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Math Help - Using Gaussian Elimination to solve for k

  1. #1
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    Using Gaussian Elimination to solve for k

    Working on this problem. Actually 2 problems, involving Gaussian elimination. Both of them say

    Find the values of k such that the system has (a) a unique solution (b) no solution and (c) infinitely many solutions

    I am required to use Gaussian Elimination

    The first problem is
    x+ky = -1
    kx+y = 1

    My solution is as follows, but not sure if correct (not sure how to do latex for this). I have seperated each entry by a comma (,)
    row 1 = 1, k, -1
    row 2 = k, 1, 1
    -k * R1 + R2 = 0, -(k2)+1, k+1

    (a) if k = -1 R3 = 0, 0 and has infinitely many solutions
    (b) if k = 1 R3 = 0, 2 and is inconsistant
    (c) if k != 1 and k != -1 the system has a unique solution

    I'm not sure if that is correct. I've gone through the problem several times, and come up with the same outcome everytime, but I'm new to this. I did really well at basic Matrix stuff, but this is the next paper after the intro one.

    The second problem is
    x-2y+3z = 2
    x+y+z = k
    2x - y + 4z = k^2

    I've gone through this a whole lot of times (I need to use Gaussian elimination) and end up with some unfactorable equation at the end. Where should I start with this problem? I don't want the answer, just some help.
    What I have been doing is laying it out in a compact matrix and then multiplying the rows to eliminate the x and y values.

    Here is my working - not sure if it is correct.
    Last edited by lukasaurus; March 2nd 2013 at 10:08 PM. Reason: updated with image of working
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  2. #2
    MHF Contributor
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    Re: Using Gaussian Elimination to solve for k

    Hey lukasaurus.

    Your first question is spot on.

    For the second one I confirmed the answer in Octave which is:

    >> B
    B =

    1 -2 3
    1 1 1
    2 -1 4

    >> det(B)
    ans = 0
    >> rref(B)
    ans =

    1.00000 0.00000 1.66667
    0.00000 1.00000 -0.66667
    0.00000 0.00000 0.00000

    This confirms no unique solution and this command:

    >> X = [2;2;4]
    X =

    2
    2
    4

    >> A = [B,X]
    A =

    1 -2 3 2
    1 1 1 2
    2 -1 4 4

    >> rref(A)
    ans =

    1.00000 0.00000 1.66667 2.00000
    0.00000 1.00000 -0.66667 0.00000
    0.00000 0.00000 0.00000 0.00000

    confirms that infinitely many solutions only occurs when k = 2 just as you have mentioned.

    If I set k = 1 for example Octave gives us:

    >> X = [2;1;1]
    X =

    2
    1
    1

    >> A = [B,X]
    A =

    1 -2 3 2
    1 1 1 1
    2 -1 4 1

    >> rref(A)
    ans =

    1.00000 0.00000 1.66667 0.00000
    0.00000 1.00000 -0.66667 0.00000
    0.00000 0.00000 0.00000 1.00000

    which is inconsistent (note the 1.0000).

    So all your answers have been confirmed with a computer program.
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  3. #3
    Junior Member
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    Re: Using Gaussian Elimination to solve for k

    Thanks so much,

    Is there a calculator that can do Matrix calculations? Last exam I had there were no calculator restrictions and I was using an old fx82 casio (because I haven't been in Uni or high school for about 15 years and it was all I had).

    And what is Octave?
    Last edited by lukasaurus; March 2nd 2013 at 10:36 PM.
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  4. #4
    MHF Contributor
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    Re: Using Gaussian Elimination to solve for k

    Octave is basically a free version of MATLAB which has a lot of the same features and is open source.

    You can also get what is known as GUIOctave which provides the front-end to make it look like MATLAB.

    GNU Octave

    Download GUI Octave 1.5.4 Free - A GUI for Octave. - Softpedia
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