Solve Equations 4x+3y=24,3x+4y-z=30,-y+4z=-24 using SOR (Successive over-relaxation) method

Solution:
We know that, for symmetric positive definite matrix the SOR method converges for values of the relaxation parameter `w` from the interval `0 < w < 2`

The iterations of the SOR method
1. Total Equations are `3`

`4x+3y-0z=24`

`3x+4y-z=30`

`0x-y+4z=-24`


2. From the above equations, First write down the equations for Gauss Seidel method
`x_(k+1)=1/4(24-3y_(k)-0z_(k))`

`y_(k+1)=1/4(30-3x_(k+1)+z_(k))`

`z_(k+1)=1/4(-24-0x_(k+1)+y_(k+1))`

3. Now multiply the right hand side by the parameter `w` and add to it the vector `x_k` from the previous iteration multiplied by the factor of `(1-w)`

`x_(k+1)=(1-w)*x_(k)+w*1/4(24-3y_(k)-0z_(k))`

`y_(k+1)=(1-w)*y_(k)+w*1/4(30-3x_(k+1)+z_(k))`

`z_(k+1)=(1-w)*z_(k)+w*1/4(-24-0x_(k+1)+y_(k+1))`

4. Initial gauss `(x,y,z) = (0,0,0)` and `w=1.25`

Solution steps are
`1^(st)` Approximation

`x_1=(1-1.25)*0+1.25*1/4[24-3(0)-0(0)]=(-0.25)*0+1.25*1/4[24]=0+7.5=7.5`

`y_1=(1-1.25)*0+1.25*1/4[30-3(7.5)+(0)]=(-0.25)*0+1.25*1/4[7.5]=0+2.34375=2.34375`

`z_1=(1-1.25)*0+1.25*1/4[-24-0(7.5)+(2.34375)]=(-0.25)*0+1.25*1/4[-21.65625]=0+-6.76758=-6.76758`

`2^(nd)` Approximation

`x_2=(1-1.25)*7.5+1.25*1/4[24-3(2.34375)-0(-6.76758)]=(-0.25)*7.5+1.25*1/4[16.96875]=-1.875+5.30273=3.42773`

`y_2=(1-1.25)*2.34375+1.25*1/4[30-3(3.42773)+(-6.76758)]=(-0.25)*2.34375+1.25*1/4[12.94922]=-0.58594+4.04663=3.46069`

`z_2=(1-1.25)*-6.76758+1.25*1/4[-24-0(3.42773)+(3.46069)]=(-0.25)*-6.76758+1.25*1/4[-20.53931]=1.69189+-6.41853=-4.72664`

`3^(rd)` Approximation

`x_3=(1-1.25)*3.42773+1.25*1/4[24-3(3.46069)-0(-4.72664)]=(-0.25)*3.42773+1.25*1/4[13.61792]=-0.85693+4.2556=3.39867`

`y_3=(1-1.25)*3.46069+1.25*1/4[30-3(3.39867)+(-4.72664)]=(-0.25)*3.46069+1.25*1/4[15.07736]=-0.86517+4.71168=3.8465`

`z_3=(1-1.25)*-4.72664+1.25*1/4[-24-0(3.39867)+(3.8465)]=(-0.25)*-4.72664+1.25*1/4[-20.1535]=1.18166+-6.29797=-5.11631`

`4^(th)` Approximation

`x_4=(1-1.25)*3.39867+1.25*1/4[24-3(3.8465)-0(-5.11631)]=(-0.25)*3.39867+1.25*1/4[12.46049]=-0.84967+3.8939=3.04424`

`y_4=(1-1.25)*3.8465+1.25*1/4[30-3(3.04424)+(-5.11631)]=(-0.25)*3.8465+1.25*1/4[15.75098]=-0.96163+4.92218=3.96056`

`z_4=(1-1.25)*-5.11631+1.25*1/4[-24-0(3.04424)+(3.96056)]=(-0.25)*-5.11631+1.25*1/4[-20.03944]=1.27908+-6.26233=-4.98325`

`5^(th)` Approximation

`x_5=(1-1.25)*3.04424+1.25*1/4[24-3(3.96056)-0(-4.98325)]=(-0.25)*3.04424+1.25*1/4[12.11833]=-0.76106+3.78698=3.02592`

`y_5=(1-1.25)*3.96056+1.25*1/4[30-3(3.02592)+(-4.98325)]=(-0.25)*3.96056+1.25*1/4[15.93899]=-0.99014+4.98093=3.9908`

`z_5=(1-1.25)*-4.98325+1.25*1/4[-24-0(3.02592)+(3.9908)]=(-0.25)*-4.98325+1.25*1/4[-20.0092]=1.24581+-6.25288=-5.00706`

`6^(th)` Approximation

`x_6=(1-1.25)*3.02592+1.25*1/4[24-3(3.9908)-0(-5.00706)]=(-0.25)*3.02592+1.25*1/4[12.02761]=-0.75648+3.75863=3.00215`

`y_6=(1-1.25)*3.9908+1.25*1/4[30-3(3.00215)+(-5.00706)]=(-0.25)*3.9908+1.25*1/4[15.98649]=-0.9977+4.99578=3.99808`

`z_6=(1-1.25)*-5.00706+1.25*1/4[-24-0(3.00215)+(3.99808)]=(-0.25)*-5.00706+1.25*1/4[-20.00192]=1.25177+-6.2506=-4.99883`

`7^(th)` Approximation

`x_7=(1-1.25)*3.00215+1.25*1/4[24-3(3.99808)-0(-4.99883)]=(-0.25)*3.00215+1.25*1/4[12.00576]=-0.75054+3.7518=3.00126`

`y_7=(1-1.25)*3.99808+1.25*1/4[30-3(3.00126)+(-4.99883)]=(-0.25)*3.99808+1.25*1/4[15.99737]=-0.99952+4.99918=3.99966`

`z_7=(1-1.25)*-4.99883+1.25*1/4[-24-0(3.00126)+(3.99966)]=(-0.25)*-4.99883+1.25*1/4[-20.00034]=1.24971+-6.25011=-5.0004`

`8^(th)` Approximation

`x_8=(1-1.25)*3.00126+1.25*1/4[24-3(3.99966)-0(-5.0004)]=(-0.25)*3.00126+1.25*1/4[12.00102]=-0.75032+3.75032=3`

`y_8=(1-1.25)*3.99966+1.25*1/4[30-3(3)+(-5.0004)]=(-0.25)*3.99966+1.25*1/4[15.99959]=-0.99991+4.99987=3.99996`

`z_8=(1-1.25)*-5.0004+1.25*1/4[-24-0(3)+(3.99996)]=(-0.25)*-5.0004+1.25*1/4[-20.00004]=1.2501+-6.25001=-4.99991`

`9^(th)` Approximation

`x_9=(1-1.25)*3+1.25*1/4[24-3(3.99996)-0(-4.99991)]=(-0.25)*3+1.25*1/4[12.00013]=-0.75+3.75004=3.00004`

`y_9=(1-1.25)*3.99996+1.25*1/4[30-3(3.00004)+(-4.99991)]=(-0.25)*3.99996+1.25*1/4[15.99997]=-0.99999+4.99999=4`

`z_9=(1-1.25)*-4.99991+1.25*1/4[-24-0(3.00004)+(4)]=(-0.25)*-4.99991+1.25*1/4[-20]=1.24998+-6.25=-5.00002`


Solution By SOR (successive over-relaxation) method.
`x=3.00004~=3`

`y=4~=4`

`z=-5.00002~=-5`

Intertions are tabulated as below

Iteration	x	y	z
1	7.5	2.34375	-6.76758
2	3.42773	3.46069	-4.72664
3	3.39867	3.8465	-5.11631
4	3.04424	3.96056	-4.98325
5	3.02592	3.9908	-5.00706
6	3.00215	3.99808	-4.99883
7	3.00126	3.99966	-5.0004
8	3	3.99996	-4.99991
9	3.00004	4	-5.00002

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