Is the vector in the space of 3 other vectors I have a set of 3 vectors
   $$ IE = {[1, 1, -3]; [2, -1, 3]; [-6, 3, -9]}$$
I want to know if the vector [1, 4, -12] , belongs (or is in the span?) to my previous set. 
So here's what I did.
$$
        \begin{matrix}
        1 & 2 & -6 & [c1]\\
        1 & -1 & 3 & [c2]\\
        -3 & 3 & -9 & [c3]\\
        \end{matrix}
=  \begin{matrix}
        1 \\
        4 \\
        -12 \\
        \end{matrix}
$$
Now I start Gauss
L2-L1
$$
        \begin{matrix}
        1 & 2 & -6 \\
        0 & -3 & 9 \\
        -3 & 3 & -9 \\
        \end{matrix}
=       \begin{matrix}
        1 \\
        3 \\
        -12 \\
        \end{matrix}
$$
L3+3L1
$$
        \begin{matrix}
        1 & 2 & -6 \\
        0 & -3 & 9 \\
        0 & 9 & -27 \\
        \end{matrix}
=       \begin{matrix}
        1 \\
        3 \\
        -9 \\
        \end{matrix}
$$
L2 / 3
and then L3-9L2
$$
        \begin{matrix}
        1 & 2 & -6 \\
        0 & 1 & -3 \\
        0 & 0 & 0 \\
        \end{matrix}
=       \begin{matrix}
        1 \\
        -1 \\
        0 \\
        \end{matrix}
$$
My Answer : 


*

*The variable C3 (3rd column) has no pivot. which means it is a free variable.

*There is no contradiction , so the vector $$ [1, 4, -12] $$ is in the space of my                                       Set.


I have 2 questions : 


*

*does the fact that I have a free variable change anything ?

*A contradiction happens when I have a line full of zeros that equals a non zero value, right ?


Like 
$$
        \begin{matrix}
        0 & 0 & 0 \\
        \end{matrix}
=       \begin{matrix}
        2 \\
        \end{matrix}
$$
Is there any other way to have a contradiction ?
Sorry if the title isn't clear, I have trouble traducing the question.. 
Thanks
 A: Your answer is correct.


*

*The fact that you have a free variable means many things. For instance, it means that your original set of three vectors were linearly dependent (one equals a linear combination of the other two). Therefore there are vectors which are outside the span of these three vectors.
We could also continue row reduction to find out just what values of $c_1,c_2,c_3$ give the linear combination we are interested in:
$$
\left[ \begin{array}{ccc|c}
1 & 2 & -6 &1 \\
0 & 1 & -3 &-1 \\
0 & 0 & 0 &0\end{array} \right] \sim
\left[ \begin{array}{ccc|c}
1 & 0 & 0 &3 \\
0 & 1 & -3 &-1 \\
0 & 0 & 0 &0\end{array} \right]$$
So $c_1=3$, and $c_2=3c_3 - 1$. Let's see if this works in general:
$$
3\left[ \begin{array}{c}
1 \\
1 \\
-3\end{array} \right] +
(3c_3-1)\left[ \begin{array}{c}
2 \\
-1 \\
3\end{array} \right]+
c_3\left[ \begin{array}{c}
-6 \\
3 \\
-9\end{array} \right]  =
\left[ \begin{array}{c}
3+2(3c_3-1)-6c_3 \\
3-(3c_3-1)+3c_3 \\
-9+3(3c_3-1)-9c_3\end{array} \right]=\left[ \begin{array}{c}
1 \\
4 \\
-12\end{array} \right]  $$
So this means you can actually find infinitely many linear combinations by taking different values for $c_3$.

*If you do row reduction, that is essentially the only way to have a contradiction. The reason it is a contradiction is because if you remember that the matrix is "shorthand" for a system of linear equations, the linear equation corresponding to a row of zeros on the left and a number like 2 on the right is of the form$$0c_1 + 0c_2 + 0c_3 = 2 \implies 0 = 2$$
which is why it is a contradiction. Any other time, if you reduce to row echelon form and you don't have a row of zeros on the left corresponding to a nonzero number in the augmented column, you essentially are using "back-substitution" to solve the system of equations.

