I've seen a couple of examples like this but there's a thing I don't understand. I have to determine a basis for the solution set of the system:
$$x-y-7=0\\ 2x-y+7=0$$
So based on the examples, this is what I know I have to do (I don't know how to format matrices, sorry for that):
$\left(\begin{matrix}1&-1&-7&0\\2&-1&7&0\end{matrix}\right)$
I know how to reduce it, so it ends up being like this:
$\left(\begin{matrix}1&0&14&0\\0&1&21&0\end{matrix}\right)$
As far as I understand it this means that:
$\left(\begin{matrix}x\\y\\7\end{matrix}\right)$
Equals:
$\left(\begin{matrix}-14\\-21\\7\end{matrix}\right)$
And I can factorize the $7$ or $-7$ (both ways the result should be a basis as far as I know, there's an infinite number of bases):
$\left(\begin{matrix}2\\3\\-1\end{matrix}\right)\times(-7)$
And this is the part where examples end it, saying that $(2, 3, -1)$ would be the basis I'm looking for, but I don't think I understand everything about it. As far as I understand the solution set of the system is only one point $(-14,-21)$, this isn't a vector space and thus can't have a basis, so I think that my understanding of 'solution set' should be wrong.
My second concern is if I need elements other than $(2, 3, -1)$ to complete the basis, because I don't understand how $(2, 3, -1)$ is even related to the equations. I also get the feel that the $7=7$ thing shouldn't even be there to begin with...
I just realized that the 7 might be a typo in the book, it should've been a z probably. Ignore everything I said then; thanks to those who helped me notice.