Linear Algebra
posted by Robert .
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

numbering the lines 14,
#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 
awesome, thanks a lot Steve, that link will come in handy!

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