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Linear algebra question

  1. Aug 3, 2009 #1

    tgt

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    1. The problem statement, all variables and given/known data
    If a linear system of n equations in n unkowns has n leading 1's in the reduced row-echelon form of its augmented matrix, then the system has exactly one solution.






    3. The attempt at a solution
    True to me but is it?
     
  2. jcsd
  3. Aug 3, 2009 #2
    Yes it is, but is this all your attempt to the solution?
     
  4. Aug 3, 2009 #3

    tgt

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    The book said False. Must be a typo?

    If there are n equstions and n leading ones, all variables must equal to a specific number. And that is the only solution in the system.
     
  5. Aug 3, 2009 #4
    Em.....unless 1's in the augmented part can be called "leading 1",then the matrix can be like this:
    1 0 2
    0 0 1
    But I have a very vague memory that the 1 in augmented part(1 in second row) can't be called "leading 1",but you should check it out.
     
  6. Aug 5, 2009 #5

    tgt

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    But there is n leading 1's with n variables. Your example have 2 leading 1's in 3 variables.

    The question does say n leading 1's in the reduced echelon matrix. The fact that its reduced echelon suggest that all columns have 0 elsewhere.
     
  7. Aug 5, 2009 #6
    Kof is right. A leading 1 is a pivot position.

    EDIT:

    Here:
    1 0 2
    0 0 1
    The 1 in the 2nd row isn't a leading 1, but here it is:
    1 0 2
    0 1 0
     
  8. Aug 5, 2009 #7
    No, my example is an augmented matrix, the 3rd column is the right hand side of the equations.
     
  9. Aug 5, 2009 #8

    tgt

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    So you are saying that if a linear system of n equations in n unkowns has n leading 1's in the reduced row-echelon form of its augmented matrix then it is inconsistent. hence the book was right.
     
  10. Aug 5, 2009 #9
    Personally I don't think 1 in augmented part can not be called leading one, and with roam's confirmation, probably it's just a typo in the book.
     
  11. Aug 6, 2009 #10

    tgt

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    The book I'm using does not draw a vertical line before the last column even for an augmented matrix. So in that case the book was right, no typo.
     
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