1. 24 Jul 2007 KKKQ 3013 PENGIRAAN BERANGKA Week 3 – Systems of Linear Equations 24 July 2007 8.00 am – 9.00 am

2. 24 Jul 2007 Week 2 Page 2 Topics Introduction Elimination Methods Decomposition Methods Matrix Inverse and Determinant Errors, Residuals and Condition Number Iteration Methods Incomplete and Redundant Systems

3. 24 Jul 2007 Week 2 Page 3 Tutorial Example 1 (WITHOUT row interchange) Forward elimination : [A].x = b [A]

4. 24 Jul 2007 Week 2 Page 4 (i) Determinant det (A) = (1).(1).(2).(0.5) = 1 det (A) = (-1) p.a 11. a 22 (1).a 33 (2).a 44 (3) = (-1) 0.(1).(1).(2).(0.5) = 1 For triangular matrix: det [A] = product of diagonal terms, a ii Triangular obtained from [A] WITHOUT any row interchange Tutorial Example 1 (WITHOUT row interchange) Alternatively,

5. 24 Jul 2007 Week 2 Page 5 (ii) LU decomposition of [A] Now, check if [L].[U] = [A] Tutorial Example 1 (WITHOUT row interchange)

6. 24 Jul 2007 Week 2 Page 6 Tutorial Example 1 (WITH row interchange) [A].x = b ORIGINAL [A][b] [A*].x = b* [A*][b*] INTERCHANGE ROWS 2-4.

7. 24 Jul 2007 Week 2 Page 7 Tutorial Example 1 (WITH row interchange) Forward elimination : [A*]

8. 24 Jul 2007 Week 2 Page 8 (i) Determinant det (A) = (-1) p.a 11. a 22 (1).a 33 (2).a 44 (3) = (-1) 1.(1).(-1).(1).(1) = 1 Tutorial Example 1 (WITH row interchange) Equals det (A) previously ! NOT det (A*) !

9. 24 Jul 2007 Week 2 Page 9 Alternatively, det (A*) = (1).(-1).(1).(1) = -1 For triangular matrix: det [ ] = product of diagonal terms, a ii Triangular obtained from [A*] Tutorial Example 1 (WITH row interchange) Hence, det (A) = (1) p.(-1) = 1

10. 24 Jul 2007 Week 2 Page 10 Tutorial Example 1 (WITH row interchange) How many row interchange ? p = ?

11. 24 Jul 2007 Week 2 Page 11 Tutorial Example 2 Refer exercise in Chapter 1 : Question #3

12. 24 Jul 2007 Week 2 Page 12 Tutorial Example 2

13. 24 Jul 2007 Week 2 Page 13 Tutorial Example 2