Q: Find all subsets of the set that forms a basis for $\displaystyle \mathbb{R}^{3}$.

$\displaystyle S=\{(1,3,-1),(-4,1,1),(-2,7,-3),(2,1,1)\}$

A: well, I did a similar one of these and got the same answer as book, but with a bunch of negative 1's unlike their strictly positive solution. I figured it was ok, because subspaces are closed under scalar multiplication (then again, we are just talking about a subset). I don't know if that’s alright.

My approach is to set up a homogenous linear combination and solve for the constants. Then I set up my new linear combination and then solve for each vector in the set S. Then that gives me all my subspaces. I have to hope that S is linearly dependent for this to work. Is that the correct method??