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anonymous

  • 5 years ago

Solve the following set of homogeneous equations by Gauss-Jordan reduction of the matrix of coefficients (without the column of zeros from the right-hand side. 5x+5y-5z=0 3x+4y-7z=0 2y-8z=0 -2y-3y+6z=0 I know to do Gauss-Jordan reduction for three equations, but not four. How do I do this?

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  1. anonymous
    • 5 years ago
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    It should work for a square matrix of any order n x n, just make sure to end up with the fourth order identity matrix on the left hand side (1's along the main diagonal). :)

  2. anonymous
    • 5 years ago
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    I write the matrix out: 5 5 -5 0 3 4 -7 0 0 2 -8 0 -2 -3 6 0 So do I just put the zeros on the left side as well?

  3. anonymous
    • 5 years ago
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    Hmm...sorry, I misread the question! At a cursory glance, what could you achieve by having 4 equations with 3 unknowns? The very structure of the question is confusing to me...I'm pretty sure that the number of equations has to equal the number of unknowns to solve using Gauss-Jordan elimination; otherwise you'd have to make an augmented matrix.

  4. anonymous
    • 5 years ago
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    Yup that's why I'm confused as well! I can't seem to find any examples online either.

  5. anonymous
    • 5 years ago
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    I think you are correct; put the constants as a fourth column on the left hand side and when you get the inverse on the right hand side, you'll just separate the right-most column as new coefficients.

  6. anonymous
    • 5 years ago
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    Found a FANTASTIC resource: http://mathrefresher.blogspot.com/2007/04/gauss-jordan-elimination.html :)

  7. anonymous
    • 5 years ago
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    I'm still trying to figure out how to get the matrix in the right form so can find what x y z are? :(

  8. anonymous
    • 5 years ago
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    I already know how to the method on the page you provided, but I can't because the matrix is not in the right form, how do I get it into the right form (thanks for the page though!)

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