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 TOL = 10 in the l∞ 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 TOL = 10−3
in the l∞ norm. 3. The linear system
2x1 − x2 + x3 = −1, 2x1 + 2x2 + 2x3 = 4, −x1 − x2 + 2x3 = −5. t has the solutin (1, 2, −√2) . 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 l∞ 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 1 0 −1 1 1 A = 2 1 − 4 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 ω = 1.2 to solve the linear systems , with a tolerance −3 (0) TOL = 10 in the l∞ 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 ] √ √ √ 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 √1 2 √ 3 7 4 √ 100x − 99.9x + x = 2.15, 5x − 6x + x − 2x = π, 2 3 4 1 √2 3 4 √ 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 TOL = 10 in the l∞ 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 4 −1 0 0 −1 0 0 0 0 0 0 0 −1 4 −1 0 0 −1 0 0 0 0 0 0 A1 = , −I , and O = . 0 −1 4 −1 0 0 −1 0 0 0 0 0 0 0 −1 4 0 0 0 −1 0 0 0 0
Form the 16 × 16 matrix A in partitioned form, A1 −I 0 0 −I A −I 0 1 A= . 0 −I A1 −I 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.
1 1 0 2 − 2 3 5 3. a. Tj = −1 0 −1 and det(λI − Tj) = λ + 4 λ. 1 1 2 2 0 √ √ 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.
0 1 − 1 2 2 2 1 1 1 c. Tg = 0 − 2 − 2 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 l∞ norm.
4. a. A is not strictly diagonally dominant.
0 0 1 b. a. Tj = 0.5 0 0.25 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) , ||r ||∞ = 0.4 × 10 . (3) t (3) −9 b. x = (0.9957894738, 0.9578947369, 0.7915789474) , ||r ||∞ = 0.1 × 10 . c. x(4) = (−0.7976470579, 2.795294120, −0.2588235305, −2.251764706)t, (4) −7 ||r ||∞ = 0.39 × 10 . d. x(4) = (−0.7534246575, 0.04109589039, −0.2808219179, 0.6917808219)t, (4) −9 ||r ||∞ = 0.11 × 10 . e. x(5) = (0.4516129032, 0.7096774197, 1.677419355, 1.741935483, 1.806451613)t, (5) −9 ||r ||∞ = 0.2 × 10 . f. x(4) = (1.000000000, 2.000000000, 1.000000000, 2.000000000, 0.9999999997, 2.0000000000)t, (4) −9 ||r ||∞ = 0.44 × 10 .
−2 8. This converges in 6 iterations with tolerance 5.00 × 10 in the l∞norm 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) ||r ||∞ = 0.046.
−2 This converges in 6 iterations with tolerance 5.00 × 10 in the l∞norm and (6) ||r ||∞ = 0.046. b. 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 0.03319706 5.94202521 -0.00099947 4.42152959 -0.00072826 3.32211694 0.02363822 4.49411603 0.00982052 4.80968966 0.00846967 3.81108707 -0.01312902
c. All tolerance lead to the same convergence specifications.