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Linear Algebra

posted by on .

Solve the following system using Gauss's algorithm

(a)
x1 + 2x2 + 4x3 + 6x4 = 3
2x1 + x2 + 3x3 = -6
-2x1 + x2 + 6x3 + 4x4 = -11
2x1 + x2 + x3 = 0

so the matrix will be:

1 2 4 6 3
2 1 3 0 -6
-2 1 6 4 -11
2 1 1 0 0

  • Linear Algebra - ,

    numbering the lines 1-4,

    #2 - 2*#1, #3 + 2*#1, #4 - 2*#1 gives

    1 2 4 6 3
    0 -3 -5 -12 -12
    0 5 14 16 -5
    0 -3 -7 -12 -6

    Follow similar steps to place zeros in the other rows off the main diagonal, to end up with

    so, (x1,x2,x3,x4) = 1/3 (-23,55,-9,-7)


    A good calculator which shows the steps can be found at

    www.gregthatcher.com/Mathematics/GaussJordan.aspx

  • Linear Algebra - ,

    awesome, thanks a lot Steve, that link will come in handy!

  • Linear Algebra - ,

    sure thing. Plus, I messed up my math. Solution is really (1,1,-3,2)

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