Page 29 - Numerical methods for chemical engineering

Elimination methods for solving linear systems 15

The augmented matrix after this operation is
 
a 11 a 12 a 13 ... a 1N b 1
(2, 1) (2, 1) (2, 1) (2, 1) 
 0 a 22 a 23 ... a 2N b 2 

 (3, 1) (3, 1) (3, 1) 
(A, b) (3, 1) =  0 a a ... a b (3,1)  (1.77)

32 33 3N 3 
 . . . . . . 

 . . . . . . . . . . . . 

...
a N1 a N2 a N3 a NN b N
where
(3, 1) (3, 1)
a b (1.78)
3 j ≡ a 3 j − λ 31 a 1 j 3 ≡ b 3 − λ 31 b 1
For the example system (1.72),
a 31 3
λ 31 = = = 3 (1.79)
a 11 1
and
 
1 1 1 4
(A, b) (3,1) =  0 −1 1 −1 
[3 − 3(1)] [1 − 3(1)] [6 − 3(1)] [2 − 3(4)]

1 1 1 4
 
=  0 −11 −1  (1.80)
0 −23 −10
In general, for N > 3, we continue this procedure until all elements of the first column are
zero except a 11 . The augmented system matrix after this sequence of row operations is
 
a 11 a 12 a 13 ... a 1N b 1
(2, 1) (2, 1) (2, 1) (2, 1) 
 0 a 22 a 23 ... a 2N b 2 

 (3, 1) (3, 1) (3, 1) 
(A, b) (N,1) =  0 a 32 a 33 ... a 3N b (3, 1)  (1.81)

3

 . . . . . . 
 . . . . . .
 . . . . . . 

(N, 1) (N, 1) (N, 1) (N, 1)
0 a a ... a b
N2 N3 NN N
We now move to the second column and perform row operations to place zeros everywhere
(2,1)
below the diagonal (2, 2) position. If a = 0, we define
22
(3, 1) (2, 1)
λ 32 = a a (1.82)
32 22
and perform the row operation 3 ← 3 − λ 32 × 2, to obtain
 
a 11 a 12 a 13 ... a 1N b 1
(2, 1) (2, 1) (2, 1) (2, 1) 
 0 a 22 a 23 ... a 2N b 2 

 
(3, 2) (3, 2) (3, 2)
(A, b) =  0 0 a 33 ... a 3N b (3, 2)  (1.83)

3

. . . . . . 
 . . . . . .

 . . . . . . 

(N, 1) (N, 1) (N, 1) (N, 1)
0 a a ... a b
N2 N3 NN N