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Homework Help: Linear Algebra - Vector Equations

  1. Jan 29, 2012 #1
    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

    Attached Files:

  2. jcsd
  3. Jan 29, 2012 #2
    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 [tex]av_1+cv_2+dv_3 = b[/tex]
  4. Jan 29, 2012 #3
    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.
  5. Jan 29, 2012 #4
    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
  6. Jan 29, 2012 #5


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    Science Advisor

    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).
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