Review Operasi Matriks

Menghitung invers matriks?

Determinan?

Matriks Singular?

Menghitung invers matriks

=

10

01

2221

1211

2221

1211

aa

aa

cc

cc

1acac

=

1

0

0

1

22221221

21221121

22121211

21121111

acac

acac

acac

acac

Determinan

� Hanya untuk square matrices

bcaddc

ba

dc

ba−≡≡

det

aaa

Jika determinan = 0 � matriks singular, tidak

punya invers

123213312132231321

321

321

321

det

cbacbacbacbacbacba

ccc

bbb

aaa

−+−+−≡

Cari invers nya"

52

42

42

21

Sistem Persamaan Linear

Simultaneous Linear Equations

Metode Penyelesaian

• Metode grafik

• Eliminasi Gauss

• Metode Gauss – Jourdan

• Metode Gauss – Seidel• Metode Gauss – Seidel

• LU decomposition

Metode Grafik

2

=

− 2

4

11

21

2

1

x

x

-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 solutionNo solutionDet [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

421 x

=

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.

=

321 1x

=

11x

⇒

=

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

Merupakan salah satu teknik paling populer

dalam menyelesaikan sistem persamaan linear

dalam bentuk:

[ ][ ] [ ]CXA =

Terdiri dari dua step

1. Forward Elimination of Unknowns.

2. Back Substitution

[ ][ ] [ ]CXA =

Forward Elimination

Tujuan Forward Elimination adalah untuk membentuk

matriks koefisien menjadi Upper Triangular Matrix

1525 1525

7.000

56.18.40

1525

−−→

112144

1864

1525

Forward Elimination

Persamaan linear

n persamaan dengan n variabel yang tak diketahui

11313212111 ... bxaxaxaxa nn =++++ 11313212111 nn

22323222121 ... bxaxaxaxa nn =++++

nnnnnnn bxaxaxaxa =++++ ...332211

. .

. .

. .

Contoh

8325

1232

7352

2232

4321

4321

4321

4321

=+−+−

=−++−

=+−+

−=−−+

xxxx

xxxx

xxxx

xxxx

matriks input

83125

12312

71352

21232

−

−−

−

−−−

8325 4321 =+−+− xxxx

Forward Elimination

83125

12312

71352

21232

−

−−

−

−−−

3216

2190

13140

92120

1211

231

−

−−

−

−−−

'

14

'

4

'

13

'

3

'

12

'

2

1'

1

5

2

2

2

RRR

RRR

RRR

RR

−=

−=

−=

=

83125 − 32

62

0 −144

5RRR −=

3216

2190

13140

92120

1211

231

−

−−

−

−−−

41599

4500

197300291

2110

1211

231

−−−−−

−

−−−

'

14

'

4

'

23

'

3

2'

2

1

'

1

219

4

2

RRR

RRR

RR

RR

+=

+=

=

=

Forward Elimination

12572

12143000

319

37100

291

2110

1211

231

−−

−−

−

−−−

'

34

'

4

3'

3

2

'

2

1

'

1

45

3

RRR

RR

RR

RR

+=

=

=

=

41599

4500

197300291

2110

1211

231

−−−−−

−

−−−

344 4RRR +=

12572

12143000

319

37100

291

2110

1211

231

−−

−−

−

−−−

1435721000

319

37100

291

2110

1211

231

−−

−

−−−

121434'

4

3

'

3

2

'

2

1

'

1

−=

=

=

=

RR

RR

RR

RR

Back substitution

29

21

121

23

432

4321

=+−

−=−−+

xxx

xxxx

4

143572

319

37

22

4

4

43

432

=

=

−=−

=+−

x

x

xx

xxx

Gauss - Jourdan

39743

22342

15231

6150

8120

15231

−−

−−−'

13

'

3

'

12

'

2

1

'

1

3

2

RRR

RRR

RR

−=

−=

=

152101

'

21

'

13RRR −=15231

142700

42110

152101

'

23

'

3

2

'

2

211

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..

Dua kemungkinan kesalahan-Pembagian dengan nol mungkin terjadi pada langkah forward

elimination. Misalkan:

771021=− xx

655

901.33099.26

7710

321

123

21

=+−

=−+

=−

xxx

xxx

xx

- Kemungkinan error karena round-off (kesalahan pembulatan)

Contoh

Dari sistem persamaan linear

−

−

−

515

6099.23

0710

2

1

x

x

x

6

901.3

7

=

−

−

−

6

901.3

7

515

6099.23

0710

− 515 3x 6

Akhir dari Forward Elimination

−

−

1500500

6001.00

0710

3

2

1

x

x

x

15004

001.6

7

=

− 6515

−

−

15004

001.6

7

1500500

6001.00

0710

Kesalahan yang mungkin terjadi

Back Substitution

99993.015005

150043 ==x

=

−

−

001.6

7

6001.00

0710 1

x

x

5.1 001.0

6001.63

2−=

−

−=

xx

3500.010

07732

1−=

−+=

xxx

=

−

15004

001.6

1500500

6001.00

3

2

x

x

Contoh kesalahan

Bandung-kan solusi exact dengan hasil perhitungan

[ ]

−=

= 1

0

2

1

x

x

X exact

[ ]

−

−

=

=

99993.0

5.1

35.0

3

2

1

x

x

x

X calculated

[ ]

−=

=

1

1

3

2

x

xX 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

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

pka

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

maksimum dari:

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

Jika nilai maksimumnya Pada baris ke p, ,npk ≤≤

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 paling

atas (pivoting element) adalah selalu paling besar

Partial Pivoting: Example

Consider the system of equations

655

901.36099.23

7710

321

21

=+−

=++−

=−

xxx

xxx

xx

655 321 =+− xxx

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

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

⇒

=

−

−

001.6

5.2

7

6001.00

55.20

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

Back Substitution

Solving the equations through back substitution

1002.6

002.63 ==x

=

−

5.2

7

55.20

0710 1

x

x

002.6

15.2

55.2 2

2 =−

=x

x

010

077 32

1 =−+

=xx

x

=

002.6

5.2

002.600

55.20

3

2

x

x

Partial Pivoting: Example

Compare the calculated and exact solution

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

the advantage of Partial Pivoting

[ ]

−=

=

1

1

0

3

2

1

x

x

x

X exact[ ]

−=

=

1

1

0

3

2

1

x

x

x

X calculated

Summary

-Forward Elimination

-Back Substitution

-Pitfalls

-Improvements

-Partial Pivoting