Question

it is not 2.77 it is 2.17 .

pls focus to number when you solving this problem

Solve the following equation with hand calculation using Power iteration scheme for outer iteration and Gauss-Seidel for matrix inversion. [3 -1 0 T&] [2.17 -1 4-1 $ 2 Lo -1 31%] koplo Assume ko and 7°as 1.000. 0 0 T61 3.36 002 O 1.951%]

Answer 1

Solve the following equation with hand calculation using Power iteration scheme for outer iteration and Gauss-Seidel for matrix inversion

Soluton,'. . feq.17 0 071 Given that To 3.36 0110, - den 4 - loi 3114) Keselo .950 Assume lot and to in 1000 consider the equation 30,-, = 4.17 -0, +4 0,5% = 3.36 -6, +30=1.95 the components of gauss-seidel integral equation dik+i)=113(2:14 +,CE)) 8,Ck+1) = 114 (3-36+,+ ) + 9CK)) Q, CE+ 1) = 11g (1.95 +0(k+) for initial iteration o(o)= [11] for k =1 0,61) = 13 (2.14 +0,00)) liz (2.17 + 1) = 1.0567 0,1); 14 (3.36 + 40,41)+ 0.CO)) = 114 (3.364 1.0564 + 1) 1.354-1 0,"") = 1/3 (1.95+ ) llz C1.95 +1.3541) = 1.1013 ni keptus ; [1.0567 1.3541 1.1013]

for k=1 0,60) = 13[4.144 ,5'3 Els [8.14 +1.3541) = 1.1747 (0) = !14 (3-36+43,4%) tape ."] = 114 (3-36+ 1.1847 +1.1013) -1.409 Azce) -413/195 +6,60)3 =l3 [1.95 +1.409] = 1.11967 kesple) = {1.1947 1.409 1.11967] for kia 0,(3) = "13 [2.14 +672)3 = !1? [14 +1409) = 1.193 0,(3) = 114 (3.36 +(3) +42¢2)] =l14 (3-36+1.193+ 1.11927) - 1.4181 $, 3) \12 (1195+ 0,633 = 1/2 (1.95 +1.4181] = 1.1827 Keres = [1.193.1.4181 1. 19243 0,64) = 1122.14+ $6313 Ell? [2.14 +1.4101] = 111960 22C4) 114 3.36+ (2,4) 42 (3)] - 114 (3.36 + 1.1960 +1.1227] =1.4196 dz(4) 113 [1.95+0,64)] llz [1.95+ 1.4196] =1,1282 Ke (A) = [1.1960 1.4196 1.1232] for k=3