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Values of a for which the system has infinite number of solutions

  1. Dec 4, 2011 #1
    Find all values a (alpha) for which the resulting system has a) no solution, b) a unique solution c) infinitely many solutions.

    x - y - az = -1
    2x +y + az = -1
    -x + y + (a^2 - 2)z = a + 2

    attempt:

    put it in matrix form and row-reduce

    1 -1 -a : -1
    2 1 a : -1
    -1 1 a^2-2 : a+2
    after row reducing I get:


    1 0 0 : 1
    0 1 -2 : 1
    0 0 a-2 : 1

    so for a = 0 the system has a unique solution (you get x=1; y=0; and z=-1/2)
    for a = 2 the system has no solution ( you get 0 0 0 : 1 on the bottom row)

    so I have a values for a unique sol. and no sol. but I still need infinitely many solutions.

    help.

    EDIT: if i'm not mistaken really you get a unique solution for any real number, a, that is not 2 (not only 0 as stated above)
    so how are you supposed to get infinitely many solutions???
     
    Last edited: Dec 4, 2011
  2. jcsd
  3. Dec 4, 2011 #2

    micromass

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    Mod note: moved to homework forums
     
  4. Dec 4, 2011 #3

    SammyS

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    If you add the first two equations together, you get 3x = -2 . So x = -2/3 no matter what the value of a is.

    If a = 0, the solution that I get is: x = -2/3, y = 1/3. z = -1/2 .

    Notice that in the third equation, z has a coefficient of (a2 - 2) , not (a - 2) .
     
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