ABSTRACT
Consider another slightly modified form of Eq. (1.165) in which bl is changed slightly from 2.0001 to 2:
Xl +Xl = 2 xI + 1.000lxl = 2
Solving Eq. (1.169) by Gauss elimination gives
(1.169a) (1.169b)
[~ I I 2]1.0001 12 Rl - RI (1.I70a) [1 I I 2]o 0.0001 I 0 (1.I70b)
Solving Eq. (1.170) yields Xl = 0 and xI = 2, which is greatly different from the solution of Eq. (1.165).
