review operasi matriks - kuliah.ftsl.itb.ac.id · jika determinan = 0 matriks singular, tidak punya...

REVIEW OPERASI MATRIKS

T E K N I K L I N G K U N G A N I T B

MENGHITUNG INVERS MATRIKS

1

0

0

1

22221221

21221121

22121211

21121111

acac

acac

acac

acac

10

01

2221

1211

2221

1211

aa

aa

cc

cc

DETERMINAN

Hanya untuk square matrices

Jika determinan = 0 matriks singular, tidak punya invers

bcaddc

ba

dc

ba

det

123213312132231321

321

321

321

det

cbacbacbacbacbacba

ccc

bbb

aaa

CARI INVERS NYA…

52

42

42

21

S I M U L T A N E O U S L I N E A R E Q U A T I O N S

SISTEM PERSAMAAN LINEAR

METODE PENYELESAIAN

• Metode grafik

• Eliminasi Gauss

• Metode Gauss – Jourdan

• Metode Gauss – Seidel

• LU decomposition

METODE GRAFIK

2

4

11

21

2

1

x

x2

-2

Det{A} 0 A

is nonsingular

so invertible

Unique

solution

SISTEM PERSAMAAN YANG TAK TERSELESAIKAN

5

4

42

21

2

1

x

x

No solution Det [A] = 0, but system is inconsistent

Then this system of

equations is not solvable

SISTEM DENGAN SOLUSI TAK TERBATAS

Det{A} = 0 A is singular

infinite number of solutions

8

4

42

21

2

1

x

x

Consistent so solvable

ILL-CONDITIONED SYSTEM OF EQUATIONS

A linear system

of equations is

said to be “ill-

conditioned” if

the coefficient

matrix tends to

be singular

ILL-CONDITIONED SYSTEM OF EQUATIONS

• A small deviation in the entries of A matrix, causes a

large deviation in the solution.

47.1

3

99.048.0

21

2

1

x

x

47.1

3

99.049.0

21

2

1

x

x

1

1

2

1

x

x

0

3

2

1

x

x

GAUSSIAN ELIMINATION

CXA

Merupakan salah satu teknik paling populer

dalam menyelesaikan sistem persamaan

linear dalam bentuk:

Terdiri dari dua step

1. Forward Elimination of Unknowns.

2. Back Substitution

FORWARD ELIMINATION

112144

1864

1525

Tujuan Forward Elimination adalah untuk

membentuk matriks koefisien menjadi Upper

Triangular Matrix

7.000

56.18.40

1525

FORWARD ELIMINATION

Persamaan linear

n persamaan dengan n variabel yang tak diketahui

11313212111 ... bxaxaxaxa nn

22323222121 ... bxaxaxaxa nn

nnnnnnn bxaxaxaxa ...332211

. .

. .

. .

CONTOH

83125

12312

71352

21232

8325

1232

7352

2232

4321

4321

4321

4321

xxxx

xxxx

xxxx

xxxx

matriks input

FORWARD ELIMINATION

83125

12312

71352

21232

32

162

190

13140

92120

12

112

31

'

14

'

4

'

13

'

3

'

12

'

2

1'

1

5

2

2

2

RRR

RRR

RRR

RR

32

162

190

13140

92120

12

112

31

41599

4500

1973002

912

110

12

112

31

'

14

'

4

'

23

'

3

2'

2

1

'

1

219

4

2

RRR

RRR

RR

RR

FORWARD ELIMINATION

12572

12143000

319

37100

291

2110

12

112

31

'

34

'

4

3'

3

2

'

2

1

'

1

45

3

RRR

RR

RR

RR

41599

4500

1973002

912

110

12

112

31

12572

12143000

319

37100

291

2110

12

112

31

1435721000

319

37100

291

2110

12

112

31

121434'

4

3

'

3

2

'

2

1

'

1

RR

RR

RR

RR

BACK SUBSTITUTION

4

143572

319

37

29

21

12

12

3

4

4

43

432

4321

x

x

xx

xxx

xxxx

GAUSS - JOURDAN

39743

22342

15231

6150

8120

15231

'

13

'

3

'

12

'

2

1

'

1

3

2

RRR

RRR

RR

142700

42110

152101

'

23

'

3

2

'

2

'

21

'

1

5

2

3

RRR

RR

RRR

6150

8120

15231

142700

42110

152101

4100

2010

1001

273'

3

'

32

'

2

'

31

'

1

21

21

RR

RRR

RRR

WARNING..

655

901.33099.26

7710

321

123

21

xxx

xxx

xx

Dua kemungkinan kesalahan -Pembagian dengan nol mungkin terjadi pada langkah

forward elimination. Misalkan:

- Kemungkinan error karena round-off (kesalahan pembulatan)

CONTOH

Dari sistem persamaan linear

515

6099.23

0710

3

2

1

x

x

x

6

901.3

7

=

Akhir dari Forward Elimination

1500500

6001.00

0710

3

2

1

x

x

x

15004

001.6

7

=

6

901.3

7

515

6099.23

0710

15004

001.6

7

1500500

6001.00

0710

KESALAHAN YANG MUNGKIN TERJADI

Back Substitution

99993.015005

150043 x

5.1 001.0

6001.6 32

xx

3500.010

077 32

1

xx

x

15004

001.6

7

1500500

6001.00

0710

3

2

1

x

x

x

CONTOH KESALAHAN

Bandung-kan solusi exact dengan hasil perhitungan

99993.0

5.1

35.0

3

2

1

x

x

x

X calculated

1

1

0

3

2

1

x

x

x

X exact

IMPROVEMENTS

Menambah jumlah angka penting

Mengurangi round-off error (kesalahan pembulatan)

Tidak menghindarkan pembagian dengan nol

Gaussian Elimination with Partial Pivoting

Menghindarkan pembagian dengan nol

Mengurangi round-off error

PIVOTING

pka ,npk

Eliminasi Gauss dengan partial pivoting mengubah tata urutan

baris untuk bisa mengaplikasikan Eliminasi Gauss secara Normal

How?

Di awal sebelum langkah ke-k pada forward elimination, temukan angka

maksimum dari:

nkkkkk aaa .......,,........., ,1

Jika nilai maksimumnya Pada baris ke p,

Maka tukar baris p dan k.

PARTIAL PIVOTING

What does it Mean?

Gaussian Elimination with Partial Pivoting ensures that

each step of Forward Elimination is performed with the

pivoting element |akk| having the largest absolute value.

Jadi,

Kita mengecek pada setiap langkah apakah angka

mutlak yang dipakai untuk forward elimination (pivoting

element) adalah selalu paling besar

PARTIAL PIVOTING: EXAMPLE

655

901.36099.23

7710

321

321

21

xxx

xxx

xx

Consider the system of equations

In matrix form

515

6099.23

0710

3

2

1

x

x

x

6

901.3

7

=

Solve using Gaussian Elimination with Partial Pivoting using five

significant digits with chopping

PARTIAL PIVOTING: EXAMPLE

Forward Elimination: Step 1

Examining the values of the first column

|10|, |-3|, and |5| or 10, 3, and 5

The largest absolute value is 10, which means, to follow the

rules of Partial Pivoting, we don’t need to switch the rows

6

901.3

7

515

6099.23

0710

3

2

1

x

x

x

5.2

001.6

7

55.20

6001.00

0710

3

2

1

x

x

x

Performing Forward Elimination

PARTIAL PIVOTING: EXAMPLE

001.6

5.2

7

6001.00

55.20

0710

3

2

1

x

x

x

Forward Elimination: Step 2

Examining the values of the first column

|-0.001| and |2.5| or 0.0001 and 2.5

The largest absolute value is 2.5, so row 2 is switched with

row 3

5.2

001.6

7

55.20

6001.00

0710

3

2

1

x

x

x

Performing the row swap

PARTIAL PIVOTING: EXAMPLE

Forward Elimination: Step 2

Performing the Forward Elimination results in:

002.6

5.2

7

002.600

55.20

0710

3

2

1

x

x

x

PARTIAL PIVOTING: EXAMPLE

002.6

5.2

7

002.600

55.20

0710

3

2

1

x

x

x

Back Substitution

Solving the equations through back substitution

1002.6

002.63 x

15.2

55.2 22

xx

010

077 32

1

xx

x

PARTIAL PIVOTING: EXAMPLE

1

1

0

3

2

1

x

x

x

X exact

1

1

0

3

2

1

x

x

x

X calculated

Compare the calculated and exact solution

The fact that they are equal is coincidence, but it does

illustrate the advantage of Partial Pivoting

SUMMARY

-Forward Elimination

-Back Substitution

-Pitfalls

-Improvements

-Partial Pivoting