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Linear dependence question

  1. Mar 19, 2013 #1
    1. The problem statement, all variables and given/known data

    For which real values of [itex]\lambda[/itex] do the following vectors form a linearly dependent set in [itex]\mathbb{R}^{3}[/itex]

    [itex]v_{1}=(\lambda ,-\frac{1}{2},-\frac{1}{2}), v_{2}=(-\frac{1}{2},\lambda ,-\frac{1}{2}), v_{3}=(-\frac{1}{2},-\frac{1}{2},\lambda )[/itex]


    3. The attempt at a solution

    I know that

    [itex]k_{1}(\lambda ,-\frac{1}{2},-\frac{1}{2})+k_{2}(-\frac{1}{2},\lambda ,-\frac{1}{2})+k_{3}(-\frac{1}{2},-\frac{1}{2},\lambda )=0[/itex]

    will have non trivial solutions if the vectors form a linearly dependent set. The problem is when I put this in matrix form with the lambdas on the diagonal I don't know how to reduce it to row echelon form.

    Is that the correct thing to do? and how can I reduce the matrix with lamdas in it if so?

    thanks any help appreciated
     
  2. jcsd
  3. Mar 19, 2013 #2

    tiny-tim

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    hi kwal0203! :smile:

    hint: if the rows of a square matrix are independent, then what can you say about the matrix? :wink:

    (btw, i assume you've already seen the one obvious value?)
     
  4. Mar 19, 2013 #3
    Is that something to do with the rank? I'm not up to studying that yet, I'm not sure what it means when the rows of a square matrix are independent. Does it mean the system will have no free variables when solving it?
     
  5. Mar 19, 2013 #4

    tiny-tim

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    first, have you done determinants (of a matrix) yet?
     
  6. Mar 19, 2013 #5
    Yes I have done work on determinants
     
  7. Mar 19, 2013 #6

    tiny-tim

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    ok, then when is the determinant zero? :wink:
     
  8. Mar 19, 2013 #7
    Lambda=x

    Using the first row of the Matrix I got:

    X^3-(3x/4)+1/4, so the determinant is 0 when that equation equals zero.

    Is that correct?

    So x=1?
     
  9. Mar 19, 2013 #8

    tiny-tim

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    that doesn't seem to fit the obvious solution :confused:

    can you show your calculations?
     
  10. Mar 19, 2013 #9
    Thanks for the help I do understand this now. The system will be linearly dependant when the determinant is equal to zero so he question is essentially asking for what values of lambda is the determinant zero.
     
  11. Mar 19, 2013 #10
    The answer in my book is lambda equals -1/2 or 1

    I'm guessing -1/2 was the obvious answer? :)
     
  12. Mar 19, 2013 #11

    tiny-tim

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    yes … you can instantly see that -1/2 makes all three vectors the same! :biggrin:

    (unfortunately, -1/2 doesn't seem to fit your equation :redface:)
    exactly! :smile:
     
  13. Mar 19, 2013 #12
    Oops it should be -1/4 in the determinant formula :)
     
  14. Mar 19, 2013 #13

    tiny-tim

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    ok, that makes sense!

    so it's x3 - 3x/4 -1/4 = 0

    the way you solve it (since it's a cubic equation, which isn't easy)

    is to notice that -1/2 is obviously a solution to the original equation, so divide by (x + 1/2),

    and you get x2 - x/2 - 1/2, which you can easily solve (what is the third solution?) :wink:
     
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