Example 1: Solving a system of equations by the Gauss-Seidel method
Use the Gauss-Seidel method to solve the system
 |
4x1 | + x2 | - x3 | = 3 |
| 2x1 | + 7 x2 | + x3 | = 19 |
| x1 | - 3 x2 | +12 x3 | = 31 |
|
| <=> | |
x1 = | | -1/4 x2 | + 1/4 x3 | + 3/4 |
| x2 = | -2/7 x1 | | - 1/7 x3 | + 19/7 |
| x3 = | -1/12 x1 | + 1/4 x2 | | + 31/12 |
|
Solution:
We have
 = |  |
0 | -1/4 | 1/4 |  |
| -2/7 | 0 | -1/7 |
| -1/12 | 1/4 | 0 | |
|
|
x1 | |
| x2 |
| x3 | |
|
|
The iteration formulas are
| <=> | |
x1(k+1) = | | -1/4 x2(k) | + 1/4 x3(k) | + 3/4 |
| x2(k+1) = | -2/7 x1(k+1) | | - 1/7 x3(k) | + 19/7 |
| x3(k+1) = | -1/12 x1(k+1) | + 1/4 x2(k+1) | | + 31/12 |
The difference between the
Gauss-Seidel method and the Jacobi method
is that here we use the coordinates x1(k),...,xi-1(k)
of x(k) already known to compute its
ith coordinate xi(k).
If we start from x1(0) = x2(0) = x3(0) = 0 and apply the iteration formulas, we obtain
| k | | x1(k) | x2(k) | x3(k) |
| 0 | | 0 | 0 | 0 |
| 1 | | 0,75 | 2,50 | 3,15 |
| 2 | | 0,91 | 2,00 | 3,01 |
| 3 | | 1,00 | 2,00 | 3,00 |
| 4 | | 1,00 | 2,00 | 3,00 |
The exact solution is: x1 = 1, x2 = 2, x3 = 3.
For instance, when k=2, we have x2(2)= 2,00:
x2(2) = -2/7 x1(2) - 1/7 x3(1) + 19/7 = -2/7 · 0,72 - 1/7 · 3,15 + 19/7 = 2,00
Go back to theory
