Metode Numerik Gauss Seidel
Nama:
Maria Ulfa Cahyani
NIM:
110210102054
Kelas:
A
1.
Soal
A=[1 1 2;2 4 1;0 2 3]
A =
1 1 2
2 4 1
0 2 3
>> b=[1;3;5]
b =
1
3
5
>> x0=[0;0;0]
x0 =
0
0
0
Jawab:
·
Metode Gauss Eliminasi
x=gauss(A,b)
n1
=
3
k
=
1
x
=
-1.8333
1.5000
0.6667
·
LU Decomposisi
x=ludec(A,b)
A
=
1
1 2
2
2 -3
0
2 3
A
=
1
1 2
2
2 -3
0
1 6
x
=
-1.8333
1.5000
0.6667
·
Jacobian
x=jacobian(A,b,x0,50)
x
=
-351.7324
-178.0725
-105.9641
·
Gauss Seidel
x=gauss_seidel(A,b,x0,10e-6,50)
fx1x2x3...
metode
iterasi gauss seidel telah konvergen
x
=
-1.8333
1.5000
0.6667
·
SOR
x=SOR(A,b,x0,1.5,10e-6,50)
C
=
0
-1 -2
-2
-4 -1
0
-2 -3
C
=
0
-1 -2
-2
0 -1
0
-2 -3
C
=
0
-1 -2
-2
0 -1
0
-2 0
x
=
1.0e+017 *
-6.4305
4.8642
-6.2350
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