Page 91 - Applied Numerical Methods Using MATLAB

80 SYSTEM OF LINEAR EQUATIONS
First, to remove the x 1 terms from equations (2.2.0.m) other than (2.2.0.a), we
subtract (2.2.0a)×a m1 /a 11 from each of them to get
(0) (0) (0) (0)
a x 1 + a x 2 + a x 3 = b (2.2.1a)
11 12 13 1
(1) (1) (1)
a x 2 + a x 3 = b (2.2.1b)
22 23 2
(1) (1) (1)
a x 2 + a x 3 = b (2.2.1c)
32 33 3
with
a (0) = a mn , b (0) for m, n = 1, 2, 3 (2.2.2a)
mn m = b m
(0)
(0)
(0)
(0)
(0)
a (1) = a (0) − (a /a )a , b (1) = b (0) − (a /a )b (0) for m, n = 2, 3
mn mn m1 11 1n m m m1 11 1
(2.2.2b)
We call this work ‘pivoting at a 11 ’ and call the center element a 11 a ‘pivot’.
Next, to remove the x 2 term from Eq. (2.2.1c) other than (2.2.1a,b), we sub-
(1) (1)
tract (2.2.1b)×a /a (m = 3) from it to get
m2 22
(0) (0) (0) (0)
a x 1 + a x 2 + a x 3 = b (2.2.3a)
11 12 13 1
(1)
(1)
a x 2 + a x 3 = b (1) (2.2.3b)
22 23 2
(2) (2)
a x 3 = b (2.2.3c)
33 3
with
(1)
(1)
(1)
(1)
(1)
a (2) = a (1) − (a /a )a , b (2) = b (1) − (a /a )b (1) for m, n = 3
mn mn m2 22 2n m m m2 22 2
(2.2.4)
We call this procedure ‘Gauss forward elimination’ and can generalize the updat-
ing formula (2.2.2)/(2.2.4) as
(k−1)
(k)
a mn = a mn − (a (k−1) /a (k−1) )a (k−1) for m, n = k + 1,k + 2,... ,M (2.2.5a)
kn
kk
mk
b (k) = b (k−1) − (a (k−1) /a (k−1) )b (k−1) for m = k + 1,k + 2,... ,M (2.2.5b)
m m mk kk k
After having the triangular matrix–vector equation as Eq. (2.2.3), we can solve
Eq. (2.2.3c) first to get
(2) (2)
x 3 = b /a (2.2.6a)
3 33
and then substitute this result into Eq. (2.2.3b) to get
(1) (1) (1)
x 2 = (b − a x 3 )/a (2.2.6b)
2 23 22
Successively, we substitute Eqs. (2.2.6a,b) into Eq.(2.2.3a) to get
3

(0) (0) (0)
x 1 = b − a x n /a (2.2.6c)
1 1n 11
n=2