Solve Equations x+y+z=3,2x-y-z=3,x-y+z=9 using Gauss Jacobi method
Solution:
Total Equations are `3`
`x+y+z=3`
`2x-y-z=3`
`x-y+z=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.
`2x-y-z=3`
`x+y+z=3`
`x-y+z=9`
From the above equations
`x_(k+1)=1/2(3+y_(k)+z_(k))`
`y_(k+1)=1/1(3-x_(k)-z_(k))`
`z_(k+1)=1/1(9-x_(k)+y_(k))`
Initial gauss `(x,y,z) = (0,0,0)`
Solution steps are
`1^(st)` Approximation
`x_1=1/2[3+(0)+(0)]=1/2[3]=1.5`
`y_1=1/1[3-(0)-(0)]=1/1[3]=3`
`z_1=1/1[9-(0)+(0)]=1/1[9]=9`
`2^(nd)` Approximation
`x_2=1/2[3+(3)+(9)]=1/2[15]=7.5`
`y_2=1/1[3-(1.5)-(9)]=1/1[-7.5]=-7.5`
`z_2=1/1[9-(1.5)+(3)]=1/1[10.5]=10.5`
`3^(rd)` Approximation
`x_3=1/2[3+(-7.5)+(10.5)]=1/2[6]=3`
`y_3=1/1[3-(7.5)-(10.5)]=1/1[-15]=-15`
`z_3=1/1[9-(7.5)+(-7.5)]=1/1[-6]=-6`
`4^(th)` Approximation
`x_4=1/2[3+(-15)+(-6)]=1/2[-18]=-9`
`y_4=1/1[3-(3)-(-6)]=1/1[6]=6`
`z_4=1/1[9-(3)+(-15)]=1/1[-9]=-9`
Equations are Divergent...
Iterations are tabulated as below
| Iteration | x | y | z |
| 1 | 1.5 | 3 | 9 |
| 2 | 7.5 | -7.5 | 10.5 |
| 3 | 3 | -15 | -6 |
| 4 | -9 | 6 | -9 |
This material is intended as a summary. Use your textbook for detail explanation.
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