Page 153 - Matrix Analysis & Applied Linear Algebra

3.10 The LU Factorization 147
Example 3.10.2

Problem 1: Use the LU factorization of A to solve Ax = b, where
   
2 2 2 12
A =  4 7 7  and b =  24  .
61822 12
Problem 2: Suppose that after solving the original system new information is
received that changes b to
 
6
˜  24  .
b =
70
˜
Use the LU factors of A to solve the updated system Ax = b.
Solution1: The LU factors of the coefficient matrix were determined in Example
3.10.1 to be
   
100 222
L =  210  and U =  033  .
341 004
The strategy is to set Ux = y and solve Ax = L(Ux)= b by solving the two
triangular systems
Ly = b and Ux = y.
First solve the lower-triangular system Ly = b by using forward substitution:
     
100 y 1 12 y 1 =12,
 210   y 2  =  24  =⇒ y 2 =24 − 2y 1 =0,
341 y 3 12 y 3 =12 − 3y 1 − 4y 2 = −24.
Now use back substitution to solve the upper-triangular system Ux = y:
     
222 x 1 12 x 1 = (12 − 2x 2 − 2x 3 )/2=6,
 033   x 2  =  0  =⇒ x 2 =(0 − 3x 3 )/3=6,
004 x 3 −24 x 3 = −24/4= −6.

˜
Solution2: To solve the updated system Ax = b, simply repeat the forward
˜
˜
and backward substitution steps with b replaced by b. Solving Ly = b with
forward substitution gives the following:
     
100 y 1 6 y 1 =6,
 210   y 2  =  24  =⇒ y 2 =24 − 2y 1 =12,
341 y 3 70 y 3 =70 − 3y 1 − 4y 2 =4.
Using back substitution to solve Ux = y gives the following updated solution:
     
222 x 1 6 x 1 =(6 − 2x 2 − 2x 3 )/2= −1,
 033   x 2  =  12  =⇒ x 2 = (12 − 3x 3 )/3=3,
004 x 3 4 x 3 =4/4=1.