Simultaneous Linear Equations and Matrix Algebra

Delve into the intricacies of solving simultaneous linear equations using matrix algebra. Explore the concept of finding the inverse of a matrix and understand the steps involved in setting up equations to find the inverse. Learn about forward elimination and back substitution in Gaussian elimination to solve linear equations efficiently.

• Mathematics
• Linear Algebra
• Matrix Operations
• Engineering Education
• Numerical Methods

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1. Simultaneous Linear Equations and Matrix Algebra Major: All Engineering Majors Author(s): Autar Kaw http://nm.MathForCollege.com Transforming Numerical Methods Education for STEM Undergraduates

2. Inverse of a Matrix

3. Find Inverse of Matrix Find inverse of 25 64 144 ?11 ?21 ?31 ?12 ?22 ?32 ?13 ?23 ?33 5 8 1 1 1 ? 1= 12 ?11 ?21 ?31 ?12 ?22 ?32 ?13 ?23 ?33 1 0 0 0 1 0 0 0 1 25 64 144 5 8 1 1 1 = 12

4. Setting up equations to find inverse ?11 ?21 ?31 ?11 ?21 ?31 1 0 0 25 64 144 5 8 1 1 1 0.04762 0.9524 4.571 = = 12 ?12 ?22 ?32 0 1 0 25 64 144 5 8 1 1 1 ?12 ?22 ?32 0.08333 1.417 5.000 = = 12 ?13 ?23 ?33 ?13 ?23 ?33 0.03571 0.4643 1.429 0 0 1 25 64 144 5 8 1 1 1 = = 12

5. Putting the solutions in matrix ?11 ?21 ?31 0.04762 0.9524 4.571 ?12 ?22 ?32 0.08333 1.417 5.000 ?13 ?23 ?33 0.03571 0.4643 1.429 = = = 0.04762 0.9524 4.571 0.08333 1.417 5.000 0.03571 0.4643 1.429 [?] 1= 1 0 0 0 1 0 0 0 1 0.04762 0.9524 4.571 0.08333 1.417 5.000 0.03571 0.4643 1.429 25 64 144 5 8 1 1 1 = 12

6. THE END

7. Nave Gauss Elimination Steps

8. Nave Gaussian Elimination A method to solve simultaneous linear equations of the form [A][X]=[C] Two steps 1. Forward Elimination 2. Back Substitution

9. Forward Elimination The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix ?1 ?2 ?3 106.8 177.2 279.2 25 64 144 5 8 12 1 1 1 = ?1 ?2 ?3 106.8 96.21 0.735 25 0 0 5 1 = 4.8 0 1.56 0.7

10. Back Substitution Solve each equation starting from the last equation ?1 ?2 ?3 106.8 96.21 0.735 25 0 0 5 1 = 4.8 0 1.56 0.7

11. THE END

12. Determinant of a Square Matrix Using Na ve Gauss Elimination

13. Theorem of Determinants If a multiple of one row of [A]nxn is added or subtracted to another row of [A]nxn to result in [B]nxn then det(A)=det(B)

14. Theorem of Determinants The determinant of an upper triangular, lower triangular or diagonal matrix [A]nxn is given by det A = ?11 ?22 ... ??? ... ??? ? = ??? ?=1

15. Forward Elimination of a Square Matrix Using forward elimination to transform [A]nxn to an upper triangular matrix, [U]nxn. ?? ? ?? ? det ? = det ?

16. Using Naive Gaussian Elimination method, find the determinant of the following square matrix. 25 64 144 5 8 1 1 1 12

17. Finding the Determinant After forward elimination steps 25 64 144 5 8 12 1 1 1 25 0 0 5 1 4.8 0 1.56 0.7 . det A = ?11 ?22 ?33 = 25 4.8 0.7 = 84.00

18. What does det(A)=0 and det(A)0 mean for [A][X]=[C] det ? = 0 implies [A][X]=[C] has no solution or infinite solutions det ? 0 implies [A][X]=[C] has a unique solution.

19. THE END

20. Nave Gauss Elimination Pitfalls

21. Pitfall#1. Division by zero 10?2 7?3= 3 6?1+ 2?2+ 3?3= 11 5?1 ?2+ 5?3= 9 ?1 ?2 ?3 0 6 5 10 2 1 7 3 5 3 11 9 =

22. Is division by zero an issue here? 12?1+ 10?2 7?3= 15 6?1+ 5?2+ 3?3= 14 5?1 ?2+ 5?3= 9 ?1 ?2 ?3 12 6 5 10 5 1 7 3 5 15 14 9 =

23. Is division by zero an issue here? YES ?1 ?2 ?3 12?1+ 10?2 7?3= 15 6?1+ 5?2+ 3?3= 14 24?1 ?2+ 5?3= 28 12 6 24 10 5 1 7 3 5 15 14 28 = ?1 ?2 ?3 12 0 0 10 0 21 7 6.5 19 15 6.5 2 = Division by zero is a possibility at any step of forward elimination

24. Pitfall#2. Large Round-off Errors ?1 ?2 ?3 20 3 5 15 10 7 3 45 = 2.249 1 1.751 9 Exact Solution ?1 ?2 ?3 1 1 1 =

25. Pitfall#2. Large Round-off Errors ?1 ?2 ?3 20 3 5 15 10 7 3 45 = 2.249 1 1.751 9 Solve it on a computer using 6 significant digits with chopping ?1 ?2 ?3 0.9625 1.05 0.999995 =

26. Pitfall#2. Large Round-off Errors ?1 ?2 ?3 20 3 5 15 10 7 3 45 = 2.249 1 1.751 9 Solve it on a computer using 5 significant digits with chopping ?1 ?2 ?3 0.99995 0.625 1.5 = Is there a way to reduce the round off error?

27. Avoiding Pitfalls Increase the number of significant digits Decreases round-off error Does not avoid division by zero

28. Avoiding Pitfalls Use Gaussian Elimination with Partial Pivoting Avoids division by zero Reduces round off error

29. THE END

30. Gauss Elimination with Partial Pivoting

31. What is Different About Partial Pivoting? At the beginning of the k th step of forward elimination, find the maximum of ???, ??+1,?,................, ??? If the maximum of these values is ??? in the ?th row, ? ? ?, then switch rows ? and ?.

32. Example (2nd step of FE) ?1 ?2 ?3 ?4 ?5 6 0 0 0 0 14 7 4 9 17 5.1 6 12 23 12 3.7 1 1 6 11 6 2 5 6 8 9 3 = 11 8 43 Which two rows would you switch?

33. Example (2nd step of FE) ?1 ?2 ?3 ?4 ?5 ?1 ?2 ?3 ?4 ?5 6 0 0 0 0 14 7 4 9 17 5.1 6 12 23 12 3.7 1 1 6 11 6 2 5 6 0 0 0 0 19 17 4 9 7 5.1 12 12 23 6 3.7 11 1 6 1 6 5 3 8 9 6 8 9 3 43 11 8 2 = = 11 8 43 6

34. Gaussian Elimination with Partial Pivoting A method to solve simultaneous linear equations of the form [A][X]=[C] Two steps 1. Forward Elimination 2. Back Substitution

35. THE END

36. Gauss Elimination with Partial Pivoting Example

37. Solve the following set of equations by Gaussian elimination with partial pivoting ?1 ?2 ?3 106.8 177.2 279.2 25 64 144 5 8 1 1 1 = 12 25 64 144 5 8 1 1 1 106.8 177.2 279.2 12

38. Forward Elimination

39. Number of Steps of Forward Elimination Number of steps of forward elimination is (n 1)=(3 1)=2

40. Forward Elimination: Step 1 Examine absolute values of first column, first row and below. 25 , 64 , 144 Largest absolute value is 144 and exists in row 3. Switch row 1 and row 3. 144 64 25 12 8 5 1 1 1 279.2 177.2 106.8 25 64 144 5 8 1 1 1 106.8 177.2 279.2 12

41. Forward Elimination: Step 1 (cont.) Divide Equation 1 by 144 and 144 64 25 12 8 5 1 1 1 279.2 177.2 106.8 64 144= 0.4444 multiply it by 64, . 279.2 0.4444 = 63.99 144 12 1 5.333 0.4444 124.1 . 64 8 1 177.2 124.1 53.10 Subtract the result from Equation 2 63.99 0 5.333 2.667 0.4444 0.5556 Substitute new equation for Equation 2 144 0 25 12 1 279.2 53.10 106.8 2.667 5 0.5556 1

42. Forward Elimination: Step 1 (cont.) Divide Equation 1 by 144 and 144 0 25 12 1 279.2 53.10 106.8 2.667 5 0.5556 1 25 144= 0.1736 multiply it by 25, . 279.2 0.1736 = 25.00 144 12 1 2.083 0.1736 48.47 . 25 2.083 2.917 5 1 106.8 48.47 58.33 Subtract the result from Equation 3 25 0 0.1736 0.8264 144 0 0 12 1 279.2 53.10 58.33 Substitute new equation for Equation 3 2.667 2.917 0.5556 0.8264

43. Forward Elimination: Step 2 Examine absolute values of second column, second row and below. 2.667 , 2.917 Largest absolute value is 2.917 and exists in row 3. Switch row 2 and row 3. 144 0 0 12 1 279.2 53.10 58.33 144 0 0 12 1 279.2 58.33 53.10 2.667 2.917 0.5556 0.8264 2.917 2.667 0.8264 0.5556

44. Forward Elimination: Step 2 (cont.) Divide Equation 2 by 2.917 and multiply it by 2.667, 2.667 2.917= 0.9143. 144 0 0 12 1 279.2 58.33 53.10 2.917 2.667 0.8264 0.5556 58.33 0.9143 = 0 0 2.917 0.8264 2.667 0.7556 53.33 0 2.667 2.667 0.5556 0.7556 0.2 53.10 53.33 0.23 . 0 0 Subtract the result from Equation 3 0 144 0 0 12 1 279.2 58.33 0.23 Substitute new equation for Equation 3 2.917 0 0.8264 0.2

45. Back Substitution

46. Back Substitution ?1 ?2 ?3 144 0 0 12 1 279.2 58.33 0.23 144 0 0 12 1 279.2 58.33 0.23 = 2.917 0 0.8264 0.2 2.917 0 0.8264 0.2 Solving for a3 0.2?3= 0.23 ?3= 0.23 0.2 = 1.15

47. Back Substitution (cont.) ?1 ?2 ?3 144 0 0 12 1 279.2 58.33 0.23 = 2.917 0 0.8264 0.2 Solving for a2 2.917?2+ 0.8264?3= 58.33 ?2=58.33 0.8264?3 2.917 =58.33 0.8264 1.15 2.917 = 19.67

48. Back Substitution (cont.) ?1 ?2 ?3 144 0 0 12 1 279.2 58.33 0.23 = 2.917 0 0.8264 0.2 144?1+ 12?2+ ?3= 279.2 ?1=279.2 12?2 ?3 =279.2 12 19.67 1.15 144 144 = 0.2917

49. Gaussian Elimination with Partial Pivoting Solution ?1 ?2 ?3 106.8 177.2 279.2 25 64 144 5 8 1 1 1 = 12 ?1 ?2 ?3 0.2917 19.67 1.15 =