Lu factorization opper matrix form problem

drdolittle · #1 $Old$ May 17th, 2009, 03:16 AM

basically i have to solve

4 -1 2 x1 = 15
-1 2 3 x2 = 5
5 -7 9 x3 =8

i know all the steps from what i gt from my lectures. can anyone of you help me just how to write in the A=LU form

this is what i have gt from my lectures

A = L U

4 -1 2 4 0 0 1 -(1/4) (1/2)
-1 2 3 -1 L(22) 0 0 1 U(23)
5 -7 9 5 L (32) L(33) 0 0 1

can you just show me how he gt the -1/4 and 1/2 in U .

the_doc · #2 $Old$ May 17th, 2009, 05:11 PM

Ok took me a while to read your matrices but turns out the answer to your question is quite simple. The product of the L and U matrices must return you original matrix A. So the two terms you are asking about are arrived at since:

$L(1,1) = A(1,1) = 4$

So then

$U(1,2) = \frac{A(1,2)}{L(1,1)} = \frac{-1}{4}$
$U(1,3) = \frac{A(1,3)}{L(1,1)} = \frac{2}{4} = \frac{1}{2}$ .

If you need any more help on LU factorisation put up another post.