# Homework Help: Linear Algebra - Vector Equations

1. Jan 29, 2012

### GreenPrint

1. The problem statement, all variables and given/known data

See attachment for the question

2. Relevant equations

3. The attempt at a solution

I don't understand...

the back of the book says that the answer is no

when I solve reduce row echelon form the augmented matrix Ab I get that

x1 = 2 - 5x3
x2 = 3 - 4x3
x3 is free
i thought thought that the answer was yes because the system is consistent and x3 can just be anything i don't understand why the answer is no

thanks for any help

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2. Jan 29, 2012

### fauboca

If the determinant is 0, they are linearly dependent. If the determinant is non-zero, then the vectors are independent.

If they are dependent, the reduce row echelon form will tell you $$av_1+cv_2+dv_3 = b$$

3. Jan 29, 2012

### DaleSwanson

I came here to ask the exact same question. Since x3 is free there are an infinite number of solutions. This has spurred a debate over on cramster, which is why I came here. I think that any of the solutions will work. For example, the values x1 = -3, x2 = -1, x3 = 1 cause the equation x1(a1) + x2(a2) + x3(a3) = b to be true.

4. Jan 29, 2012

### GreenPrint

I'm confused... is the answer yes or no and i'm not exactly sure why
although i know about linear independence and dependence and determinants and crap we haven't gotten to that part in the course yet (i prestudied the course over the break) so i think i'm suppose to come up with a sollution that dosen't involve that stuff

5. Jan 29, 2012

### Deveno

books can be wrong (it might be helpful to know what else the "back of the book" says besides just "no").

the situation here is that a3 is a linear combination of a1 and a2: (5,-6,8) = 5(1,-2,0) + 4(0,1,2). so we may as well ask if b is in the span of a1 and a2. and it is:

b = 2a1 + 3a2

= 2a1 + 3a2 + 0a3

(if x3 is free, we should be able to ignore one of the column vectors, since row rank = column rank).

(note to moderators: the link provided earlier in this thread provides a step-by-step solution to the problem posted here, and i am assuming the OP is already familiar with this solution).