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Find value h so the matrix has infinitely many solutions?

  1. Sep 16, 2005 #1

    mr_coffee

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    Hello everyone i'm confused on what i'm suppose to do to solve this matrices.
    |7 -7|5 |
    |21 h|15|

    What am I trying to make h be? I row reduced to get
    |7 -7 |5|
    |0 21+h |0|
    So if i let h = -21 will this make it have infin. solutions? because then i would have 0 0 0?
     
    mr_coffee, Sep 16, 2005
  2. jcsd
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  3. Sep 16, 2005 #2

    TD

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    Homework Helper

    Indeed, by doing so you create 2 lineair dependant equations so when reduving, one cancels out as a 0 row. You then have 1 equation in 2 unknowns, giving an infinite set of solutions.
     
    TD, Sep 16, 2005
  4. Sep 16, 2005 #3

    mr_coffee

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    Thanks! this brings up another question..
    I'm supppose to find a value of k that will make it have no solutions.
    I have:
    |1 1 4 2|
    |1 2 -4 3|
    |6 13 k 20|

    -6R2 + R3 -> R3
    |1 1 4 2|
    |1 2 -4 3|
    |0 1 24+k 2|

    now if i let k = -24 will this make it have no solutions? because 0 cannot equal 2 right?
     
    mr_coffee, Sep 16, 2005
  5. Sep 16, 2005 #4

    TD

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    Then it would say in the last row: 0 1 0 2, which is still possible since then you have y = 2. You have to reduce it further to see it, I think :smile:
    You could also compute the determinant of the coëfficiënt matrix and let it equal 0, solve for k.
     
    TD, Sep 16, 2005
  6. Sep 16, 2005 #5

    mr_coffee

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    When i keep trying to reduce it gets bad...
    I ended up with
    |1 1 4 2|
    |2 3 0 5|
    |0 0 -20-k 0|
    and now i'm stuck, anyu ideas?
     
    mr_coffee, Sep 16, 2005
  7. Sep 16, 2005 #6

    TD

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    That's strange, perhaps you made some mistakes because that doesn't seem right to me.
    Check your work again or use the other method I gave.
     
    TD, Sep 16, 2005
  8. Sep 16, 2005 #7

    fahd

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    reduce the matrix in row echelon form,and follow the rules to solve it..its simple
     
    fahd, Sep 16, 2005
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