ABSTRACT
The exact solution of Eq. (1.160) is 1.0002 0.0003 - 1.0002 1--- 0.0003 0.0003 0.0003 IX2 = - (1.I6Ia)
1.0002 - 3x2 1.0002 - 3(1/3) 0.0002 2 (1.16Ib)XI = -- -0.0003 0.0003 0.0003 3 Let's solve Eq. (1.161) using finite precision arithmetic with two to eight significant
figures. Thus,
1.00021-0.0003 x2 = -9999 and
1.0002 - 3x2 XI = 0.0003 (1.162)
The results are presented in Table 1.1. The algorithm is clearly performing very poorly. Let's rework the problem by interchanging rows I and 2 in Eq. (1.159). Thus,
XI +x2 = I
0.0003x) + 3x2 = 1.0002
Solve Eq. (1.163) by Gauss elimination. Thus,
[ I I I I ] 0.0003 3 I 1.0002 R2 - 0.0003R I
[0 1 I I I ]
2.9997 I 0.9999
(1.163a) (1.I63b)
(1.164a)
(1.164b)
0.9999 X2 = 2.9997 and ( l.164c)
Let's solve Eq. (1.164c) using finite precision arithmetic. The results are presented in Table 1.2. These results clearly demonstrate the benefits of pivoting.
