2-sistem persamaan aljabar linier-sjk-01 [compatibility mode]

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

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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)

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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 :

[email protected]

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

[email protected]

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

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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

[email protected]

⎪⎪

⎪⎪⎬

⎫

⎪⎪

⎪⎪⎨

⎧

⎪⎪

⎪⎪⎬

⎫

⎪⎪

⎪⎪⎨

⎧

⎥⎥⎥⎥⎤

⎢⎢⎢⎢⎡

=∗−−

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

[email protected]

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

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• 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 :

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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

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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

[email protected]

⎪

⎪⎬

⎫

⎪

⎪⎨

⎧=

⎪

⎪⎬

⎫

⎪

⎪⎨

⎧

⎥⎥⎤

⎢⎢⎡

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 )

[email protected]

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


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)

[email protected]

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


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

2-sistem persamaan aljabar linier-sjk-01 [compatibility mode]

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