5. Hampiran Numerik Penyelesaian Sistem Persamaan Linier
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Transcript of 5. Hampiran Numerik Penyelesaian Sistem Persamaan Linier
HAMPIRAN NUMERIK PENYELESAIAN SISTEM PERSAMAAN LINIER
Pertemuan 5
Matakuliah : METODE NUMERIK ITahun : 2008
Bina Nusantara
Bentuk umum Sistem Persamaan Linier
i
n
jjij cXa
1
nnnnnnn
nn
nn
nn
c x a x a x a xa
c x a x a x a xa
c x a x a x a xa
c x a x a x a xa
332211
33333232131
22323222121
11313212111
Untuk i = 1,2,3,…,n
nnnnnn
n
n
c
c
c
x
x
x
aaa
aaa
aaa
.
.
.
.
.
..
...
......
......
......
...
...
2
1
2
1
21
22221
11211
Bina Nusantara
Metode Penyelesaian Persamaan Linier
Eliminasi Gauss
Eliminasi Gauss-Jordan
Matriks balikan
Dekomposisi LU
Iterasi Jacobi
Iterasi Gauss-Seidel
Bina Nusantara
Eliminasi Gauss
• Sistem Persamaan dengan n persamaan dan n variabel
nnnnnnn
nn
nn
nn
c x a x a x a xa
c x a x a x a xa
c x a x a x a xa
c x a x a x a xa
332211
33333232131
22323222121
11313212111
• Nilai variabel (vektor x,) pada sistem tidak berubah jika dilakukan hal-hal berikut:
Kalikan atau bagikan suatu persamaan dengan konstanta yang tidak sama dengan nol
Ganti suatu persamaan dengan penjumlahan persamaan tersebut dengan persamaan lainnya
Bina Nusantara
Operasi Baris Elementer
nnnnnnn
nn
nn
nn
b x a x a x a xa
b x a x a x a xa
b x a x a x a xa
b x a x a x a xa
332211
33333232131
22323222121
11313212111
b’k = bk (a11) – b1 (ak1)k = 2,3,4,…,n
nnnnnn
nn
nn
nn
b xa xa xa
b xa xa xa
b xa xa xa
b x a x a x a xa
3322
33333232
22323222
11313212111
Bina Nusantara
Membentuk Segitiga Bawah
......
33333
22323222
11313212111
nnnn
nn
nn
nn
b xa
b xa xa
b xa xa xa
b x a x a x a xa
nnnnnn
nn
nn
nn
b xa xa xa
b xa xa xa
b xa xa xa
b x a x a x a xa
3322
33333232
22323222
11313212111
b’k = bk (a’ 22
) – b2 (ak2)k = 3,4,…,n
Bina Nusantara
Forward Elimination
......
33333
22323222
11313212111
nnnn
nn
nn
nn
b xa
b xa xa
b xa xa xa
b x a x a x a xa
...
...
nn
nn
a
b x
Bina Nusantara
Substitusi Kembali
...
...
n
nn
a
b x
11
211
11
131321211
)()(
a
xab
a
x a x a xabx
n
kkk
nn
...1
...,11
...1,1 nnnnnnn b x + axa
a
x - ab x
nn
nnnnn ...
1,1
...,1
...1
1
Sehingga diperoleh
......
33333
22323222
11313212111
nnnn
nn
nn
nn
b xa
b xa xa
b xa xa xa
b x a x a x a xa
Bina Nusantara
Gauss Elimination
Input A, c , n
stop
Forward Elimination
Output Solution Vector, x
Back Substitution
Gauss Elimination
Bina Nusantara
Gauss Elimination
x n = b n /a nn
sum = 0
sum = sum + a i,j *x j
Back Substitution
x i = (b i - sum)/a i,i
i = n-1, n-2, ..., 1
j = i+1, ..., n
Compute last unknown
Loop backwards over all rows except last
Loop over all columns to the right of the
current row
Compute (i) th unknown
......
33333
22323222
11313212111
nnnn
nn
nn
nn
b xa
b xa xa
b xa xa xa
b x a x a x a xa
Bina Nusantara
Gauss Elimination
m ik = a ik /a kk
a ij = a ij - m ik *a kj
k = 1, 2, ..., n-1
Forward Elimination
i = k+1, ..., n
j = k+1, ..., n
b i = b i - m ik *b k
Loop over all rows below the diagonal
position (k,k)
Loop over all columns to the right of the diagonal
position (k,k)
Compute multiplier for row (i) and column (k)
Update coefficient for row (i) and column (j)
Update right hand side for row (i) and column (j)
Loop over all rows in matrix, except last
......
33333
22323222
11313212111
nnnn
nn
nn
nn
b xa
b xa xa
b xa xa xa
b x a x a x a xa
Bina Nusantara
Contoh
1633
1023
1642
321
321
321
xxx
x xx
x xx
1633
1452
11642
321
32
321
xxx
-x x
x xx
82
5
1452
11642
32
32
321
xx
-x x
x xx
7826
2810
1642
3
32
321
x
-x x
x xx
Forward Elimination
B2+(-3/2)b1
B3+(-1/2)b1
B3+(-5)b1
Bina Nusantara
Contoh
Substitusi Balik3
26
783 x
2
)3(1028
1028 32
-
x- x
12
)3(42162
)4(16 321
x xx
Bina Nusantara
GAUSS-JORDAN ELIMINATON
nnnnnnn
nn
nn
nn
b x a x a x a xa
b x a x a x a xa
b x a x a x a xa
b x a x a x a xa
332211
33333232131
22323222121
11313212111
...
...33
...22
...11
0...000
000
000
000
nnn b x
b x
b x
b x
Bina Nusantara
Contoh:Selesaikan SPL berikut dengan eliminasi Gass-Jordan
4.71102.03.0
3.193.071.0
85.72.01.03
321
321
321
xxx
xxx
xxx
4.71102.03.0
3.193.071.0
85.72.01.03
Jawaban:Matriks perluasan(augmented matrix) dari koefisien SPL:
b’1 = b1/3
Bina Nusantara
4.71102.03.0
3.193.071.0
61667.2066667.0033333.01
b’2 = b2 – 0.1 b1
b’3 = b3 – 0.3
b1
6150.700200.10190000.00
5617.1929333.0033333.70
61667.2066667.0033333.01
b”2 = b’
2/7.033333
Bina Nusantara
0000.7100
79320.2041664.010
52356.2068063.001
0000.7100
5000.2010
0000.3001
b”’1 = b”
1 + 0.068063 b”’3
b”’2 = b’’
2 + 0.190000 b”’3
0000.7
5000.2
0000.3
3
2
1
x
x
x
Bina Nusantara
0000.7100
79320.2041664.010
52356.2068063.001
0000.7100
5000.2010
0000.3001
b”’1 = b”
1 + 0.068063 b”’3
b”’2 = b’’
2 + 0.190000 b”’3
0000.7
5000.2
0000.3
3
2
1
x
x
x
Bina Nusantara
Iterative Methods
• Consider the linear systemb Ax
nnnnnn
n
n
b
b
b
x
x
x
aaa
aaa
aaa
2
1
2
1
21
22221
11211
nnnnnn
nn
nn
b x a x a xa
b x a x a xa
b x a x a xa
2211
22222121
11212111
Bina Nusantara
Solution Methods
• Gauss Elimination – Subject to roundoff errors and ill conditioning
• Iterative methods -- Alternative to elimination method• Take initial guess of solution and then iterate to obtain
improved estimates of the solution• Jacobi and Gauss-Seidel methods• Work well for large sets of equations
Bina Nusantara
Iterative Methods
• Rearrange the equations so that an unknown is on the left-hand side of each equation:
nn
nnnnnnn
nn
nn
a
x a x a xab x
a
x a xab x
a
x a xab x
)(
)(
)(
112211
22
212122
11
121211
• Initial guess
002
01 ,,, nxxx
Bina Nusantara
Jacobi Method
nn
nnnnnnn
nn
nn
a
x a x a xab x
a
x a xab x
a
x a xab x
)(
)(
)(
011
022
0111
22
02
012121
2
11
01
021211
1
• Initial guess 002
01 ,,, nxxx
• Next approximation of the solution
Bina Nusantara
Jacobi Method
• After k iterations of this process
nn
knnn
kn
knnk
n
knn
kk
knn
kk
a
x a x a xab x
a
x a xab x
a
x a xab x
)(
)(
)(
1122111
22
2121212
11
1212111
Bina Nusantara
Example – Jacobi Method
• System of 3 equations in 3 unknowns
52
83
12
32
321
21
x x
x x x
x x
2
5
2
)(5
3
8
3
)(8
2
1
2
)(1
2213
313112
2211
kkk
kkkkk
kkk
xx x
xxxx x
xx x
• Rearrange
Bina Nusantara
Example – Jacobi Method
• Initial guess 0,0,0 03
02
01 xxx
5.22
05
2
5
667.23
008
3
8
5.02
01
2
1
021
3
03
011
2
021
1
xx
xxx
x x
1667.12
833335.15
2
5
23
)5.2(5.08
3
8
833335.12
6667.21
2
1
122
3
13
112
2
122
1
xx
xxx
xx
1,3,2 321 xxx• After 20 iterations
Bina Nusantara
Jacobi Methode
• We stop when
itolerancex
xxk
i
ki
ki
iAr allfor,~
~~
1
1
tolerancex
xx
tolerancex
xx
tolerancex
xx
kn
kn
kn
nAr
k
kk
Ar
k
kk
Ar
1
1
12
21
22
11
11
11
~
~~
~
~~
~
~~