# Linear Algebra

1. Nov 1, 2007

### will

1. The problem statement, all variables and given/known data
4) explain why S is not a basis for R^3
(a) S={(1,2,-1),(0,0,0),(1,0,1)}
(b) explain why S is not a basis for R^3
S={(2,4,5),(-1,3,6),(7,7,9),(-4,2,-4)}

2. Relevant equations all proofs

3. The attempt at a solution

(a)**The set S is a set with the 0 vector, (0,0,0). Such a set is always dependant and can therefore never be a basis for R3.

The set without the zero vector S={(1,2,-1),(1,0,1} is also no basis for R3, it contains less vectors than the dimension of R3 (which is 3).

(b)**To have a bas you need only 3 vectors by having 4 that makes this system linearly dependent and there for it doesn't represent a base.. to make it a base one has to be eliminated and check if they have linear independency

"" is there a better more elegant answer, by justifying an Axion or a theorem" my teacher is so hard" thanks guys

2. Nov 1, 2007

### verafloyd

Ahh, teachers :)

Okay, search for what "rank" means if you already don't know, then relate the number of linearly independent vectors in S to the rank of S (smaller, larger, equal...). This would be the elegant way of explaining it :)

3. Nov 1, 2007