Question

8x-3y+2z=20, 4x+11y-z=33, 6x+3y+12z=35. Solve the equations in 5 steps by Guass-Jacobi method.​

Answer 1

Answer:

X (0) = Y (0) = Z (0) = 0 and give your answer into 6dp?

8x - 3y + 2z = 20

4x - 11y -z = 33

6x + 3y + 12z = 35

Rewrite 1st equation as

8x = 20 + 3y - 2z

x = 1/8 ( 20 + 3y - 2z ) etc as an iterative process to find better and better values for x

Then we have:

x k+1 = 1/8 ( 20 + 3y k - 2z k )

y k+1 = 1/-11 ( 33 - 4x k+1 + z k )

z k+1 = 1/12 ( 35 - 6 x k+1 - 3 y k+1 )

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

1st Approximation

x₁ = 1/8 ( 20 + 3 (0) - 2 ( 0 ) )

= 1/8 ( 20 )

= 2.5

y₁ = 1/-11 ( 33 - 4 ( 2.5 ) + ( 0 ) )

= 1/-11 ( 23 )

= - 2.90909

z₁ = 1/12 ( 35 - 6 ( 2.5 ) - 3 ( - 2.90909 ) )

= 1/12 ( 35 - 15 + 6.72727

= 1/12 ( 26.272727 )

= 2.189394

2nd Approximation

x₂ = 1/8 ( 20 + 3 ( - 2.090909 ) - 2 ( 2.189394) )

= 1/8 ( 20 - 6.272727 - 4.378788 )

= 1/8 ( 9.348485 )

= 1.168561

y₂ = 1/-11 ( 33 - 4 ( 1.168561) + ( 2.189394 ) )

= 1/-11 ( 33 - 4.674244 + 2.189394 )

= 1/-11 ( 30.5155152 )

= - 2.774105

z₂ = 1/12 ( 35 - 6 ( 1.168561 ) - 3 ( - 2.774105 ) )

= 1/12 ( 35 - 7.011366 + 8.322315 )

= 1/12 ( 36.31095 )

= 3.025913

3rd Approximation

x₃ = 1/8 ( 20 + 3 ( -2.774105 ) - 2 ( 3.025913 ) )

= 1/8 ( 5.625861 )

= 0.703233

y₃ = 1/-11 ( 33 - 4 ( 0.703233 ) + ( 3.025913 ) )

= 1/-11 ( 33.212982 )

= - 3. 019362

z₃ = 1/12 ( 35 - 6 ( 0.703233 ) - 3 ( - 3. 019362 ) )

= 1/12 ( 39.83869 )

= 3.319891

I hope it's help you.

plz like and comment my answer.

Answer 2

The values after 5th iteration are x = 0.479066 , y = - 3.317098 and

z = 3.464408.

Given :

$8x-3y+2z=20\\4x+11y-z=33\\6x+3y+12z=35$

To find :

The equations in 5 steps by the Guass-Jacobi method.​

Solution:

Rewrite 1st equation as

$8x = 20 + 3y - 2z$

$x = 1/8 ( 20 + 3y - 2z )$  etc as an iterative process to find better and better values for x.

Then we have:

$x ^k^+^1 = 1/8 ( 20 + 3y ^k - 2z^ k )\\y ^k^+^1 = 1/-11 ( 33 - 4x ^k+1 + z^ k )\\z ^k^+^1 = 1/12 ( 35 - 6 x ^k^+^1 - 3 y^ k^+^1 )$

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

1st Approximation

$x_1 = 1/8 ( 20 + 3 (0) - 2 ( 0 ) )\\x_1 = 1/8 ( 20 )\\x_1 = 2.5$

$y_1 = 1/-11 ( 33 - 4 ( 2.5 ) + ( 0 ) )\\y_1 = 1/-11 ( 23 )\\y_1 = - 2.90909$

$z_1 = 1/12 ( 35 - 6 ( 2.5 ) - 3 ( - 2.90909 ) )\\z_1 = 1/12 ( 35 - 15 + 6.72727\\z_1 = 1/12 ( 26.272727 )\\z_1 = 2.189394$

2nd Approximation

$x_2 = 1/8 ( 20 + 3 ( - 2.090909 ) - 2 ( 2.189394) )\\x_2= 1/8 ( 20 - 6.272727 - 4.378788 )\\x_2= 1/8 ( 9.348485 )\\x_2= 1.168561$

$y_2 = 1/-11 ( 33 - 4 ( 1.168561) + ( 2.189394 ) )\\y_2= 1/-11 ( 33 - 4.674244 + 2.189394 )\\y_2= 1/-11 ( 30.5155152 )\\y_2= - 2.774105$

$z_2 = 1/12 ( 35 - 6 ( 1.168561 ) - 3 ( - 2.774105 ) )\\z_2= 1/12 ( 35 - 7.011366 + 8.322315 )\\z_2= 1/12 ( 36.31095 )\\z_2= 3.025913$

3rd Approximation

$x_3 = 1/8 ( 20 + 3 ( -2.774105 ) - 2 ( 3.025913 ) )\\\\x_3= 1/8 ( 5.625861 )\\\\x_3 = 0.703233$

$y_3 = 1/-11 ( 33 - 4 ( 0.703233 ) + ( 3.025913 ) )\\y_3= 1/-11 ( 33.212982 )\\y_3= - 3. 019362$

$z_3 = 1/12 ( 35 - 6 ( 0.703233 ) - 3 ( - 3. 019362 ) )\\z_3= 1/12 ( 39.83869 )\\z_3= 3.319891$

4th Approximation

$x_4 = 1/8 ( 20 + 3 ( -3.019362 ) - 2 ( 3.319891 ) )\\\\x_4 = 1/8 ( 4.302132)\\\\x_4 = 0.5377665$

$y_4 = 1/-11 ( 33 - 4 (0.5377665) + ( 3.319891) )\\\\y_4= 1/-11 ( 34.168825 )\\\\y_4= - 3.1062568$

$z_4= 1/12 ( 35 - 6 ( 0.5377665 ) - 3 ( - 3.1062568 ) )\\\\z_4= 1/12 ( 41.0921714 )\\\\z_4= 3.424347$

5th approximation.

$x_5 = 1/8 ( 20 + 3 ( -3.1062569 ) - 2 ( 3.424347) )\\\\x_5 = 1/8 ( 3.832535)\\\\x_5 = 0.479066$

$y_5 = 1/-11 ( 33 - 4 (0.479066) + ( 3.424347) )\\\\y_5= 1/-11 ( 34.5080 )\\\\y_5= - 3.137098$

$z_5= 1/12 ( 35 - 6 ( 0.473066 ) - 3 ( - 3.137098) )\\\\z_5= 1/12 ( 41.5728 )\\\\z_5= 3.464408$

Therefore, the values after the 5th iteration are x = 0.479066,

y = - 3.317098, and z = 3.464408.

#SPJ3