Question

Apply Jacobi Iteration Method to solve the equation
20x +y- 2z =17
3x+20y-Z = -18
2 x- 3 y+20z=25​

Answer 1

Given:

20x+y-2z=17  

3x+20y-z=-18

2x-3y+20z=25

To find:

Apply Jacobi Iteration Method to solve the equations.

Solution:

Total Equations are 3

20x+y-2z=17

3x+20y-z=-18

2x-3y+20z=25  

From the above equations

$x_k_+_1 = \frac{1}{20} (17 - y_k + 2z_k)\\\\y_k_+_1 = \frac{1}{20} (-18 - 3x_k + z_k)\\\\z_k_+_1 = \frac{1}{20} (25 - 2x_k + 3y_k)$

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

Solution steps are

1st Approximation

$x_1 = \frac{1}{20} (17 - (0) + 2(0)) = 0.85\\\\y_1 = \frac{1}{20} (-18 - 3(0) + (0)) = -0.9\\\\z_1 = \frac{1}{20} (25 - 2(0) + 3(0)) = 1.25$

2nd Approximation

$x_2 = \frac{1}{20} (17 - (-0.9) + 2(1.25)) = 1.02\\\\y_2 = \frac{1}{20} (-18 - 3(0.85) + (1.25)) = -0.965\\\\z_2 = \frac{1}{20} (25 - 2(0.85) + 3(-0.9)) = 1.03$

3rd Approximation

$x_3 = \frac{1}{20} (17 - (-0.965) + 2(1.03)) = 1.0012\\\\y_3 = \frac{1}{20} (-18 - 3(1.02) + (1.03)) = -1.0015\\\\z_3= \frac{1}{20} (25 - 2(1.02) + 3(-0.965)) = 1.0032$

4th Approximation

$x_4 = \frac{1}{20} (17 - (-1.0015) + 2(1.0032)) = 1.0004\\\\y_4 = \frac{1}{20} (-18 - 3(1.0012) + (1.0032)) = -1\\\\z_4= \frac{1}{20} (25 - 2(1.0012) + 3(-1.0015)) = 0.9997$

5th Approximation

$x_5 = \frac{1}{20} (17 - (-1) + 2(0.9997)) = 1\\\\y_5 = \frac{1}{20} (-18 - 3(1.0004) + (0.9997)) = -1.0001\\\\z_5= \frac{1}{20} (25 - 2(1.0004) + 3(-1)) = 1$

Solution By Gauss Jacobi Method

x = 1 ≅ 1

y = -1.0001 ≅ -1

z = 1 ≅ 1

Answer 2

Answer:

Step-by-step explanation: