Question

10x-2y-z-u-3
-2x+10y-z-u=15
-x-y+10z-2u=27
-x-y-2z+10u=-9
using Gauss-seidel method
​

Answer 1

Given:

-u+10x-2y-z=3

-u-2x+10y-z=15

-2u-x-y+10z=27

10u-x-y-2z=-9

To find:

Solve Equations using Gauss Seidel method.

Solution:

Total Equations are 4

-u+10x-2y-z=3

-u-2x+10y-z=15

-2u-x-y+10z=27

10u-x-y-2z=-9

The coefficient matrix of the given system is not diagonally dominant.

Hence, we re-arrange the equations as follows, such that the elements in the coefficient matrix are diagonally dominant.

10u-x-y-2z=-9

-u+10x-2y-z=3

-u-2x+10y-z=15

-2u-x-y+10z=27

From the above equations

$u_k_+_1 = \frac{1}{10}(-9 + x_k + y_k + 2z_k) \\\\x_k_+_1 = \frac{1}{10} (3 + u_k_+_1 + 2y_k + z_k)\\\\y_k_+_1 = \frac{1}{10}(15 + u_k_+_1 + 2x_k_+_1 + z_k)\\\\z_k_+_1 = \frac{1}{10}(27 + 2u_k_+_1 + x_k_+_1 + y_k_+_1)$

Initial gauss (u,x,y,z)=(0,0,0,0)

Solution steps are

1st Approximation

$u_1 = \frac{1}{10}(-9 + (0) + (0) + 2(0)) = -0.9\\\\x_1 = \frac{1}{10} (3 + (-0.9) + 2(0) + (0)) = 0.21\\\\y_1 = \frac{1}{10}(15 + (-0.9) + 2(0.21) + (0)) = 1.452\\\\z_1 = \frac{1}{10}(27 + 2(-0.9) + (0.21) + (1.452)) = 2.6862$

2nd Approximation

$u_2 = \frac{1}{10}(-9 + (0.21) + (1.452) + 2(2.6862)) = -0.1966\\\\x_2 = \frac{1}{10} (3 + (-0.1966) + 2(1.452) + (2.6862)) = 0.8394\\\\y_2 = \frac{1}{10}(15 + (-0.1966) + 2(0.8394) + (2.6862)) = 1.9168\\\\z_2 = \frac{1}{10}(27 + 2(-0.1966) + (0.8394) + (1.9168)) = 2.9369$

3rd Approximation

Similarly after 3rd approximation:

u₃ = -0.0371

x₃ = 0.9733

y₃ = 1.9846

z₃ = 2.9884

4th Approximation

Similarly after 4th approximation:

u₄ = -0.0065

x₄ = 0.9951

y₄ = 1.9972

z₄ = 2.9979

5th Approximation

Similarly after 5th approximation:

u₅ = -0.0012

x₅ = 0.9991

y₅ = 1.9995

z₅ = 2.9996

6th Approximation

Similarly after 5th approximation:

u₆ = -0.0002

x₆ = 0.9998

y₆ = 1.9999

z₆ = 2.9999

Solution By Gauss Seidel Method.

u  = -0.0002 ≅ 0

x = 0.9998 ≅ 1

y = 1.9999 ≅ 2

z = 2.9999 ≅ 3