Using Gauss elimination to check for linear dependence I have been trying to establish if certain vectors are linearly dependent and have become confused (in many ways). when inputting the vectors into my augmented matrix should they be done as columns or as rows ?
eg if my vectors are [ 4 -1 2 ], [-4 10 2] 
I am looking to solve a[ 4 2 2 ] + b[2 3 9]  = [0,0,0]
but am unsure how the matrix should be created
$\begin{bmatrix} 
4 & -1 & 2 \\ 
-4 & 10 & 2 \\  
\end{bmatrix}$
vs
$\begin{bmatrix} 
4 & -4\\ 
-1 & 10\\
2 & -2 
  \end{bmatrix}$
using the last method I end up with the following in row reduced echelon form (including the 0's as answers on the right)
$\begin{bmatrix} 
1 & 0 & 0\\ 
0 & 1 & 0\\
0 & 0 &0 
  \end{bmatrix}$
so to me this says a = 0, b = 0 , but I don't get what the last row means ?

If $ \: \: a \begin{pmatrix}
  4 \\
  2 \\
  2
 \end{pmatrix} $ + $ b \begin{pmatrix}
  2 \\
  3 \\
  9
 \end{pmatrix} $ = $ \begin{pmatrix}
  0 \\
  0 \\
  0
 \end{pmatrix} \: \:$ then $ \: \: \begin{pmatrix}
  4a+2b \\
  2a+3b\\
  2a+9b
 \end{pmatrix} $ = $ \begin{pmatrix}
  0 \\
  0 \\
  0
 \end{pmatrix}$.
So we have three simultaneous equations,
$ 4a+2b=0 \quad (1)\\ 2a+3b=0 \quad (2)\\ a+9b=0 \quad (3)$
We can now form a coefficient matrix,
Let $ \: A =  \begin{pmatrix}
  4 & 2 \\
  2 & 3 \\
  2 & 9
 \end{pmatrix} $
Then we put this into reduced row echelon form (RREF) using Gauss-Jordan elimination,
and we get $\; \; \begin{pmatrix}
  1 & 0 \\
  0 & 1 \\
  0 & 0
 \end{pmatrix} $.
This tells us that there is a unique solution to the simultaneous equation formed above and that is $ a=0 $ and $ b=0 $.
The reason that the bottom row is zero is because the last of the simultaneous  equations can be derived from the other two.
$ (3) = 4*(2) - \frac{5}{2}(1) $.
So it doesn't actually give us any more information.
So if the only solution is $a=0=b$, the vectors $ \begin{pmatrix}
  4 \\
  2 \\
  2
 \end{pmatrix} $ and $ \begin{pmatrix}
  2 \\
  3 \\
  9
 \end{pmatrix} $ must be linearly independent.
However, it should be clear from the 'top' elements, 4 and 2, that $b=-2a$, but the the 'middle' elements, 2 and 3, need $b=-\frac{2}{3}a$. So the vectors must be independent.
