Thursday

July 30, 2015

July 30, 2015

Posted by **Robert** on Friday, September 21, 2012 at 9:56am.

(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 -
**Steve**, Friday, September 21, 2012 at 11:22amnumbering 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 -
**Robert**, Friday, September 21, 2012 at 11:30amawesome, thanks a lot Steve, that link will come in handy!

- Linear Algebra -
**Steve**, Friday, September 21, 2012 at 11:51amsure thing. Plus, I messed up my math. Solution is really (1,1,-3,2)