Page 25 - Numerical Methods for Chemical Engineering

14 1 Linear algebra

We use the notation (A, b) (2,1) for the augmented matrix obtained after placing a zero at
the (2,1) position through the operation 2 ← 2 − λ 21 × 1,
 
a 11 a 12 ... a 1N b 1
 0 (a 22 − λ 21 a 12 ) ... (a 2N − λ 21 a 1N )(b 2 − λ 21 b 1 ) 
 
(A, b) (2,1)  a 31 a 32 ... a 3N b 3  (1.69)
= 
 . . . . 
 . . . . . . . . 

...
a N1 a N2 a NN b N
Again, it is important to note that the linear system A (2,1) x = b (2,1) has the same solution(s)
x as the original system Ax = b.
As we develop this method, let us consider the example (1.2),
x 1 + x 2 + x 3 = 4  1114 
2x 1 + x 2 + 3x 3 = 7 (A, b) =  2137  (1.70)
3x 1 + x 2 + 6x 3 = 2 3162
Since a 11 = 0,
a 21 2
λ 21 = = = 2 (1.71)
a 11 1
and the row operation 2 ← 2 − λ 21 × 1 yields
 
1 1 1 4
(A, b) (2, 1) =  [2 − (2)(1)] [1 − (2)(1)] [3 − (2)(1)] [7 − (2)(4)] 
3 1 6 2
1 1 1 4
 
=  0 −11 −1  (1.72)
3 1 6 2
We now define
(2, 1) (2, 1)
a ≡ a 2 j − λ 21 a 1 j b ≡ b 2 − λ 21 b 1 (1.73)
2 j 2
and write (A, b) (2,1) as
 
a 11 a 12 a 13 ... a 1N b 1
 (2, 1) (2, 1) (2, 1) (2, 1) 
0 a a ... a b
22 23 2N 2
 
 
(A, b) (2,1) =  a 31 a 32 a 33 ... a 3N b 3  (1.74)


 . . . . . . 

 . . . . . . . . . . . . 

...
a N1 a N2 a N3 a NN b N
We continue this process and place zeros in all elements in the first column below the
diagonal, considering next the element in row 3. If a 11 = 0,
a 31
λ 31 = (1.75)
a 11
and the row operation 3 ← 3 − λ 31 × 1 yields the new (3, 1) element

a 31
a 31 − λ 31 a 11 = a 31 − a 11 = a 31 − a 31 = 0 (1.76)
a 11