Question

Consider the linear system 11 0.5.01 21 + 12 0.5.22 + 13 0.25.13 13 0.2 -1.425 2 whose solution is (0.9,-0.8, 0.7). (b) Approximate the solution of the system by performing two iterations of the Gauss-Seidel algorithm, using (10) (0) = (0,0,0)' as the initial guess. (c) Approximate the solution of the system using one iteration of the SOR scheme, with w = 0.7 and (5) x() = (0,0,0)

Answer 1

Consider the linear system -X₂=0.2 ne 0.5X1 + Xa-0.25 X3 - X1 + 0 5X2+Xz -7.495 a ra from & we get efficient matrix AX=B 1 A = 0 D 0,5 l 1 -0.5 - -0.25 1 Here a = 1; 0,2 = 0 ; 0,3 = -1; C = 0.5; Asal Agg = -0.25 ; AB = 1 g Az a = -05 Cz 1 11 la

b) Plere lail= laialt laial ; laga 1 > 19,4 19:31 and 10331 < 19311+ 1933 80 A is not strictly diagenel dominant matrie from @ X, = [(0.2) - (0)*+x3] Xa = [-|-|us -x + 6 - 35x50 Xz = [2-X, +0.5 xa By grauss piedel Method of [0.2-X) + X ting tekst) = (-1142-XCM41) + (0.3s each x,CkH). + 0.5 x C+1) ,CK+1) ,00 Xy (left) 2- Xe

.. first X()_ 0.2 ; Xa ) -1.625 1 second iteration + (a) - (6) . ; x = 0 ; x = 0 literation [o.9-0+0] = 0.2 [-1425-0.4+0-2560) - -7.625 - (2-(0-9)+(0-5) 61.685)]- 0-9875 ġ to 0.9875 PO. 2-0 6-41.625)+ (0.9875)]- 11875 x ) = (-1425-01-1875)+(6:35) (6 5625) -3-3452 (- (1.1875) +(0.5) (-2,36563) -0.3703) 1-1875 3 x ) = -2,3656; Xce) - = -0.37031 for SoR method X{kt) = (10) x(k) + (W) [0.2 - xylk + x3 (8] (1-W) X (K) +(w) [-1.425 – x(+12460.25) xkey to Pear) - (-2) ik + (as) [ 2 - X CHAD + 0.5 x H)] YpsoX,9.05 kg ? first iteration : (1) Xi (-0.70) +(0.7)(0.2-0+0]= 0.14 (1-0.7) (0)+(0.7) 6.485-0-14+0) = -0.9975 (1-0-7)(0) + (0.7) (2-0.94 +(0.5)(-0-9975) 0.952875 = 0.14 tal- -0.9995 ; talent P: 0.952875 Answe ģ W=0.7 x, C+1) X.CO 9