1. Numerical Analysis
Lecture13

2. Chapter 3

3. Solution of Linear System of Equations and Matrix Inversion

4. Introduction Gaussian Elimination
Gauss-Jordon Elimination Crout’s Reduction
Jacobi’s Gauss- Seidal Iteration
Relaxation Matrix Inversion

5. Gauss–Seidel Iteration Method

6. It is another well-known iterative method for solving a system of linear equations of the form

7. In Jacobi’s method, the (r + 1)th approximation to the above system is given by Equations

9. Here we can observe that no element of. replaces
Here we can observe that no element of replaces entirely for the next cycle of computation.

10. In Gauss-Seidel method, the corresponding elements of
In Gauss-Seidel method, the corresponding elements of replaces those of as soon as they become available.

11. Hence, it is called the method of successive displacements
Hence, it is called the method of successive displacements. For illustration consider

12. In Gauss-Seidel iteration, the (r + 1)th approximation or iteration is computed from:

14. Thus, the general procedure can be written in the following compact form
for all and

15. To describe system in the first equation, we substitute the r-th approximation into the right-hand side and denote the result by In the second equation, we substitute and denote the result by

16. In the third equation, we substitute and denote the result by
and so on. This process is continued till we arrive at the desired result. For illustration, we consider the following example :

17. Relaxation Method

18. This is also an iterative method and is due to Southwell.
To explain the details, consider again the system of equations

19. Let
be the solution vector obtained iteratively after p-th iteration. If denotes the residual of the i-th equation of system given above , that is of

20. defined by
we can improve the solution vector successively by reducing the largest residual to zero at that iteration. This is the basic idea of relaxation method.

21. To achieve the fast convergence of the procedure, we take all terms to one side and then reorder the equations so that the largest negative coefficients in the equations appear on the diagonal.

22. Now, if at any iteration, is the largest residual in magnitude, then we give an increment to being the coefficient of xi

23. In other words, we change to
to relax
that is to reduce to zero.

24. Example Solve the system of equations
by the relaxation method, starting with the vector (0, 0, 0).

25. Solution
At first, we transfer all the terms to the right-hand side and reorder the equations, so that the largest coefficients in the equations appear on the diagonal.

26. Thus, we get
after interchanging the 2nd and 3rd equations.

27. Starting with the initial solution vector (0, 0, 0), that is taking
we find the residuals
of which the largest residual in magnitude is R3, i.e. the 3rd equation has more error and needs immediate attention for improvement.

28. Thus, we introduce a change, dx3in x3 which is obtained from the formula

29. Similarly, we find the new residuals of large magnitude and relax it to zero, and so on.
We shall continue this process, until all the residuals are zero or very small.

30. Iteration Residuals Maximum Difference Variables
number R1 R2 R3 x1 x2 x3 11 10
-15
1.875 1
9.125
8.125
1.5288 2
0.0478
6.5962

31. Iteration Residuals Maximum Difference Variables
number R1 R2 R3 x1 x2 x3 11 10
-15
15/8
=1.875 1
9.125
8.125
-9.125/(-6)
=1.5288
1.875 2
0.0478
6.5962 /7 =
1.5288 3
0.0001
2.8747/(-6) =
1.0497 4
0.4792
1.1571/8
=0.1446

32. Iteration Residuals Maximum Difference Variables
number R1 R2 R3 x1 x2 x3 5
0.3346
0.0003
-.3346/7 =
1.0497
2.0196 6
0.2881
0.0000
0.0475
-.2881/(-6)
=0.0480 7
0.048
0.1435 =
1.0017 8
0.0178
0.0659 -
2.0017

33. At this stage, we observe that all the residuals R1, R2 and R3 are small enough and therefore we may take the corresponding values of xi at this iteration as the solution.

34. Hence, the numerical solution is given by
The exact solution is

35. Numerical Analysis
Lecture13