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ΑΡΙΘΜΗΤΙΚΕΣ ΜΕΘΟΔΟΙ ΜΟΝΤΕΛΟΠΟΙΗΣΗΣ 4. Αριθμητική Επίλυση Συστημάτων Γραμμικών Εξισώσεων Gaussian elimination Gauss - Jordan 1

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Direct solution Methods Gaussian Elimination –Matrix A is transformed into an upper triangular matrix (all elements below diagonal 0) –Back substitution is used to solve the upper-triangular system Back substitution 2

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Gauss Elimination (μέθοδος απαλοιφής Gauss) 1. Gauss Elimination 2. Gauss Elimination Pitfalls 3. Gauss Elimination with Partial Pivoting 4. Determinant of a Square Matrix Using Gauss Elimination 3

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1. Gaussian Elimination A method to solve simultaneous linear equations of the form [A][X]=[C] Two steps 1. Forward Elimination 2. Back Substitution 4

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Forward Elimination The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix 5

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Forward Elimination A set of n equations and n unknowns.. (n-1) steps of forward elimination 6

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Forward Elimination Step 1 For Equation 2, divide Equation 1 by and multiply by. 7

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Forward Elimination Subtract the result from Equation 2. − _________________________________________________ or 8

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Forward Elimination Repeat this procedure for the remaining equations to reduce the set of equations as... End of Step 1 9

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Step 2 Repeat the same procedure for the 3 rd term of Equation 3. Forward Elimination.. End of Step 2 10

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Forward Elimination At the end of (n-1) Forward Elimination steps, the system of equations will look like.. End of Step (n-1) 11

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Matrix Form at End of Forward Elimination 12

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Back Substitution Solve each equation starting from the last equation Example of a system of 3 equations 13

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Back Substitution Starting Eqns.. 14

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Back Substitution Start with the last equation because it has only one unknown 15

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Back Substitution 16

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Forward Elimination For i = 1 to n-1 { // for each equation For j = i+1 to n { // for each target equation below the current For k = i+1 to n { // for each element beyond pivot column } Computational Complexity divisions multiply-add’s O(n 3 ) 17

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Computational Complexity Backward Substitution For i = n-1 to 1 { // for each equation For j = n to i+1 { // for each known variable sum = sum – A ij * x j } multiply-add’s O(n 2 ) 18

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Example 1 The upward velocity of a rocket is given at three different times Time, Velocity, 5106.8 8177.2 12279.2 The velocity data is approximated by a polynomial as: Find the velocity at t=6 seconds. Table 1 Velocity vs. time data.

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Example 1 Cont. Assume Results in a matrix template of the form: Using data from Table 1, the matrix becomes: 20

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Example 1 Cont. 1.Forward Elimination 2.Back Substitution 21

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Forward Elimination 22

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Number of Steps of Forward Elimination Number of steps of forward elimination is ( n 1) (3 1) 2 23

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Divide Equation 1 by 25 and multiply it by 64,. Forward Elimination: Step 1. Subtract the result from Equation 2 Substitute new equation for Equation 2 24

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Forward Elimination: Step 1 (cont.). Divide Equation 1 by 25 and multiply it by 144,. Subtract the result from Equation 3 Substitute new equation for Equation 3 25

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Forward Elimination: Step 2 Divide Equation 2 by −4.8 and multiply it by −16.8,. Subtract the result from Equation 3 Substitute new equation for Equation 3 26

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Back Substitution 27

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Back Substitution Solving for a 3 28

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Back Substitution (cont.) Solving for a 2 29

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Back Substitution (cont.) Solving for a 1 30

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Gaussian Elimination Solution 31

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Example 1 Cont. Solution The solution vector is The polynomial that passes through the three data points is then: 32

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2. Gauss Elimination Pitfalls 33

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Pitfall#1. Division by zero 34

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Is division by zero an issue here? 35

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Is division by zero an issue here? YES Division by zero is a possibility at any step of forward elimination 36

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Pitfall#2. Large Round-off Errors Exact Solution 37

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Pitfall#2. Large Round-off Errors Solve it on a computer using 6 significant digits with chopping 38

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Pitfall#2. Large Round-off Errors Solve it on a computer using 5 significant digits with chopping Is there a way to reduce the round off error? 39

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Avoiding Pitfalls Increase the number of significant digits Decreases round-off error Does not avoid division by zero 40

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Avoiding Pitfalls Gaussian Elimination with Partial Pivoting Avoids division by zero Reduces round off error 41

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3. Gauss Elimination with Partial Pivoting 42

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Avoiding Pitfalls Gaussian Elimination with Partial Pivoting Avoids division by zero Reduces round off error 43

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What is Different About Partial Pivoting? At the beginning of the k th step of forward elimination, find the maximum of If the maximum of the values is in the p th row,then switch rows p and k. 44

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Matrix Form at Beginning of 2 nd Step of Forward Elimination 45

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Example (2 nd step of FE) Which two rows would you switch? 46

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Example (2 nd step of FE) Switched Rows 47

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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 48

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Forward Elimination Same as Gauss elimination method except that we switch rows before each of the (n-1) steps of forward elimination. 49

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Example: Matrix Form at Beginning of 2 nd Step of Forward Elimination 50

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Matrix Form at End of Forward Elimination 51

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Back Substitution Starting Eqns.. 52

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Back Substitution 53

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Example 2 Solve the following set of equations by Gaussian elimination with partial pivoting 54

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Example 2 Cont. 1.Forward Elimination 2.Back Substitution 55

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Forward Elimination 56

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Number of Steps of Forward Elimination Number of steps of forward elimination is ( n 1 )=(3 1)=2 57

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Forward Elimination: Step 1 Examine absolute values of first column, first row and below. Largest absolute value is 144 and exists in row 3. Switch row 1 and row 3. 58

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Forward Elimination: Step 1 (cont.). Divide Equation 1 by 144 and multiply it by 64,. Subtract the result from Equation 2 Substitute new equation for Equation 2 59

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Forward Elimination: Step 1 (cont.). Divide Equation 1 by 144 and multiply it by 25,. Subtract the result from Equation 3 Substitute new equation for Equation 3 60

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Forward Elimination: Step 2 Examine absolute values of second column, second row and below. Largest absolute value is 2.917 and exists in row 3. Switch row 2 and row 3. 61

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Forward Elimination: Step 2 (cont.). Divide Equation 2 by 2.917 and multiply it by 2.667, Subtract the result from Equation 3 Substitute new equation for Equation 3 62

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Back Substitution 63

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Back Substitution Solving for a 3 64

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Back Substitution (cont.) Solving for a 2 65

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Back Substitution (cont.) Solving for a 1 66

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Gaussian Elimination with Partial Pivoting Solution 67

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Partial Pivoting: Example Consider the system of equations In matrix form = Solve using Gaussian Elimination with Partial Pivoting using five significant digits with chopping 68

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Partial Pivoting: Example Forward Elimination: Step 1 Examining the values of the first column |10|, |-3|, and |5| or 10, 3, and 5 The largest absolute value is 10, which means, to follow the rules of Partial Pivoting, we switch row1 with row1. Performing Forward Elimination 69

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Partial Pivoting: Example Forward Elimination: Step 2 Examining the values of the first column |-0.001| and |2.5| or 0.0001 and 2.5 The largest absolute value is 2.5, so row 2 is switched with row 3 Performing the row swap 70

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Partial Pivoting: Example Forward Elimination: Step 2 Performing the Forward Elimination results in: 71

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Partial Pivoting: Example Back Substitution Solving the equations through back substitution 72

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Partial Pivoting: Example Compare the calculated and exact solution The fact that they are equal is coincidence, but it does illustrate the advantage of Partial Pivoting 73

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4. Determinant of a Square Matrix Using Gauss Elimination Example 74

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Theorem of Determinants If a multiple of one row of [A] n x n is added or subtracted to another row of [A] n x n to result in [B] n x n then det(A)=det(B) 75

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Theorem of Determinants The determinant of an upper triangular matrix [A] n x n is given by 76

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Forward Elimination of a Square Matrix Using forward elimination to transform [A] n x n to an upper triangular matrix, [U] n x n. 77

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Example Using Gaussian elimination find the determinant of the following square matrix. 78

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Forward Elimination 79

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Forward Elimination: Step 1. Divide Equation 1 by 25 and multiply it by 64,. Subtract the result from Equation 2 Substitute new equation for Equation 2 80

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Forward Elimination: Step 1 (cont.). Divide Equation 1 by 25 and multiply it by 144,. Subtract the result from Equation 3 Substitute new equation for Equation 3 81

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Forward Elimination: Step 2. Divide Equation 2 by −4.8 and multiply it by −16.8,. Subtract the result from Equation 3 Substitute new equation for Equation 3 82

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Finding the Determinant. After forward elimination 83

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5. Gaussian Elimination For Tridiagonial System 84

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* Introduction of Tridiagonal System? Special Linear System Arising in Application A general tridiagonal matrix is a matrix whose nonzero elements are found only on the diagonal, subdiagonal, and superdiagonal of the matrix. if |i – j | > 1, Tridiagonal Systems I 85

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In Storage Only the diagonal, subdiagonal, and superdiagonal elements of the general tridiagonal matrix are stored. This is called tridiagonal storage mode. The elements of a general tridiagonal matrix, A, of order n are stored in three one-dimensional arrays, C, D, and E, each of length n. array C contains the subdiagonal elements, stored as follows: C = (*, a21, a32, a43,..., an,n-1) array D contains the main diagonal elements, stored as follows: D = (a11, a22, a33,..., ann) array E contains the superdiagonal elements, stored as follows: E = (a12, a23, a34,..., an-1,n, *) where "*" means you do not store an element in that position in the array Tridiagonal Systems II 86

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* Example of Tridiagonal Matrix… 2x 1 –x 2 = 1, -x 1 +2x 2 –x 3 = 0, -x 2 +2x 3 –x 4 = 0, -x 3 +2x 4 = 1. Tridiagonal Systems III 87

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Gaussian Elimination for Tridiagonal Systems 88

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1.B 1 and A n are zero. 2.This algorithm takes advantage of the zero elements that are already present in the coefficient matrix and avoids unnecessary arithmetic operations. Thus, we need to store only the new vectors a and r. Solving a Tridiagonal Systems Using the Thomas Method I 89

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Solving a Tridiagonal Systems Using the Thomas Method II Step 1 : For the first equation Step 2 : For each of the equation Step 3 : For the last equation Step 4 : by back substitution 90

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Solving a Tridiagonal Systems Using the Thomas Method III 91

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Solving a Tridiagonal Systems Using the Thomas Method IV 92

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* The required multiplications and divisions for Thomas method. For the first equation, 2divisions are needed. For each of the next n-2 equations, 2multiplications and 2 divisions are needed. For the last equation, 2 multiplications and 1 division are required. The total for elimination is 5+4(n-2). For the back substitution, n-1 multiplications are needed. Discussion of Thomas method 93

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Using the Thomas Method for a System that Would Require Pivoting for Gaussian Elimination I 94

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Using the Thomas Method for a System that Would Require Pivoting for Gaussian Elimination II 95

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Using the Thomas Method for a System that Would Require Pivoting for Gaussian Elimination III 96

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6. Gauss – Jordan Elimination 97

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Gauss-Jordan Variation of Gauss elimination Primary motive for introducing this method is that it provides a simple and convenient method for computing the matrix inverse. When an unknown is eliminated, it is eliminated from all other equations, rather than just the subsequent one. All rows are normalized by dividing them by their pivot elements Elimination step results in an identity matrix

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Introduction For inverting a matrix, Gauss-Jordan elimination is about as efficient as any other method. For solving sets of linear equations, Gauss-Jordan elimination produces both the solution of the equations for one or more right-hand side vectors b, and also the matrix inverse A −1. However, its principal weaknesses are –(i) that it requires all the right-hand sides to be stored and manipulated at the same time, and –(ii) that when the inverse matrix is not desired, Gauss-Jordan is three times slower than the best alternative technique for solving a single linear set The method’s principal strength is that it is as stable as any other direct method, perhaps even a bit more stable when full pivoting is used 99

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Graphical depiction of Gauss-Jordan

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102 Gauss-Jordan Elimination Let us consider the set of linearly independent equations. Augmented matrix for the set is:

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103 Gauss-Jordan Elimination Step 1: Eliminate x from the 2nd and 3rd equation.

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104 Gauss-Jordan Elimination Step 2: Eliminate y from the 3rd equation. 13R’ 2 +R’ 3 R’’ 3 Step 3: 0.5R’ 1 R’ 1 -R’ 2 R’’ 2 (1/168)R’’ 3 R’’’ 3

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105 Gauss-Jordan Elimination Step 4: Eliminate z from the 2 nd equation

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106 Gauss-Jordan Elimination Step 5-1: Eliminate y from the 1st equation

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107 Gauss-Jordan Elimination Step 5-2: Eliminate z from the 1st equation

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Example Given the following, determine {x} for the two different loads {c} 108

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Solution {c} T = {1 2 3} x 1 = (2)(1) + (-1)(2) + (1)(3) = 3 x 2 = (-2)(1) + (6)(2) + (3)(3) = 19 x 3 = (-3)(1) + (1)(2) + (-4)(3) = -13 {c} T = {4 -7 1) x 1 = (2)(4) + (-1)(-7) + (1)(1)=16 x 2 = (-2)(4) + (6)(-7) + (3)(1) = -47 x 3 = (-3)(4) + (1)(-7) + (-4)(1) = -23 109

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