Question

Solve 10x+y+z=12, 2x+10y+z=13, 2x+2y+10z=14 by Gauss-Seidel iteration method

Answer 1

Step-by-step explanation:

I hope it will be help full for you

Answer 2

Answer:

The values are $x=1,y=1,z=1.$

Step-by-step explanation:

Total equations are $3.$

$10x+y+z=12$

$2x+10y+z=13,$

$2x+2y+10z=14.$

From the above equations

$x_k_+_1=\frac{1}{10}(12-y_k-z_k)$

$y_k_+_1=\frac{1}{10}(13-2x_k_+_1-z_k)$

$z_k_+_1=\frac{1}{10}(14-2x_k_+_1-2y_k_+_1)$

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

Solution steps are

$1^s^t$ Approximation

$x_1=\frac{1}{10}[12-0-0]=\frac{1}{10}[12]=1.2$

$y_1=\frac{1}{10}[13-2(1.2)-0]=\frac{1}{10}[10.6]=1.06$

$z_1=\frac{1}{10}[14-2(1.2)-2(1.06)]=\frac{1}{10}[9.48]=0.948$

$2^n^d$ Approximation

$x_2=\frac{1}{10}[12-1.06-0.948]=\frac{1}{10}[9.992]=0.9992$

$y_2=\frac{1}{10}[13-2(0.9992)-0.948]=\frac{1}{10}[10.0536]=1.0054$

$z_2=\frac{1}{10}[14-2(0.9992)-2(1.0054)]=\frac{1}{10}[9.9909]=0.9991$

$3^r^d$ Approximation

$x_3=\frac{1}{10}[12-1.0054-0.9991]=\frac{1}{10}[9.9956]=0.9996$

$y_3=\frac{1}{10}[13-2(0.9996)-0.9991]=\frac{1}{10}[10.0018]=1.0002$

$z_3=\frac{1}{10}[14-2(0.9996)-2(1.0002)]=\frac{1}{10}[10.0005]=1.0001$

$4^t^h$ Approximation

$x_4=\frac{1}{10}[12-(1.0002)-1.0001]=\frac{1}{10}[9.9998]=1$

$y_4=\frac{1}{10}[13-2(1)-1.0001]=\frac{1}{10}[10]=1$

$z_4=\frac{1}{10}[14-2(1)-2(1)]=\frac{1}{10}[10]=1$

Solution By Gauss Seidel method.

$x=1,y=1,z=1.$