Can someone help me with LU factorization?

- anonymous

Can someone help me with LU factorization?

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- amistre64

lowers and uppers

- Hero

Well, at least I know what LU stands for now.

- amistre64

:)

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## More answers

- anonymous

|dw:1327272671601:dw|

- anonymous

lol that looks terrible

- amistre64

the upper is the echelon form with leading 1s

- anonymous

can u show how upper and lower will kinf od look

- anonymous

like where the 0s and ones will be

- amistre64

-2 1 ~ 1 -1/2 ~ 1 -1/2 ~ 1 -1/2
-6 4 ~ -1 4/6 ~ 0 1/6 ~ 0 1
should be the U

- anonymous

ok

- amistre64

i gotta reread up on the L part, it starts with the same column but somehow the others are changed up

- amistre64

-2 0
6 ?
is the L

- amistre64

might be the -1/6 part tho let me read up on that

- anonymous

idk it is confusing in a 2by 2 matrix

- anonymous

much simpler with 3 by 3

- amistre64

to get a 1 in the ? we multiplied by 6 ....

- anonymous

ok THANKS

- anonymous

I am gonna do it on my own and see if I get the same answer :D

- anonymous

rld you would better solve ur problem than loling in mine :P

- amistre64

-2 1
-6 4
-6(1 -1/2
6 4
-6 3
------
0 7 this one might do better for us
-2 1
-6 7 is part of the process

- anonymous

ok I thought u use the elementary row operations

- anonymous

I got a different answer

- anonymous

|dw:1327273667252:dw|

- anonymous

This is what i go t using the elementary row operations

- amistre64

i did, you wanna 1 out A first, and keep track of those pivots before you change them; they are the diagonals of your L

- amistre64

but how to go about the L after that I cant recall just yet; theres a book upstairs i might be able to read up on since thats where I learned it :)

- anonymous

amistre come help me now

- anonymous

LOL dont drive urself crazy LOL

- anonymous

Go help tomas seriously

- amistre64

ok, good luck. im going to be going thru this in a few weeks for linearA as well

- phi

when you do LU decomposition, you use elimination to find U
to find L, keep track of the multiplier you use to eliminate the "key position"
-2 1
-6 4
use -3 to get rid of the 2nd row, 1st col (scale 1st row by -3 and add to 2nd row):
-2 1
0 1
Now put the negative of -3 in the key position of the identity matrix:
1 0
3 1
You now have L and U

- anonymous

Thanks that helped out :D

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