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!)

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*} 
