REVIEW OPERASI MATRIKS T E K N I K L I N G K U N G A N I T B

REVIEW OPERASI MATRIKS

MENGHITUNG INVERS MATRIKS

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1

0

0

1

22 22 12 21 21 22 11 21 22 12 12 11 21 12 11 11

a

c

a

c

a

c

a

c

a

c

a

c

a

c

a

c

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1

0

0

1

22

21

12

11

22

21

12

11

a

a

a

a

c

c

c

c

DETERMINAN



Hanya untuk square matrices

Jika determinan = 0  matriks singular, tidak punya

invers

bc

ad

d

c

b

a

d

c

b

a



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

det

1 2 3 2 1 3 3 1 2 1 3 2 2 3 1 3 2 1 3 2 1 3 2 1 3 2 1

det

c

b

a

c

b

a

c

b

a

c

b

a

c

b

a

c

b

a

c

c

c

b

b

b

a

a

a

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CARI INVERS NYA…

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5

2

4

2

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4

2

2

1

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

METODE PENYELESAIAN

•

Metode grafik

•

Eliminasi Gauss

•

Metode Gauss – Jourdan

•

Metode Gauss – Seidel

•

LU decomposition

METODE GRAFIK

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

2

4

1

1

2

1

2 1

x

x

2 -2

Det{A}



0



A

is nonsingular

so invertible

Unique

solution

SISTEM PERSAMAAN YANG TAK

TERSELESAIKAN

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5

4

4

2

2

1

2 1

x

x

No solution

Det [A] = 0,

but system is inconsistent

Then this

system of

equations is

SISTEM DENGAN SOLUSI TAK

TERBATAS

Det{A} = 0



A is singular

infinite number of solutions

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8

4

4

2

2

1

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.

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47

.

1

3

99

.

0

48

.

0

2

1

2 1

x

x

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47

.

1

3

99

.

0

49

.

0

2

1

2 1

x

x

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1

1

2 1

x

x



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

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0

3

2 1

x

x



GAUSSIAN ELIMINATION

    

A

X



C

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

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1

12

144

1

8

64

1

5

25

Tujuan Forward Elimination adalah untuk

membentuk

matriks koefisien

menjadi Upper

Triangular Matrix

7

.

0

0

0

56

.

1

8

.

4

0

1

5

25

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FORWARD ELIMINATION

Persamaan linear

n persamaan dengan n variabel yang tak diketahui

1 1 3 13 2 12 1 11

x

a

x

a

x

...

a

x

b

a









_{n n}



2 2 3 23 2 22 1 21

x

a

x

a

x

...

a

x

b

a









_{n n}



n n nn n n n

x

a

x

a

x

a

x

b

a

_{1 1}



_{2 2}



_{3 3}



...





. .

. .

. .

CONTOH

8

3

1

2

5

1

2

3

1

2

7

1

3

5

2

2

1

2

3

2

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

8

3

2

5

1

2

3

2

7

3

5

2

2

2

3

2

4 3 2 1 4 3 2 1 4 3 2 1 4 3 2 1



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x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

matriks input

FORWARD ELIMINATION

8

3

1

2

5

1

2

3

1

2

7

1

3

5

2

2

1

2

3

2

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3

2

1

6

2

19

0

1

3

1

4

0

9

2

1

2

0

1

2

1

1

2

3

1

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







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





' 1 4 ' 4 ' 1 3 ' 3 ' 1 2 ' 2 1 ' 1

5

2

2

2

R

R

R

R

R

R

R

R

R

R

R

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









3

2

1

6

2

19

0

1

3

1

4

0

9

2

1

2

0

1

2

1

1

2

3

1

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

4

159

9

4

5

0

0

19

7

3

0

0

2

9

1

2

1

1

0

1

2

1

1

2

3

1

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

' 1 4 ' 4 ' 2 3 ' 3 2 ' 2 1 ' 1

2

19

4

2

R

R

R

R

R

R

R

R

R

R

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

FORWARD ELIMINATION

12 572 12 143 0 0 0 3 19 3 7 1 0 0 2 9 1 2 1 1 0 1 2 1 1 2 3 1                         ' 3 4 ' 4 3 ' 3 2 ' 2 1 ' 1

4

5

3

R

R

R

R

R

R

R

R

R



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

4 159 9 4 5 0 0 19 7 3 0 0 2 9 1 2 1 1 0 1 2 1 1 2 3 1                        12 572 12 143 0 0 0 3 19 3 7 1 0 0 2 9 1 2 1 1 0 1 2 1 1 2 3 1                         143 572 1 0 0 0 3 19 3 7 1 0 0 2 9 1 2 1 1 0 1 2 1 1 2 3 1                       12 143 4 ' 4 3 ' 3 2 ' 2 1 ' 1 



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R

R

R

R

R

R

R

R

BACK SUBSTITUTION

4

143

572

3

19

3

7

2

9

2

1

1

2

1

2

3

4 4 4 3 4 3 2 4 3 2 1



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

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

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

x

x

x

x

x

x

x

x

x

x

x

GAUSS - JOURDAN

39

7

4

3

22

3

4

2

15

2

3

1

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6

1

5

0

8

1

2

0

15

2

3

1



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



' 1 3 ' 3 ' 1 2 ' 2 1 ' 1

3

2

R

R

R

R

R

R

R

R



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





14

2

7

0

0

4

2

1

1

0

15

2

1

0

1



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







' 2 3 ' 3 2 ' 2 ' 2 1 ' 1

5

2

3

R

R

R

R

R

R

R

R













6

1

5

0

8

1

2

0

15

2

3

1



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

















14

2

7

0

0

4

2

1

1

0

15

2

1

0

1



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







4

1

0

0

2

0

1

0

1

0

0

1



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

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











2 7 3 ' 3 ' 3 2 ' 2 ' 3 1 ' 1

2

1

2

1

R

R

R

R

R

R

R

R











WARNING..

6

5

5

901

.

3

3

099

.

2

6

7

7

10

3 2 1 1 2 3 2 1

















x

x

x

x

x

x

x

x

Dua kemungkinan kesalahan

-Pembagian dengan nol mungkin terjadi pada langkah

forward elimination. Misalkan:

CONTOH

Dari sistem persamaan linear

             5 1 5 6 099 . 2 3 0 7 10           3 2 1 x x x           6 901 . 3 7

=

Akhir dari Forward Elimination

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

15005

0

0

6

001

.

0

0

0

7

10





















3 2 1

x

x

x

          15004 001 . 6 7

=

             6 901 . 3 7 5 1 5 6 099 . 2 3 0 7 10



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

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













15004

001

.

6

7

15005

0

0

6

001

.

0

0

0

7

10

KESALAHAN YANG MUNGKIN TERJADI

Back Substitution

99993

.

0

15005

15004

3





x

5

.

1

001

.

0

6

001

.

6

₃ 2











x

x

3500

.

0

10

0

7

7

_{2 3} 1











x

x

x



































































15004

001

.

6

7

15005

0

0

6

001

.

0

0

0

7

10

3 2 1

x

x

x

CONTOH KESALAHAN

Band

u

ng-kan solusi exact dengan hasil perhitungan

 

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























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

pk

a k pn,

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: nk k k kk

a

a

a

,

_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 |a

_kk

| 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

6

5

5

901

.

3

6

099

.

2

3

7

7

10

3 2 1 3 2 1 2 1



















x

x

x

x

x

x

x

x

Consider the system of equations

In matrix form

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









5

1

5

6

099

.

2

3

0

7

10





















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

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





































6

901

.

3

7

5

1

5

6

099

.

2

3

0

7

10

3 2 1

x

x

x



































































5

.

2

001

.

6

7

5

5

.

2

0

6

001

.

0

0

0

7

10

3 2 1

x

x

x



PARTIAL PIVOTING: EXAMPLE

                                 001 . 6 5 . 2 7 6 001 . 0 0 5 5 . 2 0 0 7 10 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

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5

.

2

001

.

6

7

5

5

.

2

0

6

001

.

0

0

0

7

10

3 2 1

x

x

x



PARTIAL PIVOTING: EXAMPLE

Forward Elimination: Step 2

Performing the Forward Elimination results in:

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002

.

6

5

.

2

7

002

.

6

0

0

5

5

.

2

0

0

7

10

3 2 1

x

x

x

PARTIAL PIVOTING: EXAMPLE

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002

.

6

5

.

2

7

002

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6

0

0

5

5

.

2

0

0

7

10

3 2 1

x

x

x

Back Substitution

Solving the equations through back substitution

1

002

.

6

002

.

6

3

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x

1

5

.

2

5

5

.

2

₂ 2

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x

x

0

10

0

7

7

_{2 3} 1

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x

x

x

PARTIAL PIVOTING: EXAMPLE

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1

1

0

3 2 1

x

x

x

X

_exact

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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