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Matrices Question

  1. Sep 10, 2005 #1
    Please help me solve this problem associated with matrices; I can't seem to figure it out.

    :eek:
     
    Last edited: Sep 10, 2005
  2. jcsd
  3. Sep 10, 2005 #2

    TD

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    If the z-coëfficiënt becomes 0 while the constant isn't 0, your system won't have a solution.

    So [tex]k^2 - k = 0 \Leftrightarrow k = 0 \vee k = 1[/tex].

    But for k = 0, the constant is 0 as well so the last equation is no longer independant, then you have a 2x3 system (which has infinite solutions).

    For k = 1, it has no solution :smile:
     
    Last edited: Sep 10, 2005
  4. Sep 10, 2005 #3
    Awesome, thanks for the help! :smile:
    Wouldn't "+1" have no solution as well?
     
  5. Sep 10, 2005 #4

    TD

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    I mean +1 lol, it has a solution for -1 :wink:

    I'll correct.

    ---

    Done.

    Since, for k = -1, the last row becomes: 0 0 2 | 1 => 2z = 1 => z = 1/2
     
  6. Sep 10, 2005 #5
    But (-1)^2 - 1 = 0 as well. So there are no solutions for "+/- 1," I suppose.
    Please correct me if I am wrong. :uhh:
     
  7. Sep 10, 2005 #6

    TD

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    Watch out, mind your minus! For k = -1, you get:

    [tex]k^2 - k \Rightarrow \left( { - 1} \right)^2 - \left( { - 1} \right) = 1 + 1 = 2[/tex]

    Which is what I wrote at the end of my previous post.
     
  8. Sep 10, 2005 #7
    Sorry, I missed that. :tongue2:
    Thanks for the help!
     
  9. Sep 10, 2005 #8

    TD

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    No problem :smile:
     
  10. Sep 11, 2005 #9
    I don't think that's true, since anything greater than 2 seems to have no solution as well.
     
  11. Sep 12, 2005 #10

    TD

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    How is that?

    Take k = 3, the last row becomes: 0 0 6 | 3 => 6z = 3 <=> z = 1/2
     
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