Page 259 - Advanced engineering mathematics

7.9 LU Factorization 239

EXAMPLE 7.32
We will solve the system AX = B, where
⎛ ⎞ ⎛ ⎞
4 3 3 −4 6 4
1 1 −1 3 4 −2
⎜ ⎟ ⎜ ⎟
A = ⎜ ⎟ and B = ⎜ ⎟ .
⎝ 2 2 −4 6 1 ⎠ ⎝ 6 ⎠
8 −2 1 4 6 1
We could solve this system by finding the reduced row echelon form of A. To illustrate LU
factorization, first factor A. Begin by finding U:

⎛ ⎞ ⎛ ⎞
4 3 3 −4 6 4 3 3 −4 6
⎜ 1 1 −1 3 4 ⎟ ⎜ 0 1/4 −7/4 4 5/2 ⎟
A = ⎜ ⎟ → ⎜ ⎟
⎝ 2 2 −4 6 1 ⎠ ⎝ 0 1/2 −11/1 8 −2 ⎠
8 −2 1 4 6 0 −8 −5 12 −6

⎛ ⎞ ⎛ ⎞
4 3 3 −4 6 4 3 3 −4 6
0 1/4 −7/4 4 5/2 01/4 −7/4 4 5/2
⎜ ⎟ ⎜ ⎟
→ ⎜ ⎟ → ⎜ ⎟ = U.
⎝ 0 0 −2 8 −7 ⎠ ⎝ 0 0 −2 0 −7 ⎠
0 0 −61 140 74 0 0 0 140 575/2
We can now form the 4 × 4matrix L by beginning with the highlighted columns and obtaining
1’s down the main diagonal:

⎛ ⎞ ⎛ ⎞
4 0 0 0 1 0 0 0
11/4 0 0 1/4 1 0 0
⎜ ⎟ ⎜ ⎟
→ = L.
⎜ ⎟ ⎜ ⎟
⎝ 21/2 −2 0 ⎠ ⎝ 1/2 2 1 0 ⎠
8 −8 −61 140 2 −32 61/2 1
Now solve LY = B. Because L is lower triangular, this is the system
y 1 = 4,
1
y 1 + y 2 =−2,
4
1
y 1 + 2y 2 + y 3 = 6
2
61
2y 1 − 32y 2 + y 3 + y 4 = 1
2
with solution
⎛ ⎞
4
−3
⎜ ⎟
Y = ⎜ ⎟ .
⎝ 10 ⎠
−408
Now solve
UX = Y.

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October 14, 2010 14:23 THM/NEIL Page-239 27410_07_ch07_p187-246