Question

= = 6.6 (3). Consider the linear system X1 +4x2 = 0 4X1 +X2 - 0 The true solution is X1 = -1/15, X2 = -1/15, X2 = 4/15. Apply the Jacobi and Gauss-Seidel methods with c0 [0,0]T to the system and find out which methods diverge more rapidly. Next, interchange the two equations to write the system as = = 4.3 +22 = 0 X1 +4.x2 = 0 [0,0]?. Iterate until ||x – 24(k)|| < 10-5. Which method = and apply both methods with x(0) converge faster?

Consider the linear system x1 +4x2 = 0 4x1 +x2 = 0 The true solution is x1 = ?1=15, x2 = 4=15. Apply the Jacobi and Gauss-Seidel methods with x(0) = [0; 0]T to the system and nd out which methods diverge more rapidly. Next, interchange the two equations to write the system as 8< : 4x1 +x2 = 0 x1 +4x2 = 0 and apply both methods with x(0) = [0; 0]T . Iterate until jjx?x(k)jj 10?5. Which method converge faster?

Answer 1

then 2 of the given solution is true a = -1/3 3/2 = 4/5 satisfy the system it should x1 +420 40, +22=0 But - 1 + =170 +6+4(stelle 467) -h 15 4 =0 15 the 15 15? x = -114 d = 4 do not satis sahishy equation 21+ 40 =0 for a = - 1 t = 4 to be the solution the 12 = 4 linear system must be 11+412 =1 421 +22=0. CamScanne CS

Interchanging the two equations we get 421 422=0 x1 +402 = 1 We first Jacobi Method. apply :21= -d2 4 422=1-21 A2 = 2 3 4 ܝܐܨ 4 Initial point is (ty, 12) =(0,0) cs Scanned with CamScanner

x () 11. 2 (0) (1) . ) 호 ㅗ 4 G 기 (2) 오. 2 0L (1) 2 an 휀 - 91 Cast 1 지 9 ㅗ 와 14. 기 he h (3) 7,42 = ) -19 Ty ti b 59 4. 261 CO 1 - ( ) b 국, 11. h hg (opt of (4) 크 아 애 9 ti 여 - -- () 2SC. ( TS EL 1 1.14 12 = 0.1835 2G G 5 = -0.06. 일송 일 00 66Y (5) -0666 = 0)834 (5) = ta 6. 006C (F) - 1커 IS6 1024. 4 6 JD - 오 4 5 = 1 - (-0.04) = 0.26 61 서 (3) 0266 S 다. = -066625 서 (3) e 16, 024584 1 G 1 00166 4. 4 f0.066252 = 0.05 4 ㅗ 02 66 25 러 (t 8.26656 { (0) 지 (③) - 02665 6 = 0.06 64 (9) 거요 T - 0.24S 6 00이654 - 형 CS Scanned with CamScanner 다 5. 1