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anonymous

  • 5 years ago

Gauss-Jordan Reduction 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 - I'm pretty confused

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  1. anonymous
    • 5 years ago
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    Hi princessanna, how do you use Gauss-Jordan reduction to solve three equations, do you use matrix and elementary row operations?

  2. anonymous
    • 5 years ago
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    Hi sklee, yes I use the matrix method. But I can't use it unless the matrix is a square, which it isn't for this question.

  3. anonymous
    • 5 years ago
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    the system of equations can be written in a matrix form: 5 5 -5 3 4 -7 0 2 -8 -2 -3 6

  4. anonymous
    • 5 years ago
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    Hi princessanna, it doesn't matter if the matrix is square or not, it is still applicable.

  5. anonymous
    • 5 years ago
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    Well I don't know how to do the method :( because I only know the way when all the 1's are diagonal?

  6. anonymous
    • 5 years ago
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    Ok, can i assume you have no problem to do the elementary row operations? As i will show you only the matrix that has been reduced, is that fine for you?

  7. anonymous
    • 5 years ago
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    Yes that would be great, thank you :)

  8. anonymous
    • 5 years ago
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    1 1 -1 0 1 -4 0 0 0 0 0 0 The operations that i have used are R_{1}(1/5), R_{3}(1/2), R_{1,2}(-3), R_{1,4}(2), R_{2,3}(-1) and R_{2,4}(1)

  9. anonymous
    • 5 years ago
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    It's pretty hard to read and understand :(

  10. anonymous
    • 5 years ago
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    From this reduced form, you can use backward substitution (that is called Gauss elimination method) or you can reduce more till it is in row reduced form. 1 0 3 0 1 -4 0 0 0 0 0 0 The operation that has been used is R_{2,1}(-1)

  11. anonymous
    • 5 years ago
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    R_{1} (1/5) means you multiply row 1 with 1/5 R_{3} (1/2) means you multiply row 3 with 1/2 R_{1,2} (-3) means you multiply row 1 with -3 and add it to row 2 to get the new row 2 etc..

  12. anonymous
    • 5 years ago
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    Is it better now? Do they helpful to you?

  13. anonymous
    • 5 years ago
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    By letting z = a, where a is any parameter, we get x = -3z = -3a and y = 4z = 4a. So the solution is (x,y,z) = (-3a,4a,a) = a(-3,4,1)

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