Tuesday, 15 Oct 2019

Slide: Invers Matriks

Slide: Invers Matriks

Slide: Invers Matriks

Topik Bahasan:

  • Definisi Invers Matriks;
  • Teknik Menentukan Invers;
    • Metode Subtitusi;
    • Metode Partisi Matriks;
    • Metode Matriks Adjoint;
    • Metode Eliminasi Gauss-Jordan;
    • Metode Perkalian Matriks Invers Elementer

Slide: Invers Matriks selengkapnya bisa Anda pelajari pada konten di bawah ini.

Slide: Invers Matriks

Slide Tunggu Sampai Slide: Invers Matriks selesai dimuat...!
Author: Dr. Ruminta

Transcript

Definisi Invers Matriks

Jika A adalah matriks ukuran n*n. Jika ada matriks B ukuran n*n sedemikian rupa sehingga :

AB = BA = I

di mana I adalah matriks identitas ukuran n*n, maka matriks A disebut non singular atau invertibeidan matriks B disebut invers dari A.

Jika matriks A tidak mempunyai invers, maka A disebut matriks singular atau non invertibei.

Notasi matriks invers : A-

1

AB = BA = I« AA= A"1 A = I

A

a,, a

11 12

a21 a22

a«1 a« 2

B=A

-1

a

1n

a

2 n

a

nn

A

-1

a11 a12

a21 a22

an1 an 2

a

1n

a

2n

a

nn

1 0

0 1

• •

• • • •

00

0 0

1

A

a11 a12 a21 a22

an1 an2

-1

a

1n

a

2n

a

nn

A

aa

11 12

a21 a22

an1 an2

a

1n

a

2n

a

nn

10 01

0 0

0 0 ••• 1

Teknik Menentukan Invers

Ada lima metode :

1.            Metode Subtitusi

2.            Metode Partisi Matriks

3.            Metode Matriks Adjoint

4.            Metode Eliminasi Gauss-Jordan

5.            Metode Perkalian Matriks Elementer

Metode Subtitusi

A

A

a

11

a

12

a

1n

a21 a22

an1 an2

a

2n

a

nn

a11 a12

a21 a22

an1 an 2

a

1n

a

2n

a

nn

1 0 0 1

0 0

0 0 l 1

Dari persamaan matriks tersebut diperoleh sebaiyak n2 persamaai simultan :

a11a11 + a12a21 + ••• + a1nan1 = 1

a11a12 + a12a22 + ••• + a1nan 2 = 0

a21a11 + a22a21 + ••• + a2 nan1 = 0       a21a12 + a22a22 + ••• + a2 nan 2 = 1

+

+ ... +

+

+ ... +

an1a11 + an 2a21 + ... + annan1 = 0          an1a12 + an 2a22 + + annan 2 = 0

a11a1n + a12a2 n + ••• + a1n«nn = 0

a21a1n + a22a2 n + ••• + a2 n®nn = 0 • • • •

• + * + + * = *

•             ?             •             I • • • I •             •

a .a + a 2a2 +... + a a = 1 1 1 2 2

Dari n2 persamaan simultan tersebut dapat diselesaikan secara subtitusi sehingga diperoleh elemen matriks invers.

A-1 =

a11 a12

a21 a22

a

1n

a

2n

an1 an2

a

nn

Contoh 1.

Tentukan invers dari matriks berikut:

                "2            3              2                              "2            3              2"            a11         a12         a13                         "1            0              0

A =         2              2              1                              2              2              1              a21         a22         a23         —           0              1              0

                1              2              2                              1              2              2              _a31      a32         a33 _                     0              0              1

                                                                                                A                                             A-1                         =                                            

2a11 + 3a21 + 2a31 = 1 2a11 + 2a21 + a31 = 0 a11 + 2a21 + 2a31 = 0

2a12 + 3a22 + 2a32 = 0

2a12 I 2a22 + ^^32 1 a12 + 2a 22 + 2a 32 = 0

2a13 + 3a23 + 2a33 = 0

I 2a23 I a33 0 a13 I 2a 23 I 2a 33 = 1

a21 I a31 1           a21         + a31 = 1              2a11       + 3a21 + 2a31 — 1

a21 2a31 1           a21         + (-2) — 1            2a11       + 3(3) + 2(-2) — 1

a31 2 / a31 2       a21         — 3        2a11       — 1 - 5 — -4 ^a11 —-2

cc 22 + — 1          +             1              + + 2a^2 — 0

-              a22 - 2a32 — 0   a22 + (1) — -1 2a12 + 3(-2) + 2(1) — 0

-              a32 — -1 ^ a32 — 1 a22 — -2       2a12 — 0 + 4 — 4 ^ a12 — 2

a23 + a33 0

cc^ 2a33 2

a23 + a33 0

2a13 + 3a23 + 2a33 — 0

a23 + (2) — 0 2a13 + 3(-2) + 2(2) — 0

a 33 — 2 / a 33 — 2 ^^23 2

2a13 — 0 + 2 — 2 ^ a13 — 1

                                2              3              2                                                              —           2              2              1

Jadi        A —       2              2              1                              A"1         —           3                              -2            - 2

                                1              2              2                                                              —           2              1              2

                                "2            3              2"            "- 2         2              1 "                                           "1 0        0"

Bukti                      2              2              1              3              -2            -              2              —           0 1          0

                                1              2              2              - 2           1              2                                              0 0          1

Contoh 2.

Tentukan invers dari matriks berikut:

A —

1              4

2              5

3

4

1 - 3 - 2

1              4

2              5

3

4

1 - 3 - 2 A

a11         a12         a13                         "1            0              0

a 21        a 22        a23         —           0              1              0

a31         a32         a33 _                     0              0              1

                A-1                         =                             1             

a11 + 4a21 + 3a31 — 1 2a11 + 5a21 + 4a31 — 0 aacn 3 a21 ^^31 0

^^12 I 4a22 I 3a32

2a12 I 5a22 I 4 1

^^12 3a22 ^^32

a 13 I 4a23 I 3a33 0 2a1315a2314a33 — 0

a13 - 3a23 - 2a33 — 1

3a21 I 21               3a21 I 21               a11         14a2113a31 — 1

7a2115a31 — 1  3(8) 12a31 — 2  a11         14(8) 13(—11) — 1

a21 — 8 a21 — 8               2a31 — -22 ^ a31 — -11 a11         — 111 — 2 ^ a11 — 2

3a22 ^^ 2a32 — 1             3^^22 ^^ 2^^32 1             ^^12 ^^ 4^^22 ^^ 3^^32 — 0 1

7a22 + 5a32 — 0                3(-5) + 2a32 — -1 a12 + 4(-5) + 3(7) — 0 j

a22 — -5 ^ a22 — -5 2a32 —14 ^ a32 — 7 a12 — 0 -1 — -1 ^ a12 — -1!

3a23 + 2a33 — 0                3a23 + 2a33 — 0                a13 + 4a23 + 3a33 — 0

7a23 + 5a33 — -1 3(2) + 2a33 — 0

a23 — 2 ^ a23 — 2 2a33 — -6 ^ a33 — -3 a13 — 0 +1 — 1 ^ a13 — 1

                1              4              3                              "2                            -1            1

Jadi A —               2              5              4              A"1 —   8                              -5            2

                1              -3            - 2                           -11                          7              - 3

                "1            4              3 "           " 2 -1      1 "                           "1            00

Bukti      2              5              4              8 - 5        2              —           0              10

                1              -3            - 2           -11 7      - 3                           0              01

Contoh 3.

Tentukan invers dari matriks berikut:

1 2 1

1              3 4

2              4 1

A

A —

1 2 1

1              3 4

2              4 1

a11         a12         a13                         1              0              0

a21         a22         a23         —           0              1              0

a31         a32         a33 _                     0              0              1

                A-1                         =                                            

0^11 I 2a21 I a31               1

an I 3 a21 I 4 ^^31             —0

2an I 4a21 I ^^31               —0

6^12 I 2^22 I 00

6^12 I 3^22 I 4    1

2a12 + 4a22 + a32 = 0

a^ I ^^23 I a^^ 00 a^ I 3a23 I 4 ^^33 0 I 4a23 I       1

^^21 3a31 1        -a21       - 3a31 = 1             a11         12a21 Ia31 — 1

a31 — 2                -a21       - 3(2) — 1            a11         12(-7) 12 — 1

a31 21 ^ a31 21  -a21       — 7 ^ a21 — -7  a11         — 1112 —13 —13

cc22 3a32 — 1    aa^22 3a32 1      a12 i 2^^22 I a32 — 0

a32 — 0                -a22 - 3(0) —-1 a12 + 2(1) + 0 — 0

a32 — 0 ^ a32 — 0 - a22 — -1 ^ a22 — 1 a12 — 0 - 2 — -2 ^ a12 — -2

a23 3 a33 — 0     a23 3 ^^33 0       ^^13 + 2^^23 + ^^33 — 0

a33 —-1               -a23 - 3(-1) — 0 a13 + 2(3) -1 — 0

a33 — 1 / ^^33 — 1 a23 3 / a23 3 ^^13 — 0 5 — 5 / ^^13 — 5

                                "1            2              1                                                              "13                         -2            - 5

Jadi        A—        1              3              4                              A"1         —           -              7              1              3

                                2              4              1                                                              2                              0              -1

                                "1            2              1"            "13         -2            -              5"                            "1 0        0"

Bukti                      1              3              4              - 7           1              3                              —           0 1          0

                                2              4              1              2              0              -              1                              0 0          1

Metoda Partisi Matriks

Mencari invers matriks dengan cara mempartisi matriks menjadi beberapa sub matriks.

A =

a11         a12         a13         a14

a 21        a22         a23         a24

a31         a32         a33         a34

_ a41     a42         a43         a44

^ A =

r An        A12

_ A21     A22 _

Ajj, AJ2, A21, dan A22 adalah sub matriks A.

Di mana : An =

22

A21 =

a11         a12         , A12 =  a13         a14

_ a21     a22 _                     _a23      a24 _

a31         a32         , A22 =  a33         a34

a41         a42                         a43         a44

Invers dari matriks partisi A adalah B, (B=A-1)

A =

nxm

q             ' An        A12

n - q       A21        A22

B =

mxk

P

m - p

' B11       B12

_ B21     B22 _

P

m - p

k-l

AB = I

" An       A12        * B11     B12                         'Ip 0 "

_ A21     A22 _     _ B21     B22 _                     _ 0 In-p _

AB =

A11B11 + A12 B 21 A11B12 + A12 B 22 A21B11 + A22 B 21 A21B12 + A22 B 22

P             p x(n - p )

(n - p )x p             n p

Perkalian matriks tersebut menghasilkan persamaan sub matriks simultan :

An Bu + A12 B21 = I (1)

A11B12 + A12 B22 = 0 (2)

A21 Bn + A22 B21 = 0 (3) A21B12 + A22 B22 = 1 ( 4)

Penyelesaian ke-4 persamaan sub matriks simultan : Dari pers. (1) An + A12 B1X B- = B-

Dari pers. (3) A221 A21 + B21B- = 0 atau B21B- = - A-1A

21 11

21 11

22 21

Karena

maka

A11 A12 A22A21 = B11

B22 =

A - A A~l A

22 ^21^11 ^12

atau

A- AA B

22 22 21 12

-1

Dari A2 jB11 + A22 B2 j = 0 ( 3 ) dan A- A21B11 + B21 = 0

Maka B12 = - A111 A12B22 atau

B = - B A A_1

12

B = - A_1 A B

21 22 21 11

Jadi A-!= B —

p

n - p

B B

11

12

B^ B

21

p

di mana B11 =

A - A A~l A

^11 12 22 21

22

n - p

-1

B22 A22 A22 A21B12

-1

B„ =

12

- BAA_1

B — - A_1 A B

21 22 21 11

atau

A-1 —

A11 A12 A221 A21 ) ( B11A12 A221 ) ( A22 A21B11 )

A~X - A- A B

22 22 21 12

-1

Contoh 1.

Tentukan invers matriks berikut:

                "2            3              2"

A =         2              2              1

                1              2              2

" An       A12

_ A21     A22 _

An =[2j A12 =[3 2]

                "2"                          "2            1"

A21 =     1              , A22 =  2              2

A- =

(A11 A12 A2>21 A21) ( B11A12 A2>21) (- A22A21B11 ) ((A22 - A22 A21B12 ).

A22 =

2 1 2 2

det A22 = 4-2 = 2, kofaktor matriks A22 : K11 = 2, K12 = -2,

K21 =-1, K22 = 2

22

K =

" Kn        K12                         " 2 - 2"   ^ Kt =    " 2 -1"

_K21      K22 _                     -1 2                         - 2 2

= Adj (A22)

A- =

22 _

AdAi) 0 A-1

det (A12 ) 22

A12 A22 A21 — [3 2]

1. B„ —

11

1 - 1

A - A A"1 A

^11 12 22 21

—1         2                             

— 2        -              2              2

-1                            * 2"       

                2                              —

1                              1             

-1

1

-1 1

/

[2]

V

l!

2

21

2

\

J

-1 f p-1

r

2. B10 —

12

- B A A — -

^11 ^12 ^22 _

[- 2l3 2]

V

1 -1/2 -1

1

v 2 j

—[2

—[- 2]

1]

3 B21 — -A22A21B11 —

-1

1

-1 1

2 1

[- 2] —

J

3 - 2

4 B22 — A22 A22A21B12 —

1

- 1

1

-1

1

1 -1

-1            1

" 3           3/'

3              / 2

- 2           -1

2 1

[2 1]

- 2 - 2" 1 2

Jadi

A- —

1 =                          B12

                B21         B22

- 2           2              1                              - 2           2              1

3              - 2           - 2           —           3              - 2           - 2

- 2           1              2                              - 2           1              2

Contoh 2.

Tentukan invers matriks berikut:

An = [1], A12 =[4 3]

                "1            4              3 1         

A —       2              5              4             

                1              -3            - 2          

" A11     A12

_ A21     A22 _

A21 =

2 1

                " 5 4 1

, A22 =  _- 3 - 2_

1

A- -

(A11 A12 A22 A21) ( B11A12 A22 ) (- A22A21B11 ) (aA22 - A22 A21B12 )

A. =

22

5 4 -3 -2

det A22 = -10 +12 = 2, kofaktor matriks A22 : Kn

= -2, K„ -

12

K21 = -4, K22

3, 5

K =

" K11      K12                         "- 2         3"            ^ Kt -     "- 2 - 4"

K21         K22 _                     - 4           5                              _ 3 5 _

= Adj (A22)

A- -

Adj(A22 ) det (A22)

^ A-1 - —

22

1              "- 2 - 4"                 "-1 - 2"

2              3 5                          3/2 5/2

A12A22A21 - [ 3]

-1 - 2 3/2 5/2

[2"                          " 1"

[1                            _ 2 _

A11 A12 A2>2> A21 - [1]

" 1"                         " 1"

_ 2 _                      _ 2 _

(A11 A12 A22 A21 ) -

1 2

-1

-[2]

1. Bn = (au AuA22 a21 ) - [2]

2. B12 =- B11A12 A- --[2][4 3]

-1 - 2

= [-1 1]

3. B21 = a- AH B11 =

1 - 2"      "2"

3/ 5/ _/2 /2 _     1

[2] =

8 -11

4. B22 AH A22 A21B12

(-1

Jadi

A-1 =

" B"        B11

_B21      B22 _

"-1 - 2"                  "-1 - 2"  "2"

V 5/ _/2 /2 _                       V V _/2 /2 _        1

-1 - 2

[-1 1]

4

-4

- 'X %

- 5 7

" 2           -1            1 "                           " 2           -1            1

8              - 5           2              —           8              -5            2

-11          7              - 3                           -11          7              -3

2

-3

Contoh 3.

Tentukan invers matriks berikut:

A —

0              1              2              2

1              1              2              3

2              2              2              3

2              3              3              3

r An        A12

_ A21     A22 _

A11 =

Ai —

21

A"1 —

(A11 A12 A2 A21) ( B11A12 A22 ) (- A22A21B11 ) (A22 - A22A21B12 ).

A22 =

2 3 33

0              1                              2              2

1              1              A12 —   2              3

2              2"                            "2            3

2              3              , A22 —                3              3

det A22 — 6-3 — -3, kofaktor matriks An : Kn — 3, K12 —-3

11

11

-12

K21 — -3, K22 — 2

K—         " Kn        K12                         " 3 - 3"   ^ Kt —  " 3 - 3"

                K21         K22                         _- 3 2 _                 _- 3 2 _

— Adj (A22)

A-1 =

22

Adj(A22 ) det (A22)

A-1 =

1

22

- 3

3 - 3 - 3 2

A A- A = 12 22 21 _

"2            2"            1              "3            - 3"         "2            2"            1              "4            6"

2              3_           - 3           - 3           2              2              3_           =3           6              6_

A - A a"1 A =

^11 12 22 21

"0            1"            -1 -3       "4            6"            1              "- 4         - 3"                         4 - 1l 3

1              1                              6              6_           = 3          _-3         - 3_                        - 1 - 1

det (at A12 A221 A21) = |-1 = 3 kofaktor matriks                A221 A21):

K11 =-1, K12 = 1 K21 = 1, K22 =

12

21

22

-4

K =

-1 1

1 -4/3

» Kt =

-1 1 1 -4 3

= Adj ( A11-A12 A22 A21)

(-A11-A12 A22 A21 ) =

1              -1 1                         -1 1                         -3 3

                _ 1 - 4/3_             = 3          _ 1 -4/3_              =            

13                                                                           _ 3 -4_

1. B„ =

(A11 - A12A22A21 )

2. B12 = -B11A12 A22 =

-3

3

1

3. B21 = -A22A21B11 =   3

22

-1 _ A -1A B

22 22 21 21

                "- 3                         3 "                                                                                                          

                _ 3                          -4                                                                                                           

                3 "           "2            2              1 "           -3            3"                            -3            2"           

                - 4           2              3              3 _          3              - 2                           4              - 2          

- 3 3 "                                     "2            2"            "- 3         3"                            "- 3         4"           

3              -              2              2              3_           _ 3          - 4                           2              - 2          

1              "- 3                         3 "           -1            "- 3         3"            "2 2"                      "-3          2

3              3                              -2            -3            3              - 2           2 3                          4              -2

-1 1

1

-2

-1 1

1 - 4

- 5 3

3

-2

Jadi

A- =

                                                - 3           3              -3            2                              - 3           3              -3            2

                B12                         3              - 4           4              - 2                           3              -4            4              -2

B21         B22 _                     - 3           4              - 5           3                              - 3           4              -5            3

                                                2              -2            3              - 2                           2              -2            3              -2

Metode Matriks Adjoint

Jika A adalah matriks ukuran n*n. Kofaktor (K) dari matriks A :

                ' K11       K12         • • • K ^              

K =          K 21 • • •             K 22        •K JV2 n • • •    di mana, Kj = (-1)1+J M„ V V

                v Kn1     Kn 2       • Knn y

Transpose dari matriks kofaktor (K) disebut matriks Adjoint A dinyatakan : Adj A

                r Ku        K12         • • • K  T              r k„         K 21        • • • K ^

Adj A = Kt =        K 21 • • •             K22         •K JV2 n • • •    =             K12 • • •              K22         •K Jvn 2 • • •

                v Kn1     Kn 2       • " Knn y                              v K1n     K 2 n      • " Knn y

Invers matriks dapat ditentukan dari matriks Adjoint (Ajd). Jika A adalah matriks ukuran n*n

dan det A^Q, maka :

r

A"1 =

r 1

\

V

det A

Adj A =

r 1

\

j

V

det A

j

K„ K

11

?12

V K1n

21

KK

22

K

2n

K

\

n1

K

2n

K

nn j

A"1 =

r Ku        K 21        Knl >

det A K12             det A K 22            det A K 2 n

det A • •              det A     det A • •

• K1n     K 2 n      • Knn

V det A det A     det A j

Contoh 1.

Tentukan inverse matriks berikut,

A =

2 3 2 2 2 1 1 2 2

Kn = (-1)

Kn = (-1)

Ku = (-1)

K 2! = (-1)3

K 22 = (-1)

2 1 2 2

21 12

22 12

3 2 22

22 12

= 4 - 2 = 2

K 23 = (-1)

= -(4 -1) = -3 K31 = (-1)

= (4 - 2) = 2 K32 = (-1)

= -(6 - 4) = -2 K33 = (-1)

= 4 - 2 = 2

23 12

32 21

22 21

23 22

= -(4 - 3) = -1

= 3 - 4 = -1

= -(2 - 4) = 2

= 4 - 6 = -2

det A = a11 K11 + a12 k12 + a13 k13 = 2(2) + 3(-3) + 2(2) = -1

4

5

6

K =

2-3 2 -2 2 -1 -1 2 -2

AdjA = K

T

( 1 ^ l-u

Adj A = -

2 -2 -1 -3 2 2 2 -1 -2

Jadi /T1 =

-2 2 1 3 -2 -2 -2 1 2

Buk+i : AAX=I

2 3 2 2 2 1 1 2 2

-2

-2

2 -2 -1 -3 2 2 2 -1 -2

2 2 1 * 3 -2 -2 2 1 2

2              1                              1              0              0

-2            -2            —           0              1              0

1              2                              0              0              1

Contoh 2.

Tentukan inverse matriks berikut,

1 4

A =

2 5 4

K,, = (-1)

K12 = ("I)3

K13 = ("1)

*21 = ("1):

K22 = ("1)'

3                             

4                             

- 2                          

5              4             

-3            -              2

2              4             

1              -2           

2              5             

1              -3           

4              3             

-3            -              2

1              3             

1              2             

= -10 +12 = 2

= -(-4 - 4) = 8

= (-6 - 5) = -11

= -(-8 + 9) = -1

= -2 - 3 = -5

K 23 = (-1)

K 3! = (-1)

K 32 = (-1)

K33 = (-1)

14          

1-            3

43          

54          

13          

24          

14          

25          

= -(-3 - 4) = 7

= 16 -15 = 1

= -(4 - 6) = 2

= 5 - 8 = -3

det A — an K11 + a12 K12 + a13 K13 — 1(2) + 4(8) + 3(-11) = 1

                2 8          -11                          2 -1 1

K —        -1 - 5      7              Adj A = Kt —      8 - 5 2

                1 2          - 3                           -11 7 -3

A- —

1 >| v 1,

Adj A =

2 8 -11

-1 -5 7

1

2

-3

Jadi A"1 =

2 -1 1 8 - 5 2 -11 7 -3

                1              4              3              2              -1            1                              1              0              0

Bukti : AA- = I     2              5              4              8              -5            2              —           0              1              0

                1              -3            - 2           -11          7              - 3                           0              0              1

Contoh 3.

Tentukan inverse matriks berikut,

A =

0              1              2              2                              1              2              3             

1              1              2              3              K,1 = (-1)2           2              2              3              = (6 +18 +18) - (18 + 9 +12) = 3

                                                                                3              3              3             

2              2              2              3                              1              2              3             

2              3              3              3              k12 = (-1)3           2              2              3              = -((6 +12 +18) - (12 + 9 +12)) = -3

K13 = (-1)

K14 = (-1)5

2 3 3 1 1 3 223 233 1 1 2 222

= (6 + 6 +18) - (12 + 9 + 6) = 3

K 21 = (-1)

2 1

2 3

33 22 23 33

= -((6 + 4 +12) - (8 + 6 + 6)) = -2

= -((6 +18 +12) - (12 + 9 +12)) = -3

                                                0              2              2                                                                                             

K 22        = (-         1)4          2              2              3              = (0 +12 +12) - (8 + 0 +12) = 4                                                                     

                                                2              3              3                                              0              2              2             

                                                                                                K 32 = (-                1)5          1              2              3              = -((0 +12 + 6) - (8 + 0 + 6))

                                                                                                                                2              3              3             

                                                0              1              2                                                                                             

K 23        = (-         -1)5        2              2              3              = -((0 + 6 +12) - (8 + 0 + 6)) = -     -4                                                           

                                                2              3              3                                              0              1              2             

                                                                                                K 33 = (-                1)6          1              1              3              = (0 + 6 + 6) - (4 + 0 + 3) = 5

                                                0              1              2                                              2              3              3             

                                                                                                                                                                               

K 24        = (-         -1)6        2              2              2              = (0 + 4 +12) - (8 + 0 + 6) = 2                                                                        

                                                2              3              3                                              0              1              2             

                                                                                                K 34 = (-                1)7          1              1              2              = -((0 + 4 + 6) - (4 + 0 + 3)) =

                                                                                                                                2              3              3             

= -4

K 3! = (-1)

1 2 2 1 2 3 3 3 3

= -3

(6 +18 + 6) - (12 + 9 + 6) = 3

K 41 = (-1)

1 2 2 1 2 3 223

-((6 +12 + 4) - (8 + 6 + 6)) = -2

4

5

K 42 = ("I)

0 2 2

1              2 3

2              2 3

K 43 = (-1)

= (0 +12 + 4) - (8 + 0 + 6) = 2

0 1 2 1 1 3 223

                0              1              2

K 44 = (-1)8         1              1              2

                2              2              2

(4 + 0 + 3))                           -3           

= (0 + 4 + 4) - (4 + 0 + 2) = 2

det A = an Kn + an Kn + au K^ + a^ K^ = 0(3) +1(-3) + 2(3) + 2(-2) = -1

K =

3              -3            3              -2

- 3           4              -4            2

3              -4            5              -3

- 2           2              -3            2

Adj A = Kt =

3              -3            3              -2

-3            4              -4            2

3              -4            5              -3

-2            2              -3            2

a"1 =

1

(

y det A

A

)

Adj A =

1

y-1)

Jadi

3              - 3           3

-3            4              -4

3              - 4           5

-2 2         -3

A4 =

-              3              3 - 3        2 3          - 4 4        - 2

-              3              4 - 5        3 2          -2 3         -2

Bukti :

-2 2

-3 2

-3 3

-3 2

3

-4

4 -2

-3 4

-5 3

0              1              2              2              - 3           3              -3            2                              1              0              0              0

1              1              2              3              3              -4            4              - 2                           0              1              0              0

2              2              2              3              - 3           4              -5            3                              0              0              1              0

2              3              3              3              2              -2            3              - 2                           0              0              0              1

2 -2 3 -2

Metode Eliminasi Gauss-Jordan

Jika A adalah matriks ukuran n*n. Eliminasi Gauss-Jordan terhadap matriks ekstensi A akan menghasilkan invers matriks A.

Matriks ekstensi (augmented matrix) A adalah matriks yang dibentuk dengan meletakan matriks identitas (I) di sebelah kanan matriks A :

r

a11 a12

a21 a22

Van1 an2

a

1n

a

2n

AI v 1 0 0 1

0 ^ 0

ann A 0 0 ••• 1 y

a11 a12

a21 a22

an1 an2

AI

a

1n

a

2n

a

10 01

nn

0 0

0 0 ••• 1

Eliminasi Gauss-Jordan terhadap matriks ekstensi A

AI

I

A

-1

a11 a12 a21 a22

an1 an 2

a

1n

a

2 n

a

1 0

0 1

nn

00

0 0

1

10 01

Eliminasi Gauss-Jordan

0 0

0              0              l               1

l11 i12 i21 a22

in1 in2

1n

2n

nn

A-1 =

*11 Z12

*21 l22

in1 in2

l1n

v2 n

nn

Di mana idj adalah enrti matriks invers yang berasal dari aijd setelah mengalami beberapa kali operasi baris elementer.

Contoh 1.

Tentukan inverse matriks berikut,

                2              3              2                              2              3              2              1              0              0

A =         2              2              1              AI=         2              2              1              0              1              0

                1              2              2                              1              2              2              0              0              1

"2 3 2 1 0 0" 2 2 10 10 1 2 2 0 0 1

b21(-1) b,i(-1/2)

2 3 2 1 0 0 0 -1 -1 -1 10 0 0.5 1 - 0.5 0 1

2              0              -1            -2            3              0              b13(2)   2              0              0              - 4           4              2

0              -1            -1            -1            1              0                              0              -1            0              -3            2              2

0              0              0.5          -1            0.5          1              b23(2)   0              0              0.5          -1            0.5          1

b12(3)

b32(1/2)

bi(1/2) b2(-1)

b3(2)

2 0 0 -4 4 2 0-10-322 0 0 0.5 -1 0.5 1

M1/2) b2(-i)

b3(2)

Jadi A~' =

-2 2 1 3 -2 -2

-2 1

2 3 2

-2

Bukti : AAT1 = /

2 2 1

1 2 2

-2

1 0 0 j-2 2 1 0 1 0 | 3 -2-2 0 0 11-21 2

2              1 "                           "1            0              0"

-2            -2            —           0              1              0

1              2                              0              0              1

Contoh 2.

Tentukan inverse matriks berikut,

A =

1              4 3

2              5 4 1 - 3 - 2

l"1 4

2 5

0 - 3 0 0

AI=

14 3 10 0 2 5 4 0 1 0 1 -3 - 2 0 0 1

3              10 0

4              0 10

1 - 3 - 2 0 0 1

1 0

M-2)

1

- 7

bsi(-l)

0 0 1

bi3(l)

b2,(-6)

1 0 0

14 3 10 0 0 - 3 - 2 - 2 10

0 - 7 - 5 -10 1

bi2(4/3) b32(-7/3)

2 -1

1

0 -3 0 -24 15 -6

00

1/ - 7/ 3 3 3

1

1 0

0 - 3 00

0

2

0 - 24

-1 15

1/ iy -7/

3 3 3

1              b2(-1/3)               1              0              0              2              -1            1

- 6                           0              1              0              8              -5            2

1              b3(-3)    0              0              1              -11          7              -3

Jadi A"1 =

2 -1 1 8 - 5 2 -11 7 -3

                "1            4              3 "           " 2           -1            1 "                           "1            0              0"

Bukti : AA- = I o 2              5              4              8              -5            2              —           0              1              0

                1              -3            - 2           -11          7              - 3                           0              0              1

Contoh 3.

Tentukan invers dari matriks berikut:

A =

2 1          2              1

4 1          2              2

12           2              2

2 1          1              2

^ AI=

2              1              2              1              1              0 0          0

4              1              2              2              0              1              0              0

1              2              2              2              0              0 1          0

2              1              1              2              0              0 0          1

2              12           110         0              0

4              1 2          2 0 1       0              0

1              2 2          2 0 0       1              0

2              1 1          2 0 0       0              1

b2i(-2)

b3i(-i/2) b4i(-l)

2 1 2 1

1 0 0 0

0 -1 - 2 0 - 2 10 0 0 1.5 1 1.5 - 0.5 0 1 0 0 0 -1 1 -1 0 0 1

1.

1.

2 12 1 1 0 0 0 0 -1 - 2 0 - 2 10 0 0 1.5 1 1.5 - 0.5 0 1 0

0 0 -11 2 0 0 1

b12(1)

b32(1.5)

-1 0 0 1 -1 10 0

0 -1 - 2 0 - 2 10 0

0 0 -2 1.5 -3.5 1.5 1 0 0 0 -1 1 -1 0 0 1

b23(-1) —

M-0.5)

2              0              0              1              -1            1              0              0              b14(-4)

0              -1            0              -1.5        1.5          -0.5        -1            0              b24(6)

0              0              -2            1.5          -3.5        1.5          1              0              -^ b34(-6)

0              0              0              0.25        0.75        -0.75      -0.5        1             

4.

5.

2 0          0              0              -4                            4              2              - 4           b1(0.5)

0 -1         0              0              6                              -5            -4            6              b2(-1)

0 0          -2            0              - 8                           6              4              - 6           -^ bs(-0.5)

0 0          0              0.25        0.75        -              0.75        -0.5        1              b4(4)

"1 0        00           -2            2              1              - 2"                                        

0 1          00           - 6           5              4              - 6                                          

0 0          1 0          4              -3            -2            3                                             

0 0          01           3 "- 2      -3 2         -2 1         4 - 2"                                     

Jadi        A- =        - 6 4 3    5 -3 -3    4 -2 -2    - 6 3 4                                   

Contoh 4.

r

Hitung invers dari matriks : A =

O 0

0 3

0 5

5

17/

0

11 0

0\ X b2

0 1 >5b3

0 1

y

V

V

1 0 1 0 1 5 0 0 1

30

0 01

>3 >3 0

1/ _ 1/ 1/ .

/6 /3 /5 y

30 b3

10 01 0 1 %

f 0 12 0 1 V0 0 1

0

13 13

\

2 0 1 _ 2 3 4

V-5 5 6y

f 2           0              1              1              0              0 >          >2 b       f 1           0              1 2          12           0 0^

_ 2          3              4              0              1              0                              _ 2          3              4              0              1 0

V_ 5       5              6              0              0              1y                           V_ 5       5              6              0              0 1y

2bl + b2 5b, + b3

01 0

b2 +b3

1 2          0              1 5y       

12           0              01           _ Xbs + b _ 53 b3 +b2

13           1 3          0             

5              _ 10        6y          

r 1           0              0              - 2           5              - 31                         r- 2         5              - 3!

0              1              0              - 8           17           -10          Jadi A-1 =             - 8           17           -10

v 0          0              1              5              -10          6 ,                           v 5          -10          6 ,

Contoh 5. Hitung invers dari matriks :

A =

0 1 4

1 2 0 3 3 8

[ A I ] =

Jadi A"1 =

0              1 2 10 0

1              0 3 0 1 0

4              - 3 8 0 0 1

1              0 3 0 1 0

0              12 10 0

0              - 3 - 4 0 - 4 1

1              0 3 0 1 0 0 1 2 1 0 0

0 0 1 3/2 - 2 1/2

-              9/2 7 - 3/2

-              2 4 -1 3/2 -2 1/2

3 0 1 0 2 1 0 0

10 01

4 -3 8 0 0 1

1 0 3 0 1                0

0 1 2 1 0                0

0 0 2 3 - 4              1

1 0 0 - 9/2 7 - 3/2"

0 10 - 2  4 -1

0 0 13/2                - 2 1/2

Metode Perkalian Matriks Invers Elementer

Jika A adalah matriks ukuran n*n. Perkalian matriks invers elementer (E) dapat menghasilkan invers matriks A.

(EnEn-1 En_2 E3E2E, )A = I

Di mana :

I = matriks identitas

A-1 = EnEn_1 En_2 l E3E2E1 ^ Invers Matriks A Et = matriks invers elementer

Et diperoleh dari transformasi matriks identitas :

E =

1 0 • 0 1 •

N, N ,

0 0 ... Nu

0 0 L Nu

Misalnya : nilai E untuk matriks ukuran 4*4.

0 0

0

^ matriks identitas (I) di mana kolom ke - i diganti oleh Nk t

E =

E =

N11                        0              0              0

N21                        1              0              0

N31                        0              1              0

N4,                         0              0              1

1              0                              N,3         0

0              1                              N23        0

0              0                              N33        0

0              0                              N43        1

E =

E =

'4

1              n12                         0              0

0              N22                        0              0

0              N32                        1              0

0              N42                        0              1

1              0              0                              N,4

0              1              0                              N24

0              0              1                              N34

0              0              0                              N44

1

N

k .i

a

1.i

a

— a

2.i

a

a

a

i.i

Nki = Normalitas vektor kolom ke-i

Misalnya : nilai Nki untuk matriks ukuran 4*4

                                " 1 "                        a1.2                        a1.3

a(i —1).i ai.i                        a1.1                        a2.2                        a33

1                              — a2.1                  >                             — a2.3

a ? i.i                     a1.1                        a2.2                        a33

a(i+1).i  Nk.1 =                   Nk.2 =   ^--          Nk.3 =  

ai.i i.i                      — a3.1                  — a3.2                  1

;                               a1.1                        a2.2                        a3.3

a( «—1).i ai.i i.i                  — a4.1 a1.1                         — a4.2 a2.2                         — a4.3 a3.3

N, =

k .4

- a

1.4

a

4.4

-a

2.4

a

4.4

-a

3.4

a

4.4

Nilai au diperoleh dari :

ak i = (Fi-1 )-1 A = Et-1 Et -2 " • E2 E1 IAi, untuk i > 2 dan I = matriks identitas

ak.i = [a1.i' a2.i' " '' ai.i' " ' ' a(n-1).i' an.i ]

Untuk i = 1 ^ ak1 = A1 atau aki = (a11 , 21, a31, , an1

) ^ elemen pada kolom 1

di mana Ai = elemen kolom ke - i pada matriks A

Nilai At adalah elemen kolom ke-i pada matriks A. Misalnya At untuk matriks ukuran 4*4.

                a11         a12         a13         a14                         a11                         a12                         a13                         a14

A =         a21         a21         a23         a24         , A, =      a21         , A2 =     a22         , A3 =     a23         , A4 =     a24

                a31         a32         a33         a34                         a31                         a32                         a33                         a34

                a41         a42         a43         a44                         _a41 _                   _a42 _                   _ a 43 _                 _ a44 _

A = [[1 A2 A3 A4]

Matriks Ft adalah identitas (I) di mana elemen kolom ke-1 ^ i diganti elemen kolom ke-1^i dari matriks A atau A. Misalnya Ft untuk matriks ukuran 4*4.

                an           0              0              0                              a11         a12         0              0                              a11         a12         a13         0                              a11         a12         a13         a14

F =          a21 a31 1 0          0 1          0 0          , F2 =     a21 a31 a22 a32 0 1          0 0          , F3=      a21 a31 a22 a32 a23 a33 0 0          , F4=         a21 a31 a22 a32 a23 a33 a24 a34

                _a41      0              0              1                              _a41      a42         0              1                              _a41      a42         a43         1                              _a41      a42         a43         a44

Langkah-langkah menghitung matriks invers :

1. Tentukan matriks Fj yaitu mengganti kolom pertama matriks identitas (In) dengan Aj. Invers dari Fj adalah (Fj)~j.

(F )-1 = E1 I = e

2.            Tentukan matriks F2 yaitu mengganti kolom kedua matriks Fj dengan A2. Invers dari F2 adalah (F2)-J.

(f2 )—1 = e2 (f )—1 = e2 E1

3.            Tentukan matriks F3 yaitu mengganti kolom ketiga matriks F2 dengan A3. Invers dari F3 adalah (F3)-J.

(f3 ) = E3 (F2 ) = E3 E2 E1

4.            Prosedur ini terus dilakukan hingga semua vektor kolom A dimasukan ke matriks Fi atau hingga Fi =A, maka akan diperoleh matriks invers :

A"1 =(F )—1 = E, L E3E2E1

Contoh 1.

Tentukan inverse matriks berikut,

A =

2 3 2 2 2 1 1 2 2

A = [[1 A2 A3 1 A

2 2 1

, A2

3 2 2

A3 =

2 1 2

                "2 0 0"                                                  

F =          2 1 0       ak.1                        = H =     

                1 0 1                                                       —

                V an                       12                           0.5

Nk.1 =   — an     =             - 2/2       =             -1

                — a31/ an                           -12                          - 0.5

1 0 0 0 1 0 0 0 1

E =

2 2 1

N

11

N

21

N

31

2 2 1

00 10 01

0.5 0 0 -1 1 0 - 0.5 0 1

(F )-1 = El —

0.5 0 0 -1 1 0 - 0.5 0 1

F2 =

2 3 0 2 2 0 1 2 1

                " 0.5       0              0"            3                              1.5"

ak2 =(F1)_14 = E14 =       -1            1              0              2              —           -1

                - 0.5       0              1              2                              0.5

                ^12/ a22                               1.5/ -1"                 "1.5"                      "1            N12        0"                            "1            1.5          0

N .2 —  1 a22      —           1/ -1       —           -1            E2 —      0              N22        0              —           0              -1            0

                ^32/a22 _                            - 0.5/ -1                                0.5                          0              N32        1                              0              0.5          1

                "1            1.5          0"            0.5          0              0"                            -1            1.5          0

(F2 )—E2 E1 —  0              -1            0              -1            1              0              —           1              -1            0

                0              0.5          1              - 0.5       0              1_                           -1            0.5          1

^3 =

2 3 2 2 2 1 1 2 2

a

k3

=(f2TA =

                ^13/^33                               -(-0,5)/0.5                           1

                — ^3 ! a33           —           -1/0.5    —           -2

                l/a33                      1/0.5                      2

{F3y=Ep2y=

1 0 1 0 1 -2 0 0 2

-1 1.5 1 -1 -1 0.5

Jadi A~l =

-2 2 1 3 -2 -2 -2 1 2

-1 1.5 0  2                              -0.5

1 -1 0     1              —           1

-1 0.5 1  2                              0.5

e3 =

1 0 M

13

0 1 N.

23

0 0 iV.

0 0 1

33

1

-2 2 3 -2 -2 -2 1 2

1 0 1 0 1 -2 0 0 2

Contoh 2.

Tentukan inverse matriks berikut,

                ~1 4        3 "                           1                              4                              3

A —       2 5          4              A = IA A2 A3 J, A =            2              , A2 =     5              , A3 =     4

                1 - 3        - 2                           1                              - 3                           - 2

                "1 0 0"                                                                                                   "1 0        0"            "1"                          "1"                         

F =          2 1 0                                       akA:                       —A        —           0 1          0              2              —           2                             

                1 0 1                                                                                                       0 0          1              1                              1                             

                V an                                       11                           1                                              " N11                     0              0"                            "1 0 0

NkA =    a21l a11                                —           - 2/1       —           - 2           E —                        N21                        1              0              —           - 2 1 0

                - a31 /a11                                            -1/1                        -1                                            N31                        0              1                              -1 0 1

                " 1 0       0"                            "1            4              0"           

(F )-1 = E —         - 2 1        0              F —        2              5              0             

                -1 0         1                              1              - 3           1             

                1 0 0"     " 4"                         "4"

°k.2 =(F1))1 A2 = E1A =  - 2 1 0    5              =             - 3

                -1 0 1     - 3                           - 7

                ^12/ a22                               " - 4/- 3 "                              " 4/3 "                   "1            Nu          0"                            "1            4/3         0

Nk .2 =  V a22     =             1 - 3        =             -13          E2 =        0              N22        0              —           0              -13          0

                - ^32/a22                             - (-7)/- 3                               - 7/3                       0              N32        1                              0              - 7/3       1

                "1            4/3         0"            " 1 0       0"                            "- v3       4/3         0

(F2 )-1 — E2 e —              0              -1/3        0              - 2 1        0              —           2/3         -1 3         0

                0              - 7/3       1              -1 0         1                              11/3       - 7/3       1

                "14 3"                    "- 53       43           0"            " 3 "                        " 13 "

F=           2 5 4       ak .3 = (F )-1 A3 =             23           -1 3         0              4              —           2/3

                1 - 3 - 2                  113         -7 3         1              - 2                           -13

N, =

k .3

ai3l a33 — a23l a33 V a33

-13

-13

- 213

-13

F )-1 = £3 F )-1 =

Jadi A- —

-13

1 0 1T- 53 0 12 23 0 0 - 3JL113 2 -11" 8 - 5 2 -11 7 -3

1                                                              "1 0        N13"                      "1            0              1

2                              E3 —                      0 1          N23        —           0              1              2

- 3                                                           0 0          N33 _                    0              0              -3

4/3                         0"                            " 2           -1            1 "                                          

-1 3                         0              —           8              - 5           2                                             

- 7/3                       1                              -11          7              -3                                           

Contoh 3.

Tentukan inverse matriks berikut,

A —

2 1 4 1 12 21

21 22 22 12

                2                              1                              2                              1                              "2            0              0              0

                4                              1                              2                              2                              4              1              0              0

^ A, —  1              , a2 —   2              , A3 —   2              , A4 —   2              F1 —      1              0              1              0

                2                              1                              1                              2                              2              0              0              1

ak.1 —

E —

                "1            0              0              0              2                              2                             

                0              1              0              0              4                              4                             

IA1 —                                                                                    —                                          

1              0              0              1              0              1                              1                             

                _0           0              0              1              _2                           2                             

Nn          0              0              0"                            12                           0              0              0

N21        1              0              0                              - 2                           1              0              0

N31        0              1              0                              -1 2                         0              1              0

N41        0              0              1                              -1                            0              0              1

NkA —

V an                       1/2                         12

- #21/°11                              - 4/2                       - 2

- 031/ au                              -1/2                        -12

- °41 / °11                             - 2/2                       -1

(F )-1 — E1 —

12 - 2

-12 -1

0 0 1 0 01 00

0 0 0 1

F =

Nk2 =

2 10 0 4 10 0 12 10 2 10 1

-              a12 / a22 1 a22

-              a32 / a22

-              a42 a22

ak .2 =(F1 )—1 A2 = E1A2 =

F )—1 = e2 Ex =

1

—1

1 -1

—1

— 0/ — 1

1 2 0 0

0 — 1 0 3/2 00

00 10 01

-1/2 —1

3/2 0 12

F =

21 41 12 21

20 20 20 11

12

—2         1

—1 2      0

—1         0

000 00 10 01

[1"                          "1/2"

1                              — 1

2                              3/2

[1                            _ 0 _

E2 =

1              N12        0              0"                            "1            12           0              0

0              N22        0              0                              0              —1         0              0

0              N32        1              0                              0              32           1              0

0              N42        0              1                              _0           0              0              1

000

—2 1 —1 2 0 —1 0

00 10 01

—1 2 1 2 0 0

0 0

0 0 1

2 —1 0 — 7/2 3/2 1

—1

ak3 ={F2 )—1 A3 =

                "—1 2    12           0              0"            "2"                          "0"

1              2              —1         0              0              2                              2

1A3 =     — 7/2    32                                                           =            

3                                              1              0              2                              — 2

                —12       0              0              1              1                              — 1

Nk.3 =

                                " - 0/- 2 "                              " 0

a23 / a33                              - 2/ - 2                   1

V a33                     1 - 2                        -12

a43 1 a33 _                          _- (-1)/- 2_                          _-12

E =

10           N13        0

0 1          N23        0

0 0          N33        0

0 0          N43        1

F )-1 = E3 (F2 )-1 =

1 0 0 1 00 00

00 10 1/2 0 1/2 1

F =

2 12 1 4 12 2 12 2 2 2 112

ak.4 =(F3 )-1 A

4

-12          12 0 0"

2              -110

-7/2        3/2 0 0

-1            0 0 1

-1 2 -3 2 7/4 3/4

-1 2 -3 2 7/4 34

12 12 -3 4 -3 4

0 1

-1 2 -1 2

0 0 0 1

12 12 -3 4 -3 4

10 01 00 00

00 10 -1 2 0 -1 2 1

00 10 -12 0 -12 1

[1"                          " 12 "                     " 0.5 "

2                              3/2                         1.5

2                              - 3/4                       - 0.75

l_2                          _ 14 _                    _ 0.25 _

NkA =

^14 /^44             

— $24 /^44        

— /$44

1/ ^44  

E4 =

1 0 0 N,

14

0 1 0 TV,

24

0 0 1 TV

34

0 0 0 TV

44

(F4y=E4(F3y =

Jadi

A~l =

-0.5/0.25 -1.5/0.25 -(-0.75)/0.25 1/0.25 10 0-2 0 10-6 0 0 1 0 0 0

10 0-2 0 10-6

0 0 1 0 0 0

-2 2 -6 5

4 -3 -2 3 -3 -2

3

4

1 -2 4 -6

3

4

3

4

-1/2 -3/2 7/4 3/4

-2 -6

3

4

1/2         0              0] [-2     2              1              -2

1/2 1      0 _ -6     5              4              -6

-3/4        -1/2        0 ~ 4       -3-2        3

-3/4        -1/2        1J |_ 3   -3-2        4

Sifat Invers Matriks

Jika A matriks invertibel hanya akan mempunyai satu matriks invers ( invers A adalah unik) dan dinyatakan oleh A-1.

AA- — I ^ A"1 A — I

Jika determinan A adalah nol (det A=0), A-1 tidak ada dan matriks A disebut matriks non invertibel atau singular.

Jika matriks A dan B adalah matriks non singular atau invertibel, maka :

(AB)"1 — A"1B"1

Jika invers dari matriks invertibei adalah invertibei maka :

(A-')-' = A

Perkalian skalar k (k^0) dengan matriks invertibei adalah invertibei, maka :

(kA)-1 = 1 A"'

Jika matriks A adalah matriks non singular atau invertibel, maka :

A )-1 =A )T

Info

Slide Matematika dan slide-slide lainnya yang ada di Site SmartStat dapat dipelajari pada tautan di bawah ini:
Daftar Slide Matematika II
Daftar Seluruh Slide

Slide lainnya bisa Anda download :di sini