# Linear Dependence/Linear Combination Question

1. Sep 9, 2015

### Tim 1234

• Member warned about posting without the homework template
S={v1, v2, v3}

v1=[1, -1], v2=[-2, 2], v3=[3, a]

a) For what value(s) a is the set S linearly dependent?
b)For what value(s) a can v3 be expressed as a linear combination of v1 and v2?

a) p=3 and m=2
3-2=1 free variable
Therefore the set has non-trivial solutions and is linearly dependent

b) I reduced the matrix to the following:

1 -2 3 0
0 0 a+3 0

Does a need to equal -3 for a linear combination to be valid?

I recognize v2=(-2)v1 - is this relevant at all?

Last edited by a moderator: Sep 9, 2015
2. Sep 9, 2015

### Staff: Mentor

You haven't answered the question -- "For what value(s) of a is the set linearly dependent?" Your answer should say something about a.
Yes, it's relevant to both question parts. Since v2 is a multiple of v1 (and vice versa), part b boils down to the question, "what values of a make v3 a scalar multiple of either v1 or v2?

3. Sep 10, 2015

### HallsofIvy

In what sense does a set have "solutions" at all?
What I think you meant to say, and what you should say, is "this set is linearly independent if and only if the only values of p, q, and r, that make p[1, -1]+ q[-2, 2]+ r[3, a]= [0, 0] are p= q= c= 0." That gives the two equations p- 2q+ 3r= 0 and -p+ 2q+ 3r= 0. But p= 2, q= 1, r= 0 will work. Therefore the set is not independent.

Personally, I dislike changing everything to matrices (and you shouldn't say "I reduced the matrix" when you haven't yet shown a matrix to begin with!). $v_3$ is a linear combination of $v_2$ and $v_3$ if and only if there exist non-zero values, p and q, such that [3, a]= p[1, -1]+ q[-2, 2] which gives the two equations p- 2q= 3 and -p+ 2q= a. What happens if you add those two equations?