The Chemistry Maths Book, Second Edition

20.7 Gauss elimination for the solution of linear equations 583

Whenλ 1 = 123 , equation (3′′) is redundant, and back substitution givesy 1 = 15 z 221 − 18 and

x 1 = 1131 − 13 z. Because zremains arbitrary, we have infinitely many solutions, one for

each possible value of z. On the other hand, no solution exists whenλ 1 ≠ 123.

0 Exercise 31

Pivoting

The elimination method involves repeated subtraction of a multiple of one equation

from another and can lead to serious differencing errors, especially for large systems

of equations. To illustrate the problem, we consider the pair of equations

(1) 0.0003x

1

1 + 1 2.513x

2

1 = 1 7.545

(2) 0.7003x

1

1 − 1 2.613x

2

1 = 1 6.167

(20.45)

of which the solution isx

1

1 = 120 ,x

2

1 = 13. To solve by the elimination method, we choose

(1) as the pivot equation and eliminatex

1

from (2) by subtracting (0.7003 2 0.0003) 1 ×

(1) from (2). If, for example, the calculation is performed using 4-figure arithmetic

then0.7003 2 0.0003 1 = 12334 , and the second equation becomes

(2′) − 5868 x

2

1 = 1 − 17600

Thenx

2

1 = 1 2.999, and equation (1) givesx

1

1 = 1 28.38. A small error inx

2

has led to a large

error inx

1

because the coefficient ofx

1

in equation (1) is small compared with that in

(2). The resulting differencing errors are avoided if equation (2) is chosen as the pivot

equation. Then, multiplication of (2) by0.0003 2 0.7003 1 = 1 0.0002484and subtraction

from (1) gives

(1′) −2.514x

2

1 = 1 −7.542

so thatx

2

1 = 1 3.000. Substitution in (2) then givesx

1

1 = 1 20.0.

The choice of equation (2) as the pivot equation is an example of partial pivoting;

that is, choosing the pivot equation in the first step as that equation in which the

coefficient ofx

1

has largest magnitude. Similarly forx

2

in the second step, and so on.

Partial pivoting is equivalent to a reordering of the equations, but is not always

sufficient. For example, equations (20.45) may have been obtained with (1) multiplied

by some large number, 3000 say:

(1) 0.9000x

1

1 + 17539 x

2

1 = 122640

(20.46)

(2) 0.7003x

1

1 − 1 2.613x

2

1 = 1 6.167

The coefficient ofx

1

in (1) is now the greater, but choosing (1) as the pivot equation

again results in a poor (but different) result. The correct procedure, called scaled

partial pivoting, is to choose as pivot equation in the first step that equation in which

the ratio of the coefficient ofx

1

to the largest other coefficient has largest magnitude.

This ratio is0.90000 275391 ≈ 1 0.0001in equation (1) and0.7003 2 2.613 1 ≈ 1 0.3in equation