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Question: 2] c) Solve the following equation by Gauss-Seidel method up to four iterations 4x-2y-z = 40, x-6y+2y = -28, x-2y+12z = -86.
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Subject:- Applied Mathematics

Marks:- 3

Mumbai Unversity>FE>Sem1>Applied Maths1

m1(81) • 9 views
 modified 4 days ago  • written 4 days ago by
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we first write the equation as

$x = ¼ [40+2y+z]$……………………….(1)

$y = 1/6 [28+x+2z]$ ……………………..(2)

$z = 1/12 [-86-x+2y]$ ………………………(3)

(i) FIRST ITERATION :-

we start with the approximation y=0 , z=0 and then we get from (1),

$∴x_1 = ¼ (40) = 10$

We use this approximation to find y i.e. put x=0, z=0 in (2)

$∴y_1= 1/6 [28+10+2(0) ]= 6.3333$

We use these values of $x_1$ and $y_1$ to find $z_1$ i.e. we put x =4 , y = 6.3333 in (3),

$∴z_1 = 1/12 [-86-10+2(6.3333) ] = -6.944$

(ii) SECOND ITERATION :-

We use latest values of y and z to find x i.e. we put y =6.3333 , z = -6.9444 in (1)

$∴x_2 = ¼ [40+2(6.3333)-6.9444] = 11.4306$

We use this approximation to find y i.e. put x=11.4306, z=-6.9444 in (2)

$∴y_2= 1/6 [28+11.4306+2(-6.9444) ]= 4.2569$

i.e. we put x =11.4306 , y = 4.2569 in (3),

$∴z_1 = 1/12 [-86-11.4306+2(4.2569) ] = -7.4097$

(iii)THIRD ITERATION :-

We use latest values of y and z to find x i.e. we put y =4.2569 , z = -7.40974 in (1)

$∴x_2 = ¼ [40+2(4.2569)-7.4097] = 10.2760$

We use this approximation to find y i.e. put x=10.2760 , z = -7.4097 in (2)

$∴y_2= 1/6 [28+10.2760+2(-7.4097) ]= 3.9094$

i.e. we put x =10.2760 , y = 3.9094 in (3),

$∴z_1 = 1/12 [-86-10.2760+2(3.9094) ] = -7.3714.$

(iv) FOURTH ITERATION:-

We use latest values of y and z to find x i.e. we put y =3.9094 , z = -7.3714 in (1)

$∴x_2 = ¼ [40+2(3.9094)-7.3714] = 10.1118$

We use this approximation to find y i.e. put x=10.1118 , z = -7.3714 in (2)

$∴y_2= 1/6 [28+10.1118+2(-7.3714) ]= 3.8948$

i.e. we put x =10.1118 , y = 3.8448 in (3),

$∴z_1 = 1/12 [-86-10.1118+2(3.8948) ] = -7.3602.$

Hence ,upto two places of decimals

x = 10.11 , y = 3.89, z = -7.36.