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Echelon form and set of solutions.

  1. Nov 12, 2012 #1
    Hello,

    I have the following system of linear equations -

    kx + 3y -z = 1
    x + 2y - z = 2
    -kx + y + 2z = -1

    I have reduced it to

    1 2 -1 : 2
    0 1 1/4 : 0
    0 0 (7-6k)/7k : (8-4k)/7

    assuming k ≠ 0.

    I would now like to be able to determine for what values of k will the system have no solutions, a single solution, and an infinite number of solutions.

    Would it be correct to say that there would be no solution for k=7/6, and a single solution for every k ≠ 7/6 (therefore, the system will never have an infinite number of solutions)?
     
  2. jcsd
  3. Nov 12, 2012 #2

    Dick

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    Sounds correct to me.
     
  4. Nov 12, 2012 #3
    What about k different than 0? I had to divide by k whilst reducing the matrix. How come it when k=0 the system still yields a solution? Is it possible that the system will never have an infinite number of solutions? (seems like something is amiss with my reduced matrix/method)
     
  5. Nov 12, 2012 #4

    Dick

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    Sure you divided by k. You still have a k in the denominator. But you can get rid of it by multiplying the last row by k. If you really want to you can also put k=0 in the original matrix and check there is no problem. To get an infinite number of solutions you need a row of zeros. Can't happen here.
     
  6. Nov 12, 2012 #5
    But what if I didn't perform the row operations properly? I went over them and yet mistakes are always possible. May you please confirm that I got it aright? I mean, that my reduced form is indeed the reduced form of that matrix?
     
  7. Nov 12, 2012 #6

    Dick

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    There's more than one reduced form, you can do it a number of different ways. But my way also showed no solutions for k=7/6.
     
  8. Nov 12, 2012 #7
    Thank you very much, Dick
     
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