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
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
3 answers
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)