# linear dependence

Printable View

• Dec 5th 2009, 07:18 AM
cookiejar
linear dependence
Find values of t for which the set of vectors
2 2t+6
-t 4t

is linearly dependent.

Please help me. I really don't know how to answer this one. Thanks!
• Dec 5th 2009, 07:40 AM
Shanks
they are linearly dependent ,is equavalent to say:
$2\times 4t=-t\times (2t+6)$
• Dec 5th 2009, 07:43 AM
HallsofIvy
Quote:

Originally Posted by cookiejar
Find values of t for which the set of vectors
2 2t+6
-t 4t

is linearly dependent.

Please help me. I really don't know how to answer this one. Thanks!

Clarification: Are these vectors written "horizontally" or "vertically". That is are they "<2, 2t+ 6> and <-t, 4t>" or "<2, -t> and <2t+6, 4t>"?

Two vectors are "dependent" if one is a multiple of the other (more generally, a set of vectors is dependent with there exist a linear combination of the vectors, with not all coefficients equal to 0, that is equal to 0. For two vectors, u and v, this says that au+ bv= 0 and then either u= (b/a)v, if a is not 0, or v= (a/b)v if it is.

If your vectors are <2, 2t+6> and <-t, 4t>, then they are dependent if and only if <2, 2t+6>= a<-5, 4t> for some a. That is saying that 2= -at and that 2t+ 6= 4at. From the first equation, a= -2/t so the second equation becomes 2t+ 6= 4(-2/t)t= -8. Solve that for t.

If your vectors are <2, -t> and <2t+6, 4t>, then they are dependent if and only if <2, -t>= a<2t+6, 4t> for some t. That is saying that 2= 2at+ 6a and -t= 4at. From the second equation, 4a= -1 so a= -1/4 and the first equation becomes 2= -(1/2)t- 3/2. Solve that for t.