Solve Equations x+y+z=7,x+2y+2z=13,x+3y+z=13 using Gauss Jacobi method
Solution:
Total Equations are `3`
`x+y+z=7`
`x+2y+2z=13`
`x+3y+z=13`
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.
`x+y+z=7`
`x+3y+z=13`
`x+2y+2z=13`
From the above equations
`x_(k+1)=1/1(7-y_(k)-z_(k))`
`y_(k+1)=1/3(13-x_(k)-z_(k))`
`z_(k+1)=1/2(13-x_(k)-2y_(k))`
Initial gauss `(x,y,z) = (0,0,0)`
Solution steps are
`1^(st)` Approximation
`x_1=1/1[7-(0)-(0)]=1/1[7]=7`
`y_1=1/3[13-(0)-(0)]=1/3[13]=4.3333`
`z_1=1/2[13-(0)-2(0)]=1/2[13]=6.5`
`2^(nd)` Approximation
`x_2=1/1[7-(4.3333)-(6.5)]=1/1[-3.8333]=-3.8333`
`y_2=1/3[13-(7)-(6.5)]=1/3[-0.5]=-0.1667`
`z_2=1/2[13-(7)-2(4.3333)]=1/2[-2.6667]=-1.3333`
`3^(rd)` Approximation
`x_3=1/1[7-(-0.1667)-(-1.3333)]=1/1[8.5]=8.5`
`y_3=1/3[13-(-3.8333)-(-1.3333)]=1/3[18.1667]=6.0556`
`z_3=1/2[13-(-3.8333)-2(-0.1667)]=1/2[17.1667]=8.5833`
`4^(th)` Approximation
`x_4=1/1[7-(6.0556)-(8.5833)]=1/1[-7.6389]=-7.6389`
`y_4=1/3[13-(8.5)-(8.5833)]=1/3[-4.0833]=-1.3611`
`z_4=1/2[13-(8.5)-2(6.0556)]=1/2[-7.6111]=-3.8056`
`5^(th)` Approximation
`x_5=1/1[7-(-1.3611)-(-3.8056)]=1/1[12.1667]=12.1667`
`y_5=1/3[13-(-7.6389)-(-3.8056)]=1/3[24.4444]=8.1481`
`z_5=1/2[13-(-7.6389)-2(-1.3611)]=1/2[23.3611]=11.6806`
`6^(th)` Approximation
`x_6=1/1[7-(8.1481)-(11.6806)]=1/1[-12.8287]=-12.8287`
`y_6=1/3[13-(12.1667)-(11.6806)]=1/3[-10.8472]=-3.6157`
`z_6=1/2[13-(12.1667)-2(8.1481)]=1/2[-15.463]=-7.7315`
Equations are Divergent...
Iterations are tabulated as below
| Iteration | x | y | z |
| 1 | 7 | 4.3333 | 6.5 |
| 2 | -3.8333 | -0.1667 | -1.3333 |
| 3 | 8.5 | 6.0556 | 8.5833 |
| 4 | -7.6389 | -1.3611 | -3.8056 |
| 5 | 12.1667 | 8.1481 | 11.6806 |
| 6 | -12.8287 | -3.6157 | -7.7315 |
This material is intended as a summary. Use your textbook for detail explanation.
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