin. dependent.
A: There's no contradiction: you reduced echelon form is wrong. Let me do it again:
\begin{align*}
&\begin{bmatrix}
2&1&7\\
3&-1&3\\
1&2&8
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&2&8\\
2&1&7\\
3&-1&3
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&2&8\\
0&-3&-9\\
0&-7&-21
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1 & 2 & 8\\
0 & 1 & 3\\
0 & 1 & 3
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1 & 0 & 2\\
0 & 1 & 3\\
0 & 0 & 0
\end{bmatrix},\\
&\text{whence the solutions: }\hspace{3.5em}\begin{bmatrix}a\\b\\c\end{bmatrix} =t\begin{bmatrix}2\\3\\-1\end{bmatrix}.
\end{align*} 
A: First note that
$$
\left[\begin{array}{ccc|c}
 2&1&7&0 \\ 
 3&-1&3&0 \\ 
 1&2&8&0 \\
\end{array}\right]
\Rightarrow
\left[\begin{array}{ccc|c}
 1&0&2&0 \\ 
 0&1&3&0 \\ 
 0&0&0&0 \\
\end{array}\right]
$$
Using matrix equations

We have
  $$
a\begin{bmatrix}
1\\
0\\
\end{bmatrix}
+b\begin{bmatrix}
0\\
1\\
\end{bmatrix}
+c\begin{bmatrix}
2\\
3\\
\end{bmatrix}
=\begin{bmatrix}
0\\
0\\
\end{bmatrix}
$$
  $$
a\begin{bmatrix}
1\\
0\\
\end{bmatrix}
+b\begin{bmatrix}
0\\
1\\
\end{bmatrix}
=-c\begin{bmatrix}
2\\
3\\
\end{bmatrix}
$$
  $$
\begin{bmatrix}
a\\
b\\
\end{bmatrix}
=-c\begin{bmatrix}
2\\
3\\
\end{bmatrix}
\neq\begin{bmatrix}
0\\
0\\
\end{bmatrix}
$$
  Therefore, the three vectors are linearly dependent. 

Using determinants

We have $$\begin{vmatrix}
 2&1&7 \\ 
 3&-1&3 \\ 
 1&2&8\\
\end{vmatrix}=
\begin{vmatrix}
 1&0&2 \\ 
 0&1&3 \\ 
 0&0&0 \\
\end{vmatrix}=0$$
  Therefore, the three vectors are linearly dependent. 

