Q4
Solve by Gauss-Seidel Method
2x+y+6z=9, 8x+3y+2z=13, X+5y+z=7
Answers
Answer:
hi i can not doing full answer am doing half answer so sorry the half answer is below
first we write the system of equation in the form
x= 1/2(9-y-6z)
y= 1/3(13-8x-2z)
z= 7-x-5y
now using the jacodi iteration formula
x^(1) =1/2(9-0-6×0) = 4.5
y^(1) = 1/3(13-8×0-2×0) = 4.333
z^(1) = 7-0-5×0= 7
as the first approximately
ye answer aage ko or bhi h par mujse itna hi solve hua
I hope isse tumari kuch help hui ho
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Answer:
Total Equations are 3
2x+y+6z=9
8x+3y+2z=13
x+5y+z=7
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.
8x+3y+2z=13
x+5y+z=7
2x+y+6z=9
From the above equations
xk+1=
1
8
(13-3yk-2zk)
yk+1=
1
5
(7-xk+1-zk)
zk+1=
1
6
(9-2xk+1-yk+1)
Initial gauss (x,y,z)=(0,0,0)
Solution steps are
1st Approximation
x1=
1
8
[13-3(0)-2(0)]=
1
8
[13]=1.625
y1=
1
5
[7-(1.625)-(0)]=
1
5
[5.375]=1.075
z1=
1
6
[9-2(1.625)-(1.075)]=
1
6
[4.675]=0.7792
2nd Approximation
x2=
1
8
[13-3(1.075)-2(0.7792)]=
1
8
[8.2167]=1.0271
y2=
1
5
[7-(1.0271)-(0.7792)]=
1
5
[5.1938]=1.0387
z2=
1
6
[9-2(1.0271)-(1.0387)]=
1
6
[5.9071]=0.9845
3rd Approximation
x3=
1
8
[13-3(1.0387)-2(0.9845)]=
1
8
[7.9147]=0.9893
y3=
1
5
[7-(0.9893)-(0.9845)]=
1
5
[5.0261]=1.0052
z3=
1
6
[9-2(0.9893)-(1.0052)]=
1
6
[6.0161]=1.0027
4th Approximation
x4=
1
8
[13-3(1.0052)-2(1.0027)]=
1
8
[7.9789]=0.9974
y4=
1
5
[7-(0.9974)-(1.0027)]=
1
5
[4.9999]=1
z4=
1
6
[9-2(0.9974)-(1)]=
1
6
[6.0053]=1.0009
5th Approximation
x5=
1
8
[13-3(1)-2(1.0009)]=
1
8
[7.9983]=0.9998
y5=
1
5
[7-(0.9998)-(1.0009)]=
1
5
[4.9993]=0.9999
z5=
1
6
[9-2(0.9998)-(0.9999)]=
1
6
[6.0006]=1.0001
6th Approximation
x6=
1
8
[13-3(0.9999)-2(1.0001)]=
1
8
[8.0002]=1
y6=
1
5
[7-(1)-(1.0001)]=
1
5
[4.9999]=1
z6=
1
6
[9-2(1)-(1)]=
1
6
[6]=1
Solution By Gauss Seidel Method.
x=1≅1
y=1≅1
z=1≅1