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Bab 2 Sistem Persamaan Aljabar Linier
Lecture 2Sistem Persamaan Aljabar Linier
METODE ELIMINASI DANMETODE ELIMINASI DAN DEKOMPOSISI
SujantokoOcean Engineering ITSOcean Engineering ITS
Bab 2 Sistem Persamaan Aljabar [email protected]
nn
nn
bxaxaxabxaxaxa
=+++=+++
..........
22222121
11212111
⎪⎪⎫
⎪⎪⎧
⎪⎪⎫
⎪⎪⎧
⎥⎥⎤
⎢⎢⎡
2
1
2
1
22221
n11211
bb
xx
a...aaa...aa
nnnnnn
nn
bxaxaxa =+++ ........
2211
22222121
⎪⎪⎭
⎪⎬
⎪⎪⎩
⎪⎨=
⎪⎪⎭
⎪⎬
⎪⎪⎩
⎪⎨
⎥⎥⎥
⎦⎢⎢⎢
⎣ n
2
n
2
nn2n1n
n22221
b...b
x...x
*
a...aa.........
a...aa
a = koefisien konstanta; x = ‘unknown’;
[ ]{ } { }bXA =aij = koefisien konstanta; xj = unknown ; bj = konstanta; n = banyaknya persamaan
Metode-Metode untuk menyelesaikan Sistem Persamaan Aljabar Linier:
1. Metode Eliminasi : Eliminasi Gauss; Gauss Jordan2 Metode Iterasi : Iterasi Jacobi; Gauss Siedel2. Metode Iterasi : Iterasi Jacobi; Gauss Siedel3. Metode Dekomposisi : Dekomposisi L-U; Cholesky.
2
M A T R I K
Operasi Matrik
⎥⎤
⎢⎡ naaa ... 11211
Kolom - j
• Penjumlahan / Pengurangan• Perkalian• Transpose
Invers Matrik⎥⎥⎥⎥
⎦⎢⎢⎢⎢
⎣ij
n
aaaa.........
... 22221
baris
-i
• Invers Matrik• Determinan
⎥⎦
⎢⎣ mnmm aaa ...21
Jenis jenis Matrikm x n
Contoh :
Jenis-jenis Matrik
• Matrik Bujur Sangkar• Matrik Diagonal • Matrik Identitas• Matrik Segitiga Atas / Bawah• Matrik Simetri• Vektor Baris
⎥⎦
⎤⎢⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
2795
B ;406813
A• Vektor Baris• Vektor Kolom
⎦⎣⎥⎦⎢⎣ 40
3
Penyelesaian: Ada, Tunggal(well condition)
Penyelesaian: Ada, Kondisi buruk(ill condition)
3x1 + 2x2 = 18-x1 + 2x2 = 2
⎬⎫
⎨⎧
⎬⎫
⎨⎧
⎥⎤
⎢⎡ 18
*23 1x
- ½ x1 + x2 = 1-2.3/5 x1 + x2 = 1.1
⎫⎧⎫⎧⎤⎡ − 111 x
x2 x2
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡− 2
*21 2
1
x⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
−
1.11
*15
3.212
1
2
1
xx
D t 3*2 ( 1)*2 8 D t 1/2 *1 ( 2 3/5)*1 0 04x1 x1
Det = 3*2 - (-1)*2 = 8 Det = -1/2 *1 - (-2.3/5)*1 = -0.04
Penyelesaian: Tak ada Penyelesaian: Tak berhingga
-½ x1 + x2 = 1-½ x1 + x2 = ½
-½ x1 + x2 = 1-1 x1 + 2x2 = 2
x2x2
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
−
211
*12
112
1
2
1
xx
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−
21
*2112
1
2
1
xx
Det = -1/2 *1 - (-1/2)*1 = 0 Det = -1/2 *2 - (-1)*1 = 0
x1x1
4
Eliminasi [email protected]
⎪⎬
⎫⎪⎨
⎧=⎪⎬
⎫⎪⎨
⎧
⎥⎥⎤
⎢⎢⎡
2
1
2
1
232221
131211
* bb
xx
aaaaaa
2
1
.......
.......EE
⎪⎭
⎬⎪⎩
⎨⎪⎭
⎬⎪⎩
⎨⎥⎥⎦⎢
⎢⎣ 3
2
3
2
333231
232221
bxaaa
Forward Elimination
3
2
....... E
⎪⎬
⎫⎪⎨
⎧⎪⎬
⎫⎪⎨
⎧⎥⎤
⎢⎡
'11
''131211
*0 bbxaaa
Forward Elimination
⎪⎭
⎪⎬
⎪⎩
⎪⎨=
⎪⎭
⎪⎬
⎪⎩
⎪⎨
⎥⎥⎥
⎦⎢⎢⎢
⎣''
3
2
3
2''
33
2322 *00
0bb
xx
aaa
'''
''33
''33
/)(/
babx =
Back Substitution
1131321211
2232322
/)(/)(
axaxabxaxabx
−−=−=
5
Proses Forward Elimination :
1. Eliminasikan x1 dari E2 dan E3Hitung: m21 = a21/a11
E’2 = E2 - m21*E1Hit / ⎪
⎪⎬
⎫
⎪
⎪⎨
⎧=
⎪
⎪⎬
⎫
⎪
⎪⎨
⎧
⎥⎥⎤
⎢⎢⎡
'2
1
2
1'23
'22
131211
*0 bb
xx
aaaaa
'2
1
..............
EE
Hitung: m31 = a31/a11E’3 = E3 – m31*E1
2. Eliminasikan x2 dari E’3
⎪⎭
⎪⎩
⎪⎭
⎪⎩⎥
⎥⎦⎢
⎢⎣
'33
'33
'32
3
0 bxaa
⎫⎧⎫⎧⎤⎡ 11131211 bxaaa
'3
2
....... E
1....... E2. Eliminasikan x2 dari E 3Hitung: m32 = a’32/a’22
E’’3 = E’3 – m32*E’2 ⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
''3
'2
1
3
2
1
''33
'23
'22
131211
*00
0bbb
xxx
aaaaaa
''3
'2
1
.......
..............
EEE
1−nnb
Proses Back Substitution :
1−= nnn
nn a
bx
1
11 −− ∑−n
ijj
iij
ii xab
1. x3 = b’’3 / a’’3
Untuk i = n-1, n-2, … , 1
11
−+== i
ii
iji a
x2. x2 = (b’2 – a’23*x3) / a’22
x1 = (b1 - a12*x2 - a13*x3) / a11 6
Algoritma Eliminasi Gauss Pivoting:i pivot = k
Forward Elimination:for k=1…n-1
for i=k+1…ni t A(i k)/A(k k)
i_pivot = kbig = |a(k,k)|for ii = k+1…n
dumy = |a(ii,k)|pivot = A(i,k)/A(k,k)for j=k…n
A(i,j) = A(i,j) - pivot * A(k,j)end
if ( dumy>big )big = dumyi_pivot = ii
end ifB(i) = B(i) - pivot * B(k)
endend
end ifend
if (i_pivot ~= k)f jj k
Back Substitution:X(n) = B(n)/A(n,n);for i=n-1…1 step-1 1
1−
= n
nn
nbx
for jj = k…ndummy = A(pivot,jj)A(i_pivot,jj)=A(k,jj)A(k,jj)=dummy;
sum = 0for j=i+1…n
sum = sum + A(I,j)*X(j)end
1−nnn
n a
11 −− ∑−n
jiij
ii xab
( ,jj) y;enddummy = C(i_pivot)C(i_pivot) = C(k)C(k) = dummyend
X(i) = (B(i)-sum) / A(i,i)end 1
1−+=∑
= iii
ijjiji
i a
xabx
C(k) = dummyEnd if
7
Contoh-1
Selesaikan sistem persamaan linier dengan metode Eliminasi Gauss. (Solusi eksak : x1 = 1, x2 = -2, x3 = 7/5 )
10 x1 + x2 – 5 x3 = 1 ……..E1-20 x1 + 3 x2 + 20 x3 = 2 ……..E2
5 x1 + 3 x2 + 5 x3 = 6 E35 x1 + 3 x2 + 5 x3 6 ……..E3
Penyelesaian:y
8
matrik bentuk disusun linier aljabar Persamaan
⎪⎪
⎪⎪⎬
⎫
⎪⎪
⎪⎪⎨
⎧
⎪⎪
⎪⎪⎬
⎫
⎪⎪
⎪⎪⎨
⎧
⎥⎥⎥⎥⎤
⎢⎢⎢⎢⎡
=∗−−
6535
21
2x1x
203205110
E1m-E2 2' 21∗=
−=−
==
E
21020
aam
11
2121
⎪⎪⎭⎪
⎪⎩⎪
⎪⎭⎪
⎪⎩⎥⎦⎢⎣ 6535 3x010*(-2)-20-
21==
15:nEliminatio Forward
010*)1(5
E1m-E3 3' 31∗=
===
E21
105
aam
11
3131
⎪⎪⎬
⎫
⎪⎪⎨
⎧
⎪⎪⎬
⎫⎪⎪⎨
⎧
⎥⎥⎤
⎢⎢⎡ −
411x
10505110
1 kolom Eleminasi
15
010*)21(-5 ==
⎪⎪⎭
⎪⎬
⎪⎪⎩
⎪⎨
⎪⎪⎭
⎪⎬
⎪⎪⎩
⎪⎨
⎥⎥⎥
⎦⎢⎢⎢
⎣
=∗
2114
3x2x
215
250
1050
E2'E3'-m 32 ∗=
===
''3E21
525
aam
22
3232
⎪⎪⎬
⎫
⎪⎪⎨
⎧
⎪⎪⎬
⎫⎪⎪⎨
⎧
⎥⎥⎤
⎢⎢⎡
∗−
411x
10505110
2 kolom Eleminasi
05*)21(-2
5 ==
9⎪⎪⎭
⎪⎬
⎪⎪⎩
⎪⎨
⎪⎪⎭
⎪⎬
⎪⎪⎩
⎪⎨
⎥⎥⎥
⎦⎢⎢⎢
⎣
=∗
274
3x2x
2500
1050
9
57
27x2
53 =⇒= 3 x
nEliminatio Backward
⎪⎬
⎫⎪⎨
⎧⎪⎬
⎫⎪⎨
⎧∴ 2
1xx1
4x10x5
522
32 =+⎪⎭
⎬⎪⎩
⎨−=⎪⎭
⎬⎪⎩
⎨∴
572
xx
3
2
24)57(10x5 2 −=⇒=+ 2 x
11)57(5)2(x10
1x5xx10
1
321
=⇒=−−+
=−+
1x)5()(1 1
10
Masalah dalam Metode Eliminasi
• Pembagian dengan NOL 2x2 + 3x3 = 84x + 6x + 7x = 3
• Kesalahan dalam pembulatan
4x1 + 6x2 + 7x3 = -32x1 + x2 + 6x3 = -5
• Sistem ILL Condition
x1 + 2x2 = 102 10
x1 = 431.1 x1 + 2x2 = 10.4
x1 + 2x2 = 10
x2 = 3
x1 = 8 (8) + 2*(1) = 101 2
1.05 x1 + 2x2 = 10.41
x2 = 1(8) 2 (1) 10
1.1*(8) + 2(1) = 10.8 ≈≈ 10.411
Solusi :
1. Penggunaan angka signifikan LEBIH BANYAK2. Pivoting
Pertukarkan baris baris sehingga elemen pivot adalahPertukarkan baris-baris sehingga elemen pivot adalah elemen terbesar
Contoh 2Contoh-2.0.0003 x1 + 3.0000 x2 = 2.0001
1.0000 x1 + 1.0000 x2 = 1.00001.0000 x1 + 1.0000 x2 = 1.00000.0003 x1 + 3.0000 x2 = 2.0001
x = 2/3 x = 2/3x2 = 2/3x1 = 2.0001 – 3*(x2)
0.0003
x2 = 2/3x1 = 1 – x2
1
Angka Sig X2 X1 Angka Sig. X2 X1Angka Sig. X2 X1
345
0.6670.66670 66667
-3.0000.00000 30000
Angka Sig. X2 X1
345
0.6670.66670.66667
0.3330.33330.333335
67
0.666670.6666670.6666667
0.300000.3300000.3300000
567
0.666670.6666670.6666667
0.333330.3333330.3333333 12
3. PenskalaanK fi i M k i d l ti b i d l h 1
Koefisien Maksimun dalam setiap baris adalah 1(dilakukan jika ada persamaan yang mempunyai koefisien terlalu besar relatif terhadap persamaan lainya)
Contoh-2. Tentukan penyelesaian sistem pers. linier dibawah ini dengan eliminasi gauss (solusi eksak : x1=1,00002 x2=0,99998)
• Dengan Penskalaan:0,00002 x1 + x2 = 1
x1 + x2 = 2
2 x1 + 100000 x2 = 100000x1 + x2 = 2
• Tanpa Penskalaan:2 x1 + 100000 x2 = 100000
x1 + x2 = 2
1 2
x1 + x2 = 2 0,00002 x1 + x2 = 1
x1 + x2 2
2 x1 + 100000 x2 = 10000049999 x2 = 49998
x1 + x2 = 2 0.99998x2 = 0,99996
x2 = 1 00x2 = 1,00x1 = 0,00
x2 = 1,00x1 = 1,00
13
Eliminasi Gauss-Jordan Invers Matrik
⎪
⎪⎬
⎫
⎪
⎪⎨
⎧=
⎪
⎪⎬
⎫
⎪
⎪⎨
⎧
⎥⎥⎤
⎢⎢⎡
2
1
2
1
232221
131211
* bb
xx
aaaaaa
⎢⎢⎢⎡
⎥⎥⎤
010001
232221
131211
aaaaaa
⎪⎭
⎪⎩
⎪⎭
⎪⎩⎥
⎥⎦⎢
⎢⎣ 33333231 bxaaa
Forward Elimination
⎢⎢
⎣ ⎥⎥⎦100333231 aaa
[A] [ I ]
⎪⎬
⎫⎪⎨
⎧
=⎪⎬
⎫⎪⎨
⎧
⎥⎥⎤
⎢⎢⎡
*2
*1
2
1
*010001
bb
xx
Forward Elimination
⎡ ⎤−−− 111001 aaa
Forward Elimination
⎪⎭
⎬⎪⎩
⎨⎪⎭
⎬⎪⎩
⎨⎥⎥⎦⎢
⎢⎣
*3
2
3
2
100010
bb
xx
NO Back Substitution⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
−−−
−−−
133
132
131
123
122
121
131211
100010001
aaaaaaaaa
*
*11
bbx =
NO Back Substitution ⎣[ I ] [A]-1
A * b*33
22
bxbx
== A * x = b
x = A-1 * b14
Algorithma Gauss-Jordan Algorithma Invers-Matrik( dengan Gauss Jordan )
Forward Elimination:
f k 1
( dengan Gauss-Jordan )
Forward Elimination:
f k 1for k=1…ndummy = A(k,k)for j=1…n+1
A(k,j) = A(k,j)/dummy
for k=1…ndummy = A(k,k)for j=1…2*n
A(k,j) = A(k,j)/dummy( ,j) ( ,j)/du yend
for i=1…n if (i<>k)
A(k,j) A(k,j)/dummyend
for i=1…n if (i<>k)if (i<>k)
dummy = A(i,k)for j=1…n+1
A(i,j) = A(i,j) – dummy * A(k,j)
if (i<>k)dummy = A(i,k)for j=1…2*n
A(i,j) = A(i,j) – dummy * A(k,j)A(i,j) A(i,j) dummy A(k,j)end
end ifend
d
A(i,j) A(i,j) dummy A(k,j)end
end ifend
dend end
15
Dekomposisi [L] [U][email protected]
⎫⎧⎫⎧⎤⎡ b
Cara Menyelesaikan Sistem Pers. Linier dengan merubah Matrik sistem A menjadi Matrik Segitiga Bawah L dan Matrik Segitiga Atas U
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
3
2
1
3
2
1
333231
232221
131211
*bbb
xxx
aaaaaaaaa
⎭⎩⎭⎩⎦⎣
[ A ] * { x } = { b }
⎪⎫
⎪⎧
⎪⎫
⎪⎧
⎥⎤
⎢⎡
⎥⎤
⎢⎡ 11131211 100 bxuul
⎪⎭
⎪⎬
⎪⎩
⎪⎨=
⎪⎭
⎪⎬
⎪⎩
⎪⎨
⎥⎥⎥
⎦⎢⎢⎢
⎣⎥⎥⎥
⎦⎢⎢⎢
⎣ 3
2
1
3
2
1
23
1312
333231
2221
11
*100
10*0bb
xxu
lllll
[ L ] * [ U ] * { x } = { b }16
Flow [email protected]
[A ] * { x} = { b}
Proses Dekomposisi Untuk memperoleh U dan L[L ]* [U ]* {x} = {b}[L ] [U ] {x} {b}
Jika [U] * {x} = {y}
[L] * {y} = {b} Proses Subs. MajuUntuk memperoleh y
[U] * {x} = {y}Proses Subs. MundurUntuk memperoleh x[U] {x} {y} p
17
Dekomposisi LU : [email protected]
Diturunkan dari proses Eliminasi Gauss, dimanaL : Elemen Pengali mij dalam proses eliminasiU : Matrik Segitiga Atas hasil dari proses eliminasi
⎪⎬
⎫⎪⎨
⎧=⎪⎬
⎫⎪⎨
⎧
⎥⎥⎤
⎢⎢⎡ 11131211
* bb
xx
aaaaaa
⎪⎭
⎬⎪⎩
⎨=⎪⎭
⎬⎪⎩
⎨⎥⎥⎦⎢
⎢⎣ 3
2
3
2
333231
232221
bb
xx
aaaaaa
[ A ] * { x } = { b }{ } { }
Proses Eliminasi Gauss
⎥⎥⎤
⎢⎢⎡
= '23
'22
131211
0 aaaaa
U⎥⎥⎤
⎢⎢⎡
= 01001
21mL⎥⎥⎦⎢
⎢⎣
''3300 a⎥
⎥⎦⎢
⎢⎣ 13231 mm
18
Dekomposisi LU : [email protected]
Matrik L dan U dicari dengan menyelesaikan persamaan [ L] * [U] = [A]
⎥⎤
⎢⎡
⎥⎤
⎢⎡
⎥⎤
⎢⎡ 1413121114131211 1000 aaaauuul
⎥⎥⎥⎥
⎦⎢⎢⎢⎢
⎣
=
⎥⎥⎥⎥
⎦⎢⎢⎢⎢
⎣⎥⎥⎥⎥
⎦⎢⎢⎢⎢
⎣ 44434241
34433231
24232221
34
2423
44434241
333231
2221
1000100
10*
000
aaaaaaaaaaaa
uuu
lllllll
ll
⎦⎣⎦⎣⎦⎣ 4443424144434241
l11=a11, l21=a21, l31=a31, l41=a41 . . . . . . li1= ai1, utk i = 1,..,n
l11*u12 = a12, l11*u13 = a13, l11*u14 = a14
u12 = a12/l11, u13 = a13/l11, u14 = a14/l11 . . . . . u1j = a1j/l11, utk j = 2,..,n
li2 = ai2-li1u12, utk i = 2,..,n u2j = (a2j-l21u1j)/l22, utk j = 3,..,n
l a l u l u utk i 3 n u (a l u l u )/l utk j 4 nli3 = ai3-li1u13-li2u23, utk i = 3,..,n u3j = (a3j-l31u1j-l32u2j)/l33, utk j = 4,..,n
li4 = ai4-li1u14-li2u24-li3u34, utk i = 4,..,n 19
Algorithma Crout for j=2…na(i,j) = a(i,j)/a(1,1)
endfor j=2…n-1
for i=j…nsum = 0
li1= ai1, utk i = 1,..,n
u1j = a1j/l11, utk j = 2,..,nfor k=1…j-1
sum = sum + a(i,k)*a(k,j)enda(i,j) = a(i,j)-sum
utk j = 2,3,…n-1
endfor k=j+1…n
sum=0for i=1..j-1
∑−
=
−=1
1
j
kkjikijij ulal
j−1
utk i = j, j+1,…,n
sum = sum + a(j,i)*a(i,k)enda(j,k) = (a(j,k) – sum)/a(j,j)
endjj
kikjiki
jk l
ulau
∑=
−= 1 utk k = j+1, j+2…,n
endsum = 0for k=1…n-1
sum = sum + a(n,k)*a(k,n)∑−
−=1n
knnknnnn ulalenda(n,n) = a(n,n) - sum
∑=1k
20
Dekomposisi LU : [email protected]
Digunakan jika Matrik Sistem [A ] adalah matrik Simetri, yaitu [A ]= [A]T
Matrik Simetri A bisa didekomposisi menjadi : [L] * [L]T = [A]⎤⎡⎤⎡⎤⎡ 000 aaaalllll
⎥⎥⎥⎥⎤
⎢⎢⎢⎢⎡
=
⎥⎥⎥⎥⎤
⎢⎢⎢⎢⎡
⎥⎥⎥⎥⎤
⎢⎢⎢⎢⎡
43433231
42322221
41312111
4333
423222
41312111
333231
2221
11
000
*000000
aaaaaaaaaaaa
lllllllll
lllll
l
⎥⎦
⎢⎣
⎥⎦
⎢⎣
⎥⎦
⎢⎣ 444342414444434241 000 aaaalllll
l11*l11 = a11, l21*l11 = a21, l31*l11 = a31, l41*l11=a41
l11 = √a11, l21 = a21/l11, l31 = a31/l11, l41 =a41/l11
l21*l21 + l22*l22 = a22, l31*l21+ l32*l22 = a32, l41*l21 + l42*l22=a42
l22 = √ (a22-l21*l21), l32= (a32 -l31*l21)/l22 , l42 = (a42-l41*l21)/l22
i
lla ∑−1
k
ii
jkjijki
ki l
llal
∑=
−= 1∑
−
=
−=1
1
2k
jkjkkkk lal
untuk i=1,2,…,k-1 21
Setelah [L] didapat maka:[email protected]
[L]*{y}={b} forward subs. didapat {y}
[L]T *{x}={y} Backward subs didapat {x}
22
Algorithma [email protected]
for k=1…nfor i=1…k-1
sum = 0sum 0for j=1…i-1
sum = sum + a(I,j)*a(k,j)end
(k i) ( (k i) )/ (i i)
i
jkjijki lla
l∑−
=
−=
1
1a(k,i) = (a(k,i)-sum)/a(i,i)
end
sum = 0
iiki l
l =
untuk i=1,2,…,k-1
for j=1…k-1sum = sum + (a(k,j))2
end(k k) √ ( (k k) ) ∑
−
−=1
2k
lala(k,k) = √ (a(k,k) - sum)end
∑=
−=1j
kjkkkk lal
23