NUMERICAL METHODS
appreciate how this would apply in (say) a computer program for a 1000-variable
case, perhaps with unforseeable zeros or very small numbers appearing on the
leading diagonal.
Solve the simultaneous equations
(a) x 1 +6x 2 − 4 x 3 =8,
(b) 3 x 1 − 20 x 2 +x 3 =12,
(c) −x 1 +3x 2 +5x 3 =3.
(27.22)
Firstly, we interchange rows (a) and (b) to bring the term 3x 1 onto the leading diagonal. In
the following, we label the important equations (I), (II), (III), and the others alphabetically.
A general (i.e. variable) label will be denoted byj.
(I) 3 x 1 − 20 x 2 +x 3 =12,
(d) x 1 +6x 2 − 4 x 3 =8,
(e) −x 1 +3x 2 +5x 3 =3.
For (j) = (d) and (e), replace row (j)by
row (j)−
aj 1
3
×row (I),
whereaj 1 is the coefficient ofx 1 in row (j), to give the two equations
(II)
(
6+^203
)
x 2 +
(
− 4 −^13
)
x 3 =8−^123 ,
(f)
(
3 −^203
)
x 2 +
(
5+^13
)
x 3 =3+^123.
Now|6+^203 |>| 3 −^203 |and so no interchange is required before the next elimination. To
eliminatex 2 ,replacerow(f)by
row (f)−
(
−^113
)
38
3
×row (II).
This gives
(III)
[ 16
3 +
11
38 ×
(−13)
3
]
x 3 =7+^1138 × 4.
Collecting together and tidying up the final equations, we have
(I) 3x 1 − 20 x 2 +x 3 =12,
(II) 38 x 2 − 13 x 3 =12,
(III) x 3 =2.
Starting with (III) and working backwards, it is now a simple matter to obtain
x 1 =10,x 2 =1,x 3 =2.
27.3.2 Gauss–Seidel iteration
In the example considered in the previous subsection an explicit way of solving
a set of simultaneous equations was given, the accuracy obtainable being limited
only by the rounding errors in the calculating facilities available, and the calcula-
tion was planned to minimise these. However, in some situations it may be that
only an approximate solution is needed. If, for a large number of variables, this is