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