Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...

3.3. LU FACTORIZATION 33
We pivot on the third row, with a multiplier of −1. Thus we get
⎛
⎞
2 1 1 3 7
⎜ 0 2 −2 1 −3
⎟
⎝ 0 0 3 7 13 ⎠
0 0 0 12 18
We have multiplied by M 3 :
We are now solving
M 3 =
⎡
⎢
⎣
1 0 0 0
0 1 0 0
0 0 1 0
0 0 1 1
⎤
⎥
⎦
M 3 M 2 M 1 Ax = M 3 M 2 M 1 b.
But we have an upper triangular form, that is, if we let
⎡
⎤
2 1 1 3
U = ⎢ 0 2 −2 1
⎥
⎣ 0 0 3 7 ⎦
0 0 0 12
Then we have
M 3 M 2 M 1 A = U,
A = (M 3 M 2 M 1 ) −1 U,
A = M 1 −1 M 2 −1 M 3 −1 U,
A = LU.
We are hoping that L is indeed lower triangular, and has ones on the diagonal. It turns out that
the inverse of each M i matrix has a nice form (See Exercise (3.6)). We write them here:
⎡
⎤ ⎡
⎤ ⎡
⎤
1 0 0 0 1 0 0 0 1 0 0 0
L = ⎢ 2 1 0 0
⎥ ⎢ 0 1 0 0
⎥ ⎢ 0 1 0 0
⎥
⎣ 3 0 1 0 ⎦ ⎣ 0 1 1 0 ⎦ ⎣ 0 0 1 0 ⎦
1 0 0 1 0 −1 0 1 0 0 −1 1
=
⎡
⎢
⎣
1 0 0 0
2 1 0 0
3 1 1 0
1 −1 −1 1
⎤
⎥
⎦
This is really crazy: the matrix L looks to be composed of ones on the diagonal and multipliers
under the diagonal.
Now we check to see if we made any mistakes:
LU =
=
⎡
⎢
⎣
⎡
⎢
⎣
1 0 0 0
2 1 0 0
3 1 1 0
1 −1 −1 1
2 1 1 3
4 4 0 7
6 5 4 17
2 −1 0 7
⎤ ⎡
⎥ ⎢
⎦ ⎣
⎤
⎥
⎦ = A.
2 1 1 3
0 2 −2 1
0 0 3 7
0 0 0 12
⎤
⎥
⎦