# 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

$$\begin{matrix} 1 & 2 & -6 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \\ \end{matrix} = \begin{matrix} 1 \\ -1 \\ 0 \\ \end{matrix}$$

1. The variable C3 (3rd column) has no pivot. which means it is a free variable.
2. There is no contradiction , so the vector $$[1, 4, -12]$$ is in the space of my Set.

I have 2 questions :

1. does the fact that I have a free variable change anything ?
2. 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

-
Note that you also multiplied L2 by -1 at some point at the end to reach your conclusion - what you have written is correct, I'm just pointing out that you did one more row-reduction step. –  Michael Boratko Sep 19 '12 at 18:04

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 $c_3$.
2. 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.