Page 27 - Numerical Methods for Chemical Engineering

16 1 Linear algebra

where

(3, 2) (3, 1) (2, 1) (3, 2) (3, 1) (2, 1)
a 3 j ≡ a 3 j − λ 32 a 2 j b 3 ≡ b 3 − λ 32 b 2 (1.84)
For our example (A, b) (3,1) (1.80) we obtain

(3, 1) (2,1)
λ 32 = a a = (−2)/(−1) = 2 (1.85)
32 22
so that

 
1 1 1 4
(A, b) (3,2) =  0 −1 1 −1 
0[(−2) − (2)(−1)] [3 − (2)(1)] [(−10) − (2)(−1)]
1 1 1 4
 
=  0 −11 −1  (1.86)
0 0 1 −8
For N > 3, we perform another N − 3 row operations to place all zeros below the principal
diagonal in the second column, yielding

 
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) 
(A, b) (N, 2) =  0 0 a 33 ... a 3N b 3 (3, 2)  (1.87)


. . . . . . 
 . . . . . .

 . . . . . . 

(N, 2) (N, 2) (N, 2)
0 0 a ... a b
N3 NN N
We then perform a sequence of row operations to place all zeros in the third column below
the (3, 3) position, then in column 4, we put all zeros below (4, 4), etc. When finished, the
augmented matrix is upper triangular, containing only zeros below the principal diagonal
a 11 , a 22 ,..., a NN :
 
a 11 a 12 a 13 a 14 ... a 1N b 1
(2, 1) (2, 1) (2, 1) (2, 1) (2, 1) 
 0 a a a ... a b

22 23 24 2N 2 
 
(3, 2) (3, 2) (3, 2)
 0 0 a 33 a 34 ... a 3N b (3, 2) 

3

(A, b) (N, N−1) =  (4, 3) (4, 3)  (1.88)
 0 0 0 a 44 ... a 4N b (4, 3) 

4

 . . . . . . 
 . . . . . .
 . . . . . . 

(N, N−1) (N, N−1)
0 0 0 0 ... a b
NN N
For our example, the final augmented matrix is
 
1 1 1 4
(A, b) (3, 2) =  0 −11 −1  (1.89)
0 0 1 −8