Aljabar matriks

ANALISIS STRUKTUR

DENGAN METODE MATRIKS

Mengapa Metode Matriks?

ALJABAR MATRIKSPertemuan Minggu I dan II

Analisis Struktur Metode Matriks

Ashar Saputra, PhD

Topik Paparan

• Aljabar matrix

• Definisi matrix

• Macam-macam matrix

• Operasi matrix

• Matrix orthogonal

• Aljabar matrix

• Solusi persamaan linier simultan

• Matrix partisi

• Program komputer untuk operasi matrix

Definisi Matriks

• Matriks adalah suatu susunan elemen dengan dobel

subscribe yang disusun dalam baris dan kolom

ij

mnm

n

n

A

aa

aa

aa

,,

,,

,,

1

221

111

A

Vektor baris

[1 x n] matrix

jn aaaaA ,, 2 ,1

Vektor kolom

i

m

a

a

a

a

A 2

1

[m x 1] matrix

Matrik bujur sangkar

B

5 4 7

3 6 1

2 1 3

Jumlah baris dan kolom sama

Macam-macam Matriks bujur sangkar

•


•


•


•

OPERASI MATRIKS

Penjumlahan

•

Sifat-sifat penjumlahan

• [A] + [B] = [B] + [A] komutatif

• [A] + [B] + [C] = ([A] + [B]) + [C]

PERKALIAN DENGAN SKALAR

3 2 5

-2 0 1

4 6 9

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9 6

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9 6 15

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9 6 15

-6

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9 6 15

-6 0

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9 6 15

-6 0 3

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9 6 15

-6 0 3

12

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9 6 15

-6 0 3

12 18

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9 6 15

-6 0 3

12 18 27

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

9 6 15

-6 0 3

12 18 27

Jeff Bivin -- LZHS

PENJUMLAHAN

(LANJUTAN)

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

=

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

+=

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

+ 26

=6 + 20

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

26 0

=+

4 - 4

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

26 0 18

=+

10 + 8

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

26 0 18

-4=+

-4 + 0

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

26 0 18

-4 12=+

0 + 12

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

26 0 18

-4 12 30=+

2 + 28

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

26 0 18

-4 12 30

16

=+

8 + 8

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

26 0 18

-4 12 30

16 40

=+

12 + 28

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

26 0 18

-4 12 30

16 40 34

=+

18 + 16

Jeff Bivin -- LZHS

Matrix Addition & Scalars

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

26 0 18

-4 12 30

16 40 34

=

Jeff Bivin -- LZHS

PERKALIAN MATRIKS

DENGAN MATRIKS LAIN

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

Jeff Bivin -- LZHS

Matrix Multiplication

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

Jeff Bivin -- LZHS

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

Jeff Bivin -- LZHS

CONTOH PERHITUNGAN

Jeff Bivin -- LZHS

Row 1 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 x 5 + 2 x 0 + 5 x 2 = 25

Jeff Bivin -- LZHS

Row 1 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 x 5 + 2 x 0 + 5 x 2 = 25

Jeff Bivin -- LZHS

Row 1 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 x 5 + 2 x 0 + 5 x 2 = 25

Jeff Bivin -- LZHS

Row 1 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25

=

Jeff Bivin -- LZHS

Row 1 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 x (-1) + 2 x 3 + 5 x 7 = 38

Jeff Bivin -- LZHS

Row 1 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 x (-1) + 2 x 3 + 5 x 7 = 38

Jeff Bivin -- LZHS

Row 1 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 x (-1) + 2 x 3 + 5 x 7 = 38

Jeff Bivin -- LZHS

Row 1 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38

=

Jeff Bivin -- LZHS

Row 1 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 x 2 + 2 x 7 + 5 x 4 = 40

Jeff Bivin -- LZHS

Row 1 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 x 2 + 2 x 7 + 5 x 4 = 40

Jeff Bivin -- LZHS

Row 1 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

3 x 2 + 2 x 7 + 5 x 4 = 40

Jeff Bivin -- LZHS

Row 1 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

=

Jeff Bivin -- LZHS

Row 2 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

-2 x 5 + 0 x 0 + 1 x 2 = -8

Jeff Bivin -- LZHS

Row 2 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

-2 x 5 + 0 x 0 + 1 x 2 = -8

Jeff Bivin -- LZHS

Row 2 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

-2 x 5 + 0 x 0 + 1 x 2 = -8

Jeff Bivin -- LZHS

Row 2 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8=

Jeff Bivin -- LZHS

Row 2 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

Jeff Bivin -- LZHS

Row 2 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

-2 x (-1) + 0 x 3 + 1 x 7 = 9

Jeff Bivin -- LZHS

Row 2 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

-2 x (-1) + 0 x 3 + 1 x 7 = 9

Jeff Bivin -- LZHS

Row 2 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

-2 x (-1) + 0 x 3 + 1 x 7 = 9

Jeff Bivin -- LZHS

Row 2 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9=

Jeff Bivin -- LZHS

Row 2 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

-2 x 2 + 0 x 7 + 1 x 4 = 0

Jeff Bivin -- LZHS

Row 2 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

-2 x 2 + 0 x 7 + 1 x 4 = 0

Jeff Bivin -- LZHS

Row 2 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

-2 x 2 + 0 x 7 + 1 x 4 = 0

Jeff Bivin -- LZHS

Row 2 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0=

Jeff Bivin -- LZHS

Row 3 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

4 x 5 + 6 x 0 + 9 x 2 = 38

Jeff Bivin -- LZHS

Row 3 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

4 x 5 + 6 x 0 + 9 x 2 = 38

Jeff Bivin -- LZHS

Row 3 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

4 x 5 + 6 x 0 + 9 x 2 = 38

Jeff Bivin -- LZHS

Row 3 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38

=

Jeff Bivin -- LZHS

Row 3 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

4 x (-1) + 6 x 3 + 9 x 7 = 77

Jeff Bivin -- LZHS

Row 3 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

4 x (-1) + 6 x 3 + 9 x 7 = 77

Jeff Bivin -- LZHS

Row 3 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

4 x (-1) + 6 x 3 + 9 x 7 = 77

Jeff Bivin -- LZHS

Row 3 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38 77

=

Jeff Bivin -- LZHS

Row 3 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

4 x 2 + 6 x 7 + 9 x 4 = 86

Jeff Bivin -- LZHS

Row 3 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

4 x 2 + 6 x 7 + 9 x 4 = 86

Jeff Bivin -- LZHS

Row 3 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

4 x 2 + 6 x 7 + 9 x 4 = 86

Jeff Bivin -- LZHS

Row 3 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38 77 86

Jeff Bivin -- LZHS


3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38 77 86

=

Jeff Bivin -- LZHS


3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38 77 86

=

Jeff Bivin -- LZHS


3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38 77 86

=

Jeff Bivin -- LZHS

Review

Jeff Bivin -- LZHS

Row 1 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25

=

Jeff Bivin -- LZHS

Row 1 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38

=

Jeff Bivin -- LZHS

Row 1 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

=

Jeff Bivin -- LZHS

Row 2 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8=

Jeff Bivin -- LZHS

Row 2 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9=

Jeff Bivin -- LZHS

Row 2 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0=

Jeff Bivin -- LZHS

Row 3 x Column 1

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38

=

Jeff Bivin -- LZHS

Row 3 x Column 2

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38 77

=

Jeff Bivin -- LZHS

Row 3 x Column 3

3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38 77 86

=

Jeff Bivin -- LZHS


3 2 5

-2 0 1

4 6 9

5 -1 2

0 3 7

2 7 4

X

25 38 40

-8 9 0

38 77 86

=

Jeff Bivin -- LZHS

Determinant dari matriks 2 x 2

a b

c d

Jeff Bivin -- LZHS

Determinant of a 2 x 2 Matrix

3 5

4 6

Jeff Bivin -- LZHS


-4 3

5 2

Jeff Bivin -- LZHS


8 4

6 5

Jeff Bivin -- LZHS

Determinant dari matriks 3 x 3

1 3 4

2 1 5

3 6 7

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

Jeff Bivin -- LZHS


1 3 4 1

2 1 5 2

3 6 7 3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


1 3 4 1 3

2 1 5 2 1

3 6 7 3 6

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

7 + 45 + 48 - 12 - 30 - 42

= 16Jeff Bivin -- LZHS


2 1 6 2

4 3 7 4

5 9 8 5

Jeff Bivin -- LZHS


2 1 6 2

4 3 7 4

5 9 8 5

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

1•1•7 + 3•5•3 + 4•2•6 - 3•1•4 - 6•5•1 - 7•2•3

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

Jeff Bivin -- LZHS


2 1 6 2 1

4 3 7 4 3

5 9 8 5 9

2•3•8 + 1•7•5 + 6•4•9 - 5•3•6 - 9•7•2 - 8•4•1

48 + 35 + 216 - 90 - 126 - 32

= 51Jeff Bivin -- LZHS


Choose a

row or

a colum

1 3 4

2 1 5

3 6 7

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

We will use the first column

to give us our cofactors

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

1 5

6 7

3 4

6 7

3 4

1 5

Notice the

alternating signs

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

1 5

6 7

3 4

6 7

3 4

1 5

Now for the minors

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

1 5

6 7

Remove the row and the

column of the cofactor

element

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

1 5

6 7

3 4

6 7



element

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

1 5

6 7

3 4

6 7

3 4

1 5



element

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

1 5

6 7

3 4

6 7

3 4

1 51(7 – 30) – 2(21 – 24) + 3(15 –

4)1(-23) – 2(–3) + 3(11) = -23 + 6 + 33 =

16

= 16Evaluate each 2x2

determinant and simplify

Jeff Bivin -- LZHS

NOW TRY A DIFFERENT

ROW OR COLUMN

Jeff Bivin -- LZHS


Choose a new

row or

a colum

1 3 4

2 1 5

3 6 7

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

This time we will use the

second row to give us our

cofactors

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

1 5

6 7

3 4

6 7

3 4

1 5

Again we have

alternating signs

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

1 5

6 7

3 4

6 7

3 4

1 5

Now for the minors

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7



element

3 4

6 7

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7



element

3 4

6 7

1 4

3 7

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7



element

3 4

6 7

1 4

3 7

1 3

3 6

Jeff Bivin -- LZHS


1 3 4

2 1 5

3 6 7

3 4

6 7

1 4

3 7

1 3

3 6-2(21 – 24) + 1(7 – 12) - 5(6 –

9)-2(-3) + 1(–5) - 5(-3) = 6 - 5 + 15 = 16

= 16Evaluate each 2x2

determinant and simplify

Jeff Bivin -- LZHS


1 -2 3 4

2 -5 1 5

2 3 -1 4

3 1 6 7

Select a row or column

to use as the co-

factors.

Jeff Bivin -- LZHS


1 -2 3 4

2 -5 1 5

2 3 -1 4

3 1 6 7

Let’s use the first row for the

co-factors

Jeff Bivin -- LZHS

1 3 4-2


1 -2 3 4

2 -5 1 5

2 3 -1 4

3 1 6 7

= 16Remember the alternating

signs.

Jeff Bivin -- LZHS

1 -2 3 4


1 -2 3 4

2 -5 1 5

2 3 -1 4

3 1 6 7

-5 1 5

3 -1 4

1 6 7

Jeff Bivin -- LZHS

+ 3 - 4



element

1 + 2


1 -2 3 4

2 -5 1 5

2 3 -1 4

3 1 6 7

-5 1 5

3 -1 4

1 6 7

Jeff Bivin -- LZHS

1 + 22 1 5

2 -1 4

3 6 7



element

+ 3 - 4


1 -2 3 4

2 -5 1 5

2 3 -1 4

3 1 6 7

-5 1 5

3 -1 4

1 6 7

Jeff Bivin -- LZHS

1 + 22 -5 5

2 3 4

3 1 7

2 1 5

2 -1 4

3 6 7



element

+ 3 - 4


1 -2 3 4

2 -5 1 5

2 3 -1 4

3 1 6 7

-5 1 5

3 -1 4

1 6 7

Jeff Bivin -- LZHS

2 1 5

2 -1 4

3 6 7

2 -5 5

2 3 4

3 1 7

2 -5 1

2 3 -1

3 1 6

1 + 2



element

+ 3 - 4


1 -2 3 4

2 -5 1 5

2 3 -1 4

3 1 6 7

-5 1 5

3 -1 4

1 6 7

Jeff Bivin -- LZHS

2 1 5

2 -1 4

3 6 7

2 -5 5

2 3 4

3 1 7

2 -5 1

2 3 -1

3 1 6

1 + 2

Now evaluate the

3x3 determinants --- more

expansion by

co-factors and minors

1(233) + 2(11) + 3(9) -

4(106)

= -142

+ 3 - 4

= -142

INVERS DAN MATRIKS

IDENTITAS

Invers Matiks Bujur Sangkar

• Untuk mencari inverse matriks dapat dipakai beberapa

metoda, antara lain metode adjoint, metode pemisahan,

Gauss-Jordan, Chelosky, dsb.

• Sebagai contoh metode Gauss-Jordan

Invers Matiks Bujur Sangkar

•

Inverses and Identities

5x = 351

51

531 x

53x

inversestivemultiplicaareand 551

identitytivemultiplicatheis1

Jeff Bivin -- LZHS

10

01

Now with Matrices

43

21

43

21

This is the

Identity Matrix

for 2 x 2 Matrices

Jeff Bivin -- LZHS

Let’s look at another example

53

87

10

01

53

87

Jeff Bivin -- LZHS

New Question

43

21

10

01

What do we

multiply a matrix

by to get the

Identity?

Jeff Bivin -- LZHS

The Inverse of a 2x2 Matrix

dc

ba

dc

ba

1

ac

bd

0dc

ba

Jeff Bivin -- LZHS


43

21

43

21

1

13

24

13

2464

1

2

1

21

23

12

Jeff Bivin -- LZHS


21

34

21

34

1

41

32

41

3238

1

11

1

114

111

113

112

Jeff Bivin -- LZHS

:' usesLet

31

25

43

21X

BAX

BAX 1A

BAXI 1

1A

BAX 1

This is our Formula!

31

25

13

24

43

21

1X2

1

914

141821X

297

79X

Jeff Bivin -- LZHS

:' usesLet

34

27

54

31X

BXA

BXA 1A1 ABIX

1A

1 ABX


14

35

34

27

54

31

1X

1532

2343

7

1X

715

732

723

743

7

1

14

35

34

27

7

1X

Jeff Bivin -- LZHS

:' usesLet

63

14

41

32

32

51X

CBAX

)( BCAX 1A

BCAXI 1

1A

BCAX 1


41

32

63

14

12

53

32

51

1X7

1

24

42

12

5371X

68

22671X

BCAX

76

78

72

726

Jeff Bivin -- LZHS

Are the two Matrices Inverses?

85

32

25

38

10

01

Jeff Bivin -- LZHS

The product of

inverse matrices

is the identity

matrix.

Identity,

therefore,

INVERSE

Matrices

Are the two Matrices Inverses?

41

23

31

24

100

010

Jeff Bivin -- LZHS

The product of

inverse matrices

is the identity

matrix.

Not the

Identity,

therefore,

NOT

INVERSE

Matrices

46

23

Jeff Bivin -- LZHS

Does the Matrix have an Inverse?

Let’s review

the definition of

the Inverse of a

2x2 Matrix


dc

ba

dc

ba

1

ac

bd

0dc

ba

Jeff Bivin -- LZHS

46

23Find the determinant!

46

23

Jeff Bivin -- LZHS

01212

Therefore, NO inverse!


144

27Find the determinant!

144

27

Jeff Bivin -- LZHS

90898

Therefore, an inverse exists!


987

654

321

Find the determinant!

Jeff Bivin -- LZHS

Therefore, NO inverse!


1 2 3 1 2

4 5 6 4 5

7 8 9 7 8

1•5•9 + 2•6•7 + 3•4•8 - 7•5•3 - 8•6•1 - 9•4•2

45 + 84 + 96 - 105 - 48 - 72

243

142

231

Find the determinant!

Jeff Bivin -- LZHS


1 3 2 1 3

2 4 1 2 4

3 4 2 3 4

1•4•2 + 3•1•3 + 2•2•4 - 3•4•2 - 4•1•1 - 2•2•3

8 + 9 + 16 - 24 - 4 - 12

Therefore, an inverse exists!

SOLUSI PERSAMAAN

LINEAR SIMULTAN

SOLUSI PERSAMAAN LINIER SIMULTAN

Persamaan Linier Simultan dengan n buah bilangan tak diketahui dapat dituliskan sebagai

berikut:

a11 x1 + a12 x2 + ⋯⋯+ a1n xn = b1

a11 x1 + a12 x2 + ⋯⋯+ a1n xn = b2

⋮ ⋮ ⋮ ⋮

an1 x1 + an2 x2 + ⋯⋯+ ann xn = bn

Secara matrix, persamaan-persamaan tersebut, bisa ditulis:

𝑎11

𝑎21

⋮𝑎𝑛1

𝑎12

𝑎22

⋮𝑎𝑛2

……

…

𝑎1𝑛

𝑎2𝑛

⋮𝑎𝑛𝑛

x1

x2

⋮xn

=

b1

b2

⋮bn

atau, secara lebih sederhana:

[A] {X} = {B}

[A] : matrix bujur sangkar koefisien persamaan linear

{X} : matrix kolo, dari bilangan yang tak diketahui

{B} : matrix kolom dari konstanta


Penyelesaian dengan metode eliminasi Gauss

Dengan cara “OPERASI BARIS”, buatlah matrix [A] menjadi “UPPER TRIANGULAR

MATRIX” (suatu matrix bujur sangkar dimana semua elemen di bawah diagonal

utama sama dengan 0)

Dengan cara eliminasi/”back substitution”, bilangan-bilangan tak diketahui dapat

diperoleh.

Contoh:

4x + 3y + z = 13

x + 2y + 3z = 14

3x + 2y + 5z = 22

4 3 11 2 33 2 5

xyz =

131422

[A] {X} = {B}

413

322

135

||| 131422

;

1

0

3

34

2

2

14

3

5

|

|

|

134

14

22

1

0

0

34

54

−14

14

114

174

|

|

|

134

434

494

;

1

0

0

34

1

−14

14

115

174

|

|

|

134

435

494

1

0

0

34

1

0

14

115

245

|

|

|

134

435

725

z = 72

5 .

5

24= 3

y = 43

5 −

11

5 . 3 = 2

x = 13

4 −

3

4 . 2 −

1

4 . 3 = 1


:' usesLet

11

7

54

23

y

x

BAU

BAU 1A

BAUI 1

1A

BAU 1


11

7

34

25

54

23

1U23

1

5

5723

1U

23

5

2357

Jeff Bivin -- LZHS

3x + 2y = 7

4x - 5y = 11

Solve the system

using inverse matrices

23

52357 , solution

:' usesLet

1

9

23

42

y

x

BAU

BAU 1A

BAUI 1

1A

BAU 1


1

9

23

42

23

42

1U8

1

25

1481U

825

47

Jeff Bivin -- LZHS

2x - 4y = 9

3x - 2y = 1

Solve the system

using inverse matrices

825

47 , solution

CONTOH GAUSS

Jeff Bivin -- LZHS

x + y + z = 6

4x – 8y + 4z = 12

2x – 3y + 4z = 3

1 1 1 6

4 -8 4 12

2 -3 4 3

Jeff Bivin -- LZHS

1 1 1 6

4 -8 4 12

2 -3 4 3

I am a 1.

Jeff Bivin -- LZHS

1 1 1 6

4 -8 4 12

2 -3 4 3

I need to

be 0.

I need to

be 0.

Jeff Bivin -- LZHS

1 1 1 6

0 -12 0 -12

0 -5 2 -9

4 - 4(1)

-8 - 4(1)

4 - 4(1)

12 - 4(6)

2 - 2(1)

-3 - 2(1)

4 - 2(1)

3 - 2(6)

12 4RR

13 2RR

1 1 1 6

0 -12 0 -12

0 -5 2 -9

I need to

be 1

Jeff Bivin -- LZHS

2121 R

1 1 1 6

0 1 0 1

0 -5 2 -9

1 1 1 6

0 1 0 1

0 -5 2 -9

I need to

be 0.

I need to

be 0.

Jeff Bivin -- LZHS

1 0 1 5

0 1 0 1

0 0 2 -4

1 - 0

1 - 1

1 - 0

6 - 1

0 + 5(0)

-5 + 5(1)

2 + 5(0)

-9 + 5(1)

21 RR

23 5RR

1 0 1 5

0 1 0 1

0 0 2 -4

I need to

be 1

Jeff Bivin -- LZHS

321 R

1 0 1 5

0 1 0 1

0 0 1 -2

1 0 1 5

0 1 0 1

0 0 1 -2

I need to

be 0.

I am a 0

Jeff Bivin -- LZHS

1 0 0 7

0 1 0 1

0 0 1 -2

1 - 0

0 - 0

1 - 1

5 – (-2)

31 RR

1 0 0 7

0 1 0 1

0 0 1 -2

x = 7

y = 1

z = -2

Reading the Solution

Jeff Bivin -- LZHS

Writing the Solution

x + y + z = 6

4x – 8y + 4z = 12

2x – 3y + 4z = 3

Jeff Bivin -- LZHS

EXAMPLE 2

Jeff Bivin -- LZHS

x + y + z = -2

2x - 3y + z = -11

-x + 2y - z = 8

1 1 1 -2

2 -3 1 -11

-1 2 -1 8

Jeff Bivin -- LZHS

1 1 1 -2

2 -3 1 -11

-1 2 -1 8

I am a 1.

Jeff Bivin -- LZHS

1 1 1 -2

2 -3 1 -11

-1 2 -1 8

I need to

be 0.

I need to

be 0.

Jeff Bivin -- LZHS

1 1 1 -2

0 -5 -1 -7

0 3 0 6

2 - 2(1)

-3 - 2(1)

1 - 2(1)

-11 - 2(-2)

-1 + 1

2 + 1

-1 + 1

8 + (-2)

12 2RR

13 RR

1 1 1 -2

0 -5 -1 -7

0 3 0 6

I would prefer to make the 3 a one in row

three rather than the -5 in row 2. Why?

Jeff Bivin -- LZHS

1 1 1 -2

0 3 0 6

0 -5 -1 -7

To

avoid

fractions!

We will

switch Row

2 and Row 3

1 1 1 -2

0 3 0 6

0 -5 -1 -7

I need to

be 1

Jeff Bivin -- LZHS

231 R

1 1 1 -2

0 1 0 2

0 -5 -1 -7

1 1 1 -2

0 1 0 2

0 -5 -1 -7

I need to

be 0.

I need to

be 0.

Jeff Bivin -- LZHS

1 0 1 -4

0 1 0 2

0 0 -1 3

1 - 0

1 - 1

1 - 0

-2 - 2

0 + 5(0)

-5 + 5(1)

-1 + 5(0)

-7 + 5(2)

21 RR

23 5RR

1 0 1 -4

0 1 0 2

0 0 -1 3

I need to

be 1

Jeff Bivin -- LZHS

31R

1 0 1 -4

0 1 0 2

0 0 1 -3

1 0 1 -4

0 1 0 2

0 0 1 -3

I need to

be 0.

I am a 0

Jeff Bivin -- LZHS

1 0 0 -1

0 1 0 2

0 0 1 -3

1 - 0

0 - 0

1 - 1

-4 – (-3)

31 RR

1 0 0 -1

0 1 0 2

0 0 1 -3

x = -1

y = 2

z = -3

Reading the Solution

Jeff Bivin -- LZHS

Writing the Solution

Jeff Bivin -- LZHS

x + y + z = -2

2x - 3y + z = -11

-x + 2y - z = 8

AX + BY = C

DX + EY = F

43

51

49

26

84

31YX

73

21

30

19

23

75YX

AX + BY = C DX + EY = F

AX = C - BY DX = F - EY

X = A-1(C – BY) X = D-1(F – EY)

A-1(C – BY) = D-1(F – EY)

A-1C – A-1BY = D-1F – D-1EY

D-1EY – A-1BY = D-1F – A-1C

(D-1E – A-1B)Y = (D-1F – A-1C)

Y = (D-1E – A-1B)-1(D-1F – A-1C)

X = X

AX + BY = C

BY = C - AX EY = F - DX

Y = B-1(C – AX) Y = E-1(F – DX)

B-1(C – AX) = E-1(F – DX)

B-1C – B-1AX = E-1F – E-1DX

E-1DX – B-1AX = E-1F – B-1C

(E-1D – B-1A)X = (E-1F – B-1C)

X = (E-1D – B-1A)-1(E-1F – B-1C)

DX + EY = F

Y = Y

43

51

49

26

84

31YX

73

21

30

19

23

75YX

Y = (D-1E – A-1B)-1(D-1F – A-1C)

X = (E-1D – B-1A)-1(E-1F – B-1C)

73359

739

73210

7350

X

73141

7329

219392

21977

Y

AX + BY = C DX + EY = F

AX = C - BY

X = A-1(C – BY)

D[ A-1(C – BY) ] + EY= F

D[A-1C – A-1BY] + EY = F

DA-1C – DA-1BY + EY = F

EY – DA-1BY = F - DA-1C

Y = (E – DA-1B)-1(F - DA-1C)

(E – DA-1B)Y = F - DA-1C

AX + BY = C

BY = C - AX

Y = B-1(C – AX)

DX + E [ B-1(C - AX ] = F

DX + E [ B-1C - B-1AX ] = F

DX + EB-1C - EB-1AX = F

DX - EB-1AX = F - EB-1C

X = (D - EB-1A)-1(F - EB-1C)

DX + EY = F

(D - EB-1A)X = F - EB-1C

43

51

49

26

84

31YX

73

21

30

19

23

75YX

X = (D - EB-1A)-1(F - EB-1C)

Y = (E – DA-1B)-1(F - DA-1C)

73359

739

73210

7350

X

73141

7329

219392

21977

Y

ALJABAR MATRIKS

(LANJUTAN)Ashar Saputra, PhD

Transpose Matriks

•

Matriks Ortogonal

•

Teori Dekomposisi MatriksBila : [A] = sebuah matrix bujur sangkar =

𝑎11 𝑎21 𝑎31

⋮𝑎𝑛1

𝑎12 𝑎22

𝑎32 ⋮

𝑎𝑛2

𝑎13

𝑎23

𝑎33

⋮ 𝑎𝑛3

… … …

…

𝑎1𝑛

𝑎2𝑛

𝑎3𝑛

⋮ 𝑎𝑛𝑛

nxn

maka matrix tersebut dapat diekspresikan dalam bentuk:

[A] = [L][U]

dimana:

[L] =

𝑙11

𝑙21

𝑙31

⋮

𝑙𝑛1

0

𝑙22

𝑙32

⋮

𝑙𝑛2

0

0

𝑙33

⋮

𝑙𝑛3

…

…

…

…

0

0

0

⋮

𝑙𝑛𝑛

=

[L] =

𝑢11

0

0

⋮

0

𝑢12

𝑢22

0

⋮

0

𝑢13

𝑢23

𝑢33

⋮

0

…

…

…

…

𝑢1𝑛

𝑢2𝑛

𝑢3𝑛

⋮

𝑢𝑛𝑛

=

“lower triangle matrix”, matrix bujur sangkar yang semua elemen di atas/di kanan diagonal utama = 0

“upper triangle matrix”, matrix bujur sangkar yang semua elemen di bawah/di kiri diagonal utama = 0

Teori Dekomposisi Matriks

Aplikasi pada solusi persamaan linier simultan

[A] {X} = {B}

[L] [U] {X} = {B} [U]{X} = {Y}

[L] {Y} = {B} {X} = [U]-1 . {Y}

{Y} = [L]-1 . {B} {X} = [U]-1 . ([L]-1 {B})

Dengan cara ini, {X} bisa diperoleh dengan cepat/mudah tanpa harus menghitung

inverse matrix [A]

(menghemat memori dan running komputer)

Dipakai pada:

Metode eleminasi Gauss

Metode Cholesky

dapat diperoleh tanpa INVERSE

dapat diperoleh tanpa INVERSE



Pada analisis struktur dengan metode matrix, akan selalu dijumpai matrix (kekakuan)

yang simetris.

Bila [A] = matrix bujur sangkar dan simetris, maka:

[A] = [L] [L]T ..... ()

𝑎11

𝑎21

⋮

𝑎𝑛1

𝑎12

𝑎22

⋮

𝑎𝑛2

𝑎13

𝑎23

⋮

𝑎𝑛3

…

…

…

𝑎1𝑛

𝑎2𝑛

⋮

𝑎𝑛𝑛

=

𝑙11

𝑙21

⋮

𝑙𝑛1

0

𝑙22

⋮

𝑙𝑛2

0

0

⋮

𝑙𝑛3

…

…

…

0

0

⋮

𝑙𝑛𝑛

𝑙11

0

⋮

0

𝑙12

𝑙22

⋮

0

𝑙13

𝑙23

⋮

0

…

…

…

𝑙1𝑛

𝑙2𝑛

⋮

𝑙𝑛𝑛

𝑎11

𝑎21

⋮

𝑎𝑛1

𝑎12

𝑎22

⋮

𝑎𝑛2

𝑎13

𝑎23

⋮

𝑎𝑛3

…

…

…

𝑎1𝑛

𝑎2𝑛

⋮

𝑎𝑛𝑛

=

𝑙11

𝑙21

⋮

𝑙𝑛1

0

𝑙22

⋮

𝑙𝑛2

0

0

⋮

𝑙𝑛3

…

…

…

0

0

⋮

𝑙𝑛𝑛

𝑙11

0

⋮

0

𝑙12

𝑙22

⋮

0

𝑙13

𝑙23

⋮

0

…

…

…

𝑙1𝑛

𝑙2𝑛

⋮

𝑙𝑛𝑛

sehingga:

𝑎11 = 𝑙112 𝑙11 = 𝑎11

𝑎21 = 𝑙11 . 𝑙21 𝑙21 = 𝑎21

𝑙11

⋮ ⋮

𝑎𝑛1 = 𝑙11 . 𝑙𝑛1 𝑙𝑛1 = 𝑎𝑛1

𝑙11

𝑎22 = 𝑙21 . 𝑙21 + 𝑙22 . 𝑙22 𝑙21 = (𝑎21 − 𝑙212)

1

2

𝑎32 = 𝑙21 . 𝑙31 + 𝑙22 . 𝑙32 𝑙32 =𝑎32− 𝑙21 . 𝑙31

𝑙22

⋮ ⋮

𝑎𝑛2 = 𝑙21 . 𝑙𝑛1 + 𝑙22 . 𝑙𝑛2 𝑙𝑛2 =𝑎𝑛2− 𝑙21 . 𝑙𝑛1

𝑙22

dan seterusnya: 𝑙𝑛3 =𝑎𝑛3− 𝑙31 . 𝑙𝑛1− 𝑙32 . 𝑙𝑛2

𝑙33

Secara umum:

𝑙𝑖𝑖 = 𝑎𝑖𝑖 − . 𝑙𝑖𝑟2𝑖−1

𝑟=1 1

2

𝑙𝑖𝑗 =𝑎𝑖𝑗− . 𝑙𝑖𝑟 . 𝑙𝑗𝑟

𝑗−1𝑟=1

𝑙𝑗𝑗 untuk i > j

𝑙𝑖𝑗 = 0 untuk i < j

Pada analisis struktur dengan metode matrix, akan selalu dijumpai matrix (kekakuan)

yang simetris.

Bila [A] = matrix bujur sangkar dan simetris, maka:

[A] = [L] [L]T ..... ()

Persamaan () dapat ditulis: [A]−1 = L L T −1

= L T −1

. L −1

= L −1 T . L −1

Bila: [M] = [L]−1

Maka: [A]−1 = [M]T . M dan L [L]−1 = [I] () [L] [M] = [I]

𝑙11

𝑙21

𝑙31

⋮

𝑙𝑛1

0

𝑙22

𝑙32

⋮

𝑙𝑛2

0

0

𝑙33

⋮

𝑙𝑛3

…

…

…

…

0

0

0

⋮

𝑙𝑛𝑛

𝑚11

𝑚21

𝑚31

⋮

𝑚𝑛1

0

𝑚22

𝑚32

⋮

𝑚𝑛2

0

0

𝑚33

⋮

𝑚𝑛3

…

…

…

…

0

0

0

⋮

𝑚𝑛𝑛

=

1

0

0

⋮

0

0

1

0

⋮

0

0

0

1

⋮

0

…

…

…

…

0

0

0

⋮

1

Sehingga:

𝑙11 .𝑚11 = 1 𝑚11 = 1

𝑙11

𝑙21 .𝑚11 + 𝑙22 .𝑚22 = 0 𝑚21 = − 𝑙21 . 𝑚11

𝑙21

𝑙31 .𝑚11 + 𝑙32 .𝑚21 + 𝑙33 .𝑚31 = 0 𝑚31 = − 𝑙31 . 𝑚11 + 𝑙32 . 𝑚21

𝑙33

⋮ ⋮

Secara umum:

𝑚𝑖𝑖 = 1

𝑙𝑖𝑖

𝑚𝑖𝑗 = . 𝑙𝑖𝑟 . 𝑚𝑟𝑗𝑖−1𝑟=1

𝑙𝑖𝑖 untuk i > j

𝑚𝑖𝑗 = 0 untuk i < j

Setelah diperoleh matrix [M], maka dengan persamaan (), inverse matrix [A] dapat

dihitung:

[A]−1 = [M]T . M Metode Cholesky

PARTIONING OF MATRICES

Suatu matrix bisa dipartisikan menjadi SUB MATRIX, dengan cara mengikutkan

hanya beberapa baris atau kolom dari matrix aslinya.

Masing-masing garis partisi harus memotong suatu baris/kolom dari matrix

aslinya.

Contoh:

[A] =

𝑎11

𝑎21

𝑎31

𝑎12

𝑎22

𝑎32

𝑎13

𝑎23

𝑎33

𝑎14

𝑎24

𝑎34

𝑎15

𝑎25

𝑎35

𝑎16

𝑎26

𝑎36

= 𝐴11

𝐴21

𝐴12

𝐴22

𝐴13

𝐴23

dimana:

𝐴11 = 𝑎11

𝑎21

; 𝐴12 = 𝑎12

𝑎22

𝑎13

𝑎23

𝑎14

𝑎24

𝐴21 = 𝑎31 ; 𝐴22 = 𝑎32 𝑎33 𝑎34

....dst!

Aturan-aturan yang dipakai untuk mengoperasikan matrix partisi persis sama

dengan mengoperasikan matrix biasa.

Contoh:

𝐴 = 5 3 14 6 2

10 3 4

3𝑥3

; 𝐵 = 1 52 43 2

3𝑥2

= 𝐵1

𝐵2

2𝑥1

= 𝐴11 𝐴12

𝐴21 𝐴22

2𝑥2

𝐴 𝐵 = 𝐴11 𝐴12

𝐴21 𝐴22

𝐵1

𝐵2

= 𝐴11 .𝐵1 + 𝐴12 .𝐵2

𝐴21 .𝐵1 + 𝐴22 .𝐵2

𝐴11 𝐵1 = 5 34 6

1 52 4

= 11 3716 44

A12 B2 = 12 3 2 =

3 26 4

A21 B1 = 10 3 1 52 4

= 16 62

A22 B2 = 4 3 2 = 12 8

𝐴 𝐵 = 14 3922 4828 70

Matrix Operations in Excel

Select the

cells in

which the

answer

will

appear

Matrix Multiplication in Excel

1) Enter

“=mmult(“

2) Select the

cells of the

first matrix

3) Enter comma

“,”

4) Select the

cells of the

second matrix

5) Enter “)”

Matrix Multiplication in Excel

Enter these

three

key

strokes

at the

same

time:

control

shift

enter

Matrix Inversion in Excel

• Follow the same procedure

• Select cells in which answer is to be displayed

• Enter the formula: =minverse(

• Select the cells containing the matrix to be inverted

• Close parenthesis – type “)”

• Press three keys: Control, shift, enter

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