Matriks dan Ruang Vektor

  O p e ra si Alja b a r Ma triks

1 Matriks dan Ruang Vektor

  M a t rik s I nve r s Definisi : Bila A.B = B.A = I, maka A dan B saling invers

• -1

  Notasi invers A adalah A Sifat-sifat Matriks Invers

Jika A dan B non singular, atau invertibel,

• 1 -1

  ( . ) A B

  

  1 = B .A

  A matriks bujur sangkar, maka :  

  A ⁿ = A.A.A. .. A n faktor



  A = I

• 1 -1 -1

  . .

     

  A A A ^{n n}

    

   = A = A .. A n faktor

  1

  1

• 1
• 1

    p A A . .

   

  1

  1 = p = 1/ p A

  A A ^{n m} . = A n + m

    A ^{n m}

  = A n.m

• 1

  Contoh : A = 1 2 3 4 A = ?

     

     

  1 Misalkan A ^

     

  

 

   

       

       

    ¹ = a b c d

  1 2 3 4 a b c d

  = 1 0 0 1 

       

        

   

  1

  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 -

• a =2

  3a + 4c =0 4c = -3 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 -

• b = -1

     

       

    _ 1/2 - 1/2

  1

  1 2 - = d c b a = ¹ A b + 2d = 0.

  2d = -b

  2

  1

  2

  1

  2  

   

   

  b d

  a ta u A ^

  1

  = 1 / A / adj (A)

  A ^    

• 3 1
>2
• 2

    ¹

  4 = 1

     

  







• 1

  2 4 - 2

• 3 1
• 2 1

    

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

  

1 A A

  . = I

OBE -1 2.

  A

     

   

  / I  I / A 

  

  1 3.

  A = 1 . adj (A) / A / t

  1

  1 = A

  4

  2

  5

  3

  6

     

     

  2

  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 .}

  3

  4

  5

  6

   

     

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

• A
Sifa t-sifa t ma triks tra nsp o se _t t 1.

  A

  = A

    _{t t t} 2.

   A + B = A B   t t 3.

  (p . A) = p . A 4. _{t t t} A . B = B A .

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

     

  2

  3

  3

  1 A dan B  = Ma ka Pe mb uktia n sifa t 1: Pe mb uktia n sifa t 2 :

     

  5 ) ( ,

       

     

    

  6

  4

  5

  5

  6

  5

  4

  6

  









  5

  4

  5

  2

  4

  1

  3

  4

  1

  3

     

        

     

     

      

  2

  1

  4

  3

  4

  3

  1

  2 = A ^{t t}

B dan

    A ^t

    

     

     

    

  4

  1

  3

  2

  4

  3

  1

  2 = A

  t t    

  2 = B A ^{t t} B A maka

     

  1

  Terbukti bahwa ^{t t t} B A B A

  2 = B A t t

  1

  3

  4

  3

  4

  2

     

  

5

  5

  4

  6

    

     

       

     ) ( Contoh pembuktian sifat 3 :

2 A 5 =

     

  5

  2 t

  1

  10

  5

     

  5 ^{t t} A maka

  3

  1

  4

  10

  15

  20

       

  10 ) , 5 (

  15

  5

  20

  10

  5

  15

  20

      

     

     

      Te rb ukti b a hwa C o nto h p e mb uktia n sifa t 4 : ^{t t} A A

  5 ) 5 ( 

  4

  12

  6

  2

  4

  1

  3

  1

  6

  3

  2 = B .

  A  

    

  19

  t m

  2

  3

    

    

    

     

    

     

     

    

    

  16

  9

  19

  8

  18

  8

  1

18 B) . (A aka

  t

adalah symetric

A - A t adalah skew symetric

  2

  3

  12

  6

  4

  3

  1

  2

  2

  1

  4

  3 .

  t ^t A B t t t

  A B B A . ) . ( 

  A

  16

  6

  Te rb ukti b a hwa Sifa t ma triks b ujur sa ng ka r A

    

    

    

     

    

     

     

    

  1

    

  

  9

  8

  19

  18

  8

• A

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

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

  A

  ( + A )

  t

  ( A ) So a l La tiha n :

• ma triks ske w syme tric C = 1/ 2 A

  1.

       _t

       _t

       

  1 =  

  

3

1 - 2 - 1 -

  1 3 - 1 4 -

  2

  4

  A maka A ..... : ,

1 -

       

  2.

  1 =

  2

  2

  3

  1

  1

  1

  ..... : ,

  A maka A

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

  De finisi : d ise b ut ma triks te re d uksi = adj   m x m 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)

  Ma triks Ese lo n (Elimina si G a uss)

























  5

  2

  3

  4

  5

  6

  1

  2

  3

  4

  1

             

  2

  3

  4

  1

  2

  3

  1

  2

  1

  1 Ma triks Ese lo n Te re d uksi (Elimina si G a uss Jo rd a n):         

          

  1

  1

  1

  1

  1

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

  1 2 4 0 1 7 0 0 1

      

      

  1 0 0  

  O p e rasi Baris Ele m e nte r (O BE) De finisi : b = me nuka r b a ris ke i d e ng a n b a ris ke j _ij b (p ) = me ng a lika n b a ris ke i d e ng a n p _i b (p ) = b + p .b _{ij i 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 .

  3 = 0 20 28

      

  2 = 3 6 9 4b

       b

     

  4 5 6 3 26 37 0 5 7 

    ( ) .

      

      

      

      

      

  3

  C o nto h :

  2

  2

  23

  12 2(3)

  3 6 9 0 5 7 b b b b

  1 2 3 0 5 7 b 4 5 6

  4 2 3 4 5 6 0 5 7 b 5 6

  4

  4

  1

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 :

  I = 1 0 0 0 1 0 0 0 1

  E = 1 0 0 0 5 0 0 0 1 ^{3 b 2} 

     

       

     

      ^{( ) 5 b 3 2} I =

  1 0 0 0 1 0 ^{( / ) 1 5}       

    I = 1 0 0 0 1 0 0 0 1

  = 0 1 0 1 0 0 0 0 1 ^{3 b 12} 

       

  4

      

     

  ( )     

  I = 1 0 0 0 1 0 0 0 1 ^{32 3 3 2 3} ( )

  4 b b b b

  4

  ( ) .

     

        

     

     

  E = 1 0 0 0 1 0 0 4 1 ^{3 32 3 3 2} 

  I = 1 0 0 0 1 0 0 0 1 b b b b

     

       

  1 0 0 0 1 0 0 0 1   

      ^{E b 3 12} I =

  













         

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

yang terjadi bila OBE tersebut dilakukan pada matriks A C o nto h :

  

  1 2  

  3 4 

  b

  A ¹²    

•  

  3

  4

  1

  2     1 0 0 1

     

  b ₁₂

   I = E =

  2

        0 1 1 0    

  1 2  

  3 4  E.A = =

   1  

       

  1

  3

  4

  1

  2       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 _-1 ma ka E 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.

• 1

  (AI) OBE (I A ) C o nto h 1:

   

  1

  2

  

  1 A maka A 

  = , : ?  

  3

  4 _{So lusi :}  

  1  b ( ) ₂

  1

  2

  1

  1

  2

  1     b (-3) ₂₁

  2    

         

  3

  4

  1 - -

  2

  3

  1        

  1

  2

  1

  1

  2 1 - b (-2) ₁₂

         

  1

  1

  1

  1 Ja d i    

     

  

• 2

  2

  1

  1

  1

  1 2 - =

1 A

  )

  6

  (B  I) OBE ( I B

   B maka B

       

     

  1  

  2

  

6

  C o nto h 2 : So lusi :

  2

  8

  

6

  2

  8

  

8

  ? ,

• 1

  31

  1

  1

  1     

  







     _{ ) 2 ( b ) 2 / 1 ( b 21 1} 1/2

  3

  3

  

1

  1

  6

  6

  2

  

  1

  3 3 1/2  

  1

  3 3 1/2  _{b ( 12  3 ) b ( 1 / 2 )}       

        2 -

  1 ² 1 1/2 1/2 _{b ( 32  2 )}         

• 1

     

  2 2 -

  1

  1

  2 2 -

  1

  1     ¹

  

  1

  3

  2

• 

  1 ₂   _{b ( 1 / 2 )} ³

•      1 1/2 1/2 

     2 -

  1

  1  

   

  1

  3

  2

• 1

  1

  2 

   _ b ( ₁₃ 3 )

  1

  1

      

  2

  2

  1

  1

  1 1 -

  2 = B

  Jadi

  1 -

      

• 2

  1

  1

  2

  1

  2

  1

  2

  1

  2

  1

  1 0 0 2 0 -1 1 0 - 0 0 0 1 0 -

  33

  1

• 1
• 2

  2

  2

          

          

  I

3 B

  Ma triks ya ng tid a k me mp unya i inve rs _{C o nto h :}      

1 B

  

  2

  ) 1 ( b ) 2 ( b _b

  1

  1

  2

  1

  2

  1 1 -

  1

  1

  1

  1

  3

     

    _{  23 31 21}

       

   

     

  1

  2

  2

  1 1 -

  1

  2

  1 2 - 3 - 3 -

      

  1

  )

3 (

)

1 (

_{32 12 b} ^b

  1

  1

  2

  1

  1

  1

  1 1 -

  1 2 - 3 - 3 -

  1

      

  2

  1 1 -

  1

  1 5 - 1 1 -

  3

  



      

      

         

  

Sebelah kiri bukan matriks identitas, maka Matriks

B tak mempunyai invers.

  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

  A = 2 1 1

      

      

• 1 2 1 1 -1 2

  A = 3 4 -1 1 0 3 

      