Matriks dan Ruang Vektor

O p e ra si Alja b a r Ma triks

1

Definisi :

Bila A.B = B.A = I, maka A dan B saling

invers

Notasi invers A adalah A

-1

Sifat-sifat Matriks Invers

Jika A dan B non singular, atau invertibel,

maka:

A.B juga non singular

3

( . )

A B



1

= B .A

-1

-1

A matriks bujur sangkar, maka :





A

n

= A.A.A. .. A n faktor



A

0

= I

 





A

n

= A

-1 n

= A

-1

.

A

1

.

A

1

.. A

-1



n faktor

4



p A

.



1

= p

-1

.

A

1

= 1/ p A

-1

A

n

.

A

m

= A

n + m

 

A

n m

= A

n.m

Contoh : A =

1 2

3 4

A = ?

-1











 

5

Misalkan

A 



 



  



 

  



 



 



 

 

1 ₌ a b

c d

1 2 3 4

a b

c d =

1 0 0 1

   

  

  

 

 

 

1 0

0 1 = 4d 3b

4c 3a

2d b

2c a

a+2c = 1 b+2d = 0 3a+4c= 0 3b+4d= 1

a+2c =1 x2 2a+4c =2

3a+ 4c=0 x1 3a+4c =0

3a + 4c =0

4c = -3

6

4

)

2

(

3

4

3











a

c

2

1

1

2

3

_



c

b+2d =0 x2 2b+4d =0

3b+4d =1 x1 3b+4d =1

7

   

     

 



1/2 1/2 1

1 2

= d c

b a = 1

A

b + 2d = 0. 2d = -b

2

1

2

1

2













b

a ta u

8

A

1

= 1

/ A /

adj (A)

A 



 

 

1 _{= 1} 4

-2

- 2 -3 1















- 1

2

4 - 2

-3 1















- 2 1

1 1/ 2 -1/ 2

1.

Rumus p e nye le sa ia n Ma triks Inve rs

2.

3.

9

A A

.

1

= I



_/

_I



_

OBE

_{ }

_



_I

_/

_A

-1



A

A

1

= 1

Matriks Transp o se

Ma triks tra nsp o se d ip e ro le h d e ng a n me nuka r e le me n-e le me n b a ris me n-ja d i e le me n-e le me n ko lo m d a n se -b a liknya .

C o nto h :

Tra nsp o se d a ri A a d a la h :

10













6

5

4

3

2

1

=

A





















6

3

5

2

Sifa t-sifa t ma triks tra nsp o se

1.

2.

3.

4.

C o nto h p e mb uktia n sifa t ma triks tra nsp o se :

11

 

t t

=

A

A



A + B = A



t t _ Bt

(p . A) = p . A

t t



A . B = B A



t t _. t



























2

4

1

3

4

1

3

2

=

dan

B

13

Contoh pembuktian sifat 3 :

Te rb ukti b a hwa

Sifa t ma triks b ujur sa ng ka r A

15































































9

8

19

18

8

1

6

2

16

3

12

6

4

3

1

2

2

1

4

3

.

t t

A

B

t t

t

A

B

B

A

.

)

.

(



A

+ A

t

adalah symetric

3. A d a p a t d itulis se b a g a i jumla h d a ri sua tu

ma triks syme tric B = 1/ 2 d a n sua tu

ma triks ske w syme tric C = 1/ 2

So a l La tiha n :

Te ntuka n Tra nsp o se Sua tu Ma triks d ib a wa h ini !

16

)

A

+

(

t

A

)

A

-(

t

Matriks Ese lo n dan Matriks Ese lo n te re duksi

De finisi : d ise b ut ma triks te re d uksi b ila me me nuhi :

1. Bila a d a b a ris ya ng ta k se mua no l, ma ka

e le me n p e rta ma ya ng  0 ha rus b ila ng a n 1

2. Ele me n p e rta ma ya ng  0 p a d a b a ris

d ib a wa hnya ha rus d ise b e la h ka na n 1

3. Ba ris ya ng se mua no l ha rus p a d a b a g ia n

b a wa h (b a ris-b a ris b a wa h)

18

 

C o nto h Ma triks Ese lo n

C o nto h Ma triks Ese lo n Te re d uksi

21

1 2 4

0 1 7

0 0 1





















1 0 0

0 1 0

0 0 1













O p e rasi Baris Ele m e nte r (O BE)

De finisi :

b_ij = me nuka r b a ris ke i d e ng a n b a ris ke j

b_i(p ) = me ng a lika n b a ris ke i d e ng a n p

b_ij (p ) = b_i + p .b_j

G a nti b a ris ke i d e ng a n b a ris b a ru ya ng

me rup a ka n b a ris ke i d ita mb a h d e ng a n b a ris ke j ya ng d ika lika n d e ng a n p .

C o nto h :

23

1

4

4

4

2 3

4 5 6

0 5 7

b

5 6

1 2 3

0 5 7

b

4 5 6

3 6 9

0 5 7

b

b

b

b

12 2(3) 23

2 2 3

































































( )

.

4 5 6 3 26 37 0 5 7



   



   

b

₂

= 3 6 9

4b

₃

= 0 20 28

Ma triks Ele me nte r d a n sifa t-sifa tnya :

De finisi :

A _nxn d ise b ut ma triks e le me nte r, b ila d e ng a n se ka li

me la kuka n O BE te rha d a p I_n d i p e ro le h A_nxn

C o nto h :

24

I =

1 0 0 0 1 0 0 0 1

E =

1 0 0 0 5 0 0 0 1

3 b2



   



   





   



   

( )5

b

3

2 I =

1 0 0 0 1 0 0 0 1

( / )1 5





   



25 _{I =} 1 0 0_{0 1 0}

E = Matriks elementer, maka E.A = matriks baru

yang terjadi bila OBE tersebut dilakukan pada

matriks A

A



OBE

 



_{= E.A}

Se tia p Ma triks Ele me nte r a d a la h ma triks ta k sing ula r.

Inve rs ma triks e le me nte r jug a ma triks e le me nte r.

I O BE E

ma ka E-1 _{jug a e le me nte r}

C a ra p e nye le sa ia n inve rs ma triks d e ng a n O BE.

27

Ja d i

29



















2

1

2

1

1

1

2

-=

C o nto h 2 :

So lusi :

30

?

,

8

8

2

6

8

2

6

6

2

1

_

























B

maka

B

33

1

0

0 0 2 0 -1

1 0 - 0

0 0 1 0 -

1 2 1

2 21 1

2 21





















I

₃

_B

-1

Jadi





































2

1

2

1

0

0

2

1

2

1

-2

1

1

0

2

35

So a l la tiha n :

1) C a ri inve rs ma triks d a ri

2) C a ri inve rs ma triks d a ri

36

A =

2 1 1

-1 2 1

1 -1 2





















A =

3 4 -1

1 0 3

2 5 - 4











