Home > Matrix & Vector calculators > Solving systems of linear equations using Gauss Seidel method example

5. Gauss Seidel method example ( Enter your problem )
  1. Example `2x+y=8,x+2y=1`
  2. Example `2x+5y=16,3x+y=11`
  3. Example `2x+y+z=5,3x+5y+2z=15,2x+y+4z=8`
  4. Example `x+y+z=7,x+2y+2z=13,x+3y+z=13`
Other related methods
  1. Inverse Matrix method
  2. Cramer's Rule method
  3. Gauss-Jordan Elimination method
  4. Gauss Elimination Back Substitution method
  5. Gauss Seidel method
  6. Gauss Jacobi method
  7. Elimination method
  8. LU decomposition using Gauss Elimination method
  9. LU decomposition using Doolittle's method
  10. LU decomposition using Crout's method
  11. Cholesky decomposition method
  12. SOR (Successive over-relaxation) method
  13. Relaxation method

2. Example `2x+5y=16,3x+y=11`
(Previous example)
4. Example `x+y+z=7,x+2y+2z=13,x+3y+z=13`
(Next example)

3. Example `2x+y+z=5,3x+5y+2z=15,2x+y+4z=8`





Solve Equations 2x+y+z=5,3x+5y+2z=15,2x+y+4z=8 using Gauss Seidel method

Solution:
Total Equations are `3`

`2x+y+z=5`

`3x+5y+2z=15`

`2x+y+4z=8`


From the above equations
`x_(k+1)=1/2(5-y_(k)-z_(k))`

`y_(k+1)=1/5(15-3x_(k+1)-2z_(k))`

`z_(k+1)=1/4(8-2x_(k+1)-y_(k+1))`

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

Solution steps are
`1^(st)` Approximation

`x_1=1/2[5-(0)-(0)]=1/2[5]=2.5`

`y_1=1/5[15-3(2.5)-2(0)]=1/5[7.5]=1.5`

`z_1=1/4[8-2(2.5)-(1.5)]=1/4[1.5]=0.375`

`2^(nd)` Approximation

`x_2=1/2[5-(1.5)-(0.375)]=1/2[3.125]=1.5625`

`y_2=1/5[15-3(1.5625)-2(0.375)]=1/5[9.5625]=1.9125`

`z_2=1/4[8-2(1.5625)-(1.9125)]=1/4[2.9625]=0.7406`

`3^(rd)` Approximation

`x_3=1/2[5-(1.9125)-(0.7406)]=1/2[2.3469]=1.1734`

`y_3=1/5[15-3(1.1734)-2(0.7406)]=1/5[9.9984]=1.9997`

`z_3=1/4[8-2(1.1734)-(1.9997)]=1/4[3.6534]=0.9134`

`4^(th)` Approximation

`x_4=1/2[5-(1.9997)-(0.9134)]=1/2[2.087]=1.0435`

`y_4=1/5[15-3(1.0435)-2(0.9134)]=1/5[10.0429]=2.0086`

`z_4=1/4[8-2(1.0435)-(2.0086)]=1/4[3.9045]=0.9761`

`5^(th)` Approximation

`x_5=1/2[5-(2.0086)-(0.9761)]=1/2[2.0153]=1.0077`

`y_5=1/5[15-3(1.0077)-2(0.9761)]=1/5[10.0248]=2.005`

`z_5=1/4[8-2(1.0077)-(2.005)]=1/4[3.9797]=0.9949`

`6^(th)` Approximation

`x_6=1/2[5-(2.005)-(0.9949)]=1/2[2.0001]=1.0001`

`y_6=1/5[15-3(1.0001)-2(0.9949)]=1/5[10.01]=2.002`

`z_6=1/4[8-2(1.0001)-(2.002)]=1/4[3.9979]=0.9995`

`7^(th)` Approximation

`x_7=1/2[5-(2.002)-(0.9995)]=1/2[1.9985]=0.9993`

`y_7=1/5[15-3(0.9993)-2(0.9995)]=1/5[10.0033]=2.0007`

`z_7=1/4[8-2(0.9993)-(2.0007)]=1/4[4.0008]=1.0002`

`8^(th)` Approximation

`x_8=1/2[5-(2.0007)-(1.0002)]=1/2[1.9991]=0.9996`

`y_8=1/5[15-3(0.9996)-2(1.0002)]=1/5[10.0009]=2.0002`

`z_8=1/4[8-2(0.9996)-(2.0002)]=1/4[4.0007]=1.0002`


Solution By Gauss Seidel Method.
`x=0.9996~=1`

`y=2.0002~=2`

`z=1.0002~=1`

Iterations are tabulated as below
Iterationxyz
12.51.50.375
21.56251.91250.7406
31.17341.99970.9134
41.04352.00860.9761
51.00772.0050.9949
61.00012.0020.9995
70.99932.00071.0002
80.99962.00021.0002



This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here



2. Example `2x+5y=16,3x+y=11`
(Previous example)
4. Example `x+y+z=7,x+2y+2z=13,x+3y+z=13`
(Next example)





Share this solution or page with your friends.


 
Copyright © 2023. All rights reserved. Terms, Privacy
 
 

.