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Gaussian elimination system of Eqs, the case of no solution

  1. Nov 21, 2016 #1
    1. The problem statement, all variables and given/known data

    system of equations is as follows
    4x +2y -2z = 0
    2x + y -z= 1
    3x +y -2z = 1

    2. Relevant equations


    3. The attempt at a solution


    Using gaussian elimination

    we can multiply mid-eq, by (-2) [[[actually... it is simply a basic equation procedure]]]
    2x+y-z=1 |*(-2)
    =
    -4x -2y +2z = -2

    further with using gaussian elimination (?) we sum together
    4x+2y-2z=0
    -4x-2y+2z=-2
    result
    0=-2


    ´my teacher said something to the effect that, it is concluded that the system of equations does not have solutions... because a non-sensical result came out of the procedure.

    Now, my teacher gave me some kind of proof for the Gaussian elimination method, but I'm still little bit uncertain why the "untrue equation midresult" causes the entire system of eqiuation to not have solution...
    Our course ended today also, so I can't ask my teacher except by email.

    perhaps it's a dumb question but anyway...

    1.) By what reasoning is it arrived to this conclusion that when a "non-sensical equation result" comes out from the procedure, that this "mid-result" if you allow me to call it that, causes the system of equations to not have solution (Therefore, you don't have to calculate any further using Gaussian elimination?)

    sorry if I failed to think about the problem rigorously enough as required by homework forum rules.
     
  2. jcsd
  3. Nov 21, 2016 #2

    BvU

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    If you look at the first two equations: an ##\ (x,y,z) \ ## that satisfies equation two would satisfy ##\ 4x +2y-2z=2\ ## ( a simple multiplication by 2) which makes it impossible to satisfy the first equation.

    The wording is: equations 1 and 2 are not independent.
     
  4. Nov 21, 2016 #3
    I can see how when you compare the equations....

    One of the equations equals zero
    And the same equation equals two... seems difficult to find an (x, y, z)

    What does that even mean equations are not independent?
     
  5. Nov 21, 2016 #4

    Ray Vickson

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    If you set ##2x+y-z = w## the first two equations say ##2w = 0## and ##w = 1##. There is very obviously NO possible solution.
     
  6. Nov 25, 2016 #5

    BvU

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    Two equations are dependent if a number can be found such that you get the other when multiplying the first with that number.
    Like ##2x + 2y = 4## and ##x + y = 2##: basically only one equation instead of 2.
    For more equations: if you can manipulate a group of one or more equations to get one of the remaining equations.
    (manipulate: mutiply all terms with a number, add, etc.)​

    Maybe my classification wasn't correct: when you multiply the first by 1/2 you don't exactly get the second equation, only the same coefficients for ##x,y## and ##z##.
    Perhaps the term for ##2w = 0## and ##w = 1## is 'conflicting' or 'contradictory' . Can a native english purist mathemagician help me out?
     
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