AL Modul 2 Invers Matrik

7/26/2019 AL Modul 2 Invers Matrik

1/24

MODUL 2INVERS MATRIK

PRAYUDI STT PLN


2/24

PENGERTIAN INVERS MATRIK

Matrik bujur sangkar A dikatakan !"un#ai in$!rs% jika t!rda"at

atrik & s!d!ikian ru"a s!'ingga (

A& ) &A ) I

diana I atrik id!ntitas

& dikatakan in$!rs atrik A ditu*is A+,% aka% AA+, ) A+,A ) I

A dikatakan in$!rs atrik & ditu*is &+,% aka% &+,&) &&+,) I

-.nt.' / A& ) &A ) I

,,,

201,02

30

40202,

=

,,,

201,02

=

,11

1,111,

30

40202,


3/24

TEKNIK MENGHITUNG INVERS

M!t.d! Adj.int atrik

M!t.d! ."!rasi !*!!nt!r baris

M!t.d! P!rka*ian In$!rs Matrik E*!!nt!r

M!t.d! "artisi atrik

Pr.gra K."ut!r + MAT-ADS% MATLA&

5S O6I-E E7-ELL


4/24

M!t.d! Adj.int Matrik

Andaikan A atrik bujur sangkar b!r.rd. 8n9n:% -ij)8;,:i


5/24

Kasus% n ) 0

=

00020,

20222,

,0,2,,

aaa

aaa

aaa

A

=

=

0020,0

0222,2

0,2,,,

0020,0

0222,2

0,2,,,,;

MM;M

M;MM;

MM;M

d!t8A:

,

---

---

---

d!t8A:

,A

CONTOH :

=

334340

402

A d!t8A:) ,

=

=

=

,;2,;

2;3

,;33;

@:;8,2:;8,1;,:;8,3

,2:;8,1;:,;,1821:;8,3;

,:;8,321:;8,3;23:;821

40

02

34

02;

34

40

30

42;

34

42

34

30;

34

40

33

40;

33

34

8,:,A ,;


6/24

KASUS ( n ) 4

=

=

440424,4

400020,0

420222,2

4,0,2,,,

,;

4440424,

0400020,

2420222,

,4,0,2,,

MM;MM;

M;MM;MMM;MM;

M;MM;M

d!t8A:, A

aaaa

aaaaaaaa

aaaa

A

-ONTOB (

Bitung*a' in$!rs atrik b!rikut ini (

Eks"ansi baris ;, (

d!t8a:)M,,;2M,2


7/24

INVERS : OPERASI ELEMENTER BARIS

O"!rasi E*!!nt!r baris #ang

digunakan ada*a' (

8,:? BjkBj

82:? BjBi

80:? BjBj < kBj

,111

???,??????

11,1

111,

???

?????????

???

???

2,

2222,

,,2,,

nnnn

ii

n

n

aaa

a

aaa

aaa

Langka';*angka' s!bagai b!rikut

8,:? &!ntuk atrik *!ngka" A%I

82:? D!ngan s!rangkain ."!rasi

!*!*!t!r baris r!duksi*a' A%I

!njadi atrik b!rb!ntuk I%&

80:? A1 ) &

nnnn

ii

n

n

bbb

b

bbb

bbb

???

?????????

???

???

,111

???,??????

11,1

111,

2,

2222,

,,2,,

nnnn

ii

n

n

bbb

b

bbb

bbb

???

?????????

???

???

AFadi%

2,

2222,

,,2,,

,;O"!rasi !*!!nt!r baris

Ga.uss;F.rdan


8/24

CONTOH :

M?Asa*

2 0 4 , 1 1

0 4 3 1 , 1

4 3 3 1 1 ,

It!rasi;,

, ,?3 2 1?3 1 1 B,)8,a,,:B,

1 ;1?3 ;, ;,?3 , 1 B2)B2;8a2,a,,:B,1 ;, ;0 ;2 1 , B0)B0;8a0,a,,:B,

It!rasi;2

, ,?3 2 1?3 1 1

1 , 2 0 ;2 1 B2)8,a22:B21 1 ;, , ;2 , B0)B0;8a02a22:B2

It!rasi;0

, ,?3 2 1?3 1 1

1 , 2 0 ;2 1

1 1 , ;, 2 ;, B0)8,a00:B0


9/24

It!rasi;4

, ,?3 1 2?3 ;4 2 B,)B,;8a,0a00:B01 , 1 3 ; 2 B2)B2;8a20a00:B0

1 1 , ;, 2 ;,

It!rasi;3

, 1 1 ;3 3 ;, B,)B,;8a,2a22:B2

1 , 1 3 ; 2

1 1 , ;, 2 ;,

Lanjutan (

= ,;2,;

2;3

,;33;

AFadi% ,;


10/24

Matrik AHa*

, 2 0 4 , 1 1 1

2 0 3 3 1 , 1 1

0 3 C 4 1 1 , 10 1 1 1 ,It!rasi ; ,

, 2 0 4 , 1 1 1 B,)8,a,,:B,

1 ;, ;, ;0 ;2 , 1 1 B2)B2;8a2,a,,:B,

1 ;, ;2 ; ;0 1 , 1 B0)B0;8a2,a,,:B,1 1 ;, ; ;0 1 1 , B4)B4;8a4,a,,:B,

It!rasi ; 2

, 2 0 4 , 1 1 1

1 , , 0 2 ;, 1 1 B2)8,a22:B2

1 1 ;, ;3 ;, ;, , 1 B0)B0;8a02a22:B21 1 ;, ; ;0 1 1 , B4)B4;8a42a22:B2

It!rasi;4

, 2 0 4 , 1 1 1

1 , , 0 2 ;, 1 1

1 1 , 3 , , ;, 1 B0)8,a00:B01 1 1 ;, ;2 , ;, , B4)B4; a40a00 B0


11/24

It!rasi;3

, 2 0 4 , 1 1 1

1 , , 0 2 ;, 1 1

1 1 , 3 , , ;, 1

1 1 1 , 2 ;, , ;, B4)8,a44:B4It!rasi;

, 2 0 1 ;C 4 ;4 4 B,)B,;a,4B4

1 , , 1 ;4 2 ;0 0 B2)B2;a24B4

1 1 , 1 ;@ ; 3 B0)B0;a04B4

1 1 1 , 2 ;, , ;,It!rasi;C

, 2 1 1 21 ;,4 ,4 ;,, B,)B,;a,0B0

1 , 1 1 3 ;4 0 ;2 B2)B2;a20B0

1 1 , 1 ;@ ; 3

1 1 1 , 2 ;, , ;,It!rasi;

, 1 1 1 ,1 ; ;C B,)B,;a,2B2

1 , 1 1 3 ;4 0 ;2

1 1 , 1 ;@ ; 3

1 1 1 , 2 ;, , ;,

=

,;,,2

3;@;

2;04;3

C;;,1

AFadi%

,;


12/24

PERKALIAN MATRIK ELEMENTER

8,:? Matrik !*!!nt!r ada*a'

atrik #ang di"!r.*!' dari."!rasi !*!!nt!r #ang

dik!nakan "ada atrik

id!ntitas?

82:? S!tia" atrik !*!!nt!r

!"un#ai in$!rs% dan s!tia"

atrik bujur sangkar b!r.rd.

8n9n: #ang !"un#ai in$!rs

!ki$a*!n baris t!r'ada" atrik

id!ntitas I?

80:? Akibatn#a% jika (

EkEk1Ek2JE2E1A ) I%

aka%

A1) EkEk1Ek2JE2E1

Matrik !*!!nt!r E di"!r.*!' dari

trans=.rasi atrik id!ntitas diana

"ada k.*. k!;I diganti d!ngann.ra*itas $!kt.r k.*. (

=

,??????11

???????????????????

1??????11???????????????????

1??????,1

1??????1,

?

?

?

?

ik

ik

ik

ik

i

N

N

N

N

E

=

iiin

aii

iii

ik

aa

aa

N

???

,

???

%

%,

%

iiiik IAEEEN ,2,% ???

(diana


13/24

CONTOHBitung in$!rs atrik A

Jawab :

M!ng'itung E,

=

334

340

402

A

=

=

=

3

3

4

A/

3

4

0

A/

4

0

2

A 02,

=

=

,12;

1,,?3;

111?3

,1aa;

1,aa;

11,a

E

,,0,,,2,

,,

,

M!ng'itung E2

=

=

=

=

=

=

,2;,

12;0

104;

,12

1,,?3;

111?3

,2;1

12;1

10,

EE

,2;112;1

10,

,:8;,:8;1?3;111?3;,1

11?3;,?3;,

E

,;1?3;

,?3

34

0

,12;1,,?3;

111?3

AEN

,2

2

2,2


14/24

Meng!"#ng E$ %an In&e'( Ma"'!k

,;2,;2;3

,;33;

,2;,12;0

104;

,;112,1

,;1,

EEE

,;112,1

,;1,

,G8;,:1182:G8;,:;,1

8;,:G8;,:;1,

E

,;2

,;

33

4

,2;,12;0

104;

AEEN

,20

0

0,20

Fadi In$!rs Matrik

=

,;2,;

2;3

,;33;

A ,;


15/24

CONTOHBitung in$!rs atrik A

Jawab :

M!ng'itung E,

=

0

4C30

3302

402,

A

=

=

,110;

1,10;

11,2;

111,

E

0

0

2

,

AN

,

,,

M!ng'itung E2

=

=

=

=

=

=

,110;

1,,;,;

11,;2

1120;

,110;

1,10;

11,2;

111,

,111

1,,;1

11,;1

112,

EE

,111

1,,;1

11,;1

112,

,118;,:;1

1,8;,:8;,:;1

11,;,1

1128;,:;,

E

1,;

,;

2

3

0

2

,110;1,10;

11,2;

111,

AEN

,2

2

2,2


16/24

M!ng'itung E0

=

=

=

=

=

=

,,;,2;

1,;,,

1,2;,

1,,0;

,110;

1,,;,;

11,;2

1120;

,111

1,11

11,1

111,

EEE

,,;11

1,;11

1,,1

1,1,

,8;,:8;,:;11

1,8;,:11

1,8;,:;,1

1,8;,:;1,

E

,;

,;,

,

C3

0

,110;

1,,;,;11,;2

1120;

AEEN

,20

0

0,20


17/24

M!ng'itung E4 dan In$!rs Matrik

=

=

=

=

=

=

=

,;,,;2

3;@;

2;04;3

C;;,1

,,;,2;

1,;,,

1,2;,

1,,4;

,;111

3,11

2;1,1

C;11,

EEEEA

,;111

3,112;1,1

C;11,

,8;,:111

38;,:;,118;2:8;,:;1,1

8;C:8;,:;11,

E

,;

3

2;

C;

4

3

4

,,;,2;

1,;,,

1,2;,

1,,0;

AEEEN

,204,;

0

4,204


18/24

INVERS : PARTISI MATRIK )1*Partisi atrik A #ang b!r.rd. 89n:

ada*a' sub atrik;sub atrik #angdi"!r.*!' dari A d!ngan ara

!b!rikan batasan;batasan garis

'.ris.nta* diantara dua baris dan

atau !b!rikan batasan;batasan

garis $!rtika* diantara dua k.*.?

CONTOH

=

0

4C303302

402,

A

=

=

=

=

4C A

0

30A

3340 A

022,A

(ada*a'AatrikPartisi

222,

,2,,

=

0,334

,00403002

20440

04302

A

CONTOH


19/24

INVERS : PARTISI MATRIK )2*Andaikan A atrik bujur sangkar

b!r.rd. 8n9n: #ang !"un#aiin$!rs% #aitu ( A1) &% dan

"artisin#a asing;asing ada*a' (

Kar!na% A&)&A)I aka

di"!r.*!' (

=

=

222,

,2,,

222,

,2,,

&&

&&&/

AA

AAA

=

=

I1

1I

AA

AA

&&

&&

I1

1I

&&

&&

AA

AA

222,

,2,,

222,

,2,,

222,

,2,,

222,

,2,,

Dari "!rka*ian atrik di"!r.*!' 'asi* (

8,:? A11&11< A12&21) I

82:? A11&12< A12&22) 1

80:? &21A11< &22A21) 1

84:? &21A12< &22A22) I

D!ngan asusi% A111 ada% dan

&22) L1ada

Maka ruus untuk !ng'itung in$!ratrikn#a ada*a' (

8,:? &12 ) +8A 111A12:L

1

82:? &21) + L18A21A11

1:

80:? &11 ) A111


20/24

CONTOH :Kasus n)4? Bitung*a' in$!rs

atrik b!rikut ini

FaHab (

=

0

4C30

3302

402,

A

=

=

=

=

4C A

0

30A

33

40 A

02

2,A

(ada*a'AatrikPartisi

222,

,2,,

Meng!"#ng L

=

=

=

=

=

=

=

=

=

=

,;,

3;

,;8;,:;

8;3:;;

3;

,LFadi%

;,;

3;,;

,2@

@;

4C

0,

2;,

0

30;

4C

AAAAL

10

,,

,;2

20;

0

30AA

0,

2;,

33

40

,;2

20;AA

,;2

20;

,2;

2;0

:408

,A

,;

,2,;

,,2,22

,;

,,2,

,2,;

,,

,;

,,


21/24

Meng!"#ng In&e'( Ma"'!k

=

=

=

=

=

=

=

=

=

4;3

;,0

0;0

;,0

,;2

20;

,2;

;@

0,

2;,

,;2

20;

:A8AL:AA8A&

,;2

@;

10

,,

,;,

3;;

:A8A;L&

2;0

C;

,;,

3;

0,

2;,;

L:A;8A&

,;,,2,

,;,2

,;,,

,;,,,,

,;,,2,

,;2,

,;,2

,;,,,2

=

=

,;,,2

3;@;

2;04;3

C;;,1

&&

&&A

222,

,2,,,;

CONTOH :

Bitung in$!rs atrik A b!rikut (

FaHab ( Partisi atrik A

=

0,334

,0040

3002

20440

04302

A

=

=

=

=

0,3

,00

30

A

34

40

02

A

204

043 A

40

02A

222,

,2,,


22/24

Meng!"#ng L

2,;

,1

1,

2;0

04;

34

40

02

AA

3C

;C;;

204

043

2;0

04;AA

2;0

04;

20;

0;4

@

,A

,;,,2,

,2,;

,,

,;,,

=

=

=

=

=

,;,;,

,;2;1

,1,;

,:;812:8;,;1:;8,

2:8;,;4:;822:8;2;

1:;8,2:8;2;,:;81

,

,L

Fadi%

2,;2

,;1,;

2,;,

,20

204

043

;

0,3

,00

30

3C;C;;

34

40

02

;

0,3

,00

30

AAAAL

,;

,2,;

,,2,22


23/24

Meng!"#ng In&e'( Ma"'!k

=

=

=

=

=

=

=

=

=

2C;3

00C;

23;2

010;

2;0

04;

0;2

4;,

22;

3C;C;;

2;004;

:A8AL:AA8A&

02;

4,;

2;2

2,;

,1

1,

,;,;,

,;2;1

,1,;

;

:A8A;L&

4,C2

3;21;2;

,;,;,,;2;1

,1,;

3C

;C;;

;

L:A;8A&

,;,,2,

,;,2

,;,,

,;,,,,

,;,,2,

,;2,

,;,2

,;,,,2

=

=

,;,;,02;

,;2;14,;

,1,;2;2

4,C22C;3

3;21;2;00C;

&&&&AFadi%

222,,2,,,;


24/24

SOAL TUGAS IV

Bitung in$!rs atrik A b!rikut ini d!ngan ara (

=

,,,4,,2

22,

,,,

bbaabbaa

aabb

aabb

A

a? M!t.d! Adj.int

b? P!rka*ian atrik !*!!nt!r

? O"!rasi !*!!nt!r baris

d? M!t.d! "artisi atrik

=

aaabb

aaabb

aaabbbbbaa

bbbaa

A

2,

2,

220,2,2,,

,2,2

Bitung in$!rs atrik A b!rikut ini d!ngan 2 ara "artisi #ang b!rb!da(

=

,0,,2

0,,20

,,204402,

02,,2

aaabb

aaabb

aaabbbbbaa

bbbaa

A

AL Modul 2 Invers Matrik

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Transcript of AL Modul 2 Invers Matrik