Question

Consider the linear system 5x1 - 21 + X1 - 22 + x3 = 1 5.22 - 23 = 2 22 5 5x3 = 3 (a) Discuss the convergence of the iterative solutions of this system generated by the Jacobi and Gauss-Seidel methods, by considering their iterative matrices. (b) If both methods converge, which one of them converges faster to the exact solution of this system? (c) Starting with the initial approximation x(0) = [0,0,0], find the number of iterations needed to achieve an approximation accurate to within 10% using the Jacobi method. (d) If the approximate solution x(10) is computed by the Gauss-Seidel method, estimate the error ||x - x(10) ||0o, that is find an upper bound for this error.

Answer 1

Gauss Jacobi and gauss Seidel methods are used to solve these equations.

Gauss Seidel method is an advancement of gauss Jacobi method.

In Jacobi method we find next iteration using the values from previous iteration,

while in Seidel method we put the values that have been calculated from the same iteration and the previous iteration for faster convergence to solution.

The system should be diagonally dominant for the convergence i.e. absolute value of diagonal element must be greater than absolute values sum of other entries.

If not so, we adjust the equations by placing them above or below.

Given 5x, - xy + xy = 1 x + 5% - x₂ = 2 x - x₂ +503 = 3 GAUSS- JACOBI METHOD: AN ооо x + ) - 2 - x + x + 5 x enti) = [ 3-xcm) + (m)] 15 Taking x lol = 0 mo M₂ (0) = 0 X3 (0) = 0 => x = (2+0 -0)/5 = (-2) my (0) = (2-0+0)/5 = (04) xx (0) = (3-0 +0)/5 = (6) >(3) = (1 + (ou) – (06)]/5 = 63)/5 = '016 12 (2) = (2 - (-2) + (-6))/5 (2.4)/5 = 48 M3 ) = (3 + (--2) + (04)= (3:2)/5 = .64 5 .168. , (3) = (1 + (.48) - (64)/5 = 423) = (2-(16) + (64))/s = x. (3) (3 - (16) + (045)/5 = .496 664 Scanned with Cam Scanner

=> 1 = (1 + (496) - (6664)/5 = /AIGG4 Y ME (2 - (168) + (604)) = -4992 rz 2011 (3. - (168) + (496)) = .6656 (s) = (1 + (4992) - (6656] [5 = 1.1668 > (S) = [2 -(ccm) + folia lls = .49984 Y3 (5) = (3-(1904) + (4942) ] ] = "GO656 =) X, 6) = (1 + (49984] - (GGGSC)/5 = %.166656 Y C6) = (2- (1668) + (-66656))/= .499952 D3 (6) = 6 -(1968) + (049984))/5 = 1,666608) = 7) = [2 + (49952.) - (.666608)] Is = /1666688 *21*1= € - (166656) + (.coccos) ]/s = y qaqqou m2 (+) 13 - (166656) + (499952)/5 = 1.6666592) >y (8) = (1 + (uaggaou) - (.6666 542)/5 = 7:16666624 (8) = 12 -6.1666688) + (.6666 592))/5 = .49999808 ry (15) = (3-(1666688) + (4999904))/5 = 1.66666432 Hence from 7th and 8th iteration we can approximate ( 1 ) = 1.16666 | correct upto uggag s-decimal 1.666667 places: CS Scanne dati took 8 - stepdans (c.) CamScanner

Gauss seidel Method: sc (M+1) = (2 + schms - x M) /s v, (n+1) = (2 - x +1) + x (%) /s x (M+1) = (3 *- anti) + H (Mt) ) 15 Jaking (1 M Xg] = [0,0,0] = 4,() = (1 +0 -0 S5 = 1.2 | x2 (1) = (2 - (62) + (0) 15 = 1.36 ME (I) = (3-(2) + (-36) )/5 = 1.632) = x (2) = (1 + (036) - (+632)] /5 = 1.1456 x2 (2) = ( 2 - (1456) +(+632)]ls = .49728 23 (2) = ( 3 -(1456) + (.49+28)] /s = 1.670336 M = , 3) = (2 + (49728) -(670336)] /s = /1653888| x2 (3) = (2 - (1653888) + (5670336)]/s = .500989 x3 (3) = ( 3 - (11653888) + (500989)]/s = .66712004) (a) = (1 + (1500989) - (.66712)] Is = 1.166773 (4) = (2- (+166773) + (06642)] 15 = 1.5000694 (4) = (3 - ( - 6 6 1+3) + ( Soo06 4) 3/s = .66 G6 S4 2 2,65) = ( 4 +0.50006а) - (.666659)]/s = 7.166682 x2 (s) = (2 - (.166682) + (.666659)] /s = .49999 54 12 (5) - [3 - ( 166682) + (1 4999954)/5 = 1.666662) =

(6) = (1 + (-4aa995) – 6.666662)] /s = 4.166666 | (6) = (2-6.166666} +6.666662)] /s = . yaa 9992 (6) = ( 3 – (166666) + (yaaaaa) ]/s =1.6666666/ and o we can conclude: - (16 Hz Hz) Z ( 6 1 2 1 2 1 2 1 2 (exact solution) ( (exact (6) Hence we can guess the exact answer by both methods but rate of convergence of Seidel method is faster than Jacobi method Seidel method converges faster to the exact solution. (do we notice that at the 6th step only Seidel method is very close to the exact solution (error <106) i ll x - x 10) los ie error between exact solution and approximate scio) almost reaches "". (correct upto about 8 decimal places). so llx-se la llao - 0 ans (about 10 20) Cs