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

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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

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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

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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

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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

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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

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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

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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

~

~~

~

~~

~

~~