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Department of Mathematics MTL107: Numerical Methods And Department of Mathematics MTL107: Numerical methods and Computations MATLAB PRACTICE SHEET-3: Iterative Techniques in Matrix Algebra-Jacobi, Gauss-Siedel Iterative Techniques, Successive Over-Relaxation(SOR) Techniquesfor Solving Linear Systems, Condition Number, Well-conditioning, Ill-conditioning, The Conjugate Gradient Method. −3 1. Use the Jacobi method to solve the linear systems , with T OL = 10 in the l1 norm. using x(0) = 0: a. 3x1 − x2 + x3 = 1; b. 10x1 − x2 = 9; 3x1 + 6x2 + 2x3 = 0; −x1 + 10x2 − 2x3 = 7; 3x1 + 3x2 + 7x3 = 4: −2x2 + 10x3 = 6: c. 10x1 + 5x2 = 6; d. 4x1 + x2 + x3 + x5 = 6; 5x1 + 10x2 − 4x3 = 25; −x1 − 3x2 + x3 + x4 = 6; −4x2 + 8x3 − x4 = −11:6; 2x1 + x2 + 5x3 − x4 − x5 = 6: −x3 + 5x4 = −11; −x1 − x2 − x3 + 4x4 = 6; 2x2 − x3 + x4 + 4x5 = 6: 2. Use the Gauss-Seidel method to solve the linear systems in Exercise 1, with T OL = 10−3 in the l1 norm. 3. The linear system 2x1 − x2 + x3 = −1; 2x1 + 2x2 + 2x3 = 4; −x1 − x2 + 2x3 = −5: t has the solutin (1; 2; −p2) : 5 a. Show that ρ(Tj) = 2 > 1: b. Show that the Jacobi method with x(0) = 0 fails to give a good approximation after 25 iterations. 1 c. Show that ρ(Tg) = 2 : d. Use the Gauss-Seidel method with x(0) = 0 to approximate the solution to the linear −5 system to within 10 in the l1 norm. 4. The liear system x1 − x3 = 0:2; 1 1 − 2 x1 + x2 + 4 x3 = −1:425; 1 x1 − 2 x2 + x3 = 2: has the solutin (0:9; −0:8; 0:7)t: a. Is the coefficient matrix 21 0 −13 1 1 A = 4 2 1 − 4 5 1 1 − 2 1 strictly diagonally dominant ? b. Compute the spectral radius of the Gauss-Seidel matrix Tg. c. Use the Gauss-Seidel iterative method x(0) = 0 to approximate the solution to the linear system with a tolerance of 10−2 and a maximum of 300 iterations. d. What happens in part(c) when the system is changed to x1 − 2x3 = 0:2; 1 1 − 2 x1 + x2 − 4 x3 = −1:425; 1 x1 − 2 x2 + x3 = 2: 5. Use the SOR method with ! = 1:2 to solve the linear systems , with a tolerance −3 (0) T OL = 10 in the l1 norm. using x = 0: a. 3x1 − x2 + x3 = 1; b. 10x1 − x2 = 9; 3x1 + 6x2 + 2x3 = 0; −x1 + 10x2 − 2x3 = 7; 3x1 + 3x2 + 7x3 = 4: −2x2 + 10x3 = 6: c. 10x1 + 5x2 = 6; d. 4x1 + x2 + x3 + x5 = 6; 5x1 + 10x2 − 4x3 = 25; −x1 − 3x2 + x3 + x4 = 6; −4x2 + 8x3 − x4 = −11:6; 2x1 + x2 + 5x3 − x4 − x5 = 6: −x3 + 5x4 = −11; −x1 − x2 − x3 + 4x4 = 6; 2x2 − x3 + x4 + 4x5 = 6: 6. (i) Use Gaussian elimination and 3-digit rounding arithmetic to approximate the solution to the following linear systems. (ii) Then use one iteration of iterative refinement to im- prove the approximation, and compare the approximations to the actual solutions. a. 0:03x1 + 58:9x2 = 59:2; b. 3:03x1 − 12:1x2 + 14x3 = −119; 5:31x1 − 6:10x2 = 47:0; −3:03x1 + 12:1x2 − 7x3 = 120; t Actual solution [10; 1] 6:11x1 − 14:2x2 + 21x3 = −139: 1 t Actual solution [0; 10; 7 ] p p p c. 1:19x1 + 2:11x2 − 100x3 + x4 = 1:12; d. πx1 − ex2 + 2x3 − 3x4 = 11; 14:2x − 0:122x + 12:2x − x = 3:44; π2x + ex − e2x + 3 x = 0; 1 2 3 4 p1 2 p 3 7 4 p 100x − 99:9x + x = 2:15; 5x − 6x + x − 2x = π; 2 3 4 1 p2 3 4 p 3 2 1 15:3x1 + 0:110x2 − 13:1x3 − x4 = 4:16; π x1 − e x2 − 7x3 + 9 x4 = 2; Actual solution [0:176; 0:0126; −0:0206; −1:18]t: Actual solution [0:788; −3:12; 0:167; 4:55]t: −3 −1 7. Perform the conjugate gradient method with T OL = 10 in the l1 norm, (C = C = I ) on each of the following linear systems. a. 3x1 − x2 + x3 = 1; b. 10x1 − x2 = 9; −x1 + 6x2 + 2x3 = 0; −x1 + 10x2 − 2x3 = 7; x1 + 2x2 + 7x3 = 4: −2x2 + 10x3 = 6: c. 10x1 + 5x2 = 6; d. 4x1 + x2 − x3 + x4 = −2; 5x1 + 10x2 − 4x3 = 25; x1 + 4x2 − x3 − x4 = −1; −4x2 + 8x3 − x4 = −11; −x1 − x2 + 5x3 + x4 = 0: −x3 + 5x4 = −11; x1 − x2 + x3 + 3x4 = 1; e. 4x1 + x2 + x3 + x5 = 6; f. 4x1 − x2 − x4 = 0; x1 + 3x2 + x3 + x4 = 6; −x1 + 4x2 − x3 − x5 = 5; x1 + x2 + 5x3 − x4 − x5 = 6; −x2 + 4x3 − x6 = 0: x2 − x3 + 4x4 = 6; −x1 + 4x4 − x5 = 6: x1 − x3 + 4x5 = 6: −x2 − x4 + 4x5 − x6 = −2: −x3 − x5 + 4x6 = 6: 8. Given 2 3 2 3 2 3 4 −1 0 0 −1 0 0 0 0 0 0 0 6−1 4 −1 0 7 6 0 −1 0 0 7 60 0 0 07 6 7 6 7 6 7 A1 = 6 7 ; −I 6 7 ; and O = 6 7 : 4 0 −1 4 −15 4 0 0 −1 0 5 40 0 0 05 0 0 −1 4 0 0 0 −1 0 0 0 0 Form the 16 × 16 matrix A in partitioned form, 2 3 A1 −I 0 0 6−I A −I 0 7 6 1 7 A=6 7. 4 0 −I A1 −I 5 0 0 −I A1 Let b = (1; 2; 3; 4; 5; 6; 7; 8; 9; 0; 1; 2; 3; 4; 5; 6)t: a. Solve Ax = b using the conjugate gradient method with tolerance 0.05. b. Solve Ax = b using the preconditioned conjugate gradient method with C −1 = D−1=2 and tolerance 0.05. c. Is there any tolerance for which the methods of part(a) and part(b) require a different number of iterations ? ANSWERS 1. Jacobi's Algorithm gives the following results: a. x(10) = (0:03507839; −0:2369262; 0:6578015)t b. x(6) = (0:9957250; 0:9577750; 0:7914500)t c. x(22) = (−0:7975853; 2:794795; −0:2588888; −2:251879)t d. x(14) = (−0:7529267; 0:04078538; −0:2806091; 0:6911662)t: 2. The Gauss-Seidel Algorithm gives the following results: a. x(6) = (0:03535107; −0:2367886; 0:6577590)t b. x(4) = (0:9957475; 0:9578738; 0:7915748)t c. x(10) = (−0:7973091; 2:794982; −0:2589884; −2:251798)t d. x(7) = (0:7866825; −1:002719; 1:866283; 1:912562; 1:989790)t: 2 1 1 3 0 2 − 2 3 5 3. a. Tj = 4−1 0 −15 and det(λI − Tj) = λ + 4 λ. 1 1 2 2 0 p p 5 5 Thus, the eigenvalues of Tj are 0 and ± 2 i; so ρ(Tj) = 2 > 1: b. x(25) = (−20:827873; 2:0000000; −22:827873)t: 20 1 − 1 3 2 2 2 1 1 1 c. Tg = 40 − 2 − 2 5 and det(λI − Tg) = λ λ + 2 1 0 0 − 2 1 1 Thus, the eigenvalues of Tg are 0, − 2 ; and − 2 ; and ρ(Tj) = 1=2: (23) t −5 d. x = (1:0000023; 1:9999975; −1:0000001) is within 10 in the l1 norm. 4. a. A is not strictly diagonally dominant. 2 0 0 1 3 b. a. Tj = 40:5 0 0:255 and ρ(Tj) = 0:97210521: −1 0:5 0 Since Tj is convergent, the Jacobi method will converge. c. With x(0) = (0; 0; 0)t; x(187) = (0:90222655; −0:79595242; 0:69281316)t: d. ρ(Tj) = 1:39331779371: Since Tj is not convergent, the Jacobi method will not converge. 5. The SOR Algorithm gives the following results. a. x(12) = (0:03488469; −0:2366474; 0:6579013)t b. x(7) = (0:9958341; 0:9579041; 0:7915756)t c. x(8) = (−0:7976009; 2:795288; −0:2588293; −2:251768)t d. x(7) = (−0:7534489; 0:04106617; −0:2808146; 0:6918049)t: e. x(10) = (0:7866310; −1:002807; 1:866530:1:912645; 1:989792)t f. x(7) = (0:9999442; 1:999934; 1:000033; 1:999958; 0:9999815; 2:000007)t: 6. Gaussian elimination and iterative refinement give the following results. a. (i) (10:0; 1:01)t; (ii) (10:0; 1:00)t b. (i) (12:0; 0:499; −1:98)t; (ii) (1:00; 0:500; −1:00)t c. (i) (0:185; 0:0103; −0:0200; −1:12)t; (ii) (0:177; 0:0127; −0:0207; −1:18)t d. (i) (0:799; −3:12; 0:151; 4:56)t; (ii) (0:758; −3:00; 0:159; 4:30)t (3) t (3) −9 7. a. x = (0:06185567013; −0:1958762887; 0:6185567010) ; jjr jj1 = 0:4 × 10 : (3) t (3) −9 b. x = (0:9957894738; 0:9578947369; 0:7915789474) ; jjr jj1 = 0:1 × 10 : c. x(4) = (−0:7976470579; 2:795294120; −0:2588235305; −2:251764706)t; (4) −7 jjr jj1 = 0:39 × 10 : d. x(4) = (−0:7534246575; 0:04109589039; −0:2808219179; 0:6917808219)t; (4) −9 jjr jj1 = 0:11 × 10 : e. x(5) = (0:4516129032; 0:7096774197; 1:677419355; 1:741935483; 1:806451613)t; (5) −9 jjr jj1 = 0:2 × 10 : f. x(4) = (1:000000000; 2:000000000; 1:000000000; 2:000000000; 0:9999999997; 2:0000000000)t; (4) −9 jjr jj1 = 0:44 × 10 : −2 8. This converges in 6 iterations with tolerance 5:00 × 10 in the l1norm and a. Solution Residual 2.55613420 0.00668246 4.09171393 -0.00533953 4.60840390 -0.01739814 3.64309950 -0.03171624 5.13950533 0.01308093 7.19697808 -0.02081095 7.68140405 -0.04593118 5.93227784 0.01692180 5.81798997 0.04414047 5.85447806 0.03319707 5.94202521 -0.00099947 4.42152959 -0.00072826 3.32211695 0.02363822 4.49411604 0.00982052 4.80968966 0.00846967 3.81108707 -0.01312902 (6) jjr jj1 = 0:046: −2 This converges in 6 iterations with tolerance 5:00 × 10 in the l1norm and (6) jjr jj1 = 0:046: b.
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