ΑΡΙΘΜΗΤΙΚΕΣ ΜΕΘΟΔΟΙ ΜΟΝΤΕΛΟΠΟΙΗΣΗΣ 4. Αριθμητική Επίλυση Συστημάτων Γραμμικών Εξισώσεων • Gaussian elimination • Gauss - Jordan 1 Direct solution Methods • Gaussian Elimination a11 ai1 an1 a1i aii ani a1n x1 b1 ain xi bi ann xn bn a11 0 0 a1i a1n x1 b1 ~ a~ii a~in xi bi ~ 0 a~nn xn bn Back substitution – Matrix A is transformed into an upper triangular matrix (all elements below diagonal 0) – Back substitution is used to solve the upper-triangular system 2 Gauss Elimination (μέθοδος απαλοιφής Gauss) 1. 2. 3. 4. Gauss Elimination Gauss Elimination Pitfalls Gauss Elimination with Partial Pivoting Determinant of a Square Matrix Using Gauss Elimination 3 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 Forward Elimination The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix 25 5 1 x1 106.8 64 8 1 x 177.2 2 144 12 1 x3 279.2 5 1 x1 106.8 25 0 4.8 1.56 x 96.21 2 0 0.7 x3 0.735 0 5 Forward Elimination A set of n equations and n unknowns a11x1 a12 x2 a13 x3 ... a1n xn b1 a21x1 a22 x2 a23 x3 ... a2 n xn b2 . . . . . . an1x1 an 2 x2 an3 x3 ... ann xn bn (n-1) steps of forward elimination 6 Forward Elimination Step 1 For Equation 2, divide Equation 1 by a11 and multiply by a21 . a21 a (a11x1 a12 x2 a13 x3 ... a1n xn b1 ) 11 a21 a21 a21 a21x1 a12 x2 ... a1n xn b1 a11 a11 a11 7 Forward Elimination Subtract the result from Equation 2. a21x1 a22 x2 a23 x3 ... a2 n xn b2 a21 a21 a21 − a21x1 a a12 x2 ... a a1n xn a b1 11 11 11 _________________________________________________ a21 a21 a21 a22 a12 x2 ... a2 n a1n xn b2 b1 a11 a11 a11 or a x ... a x b ' 22 2 ' 2n n ' 2 8 Forward Elimination Repeat this procedure for the remaining equations to reduce the set of equations as a11x1 a12 x2 a13 x3 ... a1n xn b1 ' ' a22 x2 a23 x3 ... a2' n xn b2' ' ' a32 x2 a33 x3 ... a3' n xn b3' . . . . . . . . . ' an' 2 x2 an' 3 x3 ... ann xn bn' End of Step 1 9 Forward Elimination Step 2 Repeat the same procedure for the 3rd term of Equation 3. a11x1 a12 x2 a13 x3 ... a1n xn b1 ' ' a22 x2 a23 x3 ... a2' n xn b2' " a33 x3 ... a3" n xn b3" . . . . . . " an" 3 x3 ... ann xn bn" End of Step 2 10 Forward Elimination At the end of (n-1) Forward Elimination steps, the system of equations will look like a11 x1 a12 x2 a13 x3 ... a1n xn b1 ' ' a22 x2 a23 x3 ... a2' n xn b2' " a33 x3 ... a3"n xn b3" . . . . . . n 1 n 1 ann xn bn End of Step (n-1) 11 Matrix Form at End of Forward Elimination a11 a12 0 a' 22 0 0 0 0 a1n x1 b1 ' ' a 2 n x2 b2 a"3n x3 b3" (n1 ) xn bn(n-1 ) 0 ann a13 a ' 23 " 33 a 0 12 Back Substitution Solve each equation starting from the last equation 5 1 x1 106.8 25 0 4.8 1.56 x 96.21 2 0 0.7 x3 0.735 0 Example of a system of 3 equations 13 Back Substitution Starting Eqns a11 x1 a12 x2 a13 x3 ... a1n xn b1 ' ' a22 x2 a23 x3 ... a2' n xn b2' " a33 x3 ... an" xn b3" . . . n 1 . . . n 1 ann xn bn 14 Back Substitution Start with the last equation because it has only one unknown ( n 1) n ( n 1) nn b xn a 15 Back Substitution ( n 1) n ( n 1) nn b xn a xi bii 1 ai,ii11 xi 1 ai,ii12 xi 2 ... ai,in1 xn i 1 aii i 1 xi bi aiji 1 x j for i n 1,...,1 n j i 1 i 1 ii a for i n 1,...,1 16 Computational Complexity • Forward Elimination For i = 1 to n-1 { For j = i+1 to n { // for each equation // for each target equation below the current A ji n 1 n2 M ji , A ji 0 (n i ) divisions A 2 1 For k = i+1 to nii { // for each elementi beyond pivot column } } } A jk A jk M ji Aik 3 O(n ) n 1 2 3 (n i ) n 3 i 1 2 multiply-add’s 17 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 – Aij * xj } } n 1 n2 (n i ) multiply-add’s 2 i 1 2 O(n ) 18 Example 1 The upward velocity of a rocket is given at three different times Table 1 Velocity vs. time data. Time, t s 5 8 12 Velocity, v m/s 106.8 177.2 279.2 The velocity data is approximated by a polynomial as: vt a1t 2 a 2 t a3 , Find the velocity at t=6 seconds . 5 t 12. Example 1 Cont. Assume vt a1t 2 a2t a3 , 5 t 12. Results in a matrix template of the form: t12 2 t 2 t32 t1 t2 t3 1 1 1 a1 v1 a v 2 2 a3 v3 Using data from Table 1, the matrix becomes: 25 5 1 a1 106.8 64 8 1 a 177.2 2 144 12 1 a3 279.2 20 Example 1 Cont. 25 5 1 a1 106.8 25 5 1 106.8 64 8 1 a2 177.2 64 8 1 177.2 144 12 1 a3 279.2 144 12 1 279.2 1. Forward Elimination 2. Back Substitution 21 Forward Elimination 22 Number of Steps of Forward Elimination Number of steps of forward elimination is (n1)(31)2 23 Forward Elimination: Step 1 25 5 1 106.8 64 8 1 177.2 144 12 1 279.2 25 Divide Equation 1 by 25 and 64 2.56 . multiply it by 64, 25 5 1 106.8 2.56 64 12.8 2.56 273.408 . Subtract the result from Equation 2 Substitute new equation for Equation 2 64 64 0 177.2 8 1 12.8 2.56 273.408 4.8 1.56 96.208 5 1 106.8 25 0 4.8 1.56 96.208 144 12 1 279.2 24 Forward Elimination: Step 1 (cont.) 5 1 106.8 25 Divide Equation 1 by 25 and 0 4.8 1.56 96.208 144 144 12 1 279.2 multiply it by 144, 25 5.76 . 25 5 1 106.8 5.76 144 28.8 5.76 615.168 12 1 Subtract the result from 144 144 28.8 5.76 Equation 3 0 16.8 4.76 . 279.2 615.168 335.968 5 1 106.8 25 Substitute new equation for 0 4 . 8 1 . 56 96 . 208 Equation 3 0 16.8 4.76 335.968 25 Forward Elimination: Step 2 5 1 106.8 25 0 4.8 1.56 96.208 0 16.8 4.76 335.968 0 Divide Equation 2 by −4.8 and multiply it by −16.8, 16.8 3.5 . 4. 8 4.8 1.56 96.208 3.5 0 16.8 5.46 336.728 Subtract the result from Equation 3 Substitute new equation for Equation 3 0 0 0 16.8 4.76 335.968 0.7 0.76 16.8 5.46 336.728 0 5 1 106.8 25 0 4.8 1.56 96.208 0 0 0.7 0.76 26 Back Substitution 27 Back Substitution 5 1 106.8 25 5 1 a1 106.8 25 0 4.8 1.56 96.2 0 4.8 1.56 a 96.208 2 0 0 0.7 0.7 0 0 0.7 a3 0.76 Solving for a3 0.7 a3 0.76 0.76 a3 0 .7 a3 1.08571 28 Back Substitution (cont.) 5 1 a1 106.8 25 0 4.8 1.56 a 96.208 2 0 0 0.7 a3 0.76 Solving for a2 4.8a2 1.56a3 96.208 96.208 1.56a3 a2 4.8 96.208 1.56 1.08571 a2 4.8 a2 19.6905 29 Back Substitution (cont.) 5 1 a1 106.8 25 0 4.8 1.56 a 96.2 2 0 0 0.7 a3 0.76 Solving for a1 25a1 5a2 a3 106.8 106.8 5a2 a3 a1 25 106.8 5 19.6905 1.08571 25 0.290472 30 Gaussian Elimination Solution 25 5 1 a1 106.8 64 8 1 a2 177.2 144 12 1 a3 279.2 a1 0.290472 a 19.6905 2 a3 1.08571 31 Example 1 Cont. a1 0.290472 a 19.6905 2 a3 1.08571 Solution The solution vector is The polynomial that passes through the three data points is then: vt a1t 2 a2t a3 0.290472t 2 19.6905t 1.08571, 5 t 12 v6 0.2904726 19.69056 1.08571 2 129.686 m/s . 32 2. Gauss Elimination Pitfalls 33 Pitfall#1. Division by zero 10 x2 7 x3 3 6 x1 2 x2 3x3 11 5 x1 x2 5 x3 9 0 10 7 x1 3 6 2 3 x2 11 5 1 5 x3 9 34 Is division by zero an issue here? 12 x1 10 x2 7 x3 15 6 x1 5 x2 3x3 14 5 x1 x2 5 x3 9 12 10 7 x1 15 6 5 3 x2 14 5 1 5 x3 9 35 Is division by zero an issue here? YES 12 x1 10 x2 7 x3 15 6 x1 5 x2 3x3 14 24 x1 x2 5 x3 28 12 10 7 x1 15 6 5 3 x2 14 24 1 5 x3 28 12 10 7 x1 15 0 0 6.5 x2 6.5 12 21 19 x3 2 Division by zero is a possibility at any step of forward elimination 36 Pitfall#2. Large Round-off Errors 15 10 x1 45 20 3 2.249 7 x 1.751 2 1 3 x3 9 5 Exact Solution x1 1 x 1 2 x3 1 37 Pitfall#2. Large Round-off Errors 15 10 x1 45 20 3 2.249 7 x 1.751 2 1 3 x3 9 5 Solve it on a computer using 6 significant digits with chopping x1 0.9625 x 1.05 2 x3 0.999995 38 Pitfall#2. Large Round-off Errors 15 10 x1 45 20 3 2.249 7 x 1.751 2 1 3 x3 9 5 Solve it on a computer using 5 significant digits with chopping x1 0.625 x 1.5 2 x3 0.99995 Is there a way to reduce the round off error? 39 Avoiding Pitfalls Increase the number of significant digits • Decreases round-off error • Does not avoid division by zero 40 Avoiding Pitfalls Gaussian Elimination with Partial Pivoting • Avoids division by zero • Reduces round off error 41 3. Gauss Elimination with Partial Pivoting 42 Avoiding Pitfalls Gaussian Elimination with Partial Pivoting • Avoids division by zero • Reduces round off error 43 What is Different About Partial Pivoting? At the beginning of the kth step of forward elimination, find the maximum of akk , ak 1,k ,................, ank If the maximum of the values is a pk in the p th row, k p n, then switch rows p and k. 44 Matrix Form at Beginning of 2nd Step of Forward Elimination a11 a12 0 a' 22 ' 0 a32 0 a'n 2 a13 ' 23 ' 33 an' 3 an' 4 a a a1n x1 b1 ' ' a 2 n x2 b2 a3' n x3 b3' ' xn bn' ann 45 Example (2nd step of FE) 6 14 5.1 3.7 6 x1 5 0 7 6 1 2 x2 6 4 12 1 11 x3 8 0 9 23 6 8 x4 9 0 0 17 12 11 43 x5 3 Which two rows would you switch? 46 Example (2nd step of FE) 6 14 5.1 3.7 6 x1 5 0 17 12 11 43 x 3 2 4 12 1 11 x3 8 0 9 23 6 8 x4 9 0 0 7 6 1 2 x5 6 Switched Rows 47 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 Forward Elimination Same as Gauss elimination method except that we switch rows before each of the (n-1) steps of forward elimination. 49 Example: Matrix Form at Beginning of 2nd Step of Forward Elimination a11 a12 0 a' 22 ' 0 a32 0 a'n 2 a13 a'23 ' 33 a a ' n3 a ' n4 a1n x1 b1 a'2 n x2 b2' ' ' a3n x3 b3 ' ' ann xn bn 50 Matrix Form at End of Forward Elimination a11 a12 0 a' 22 0 0 0 0 a1n x1 b1 a'2 n x2 b2' " " a3n x3 b3 (n1 ) (n-1 ) 0 ann xn bn a13 a'23 " 33 a 0 51 Back Substitution Starting Eqns a11 x1 a12 x2 a13 x3 ... a1n xn b1 ' ' a22 x2 a23 x3 ... a2' n xn b2' " a33 x3 ... an" xn b3" . . . n 1 . . . n 1 ann xn bn 52 Back Substitution ( n 1) n ( n 1) nn b xn a i 1 xi bi n i 1 aij x j j i 1 i 1 ii a for i n 1,...,1 53 Example 2 Solve the following set of equations by Gaussian elimination with partial pivoting 5 1 a1 106.8 25 64 8 1 a 2 177.2 144 12 1 a3 279.2 54 Example 2 Cont. 25 5 1 a1 106.8 25 5 1 106.8 64 8 1 a 177.2 64 8 1 177 . 2 2 144 12 1 a3 279.2 144 12 1 279.2 1. Forward Elimination 2. Back Substitution 55 Forward Elimination 56 Number of Steps of Forward Elimination Number of steps of forward elimination is (n 1)=(31)=2 57 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. 25 5 1 106.8 144 12 1 279.2 64 8 1 177.2 64 8 1 177.2 144 12 1 279.2 25 5 1 106.8 58 Forward Elimination: Step 1 (cont.) 144 12 1 279.2 64 8 1 177.2 25 5 1 106.8 144 Divide Equation 1 by 144 and 64 0.4444 . multiply it by 64, 144 12 1 279.2 0.4444 63.99 5.333 0.4444 124.1 . Subtract the result from Equation 2 64 63.99 0 8 1 177.2 5.333 0.4444 124.1 2.667 0.5556 53.10 1 279.2 Substitute new equation for 144 12 0 2.667 0.5556 53.10 Equation 2 25 5 1 106.8 59 Forward Elimination: Step 1 (cont.) 12 1 279.2 Divide Equation 1 by 144 and 144 0 2.667 0.5556 53.10 25 multiply it by 25, 0.1736. 144 25 5 1 106.8 144 12 1 279.2 0.1736 25.00 2.083 0.1736 48.47 . Subtract the result from Equation 3 Substitute new equation for Equation 3 25 25 0 5 1 106.8 2.917 0.8264 58.33 2.083 0.1736 48.47 12 1 279.2 144 0 2.667 0.5556 53.10 0 2.917 0.8264 58.3360 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. 1 279.2 1 279.2 144 12 144 12 0 2.667 0.5556 53.10 0 2.917 0.8264 58.33 0 2.917 0.8264 58.33 0 2.667 0.5556 53.10 61 Forward Elimination: Step 2 (cont.) 12 1 279.2 144 0 2.917 0.8264 58.33 0 2.667 0.5556 53.10 0 Divide Equation 2 by 2.917 and multiply it by 2.667, 2.667 0.9143. 2.917 2.917 0.8264 58.33 0.9143 0 2.667 0.7556 53.33 Subtract the result from Equation 3 0 0 0 Substitute new equation for Equation 3 12 1 279.2 144 0 2.917 0.8264 58.33 0 0 0.2 0.23 . 2.667 0.5556 2.667 0.7556 0 53.10 53.33 0.2 0.23 62 Back Substitution 63 Back Substitution 1 279.2 1 144 12 144 12 0 2.917 0.8264 58.33 0 2.917 0.8264 0 0 0 0.2 0.23 0 0.2 a1 279.2 a 58.33 2 a3 0.23 Solving for a3 0.2a3 0.23 0.23 a3 0.2 1.15 64 Back Substitution (cont.) 1 144 12 0 2.917 0.8264 0 0.2 0 a1 279.2 a 58.33 2 a3 0.23 Solving for a2 2.917 a2 0.8264a3 58.33 58.33 0.8264a3 a2 2.917 58.33 0.8264 1.15 2.917 19.67 65 Back Substitution (cont.) 1 144 12 0 2.917 0.8264 0 0.2 0 a1 279.2 a 58.33 2 a3 0.23 Solving for a1 144a1 12a2 a3 279.2 279.2 12a2 a3 a1 144 279.2 12 19.67 1.15 144 0.2917 66 Gaussian Elimination with Partial Pivoting Solution 25 5 1 a1 106.8 64 8 1 a 177.2 2 144 12 1 a3 279.2 a1 0.2917 a 19.67 2 a3 1.15 67 Partial Pivoting: Example Consider the system of equations 10 x1 7 x2 7 3x1 2.099 x2 6 x3 3.901 5 x1 x2 5 x3 6 In matrix form 7 0 x1 7 10 3.901 3 2.099 6 x 2 = 6 5 1 5 x 3 Solve using Gaussian Elimination with Partial Pivoting using five significant digits with chopping 68 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 7 0 x1 7 10 3 2.099 6 x 3.901 2 5 1 5 x3 6 7 0 x1 7 10 0 0.001 6 x 6.001 2 0 2.5 5 x3 2.5 69 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 7 0 x1 7 10 0 0.001 6 x 6.001 2 0 2.5 5 x3 2.5 7 0 x1 7 10 0 x 2.5 2 . 5 5 2 0 0.001 6 x3 6.001 70 Partial Pivoting: Example Forward Elimination: Step 2 Performing the Forward Elimination results in: 0 x1 7 10 7 0 2.5 x 2.5 5 2 0 0 6.002 x3 6.002 71 Partial Pivoting: Example Back Substitution Solving the equations through back substitution 0 x1 7 10 7 0 2.5 x 2.5 5 2 0 0 6.002 x3 6.002 6.002 x3 1 6.002 2.5 5 x3 x2 1 2.5 7 7 x 2 0 x3 x1 0 10 72 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 x1 0 X calculated x2 1 x3 1 X exact x1 0 x 2 1 x3 1 73 4. Determinant of a Square Matrix Using Gauss Elimination Example 74 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) 75 Theorem of Determinants The determinant of an upper triangular matrix [A]nxn is given by det A a11 a22 ... aii ... ann n a ii i 1 76 Forward Elimination of a Square Matrix Using forward elimination to transform [A]nxn to an upper triangular matrix, [U]nxn. Ann U nn det A det U 77 Example Using Gaussian elimination find the determinant of the following square matrix. 5 1 25 64 8 1 144 12 1 78 Forward Elimination 79 Forward Elimination: Step 1 25 5 1 64 8 1 144 12 1 25 Divide Equation 1 by 25 and 64 2.56 . multiply it by 64, 25 5 1 2.56 64 12.8 2.56 . Subtract the result from Equation 2 Substitute new equation for Equation 2 64 64 0 8 12.8 1 2.56 4.8 1.56 5 1 25 0 4.8 1.56 144 12 1 80 Forward Elimination: Step 1 (cont.) 5 1 25 0 4.8 1.56 Divide Equation 1 by 25 and multiply it by 144, 144 5.76 . 144 12 1 25 25 5 1 5.76 144 28.8 5.76 . Subtract the result from Equation 3 Substitute new equation for Equation 3 144 12 144 28.8 0 16.8 1 5.76 4.76 5 1 25 0 4.8 1.56 0 16.8 4.76 81 Forward Elimination: Step 2 5 1 25 0 4.8 1.56 0 16.8 4.76 0 Divide Equation 2 by −4.8 and multiply it by −16.8, 16.8 3.5 . 4. 8 4.8 1.56 3.5 0 16.8 5.46 . Subtract the result from Equation 3 Substitute new equation for Equation 3 0 0 0 16.8 4.76 16.8 5.46 0 0 .7 5 1 25 0 4.8 1.56 0 0 0.7 82 Finding the Determinant After forward elimination 5 1 25 5 1 25 64 8 1 0 4.8 1.56 0 0.7 144 12 1 0 . det A u11 u22 u33 25 4.8 0.7 84.00 83 5. Gaussian Elimination For Tridiagonial System 84 Tridiagonal Systems I * 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 , 85 Tridiagonal Systems II 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 86 Tridiagonal Systems III * Example of Tridiagonal Matrix… 2x1 –x2 = 1, -x1 +2x2 –x3 = 0, -x2 +2x3 –x4 = 0, -x3 +2x4 = 1. 2 1 0 0 1 2 1 0 0 1 2 1 0 0 1 2 | 1 | 0 | 0 | 1 87 Gaussian Elimination for Tridiagonal Systems 2x1 – x2 1, - x1 2x2 – x3 0, x1 – - x2 2x3 – x4 0, - x3 2x4 1. x4 1; 1 3 (– )(1) 1; 4 4 1 2 x2 (– )(1) 1; 3 3 1 1 x1 (– )(1) 1. 2 2 x3 1 1 x2 , 2 2 3 1 x2 – x3 , 2 2 - x2 2x3 – x4 0, - x3 2x4 1. x1 – 1 x2 2 x2 – 1 , 2 1 , 3 1 , 4 1. 2 x3 3 x3 – 3 x4 4 x4 88 Solving a Tridiagonal Systems Using the Thomas Method d 1 x1 a1 x 2 I r1, b 2 x1 d 2 x 2 a 2 x3 .... r 2, bn 1 xn 2 dn 1 xn 1 an 1 xn rn 1, bnxn 1 dnxn rn. 1. B1 and An 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. 89 Solving a Tridiagonal Systems Using the Thomas Method II Step 1 : For the first equation a1 r1 a1 , r1 d1 d1 Step 2 : For each of the equation ai ri biri 1 ai , ri di biai 1 di biai 1 Step 3 : For the last equation rn bnrn 1 rn dn bnan 1 xn rn, Step 4 : by back substitution xi ri aixi 1, i n 1, n 2, n 3,....,2,1. 90 Solving a Tridiagonal Systems Using the Thomas Method 2 x1 x 2 III 1, x1 2 x 2 x 3 0, x 2 2 x3 x 4 0, x3 2 x 4 1. d (2,2,2,2); a (1,1,1,0); b (0,1,1,1); r (1,0,0,1). 91 Solving a Tridiagonal Systems Using the Thomas Method a1 IV a1 1 r1 1 , r1 . d1 2 d1 2 a2 a2 1 2 , d 2 b 2 a1 2 ( 1)( 1 / 2) 3 r2 r 2 b 2 r1 0 ( 1)(1 / 2) 1 . d 2 b 2 a1 2 ( 1)( 1 / 2) 3 x 4 r 4 1, x 3 r 3 a 3 x 4 1 / 4 ( 3 / 4) 1, x 2 r 2 a 2 x 3 1 / 3 ( 2 / 3)(1) 1, x1 r 1 a1 x 2 1 / 2 ( 1 / 2)(1) 1. 92 Discussion of Thomas method * 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. * The Thomas algorithm requires that d1 0 and di - biai - 1 0 for each i. 93 Using the Thomas Method for a System that Would Require Pivoting for Gaussian Elimination I 2 x1 x 2 1, x1 2 x 2 x3 0, 2 x 2 x3 x 4 3 x3 2 x 4 x5 4 , 3 0, x 4 2 x5 x 6 0, x 5 2 x 6 1. x1 1 x2 2 1 , 2 1 , 3 1, x2 2 x3 3 x4 x3 2 x 4 x5 0, x 4 2 x 5 x 6 0, x 5 2 x 6 1. 94 Using the Thomas Method for a System that Would Require Pivoting for Gaussian Elimination II x1 1 x2 2 1 , 2 1 , 3 1( solve), x2 2 x3 3 x4 x3 2 x 4 x5 2 x5 x 6 0( skip for now), 1(u sin gx 4 1), x 5 2 x 6 1. 95 Using the Thomas Method for a System that Would Require Pivoting for Gaussian Elimination III x 6 1, x1 1 x2 2 x2 1 , 2 1 , 3 1, 2 x3 3 x4 x3 2 x 4 x5 0, 1 1 x5 x6 , 2 2 x 6 1. 1 1 x 6 1, 2 2 x 4 1, x5 x 3 2 x 4 x 5 1, 1 2 x 3 1, 3 3 1 1 x1 x 2 1. 2 2 x2 96 6. Gauss – Jordan Elimination 97 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 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 Graphical depiction of Gauss-Jordan a11 a12 ' a a 22 21 a31 a32 a13 | c1 ' a23 | c2' '' '' a33 | c3 1 0 0 | c1n n 0 1 0 | c 2 0 0 1 | c3 n Graphical depiction of Gauss-Jordan a11 a12 ' a a 22 21 a31 a32 a13 | c1 ' a23 | c2' '' '' a33 | c3 1 0 0 | c1n n 0 1 0 | c 2 0 0 1 | c3 n 1 0 0 | c1n n 0 1 0 | c 2 0 0 1 | c3 n c1n x1 x2 c2 n x3 c3 n Gauss-Jordan Elimination Let us consider the set of linearly independent equations. 2 x 4 y 5z 36 3x 5 y 7 z 7 5 x 3 y 8 z 31 Augmented matrix for the set is: 36 2 4 5 3 5 7 7 3 8 31 5 102 Gauss-Jordan Elimination Step 1: Eliminate x from the 2nd and 3rd equation. 36 2 4 5 3 5 7 7 3 8 31 5 5 36 2 4 0 1 14.5 61 0 13 20.5 121 2 x 4 y 5z 36 3x 5 y 7 z 7 5x 3 y 8z 31 2 x 4 y 5z 36 y 14.5z 61 13 y 20.5z 121 103 Gauss-Jordan Elimination Step 2: Eliminate y from the 3rd equation. 13R’2+R’3 R’’3 36 2 4 5 0 1 14.5 61 0 0 168 672 2 x 4 y 5 z 36 y 14.5 z 61 168 z 672 Step 3: 1 2 2.5 18 0 1 14.5 61 -R’2 R’’2 1 4 (1/168)R’’3 R’’’3 0 0 0.5R’1 R’1 x 2 y 2.5z 18 y 14.5z 61 z4 104 Gauss-Jordan Elimination Step 4: Eliminate z from the 2nd equation 1 2 2.5 18 0 1 14.5 61 1 4 0 0 x 2 y 2.5z 18 y 14.5z 61 z4 ( Row 3) (14.5) ( Row 2) 1 2 2.5 18 0 1 0 3 0 0 1 4 x 2 y 2.5 z 18 y 3 z4 105 Gauss-Jordan Elimination Step 5-1: Eliminate y from the 1st equation 1 2 2.5 18 0 1 0 3 0 0 1 4 x 2 y 2.5 z 18 y 3 z4 ( Row 2) (2) ( Row 1) New Row 1 1 0 2.5 12 0 1 0 3 0 0 1 4 x 2.5 z 12 y 3 z4 106 Gauss-Jordan Elimination Step 5-2: Eliminate z from the 1st equation 1 0 2.5 12 0 1 0 3 0 0 1 4 x 2.5 z 12 y 3 z4 ( Row 3) (2.5) ( Row 1) New Row 1 1 0 0 2 0 1 0 3 0 0 1 4 x2 y 3 z4 107 Example Given the following, determine {x} for the two different loads {c} Ax c 2 1 1 1 A 2 6 3 3 1 4 cT 1 T c 4 2 3 7 1 108 Solution Ax c 2 1 1 1 A 2 6 3 3 1 4 c 1 T c 4 T 2 3 7 1 {c}T = {1 2 3} x1 = (2)(1) + (-1)(2) + (1)(3) = 3 x2 = (-2)(1) + (6)(2) + (3)(3) = 19 x3 = (-3)(1) + (1)(2) + (-4)(3) = -13 {c} T= {4 -7 1) x1 = (2)(4) + (-1)(-7) + (1)(1)=16 x2 = (-2)(4) + (6)(-7) + (3)(1) = -47 x3 = (-3)(4) + (1)(-7) + (-4)(1) = -23 109