Question

Question 11

In Exercises 9-12, show that the Gauss-Seidel method diverges for the given system using the initial approximation (x1, x2,...,x) = (0,0,...,0). 9. x– 2x2 = -1 2xy + x2 = 3 11. 2x, – 3x2 = -7 x1 + 3x2 – 10x3 = 9 3x + x3 = 13 10. - x + 4x, = 1 3xı – 2x2 = 2 12. x, + 3x, – x3 = 5 3x1 - x2 = 5 x2 + 2x3 = 1

Answer 1

.No.1 2x - 3x2 - 7 x + 3x2 - 102 3 = 9 3x + x3 = 13 & (21,22, 12n) = (0, 0 04 matrix System is The coefficient mahix of given not diagonally dominant : ke arrange the egn in in the coefficient mabix are such that diagonally the elements demnant: So, we 30 tox2 + Hz - 13 2x - 302 - 0x3 = -7 x +32 - 10x3 -9 get x = ½ (13 - 0x2 - x3) X, -1/3 (-7-22,-03) = -1/10 (9-x - 3x21 then 1st approximahen : 13 113-0-01 - 7313] = 4.3333 x = -1/3 (-1-2 (4-3333) -01 : 1/3 (-15-66677 = 5.2222 x . -4/10 (9 - (4 : 3333) – 3 [5-2222]] = -410 [11] = 11 X2 2nd applex mahon 13.113 - 0 -1.1: V3 11.91 = 3.9661 12) - 13.-7-2(3-96666)-0] =-13-14-9333-4.9779 (*) . - [ 3 - (3-9 6 6 6 ) - 3 ( 4-917 3) = - to f-9-91 - 0 - 94

327 Approximation : 4th X = 1/3 (13 - 0 - (0.69)] = 73 [12.01] = 4.0033 2,5) = -4/3 (-7-214-0033) - 0 1 - - 13 [-15.0061] - 5-0022 X3% -10 (9- (4.0033) -3(5.02 22)] = -80(-10-01): 1.00 approximation: X, 4) = 1/3 (13 - 0 - (1-001) 43 (1.999] 3-9996 X24) = -4/3(-7-2(3-9996) - 01--1/3[-14.99931 = 4.9997 X34) = -1/0 (9-( 3.9996) - 3(4.9997)) = -4/10 (-9.999)-0.9999 5th approximation CST 13 (13.-0 - (0.9999) 7= 13 [12:0001 - 4000003 2,5) = -13 (-7 - 2(4.00003)-01 : -13(-15.00006)-5-00002 x3 = -1/10 (9- (4.00003) -3(5.0002)]-410(-10.0001] =-0000! de table given in such : 262 26, Ileration 203 4.3333 3.9666 4·0033 3.9996 4.00003 5.2222 4.9778 5.0022 4-9997 5.00002 1 0.99 1.001 0.9999 1.00001 As we the & give can value same see of for x in 2 3rd 4th f 5th iteration 3.966641.00333.9996 < 4.00003 x2 and az. So next Increassing iterations then show decreassing that given and so on for system is diverge.
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