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Simulataneous linear equations

  1. Dec 3, 2009 #1
    The simultaneous linear equations
    mx+12y=24---------------1
    3x+my=m------------------2

    have a unique solution for m= ?
    The answer should be m Є R\{-6,6}


    what I tried doing was substituting the 2nd equation into m, expanded, grouped them and factorised but I struck a dead end.
     
  2. jcsd
  3. Dec 3, 2009 #2

    berkeman

    User Avatar

    Staff: Mentor

    That's the whole question? Seems like you have 3 unknowns and 2 equations... ?
     
  4. Dec 3, 2009 #3
    Yes, that is the whole question.
     
  5. Dec 3, 2009 #4

    Mark44

    Staff: Mentor

    The problem isn't to solve for m, but to state conditions for which there will be a unique solution for m.

    The two equations can be put in an augmented matrix form, like so.
    [m 12 | 24]
    [3 m | 12]

    Do you know how to use matrix operations to row-reduce this augmented matrix?
     
  6. Dec 3, 2009 #5
    I've touched on a little bit of matrix operations.
    shouldn't it be
    [m 12 | 24]
    [3 m | m] ?? and from there i wouldn't know where to go.
     
  7. Dec 3, 2009 #6

    Mark44

    Staff: Mentor

    Right. That 3rd entry in the 2nd row should be m, not 12 as I had.

    The basic row operations are
    • Exchange two rows
    • Replace a row by a multiple of itself
    • Add a multiple of one row to another row

    I think this is the direction they want you to go in this problem.
     
  8. Dec 3, 2009 #7
    okkkk, but i am not very familiar with how to do that.
     
  9. Dec 3, 2009 #8

    Mark44

    Staff: Mentor

    So going back to the system of equations,

    mx + 12y = 24
    3x + my = m

    is there some multiple of the first equation you could add to the second equation to eliminate a variable? The operations you can apply to your system of equations are exactly the same operations as I listed in my previous post
     
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