The Chemistry Maths Book, Second Edition

582 Chapter 20Numerical methods

Step 2. Elimination ofy

Equation (2′) is the new pivot equation, and yis eliminated from subsequent equations

by addition or subtraction of appropriate multiples of (2′). In the present case,

subtraction of 4 1 × 1 (2′) from (3′) gives

(1) x 1 + 12 y 1 + 13 z 1 = 126

(2′) −y 1 − 15 z= 1 − 18 (20.44)

(3′′)12z= 1 33

The equations are in triangular or echelon form, and the solution is completed by

back substitution.

Step 3. Back substitution

The equations (20.44) are solved in reverse order. Equation (3′′) gives z 1 = 11124 ,

substitution for this in (2′) givesy 1 = 11724 , and substitution for yand zin (1) gives

x 1 = 13724.

0 Exercises 27–30

The following example demonstrates that Gauss elimination applies to any system of

linear equations, even when there is no unique solution or when there is no solution.

EXAMPLE 20.16Use Gauss elimination to solve the equations (see Section 17.4)

(1) 2x 1 + 12 y 1 + 1 z 1 = 110

(2) x 1 + 12 y 1 − 12 z 1 = 1 − 3

(3) 3x 1 + 12 y 1 + 14 z 1 = 1 λ

where λis a number.

Step 1.Subtract from (2) and from (3):

(1) 2x 1 + 12 y 1 + 1 z 1 = 110

(2′)

(3′)

Step 2.Add (2′) to (3′):

(1) 2x 1 + 12 y 1 + 1 z 1 = 110

(2′)

(3′′)0 1 = 1 λ 1 − 123

yz−=−

5

2

8

−+ = −yz

5

2

λ 15

yz−=−

5

2

8

3

2

()× 1

1

2

()× 1