<s>[INST] 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!)
 [/INST] 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. 

</s>