materi matriks smk ap

1

MATRIKS

Matrix asalah susunan bilangan berbentuk segi-4 yang terdiri atas baris dan kolom yang ditulis dalam sepasang tanda kurung.

DEFINISI

NOTASI OF MATRIKS

mnmm

n

n

m aaa

aaaaaa

a

aa

A

...............

...

...

...

32

22322

11312

1

21

11

Nama Matriks

Amxn

ELEMENT MATRIKS

mnmm

n

n

m aaa

aaaaaa

a

aa

A

...............

...

...

...

32

22322

11312

1

21

11elementbaris

1

Letak elemenElemen kolom 1

ORDO

mnmm

n

n

m aaa

aaaaaa

a

aa

A

...............

...

...

...

32

22322

11312

1

21

11 Baris 1

Baris 2

Baris m

Kolom1 Kolom2

Kolom 3 Kolom n

Ordo m x n Notasi : A m x n

53

12

643970182

Ζ

1. Apakah nama matriks di atas?

Contoh:

53

12

643970182

Ζ

2. Sebutkan elemen baris 3 dan kolom 4!

53

12

643970182

Ζ

3. Sebutkan elemen baris ke-2!

53

12

643970182

Ζ

3. Sebutkan ordo matriks di atas dan notasinya!

Ordo 3 x 4

Notasi : Z 3 x 4

JENIS-JENIS MATRIKS

10

MATRIKS BARIS

4991N

MATRIKS KOLOM

603

S

MATRIKS DIAGONAL

100010004

M

MATRIKS IDENTITAS

000

000

W

100010001

W

Penjumlahan Perkalian

Matriks 0

1001

W

MATRIKS SEGITIGA

14002110341

W

Segitiga Atas

Segitiga bawah

28472030600290001

W

TRANSPOS MATRIKS

Transpos matriks A terjadi jika setiap baris pada matriks tersebut berubah menjadi kolom . Transpose matriks A ditulis A’ atau At. Sehingga A m x n menjadi A’ n x m.

Elemen baris 1 matriks A Kolom 1 matriks A’

Elemen baris 2 matriks A Kolom 2 matriks A’

dst

TRANSPOS MATRIKS

93

214

105

21'A

A 4 x 2

921102

3451

A

A’ 2 x

4

93

214

105

21tA

Tentukan transpose matriks berikut!

125

A

413221130

B

Tentukan transpose matriks berikut!

125

tA 125 A

Jawab

421123310

B

413221130

tB

413221130

B

hgc

fe

Bdb

aA

a = e

d = h

b = fc = g

PERSAMAAN MATRIKS

zx

2-1

B32-

1A

4

If A = B, tentukan nilai x dan z!

y4x2x52

Bz6yx2

A

Jika A = B, tentukan nilai x, y dan z!

y4x2x52

Bz6yx2

A

2 = 2

z = 4x - y

x + y = -56 = 2x

2 = 2

z = 4x – y

= 4.3 – (-8)

z = 12 + 8 = 20

x + y = -5

3 + y = -5

y = -5 - 3 = -8

6 = 2x

x = 6/2 = 3

205

652

B206

2A

2.36

20 4.3 – (-8)12 + 820

3 + (-8)-5

y4x2x52

Bz6yx2

A

1. PENJUMLAHAN DAN PENGURANGAN MATRIKS

Dua atau lebih matriks dapat dijumlahkan atau dikurang kan jika :

a. Matriks tersebut berordo sama

b. Yang dioperasikan elemen yang seletak

Contoh:

215

49

942

7010

8122

536

CBA

Dapatkah A dan C dijumlahkan?

Jika

Dapatkah A dan B dijumlahkan?

1164

12316

942

7010

8122

536

A + B = …

Untuk

942

7010

8122

536

BA

1780

234

8122

536

942

7010

B - A = …

2. PERKALIAN MATRIKS

a. Perkalian 2 buah matriks

=

tsrqpo

nm

lkj

ihg

fed

cba

CBA321

3 x 3 3 x 2 2 x 4

1. Dapatkah A dan C dikalikan?

2. Dapatkah A dan B dikalikan?

Contoh

05

20

12/1

43

802

536

CA

Dapatkah A dan C dikalikan?

Diberikan

A 3 x 2 C2 x 4

=

C2 x 4

Z3 x 4

05

20

12/1

43

802

536

CA

A x C = …

Untuk

A 3 x 2 C2 x 4

=

K1C K2C K3C K4C

B1A

B2A

B3A

K1C K2C K3C K4C

B1A

B2A

B3A

05

20

12/1

43

802

536

CAa = (6x3)+(2x4)

= 18 + 8

= 26

K1C K2C K3C K4C

B1A

B2A

B3A

05

20

12/1

43

802

536

CAa = (-3x3)+(0x4)

= -9 + 0

= -9

26

K1C K2C K3C K4C

B1A

B2A

B3A

05

20

12/1

43

802

536

CAa = (5x5)+(0x-8)

= 25 + 0

= 25

26

-9

K1C K2C K3C K4C

B1A

B2A

B3A

05

20

12/1

43

802

536

CA

26

-925

26

-925

05

20

12/1

43

802

536

CA

-17 10,5

-1,51 -4

016

30

-15A.C =

Kerjakan soal berikut!

232

140

421

42

53

21

ZX Y

Diberikan

Tentukanlah matriks :

1.X.Y

2.Z.X

1.

b. Perkalian Matriks dengan skala

Multiplication a real number with matrix A is multipilcation each elements of matrix A by that real number

k.A = [k.amn]

Example

802

536

A

Determine 2 x A if

Answer

2 x 82 x 02 x 2

5 x 23 x2

6 x 2 2.A =

1604

10612

=

DETERMINANT

Determinant of matrix

a.Only used in square

b.are substraction with elements 1st diagonal and 2nd diagonal, where each elements enclosed

a. DETERMINANT ORDO 2 X 2

If

dcba

A

than|A| = ad - bc

Example

Determine value of determinant matrix below

61-105

A

Answer:

|A| = 5.6 – 10.-1 = 30 + 10 = 40

DETERMINAN ORDO 3 x 3

If given

ihgfedcba

A

than |A| =

heb

gda

ihgfedcba

A

DETERMINAN ORDO 3 x 3

|A| =

heb

gda

ihgfedcba

A

= (a.e.i + b.f.g + c.d.h) –(c.e.g + a.f.h + b.d.i)

Example

Determine determinat of

531312740

A

Answer: = (0.1.5 + 4.-3.-1 + 7.2.3) –(-1.1.7 +3.-3.0 + 5.2.4)

= (0+12+42) – (-7+0+40)

= 54 – 33 = 21

314

120

531312740

A

4. ADJOINAdjoin matrix A is the result transpose from kofaktor matriks A.

Matrix A

Adjoin Matrix A

Minor Matrix A

Kofaktor Matrix A

Minor Jika maka minor

61-105

A

M11 = 6

61-105

AM12 = -1

61-105

A

M21 = 10

61-105

AM22 = 5

61-105

A

5101-6

A

a. Ordo 2 x 2

Kofactor

If than kofactor

61-105

A

M11 = 6 .-11+1 = 6M12 = -1. -11+2 = -1.-1 =1M21 = 10. -12+1 = 10. -1 = -10M22 = 5. -1 2+2 = 5.1 = 5

510-16

A

5101-6

A

-

-

Adjoin

If than Adjoin matrix A

Resulted from the its kofactor

61-105

A

510-16

Akofaktor

5101-6

AMinor

5110-6

A Adjoin

205321211

A

M11 = 2.-2 – (0.3)

= -4- 0

= -4

205321211

A

M12 = 1.-2 – (-5.3)

= -2 – (-15)

= 13

205321211

A

M13 = 1.0 – (-5.2)

= 0 – (-10)

= 10

If , minor matrix A showed

next

205321211

A

b. Ordo 3 x 3

205321211

A

M21 = 1.-2 – 0.2

= -2- 0

= -2

205321211

A

M22 = 1.-2 – (-5.2)

= -2 – (-10)

= 8

205321211

A

M23 = 1.0 – (-5.1)

= 0 – (-5)

= 5

205321211

A

M21 = 1.3 – (2.2)

= 3 - 4

= -1

205321211

A

M22 = 1.3 – (1.2)

= 3 – 2

= 1

205321211

A

M23 = 1.2 – (1.1)

= 2 – (1)

= 1

Kofactor

11158210134-

A

:Minor

A

11-15-82

1013-4-A

:Kofactor

Adjoin

111582111-

:AMinor

11-15-8211-1-

:AKofactor

15-11-81-1-21-

:A Adjoin

205321211

A

given If

Inverse matrix A AAdjoin.|A|

1A 1 0 A ,

5. INVERSE

a. Inverse ordo 2 X 2

,

dcba

AIf

acbd

|A|1Aor 1

AAdjoin.|A|

1A 1

Answer

dcba

Aif

acbd

|A|1Ao 1r

AAdjoin.|A|

1A , 1

Contoh:

61-105

A

Determine inverse from

Answer :

51106

|)10.1(6.5|11A

51106

|1030|11A

51106

4011A

.adjA|A|

1A 1

40/540/140/1040/61A

8/140/14/120/31A

II. MATRIX APPLICATIONUsing to determine variabel value of linear equation. If the equation have variabel x dan y, than ..

|A||X|x

|A||Y|y

Example Determine value of x dan y from the next

equations2x + 3y = 7x - 2y = 7

7

72132

yx

734(1.3)22.|A|21

32A

35211

4(7.3)27.|X|27

37X

774(1.7)2.|Y|1

72Y

17

7

5735

|A||X|x

17 -7|A|

|Y|y

Competence Check1. Given

(A.B)-1 = ….

455-6-

Ba4321

A nd

42-3-1

b.

2-1211

21

e.

21211-

21

c.

21-211-

21

d.

1234

a.

2. Determine solution set from the next l

are ….

54

yx

2132

)2,1( d.)2,1(b.

)1,2(e.)2,1( c.)2,1(a.

204321301

A

b. Find determinan and adjoint from the next matrix

3-1-12

B

3-1-12

B

Minor Jika maka minor

3-1-12

A

M11 = -3

3-1-12

AM12 = -1

3-1-12

A

M21 = 1

3-1-12

AM22 = 2

3-1-12

A

211-3-

A

Kofactor

If than kofactor

3-1-12

A

M11 = -3 .-11+1 = -3M12 = -1. -11+2 = -1.-1 =1M21 = 1. -12+1 = 1. -1 = -1M22 = 2. -1 2+2 = 2.1 = 2

2-1-13-

A

211-3-

A

-

-

SOAL

3-1-12

A

2-1-13-

A

211-3-

A

2

31

1-

MINOR

KOFACTOR

ADJOINT