Gabungan Bab II Matrik

8/18/2019 Gabungan Bab II Matrik

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1

MATRIK

(KALKULUS III)

Definisi Matrik

Yaitu suatu elemen yang disusun dengan adanya baris dan kolom aij , dimana

i = nomor baris dan j = nomor kolom.

1. Penjumlahan / Penuranan

!"nt"h#

Jika : A =

11

32

, B =

14

53

it : a. A ! B

b. A " B

Pen$elesaian #

A ! B =

11

32

!

14

53

=

++

++

1141

5332

=

25

#5

A $ B =

11

32

$

14

53


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2

=

−−

−−

1141

5332

=

−

−−

%3

21

%. Perkalian matrik

&isal : A&% ' %& !%%

Baris ' (olom = )erkalian &atrik

!"nt"h#

Jika : A =

111

132

, B =

11

41

32

it : A * B

Pen$elesaian#

A * B =

111

132

*

11

41

32

=

++++

++++

1.14.13.11.11.12.1

1.14.33.21.11.32.2

=

#4

1+#

A ' ≠ ' A

!"nt"h #

1


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3

Jika : A =

42

31

, B =

22

13

it : A * B B * A

)enyelesaian :

A * B =

42

31

*

22

13

=

++++

2.41.22.43.2

2.31.12.33.1

=

1%14

-+

B * A =

22

13

*

42

31

B * A =

++++

4.23.22.21.2

4.13.32.11.3

=

14

135

Jadi, terbukti matrik A * B ≠ B * A

&. Trans*"se Suatu Matrik

/rans0ose suatu matrik adala matrik yang meruba kolom menjadi

baris dan baris menjadi kolom.

!"nt"h #


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4

Jika : A =

41123

154

it : a. At

b. A ! At

. Att = A

Pen$elesaian #

a. A =

411

23

154

At =

421

15

134

b. A ! At =

411

23

154

!

421

15

134

=

#32

312#

2##

A ! At = matrik simetris, bukti a12 = a21

. (At) t A

A =

411

23

154


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5

t

42115

134

=

41123

154

Jadi /erbukti Att = A

(A + ) t At + t

!"nt"h #

Jika : A =

123

3-5

54

, B =

121

43

543

it : A ! Bt = At ! Bt

)enyelesaian :

At =

135

2-

354

B t =

145

234

13

A ! B =

123

3-5

54

!

121

43

543

=

244

-1%11

1%1%-


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A ! Bt =

t

244

-1%11

1%1%-

=

2-1%

41%1%

411-

A t ! B t =

1352-

354

!

145234

13

=

2-1%

41%1%

411-

Jadi /erbukti A ! Bt = A t ! B t

(A ' )t t ' A t , At ' t

!"nt"h #

Jika : A =

11

425

234

, B =

223

152

421

it : A * Bt = Bt * At At

)enyelesaian :

A t =

142

123

54


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-

B t =

214

252

321

A * B =

11

425

234

*

223

152

421

=

++++++++++++

++++++

2.11.14.2.15.12.3.12.11.

2.41.24.52.45.22.53.42.21.5

2.21.34.42.25.32.43.22.31.4

=

2-1+11

3%2#21

232-1

A * Bt =

2-3%23

1+2#2-11211

B t * A t =

214

252

321

*

142

123

54

B t * A t =

++++++++++++++++++

1.21.1.44.22.15.42.23.14.4

1.21.5.24.22.55.22.23.54.2

1.31.2.14.32.25.12.33.24.1


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#

=

2-3%231+2#2-

11211

At * Bt

142

123

54

*

214

252

321

++++++++++++++++++

2.12.43.21.15.42.24.12.41.2

2.12.23.31.15.22.34.12.21.3

2.2.53.41.5.52.44.2.51.4

=

12514

151-11

343+3#

Jadi terbukti A * Bt = B t * A t At * Bt

-. Matrik Simetris

aij a ji

"nt"h #

−−53

542

321

#3

#2

324

#1

132

24

Kunin$a # dikatakan matrik simetris jika :

a1% a%1

a1& a&1

a&% a%&


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+

. Determinan

ara menari determinan dari suatu matrik ada bebera0a ara, yaitu :

• ara ko6aktor

• ara sarrus

ara (o6aktor

A =

2221

1211

aa

aa

0et A A

a11 a%% a%1 a1%

!"nt"h #

Jika : A =

-1

32

it : det A

Pen$elesaian #

det A = A

=

-1

32

= 2.- " 3.1

= 14 " 3

= 11

!"nt"h #


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1%

Jika : A =

524333

151

it : det A

Pen$elesaian #

ara 1 :

det A = A

=

524333

151

= 1

52

33

$ 5

54

33

! 1

24

33

= 1 3.5 " 3.2 " 5 3.5 " 3.4 ! 1 3.2 " 4.3

= 1 15 " " 5 15 " 12 ! 1 " 12

= + " 15 "

= $12

ara 2 :

7umus eterminan

A

=

131312121111 K a K a K a +−

A =

524

333

151

(olom 1 8 &inor dari baris 1 kolom 1


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11

( 11 det= $11!1

52

33

= 1 3.5 " 2.3

= 15 "

= +

(olom 28 &inor dari baris 1 kolom 2

( 12 det= $11!2

54

33

= $1 3.5 " 4.3

= $1 15$12

= 3

(olom 3 8 &inor dari baris 1 kolom 3

( 13 det= $11!3

24

33

= 1 3.2 " 4.3

= 1 $12

= $

Jadi A

= 1 + $ 5 3 ! 1 $

= + " 15 "

= $12


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12

ara 3 : ara 9arrus

!"nt"h #

Jika : A =

524

333

151

it : det A

)enyelesaian :

A

=

24

33

51

524

333

151

= 1.3.5 ! 5.3.4 ! 1.2.3 " 5.3.5

= 15 ! % ! " 12 " " -5

= $12

2. In3ers Matrik

A41 A

1

. At kofaktor

!"nt"h #

Jika : A =

%41

.14

532

it : A$1

Pen$elesaian #

A$1 = A

1

. At kofaktor


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13

A$1 = A

A kofaktor t

A$1 = A

adjA

dimana = Adj A = At kofaktor

imana A (o6aktor =

333231

232221

131211

K K K

K K K

K K K

A =

%41

14

532

(olom 1 8 &inor dari baris 1 kolom 1 a11 = 2

( 11 = $11!1

%4

1

= 1 1.% " 4.

= $24

8 &inor dari baris 2 kolom 1 a21 = 4

( 21 = $12!1

%4

53

= $1 3.% " 5.4

= 2%

8 &inor dari baris 3 kolom 1 a31= 1


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14

( 31 = $13!1

1

53

( 31 = 1 3. " 5.1

= 13

(olom 28 &inor dari baris 1 kolom 2 a12 = 3

( 12 = $11!2

%14

= $1 4.% " 1.

=


( 22 = $12!2

%152

= 1 2.% " 5.1

= $5


( 32 = $13!2

452

= 1 2. " 4.5

= #

(olom 38 &inor dari baris 1kolom3 a13 = 5


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15

( 13 = $11!3

41

14

= 1 4.4 " 1.1

= 15

8 &inor dari baris 2 kolom 3 a23 =

( 23 = $12!3

41

32

= $1 2.4 " 3.1

= $5

8 &inor dari baris 3 kolom 3 a33 = %

( 33 = $13!3

14

32

= 1 2.1 " 4.3

= $1%

A (o6aktor =

−−−

−

1%#13

552%

1524

At

ko6aktor =

−−−

−

1%515

#5

132%24


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1

A= A11 ( 22 ! A12 ( 12 ! A13

= 2 $24 ! 3 ! 5 15

A= $4# ! 1# ! -5

= 45

A$1 = A

A kofaktor t

A$1 = 45

1

−−

−

−

1%515

#5.

132%24

A$1 =

−−

−

−

45;1%45;545;15

45;#45;545;

45;1345;2%45;24

A$1 =

−−

−

−

+;2+;115;5

45;#+;115;2

45;13+;415;#

5. A0j"int Matrik ujur Sankar

/eta0kan A = ij A suatu matrik bujur sangat n ' n dan ' ij adala 6aktor aij,

maka menurut de6inisi:

A0j"int A A0j A =

321

22212

12111

...

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

...

...

nnn

n

n

K K K

K K K

K K K

At K"fakt"r


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1-

&isal : A =

433

232

321

=

333231

232221

131211

aaaaaa

aaa

!"nt"h #

Jika A =

433

232

321

it : Adj A dan A$1

Pen$elesaian #


( 11 = $11!1

43

23

= 1 3.4 " 3.2

=


( 21 = $12!1

43

32

= $1 2.4 " 3.3

= 1



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1#

( 31 = $13!1

23

32

= 1 2.2 " 3.3

= $ 5


( 12 = $11!2

43

22

= $1 2.4 " 3.2

= $ 2


( 22 = $12!2

43

31

= 1 1.4 " 3.3

= $ 5


( 32 = $13!2

22

31

= $1 1.2 " 2.3

= 4


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1+


( 13 = $11!3

33

32

= 1 2.3 " 3.3

= $ 3


( 32 = $12!3

33

21

= $1 1.3 " 2.3

= 3


( 33 = $13!3

32

21

= 1 1.3 " 2.2

= $ 1

Adj A =

333231

232221

131211

K K K

K K K

K K K


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2%

Adj A =

−−−

−−

145

351

32


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21

it : >igen ?alue dan >igen ?ektor

Pen$elesaian #

A I −λ =

λ

λ

%

%

$

32

41

=

−−−−

32

41

λ

λ

A I −λ =

−−−−

32

41

λ

λ

= [ ]423 −−−−− λ λ I

=2

λ $4λ ! $#

=2

λ $ 4λ $ 5

= λ $5 λ ! 1 λ = 5 dan λ = $1

>igen alue adala akar$akar dari λ maka, eigen aluenya adala λ = 5, λ = $1

>igen ektornya :

• λ

= 5 kita subsitusikanλ

= 5 ke dalam matrik karakteristik yang diketaui

dan dengan menyelesaikan 0ersamaan linier omogen yang di da0at.

A I −λ =

−−−−

32

41

λ

λ

=

−−−−352

415


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22

A I −λ =

−

−

22

44

−

−22

44

y

x

=

%

%

4* $ 4Y = % * = 1

$2* ! 2Y = % Y = 1

>igen ?ektornya 0ada λ = 5 adala 1,1

• λ = $1

A I −λ =

−−−−

32

41

λ

λ

=

−−−

−−−312

411

=

−−−−42

42

−−

−−

42

42

y

x

=

%

%

2' " 4y = %

$2' " 4y = %

' = 2, y = $1

Jadi, >igen ?ektornya 0ada λ = $1 adala 2,$1


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23

9. Tuas

9.1. Penjumlahan / Penuranan Matrik

1. Jika A =

[2 1 1

3 4 1

] B =

[3 4 2

1 1 1

]a. A ! B b. A " B

2. Jika A ¿[2 7 5

4 3 0

5 2 4]B=[

1 0 4

3 2 1

6 5 1]

a. A ! B

b. A " B


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24

9.%. Perkalian Matrik

3. Jika A = [3 4 21 1 1] , B = [3 2

4 1

1 1]

A @ B

4. Jika A = [2 3 14 3 2] , B= [3 3

2 3

1 3]

A @ B

5. Jika A = [2 2 24 3 1] , B = [2 4

3 2

2 2]A @ B B @ A

. Jika A = [3 1 2

4 1 3

2 1 1] B=[

3 2 1

1 2 1

3 1 1]

A @ B B @ A

+.3. /rans0ose &atrik

-. Jika A = [2 2 7

1 4 4

8 6 1]

a. At

b. A ! At = &atrik 9imetris

. At

¿ ¿t = A

#. Jika A = [2 2 5

3 3 7

2 1 4]

a. At


. A

t

¿ ¿t = A


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2

15. Jika A =

[4 2 5

2 2 2

4 1 3

]a. &enggunakan ara ko6aktor b. &enggunakan ara sarrus

1. Jika A = [4 2 1

2 2 1

2 3 2]

a. &enggunakan ara (o6aktor

b. &enggunakan ara 9arrus

9.2. In3ers Matrik

1-. Jika A = [2 2 4

1 2 1

4 2 4] A

−1

1#. A = [2 2 3

1 2 3

4 1 1]

A−1

+.-. Adjoint &atrik

1+. Jika A = [1 2 1

2 4 1

1 1 1]

Adj A

9.6. 7ien 8alue 0an 7ien 8ekt"r


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2-

2%. Jika diketaui A = [1 24 3] dan < = [1 00 1 ]/entukan >igen ?alue dan >igen ?ektor

21. Jika diketaui A = [1 −21 4 ] dan < = [1 00 1 ]/entukan >igen?alue dan >igen?ektor

1:. Rankuman

Definisi matrik

Yaitu suatu elemen yang disusun dengan adanya baris dan kolom aijimana : i= nomor baris dan j = nomor kolom.

1. Penjumlahan 0an Penuranan Matrik

)enjumlaan dan 0engurangan antara dua matrik dilakukan 0ada anggota

bariskolom yang sama.

%. Perkalian Matrik

&isal, a&% ;


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2#

/rans0ose suatu matrik adala matrik yang meruba kolom menjadi

baris dan baris menjadi kolom.

A + At Matrik Simetris= ukti A1% A%1

(At) t A

(A + ) t At + t

(A ' ) t t ' A t

-. Matrik Simetris

aij a ji

kunin$a # dikatakan matrik simetris jika:a1% a%1

a1& a&

a&% a%&

. Determinan

ara menari determinan dari suatu matrik ada bebera0a ara, yaitu :

•

ara ko6aktor • ara sarrus

2. In3ers matrik

A$1 = A

1

A A$1 = A

A kofaktor t

A$1 = A

adjA

dimana = Adj A = Atko6aktor

imana A (o6aktor =

333231

232221

131211

K K K

K K K

K K K


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2+

5. A0j"int Matrik ujur Sankar

Adjoint A = Adj A =

321

22212

12111

...

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

...

...

nnn

n

n

K K K

K K K

K K K

= Atko6aktor

6. 7ien 8alue 0an 7ien 8ekt"r

>igen alue adala akar$akar dari λ , dengan ara manari arga A I −λ

,dimana i =

1%

%1

dan i = matrik identitas. sedangkan eigen ektor

dida0atkan dari asil menari eigen alue, lalu dari nilai$nilai λ dida0atkan arga

' dan y nya itula eigen ektornya.

35


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3%

11. S"al 0an Pem


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32/56

32

A @ B =

[3 4 2

1 1 1

] @

[3 2

4 1

1 1

] = [3.3+ 4.4+ 2.1 3.2+4.1+2.11.3+1.4+1.1 1.2+1.1+1.1]A @ B = [9+16+2 6+4+23+4+1 2+1+1 ] = [27 128 4 ]

4. Jika A = [2 3 14 3 2] , B= [3 3

2 3

1 3]

itanya: A @ B

Pen$elesaian #

A @ B = [2 3 14 3 2] @

[

3 3

2 3

1 3

]= [2.3+3.2+1.1 2.3+3.3+1.34.3+3.2+2.1 4.3+3.3+2.3]= [ 6+6+1 6+9+312+6+2 12+9+6]= [13 1820 27]

5. Jika A =

[2 2 24 3 1]

, B =

[2 4

3 2

2 2]itanya : A @ B B @ A

Pen$elesaian #

A @ B = [2 2 24 3 1] @ [2 4

3 2

2 2]


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33

A @ B = [2.2+2.3+2.2 2.4+2.2+2.24.2+3.3+1.2 4.4+3.2+1.2]= [4+ 6+4 8+4 +48+ 9+2 16+ 6+2]= [14 1619 24 ]

B @ A = [2 4

3 2

2 2]×[2 2 24 3 1]

= [2.2+ 4.4 2.2+4.3 2.2+4.13.2+ 2.4 3.2+2.3 3.2+2.12.2+ 2.4 2.2+2.3 2.2+2.1]

= [20 16 8

14 12 8

12 10 6]

Jadi, terbukti A @ B B @ A = [14 1619 24 ] [20 16 8

14 12 8

12 10 6]

. Jika A = [3 1 2

4 1 3

2 1 1] B=[

3 2 1

1 2 1

3 1 1]

itanya # A @ B B @ A

Pen$elesaian #

A @ B =

[3 1 2

4 1 3

2 1 1

]×

[3 2 1

1 2 1

3 1 1

] = [

3.3+1.1+2.3 3.2+1.2+2.1 3.1+1.1+2.14.3+1.1+3.3 4.2+1.2+3.1 4.1+1.1+3.12.3+1.1+1.3 2.2+1.2+1.1 2.1+1.1+1.1]

A @ B = [ 9+ 1+ 6 6 +2+2 3+1+212+1+6 8+ 2+ 3 4+1+36 +1+ 3 4+2+1 2+1+1 ]


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34

¿

[16 10 6

22 13 8

10 9 4

]B ' A = [

3 2 1

1 2 1

3 1 1]×[

3 1 2

4 1 3

2 1 1]

=

[3.3+ 2.4+1.2 3.1+2.1+1.1 3.2+ 2.3+1.11.3+ 2.4+1.2 1.1+2.1+1.1 1.2+ 1.3+1.13.3+ 1.4+1.2 3.1+1.1+1.1 3.2+ 1.3+1.1

]= [

9+8+2 3+2+1 6+6+13+8+2 1+2+1 2+6+19+4+2 3+1+1 6+3+1]

= [19 7 13

13 4 9

15 5 10]

/erbukti A @ B B @ A = [13 1820 27] [19 7 13

13 4 9

15 5 10]

11.3. /rans0ose &atrik

-. Jika A = [2 2 7

1 4 4

8 6 1]

itanya :a. A

t


. A

t

¿ ¿t = A

Pen$elesaian #

a. At =

[

2 2 7

1 4 4

8 6 1

]


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35

At =

[2 1 8

2 4 6

7 4 1

] b. A ! A

t = [2 2 7

1 4 4

8 6 1] ! [

2 1 8

2 4 6

7 4 1]

= [2+2 2+ 1 7+ 81+2 4+ 4 4 +68+7 6+ 4 1+ 1]

=

[ 4 3 15

3 8 1015 10 2 ]Jadi, terbukti A ! A t = &atrik 9imetris

. A

t

¿¿t = A

A = [2 2 71 4 48 6 1

]t

= [2 1 82 4 67 4 1

]t

= [2 2 7

1 4 4

8 6 1]

Jadi, terbukti A

t

¿ ¿t = A

#. Jika A = [2 2 5

3 3 7

2 1 4]

itanya :

a. At


. A

t

¿ ¿t = A

Pen$elesaian #

a. At = [

2 2 5

3 3 7

2 1 4] t

At = [

2 3 2

2 3 1

5 7 4]


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3

b. A ! At =

[2 2 5

3 3 7

2 1 4

] !

[2 3 2

2 3 1

5 7 4

]= [2+2 2+3 5+23+2 3+3 7+12+5 1+7 4+4 ]

A ! At = [

4 5 7

5 6 8

7 8 8]

Jadi, terbukti A ! At = &atrik 9imetris

. A

t

¿ ¿t = A

A = [2 2 53 3 72 1 4

]t

= [2 3 22 3 15 7 4

]t

= [2 2 5

3 3 7

2 1 4]

Jadi, terbukti jika A

t

¿ ¿t = A

+. Jika A = [5 −3 −14 −1 23 2 2

] B = [4 3 2

5 1 1

5 2 2]

itanya : A ! Bt = At ! At

Pen$elesaian#

At = [ 5 4 3

−3 −1 2−1 2 2] Bt = [

4 5 5

3 1 2

2 1 2]

A!B = [5 −3 −14 −1 23 2 2

] ! [4 3 2

5 1 1

5 2 2]


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3-

=

[5+ 4 −3+3 −1+24+5 −1+1 2+13+5 2+2 2+2

] = [

9 0 1

9 0 3

8 4 4]

A ! Bt =

[9 9 8

0 0 4

1 3 4

] At!Bt = [

5 4 3

−3 −1 2−1 2 2] ! [

4 5 5

3 1 2

2 1 2]

= [ 5+ 4 4+5 3+5−3+3 −1+1 2+2−1+2 2+1 2+2]

[9 9 8

0 0 4

1 3 4]

Jadi terbukti A!Bt = At ! Bt = [9 0 1

9 0 3

8 4 4]=[

9 9 8

0 0 4

1 3 4]

1%. Jika A = [ 5 −3 2−3 −2 83 1 −2] , B = [

5 4 2−3 −2 8

2 1 −2]itanya : ( A+B)t = A t + Bt

Pen$elesaian #

At = [

5 4 3

−3 −2 12 8 −2] Bt = [

5 −3 2−3 −2 8

2 1 −2]


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3#

A ! B =

[ 5 −3 2−3 −2 8

3 1 −2

] !

[ 5 4 2

−3 −2 82 1 −2

] = [

5+5 −3+ 4 2+24+(−3) −2+(−2) 8+8

3+ 2 1+1 −2+(−2)]A ! B = [

10 1 4

1 −4 165 2 −4 ]

( A+B)t =

[10 1 5

1 −4 24 16 −4 ]

At + Bt = [

5 4 3

−3 −2 12 8 −2] ! [

5 −3 2−3 −2 8

2 1 −2] = [

5+5 4+(−3) 3+2−3+4 −2+(−2) 1+1

2+2 8+8 −2+(−2)] = [

10 1 5

1 −4 24 16 −4 ]

Jadi, terbukti ( A+B)t = A t +Bt = [10 1 5

1 −4 24 16 −4 ]

11. Jika A = [1 2 1

2 3 1

3 2 2] B = [

4 3 2

2 3 3

2 1 1]

itanya : ( A × B)t =Bt × A t ≠ A t × Bt

Pen$elesaian #

At = [

1 2 3

2 3 2

1 1 2]Bt = [

4 2 2

3 3 1

2 3 1]


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3+

A × B =

[1 2 1

2 3 1

3 2 2

]×

[4 3 2

2 3 3

2 1 1

]= [

4+4+2 3+6+1 2+6+18+6+2 6+9+1 4+9+1

12+4+4 9+6+2 6+6+2]

=

[10 10 9

16 16 14

20 17 14

]( A × B)t = [

10 16 20

10 16 17

9 14 14]

Bt

× At =

[

4 2 2

3 3 1

2 3 1

]×

[

1 2 1

2 3 1

3 2 2

] = [4+4+2 8+6+2 12+4+43+6+1 6+9+1 9+6+22+6+1 4+9+1 6+6+2 ]

= [10 16 20

10 16 17

9 14 14]

At

× Bt = [

4 6 23 5 2

2 1 4 ]×[2 3 17 4 3

1 6 4] = [

4 +6+6 2+ 6+9 2+2+38+9+ 4 4+9+ 6 4+3+ 24+ 3+4 2+3+6 2+1+2]

=

[

16 17 7

21 19 9

11 11 5

]


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4%

Jadi, terbukti ( A × B)t =Bt × A t ≠ A t × Bt

[10 16 20

10 16 17

9 14 14]=[

10 16 20

10 16 17

9 14 14]≠[

16 17 7

21 19 9

11 11 5]

12. Jika A = [1 2 4

2 5 1

3 2 2] , B = [

4 3 2

5 2 4

6 1 1]

itanya : ( A x B)t = B t × A t A t @ B t

Pen$elesaian #

At = [

1 2 3

2 5 2

4 1 2]Bt = [

4 5 6

3 2 1

2 4 1]

A @ B = [1 2 4

2 5 1

3 2 2] @ [

4 3 2

5 2 4

6 1 1]

=

[1.4 +2.5+ 4.6 1.3+2.2+ 4.1 1.2+2.4+ 4.12.4

+5.5

+1.6 2.3

+5.2

+1.1 2.2

+5.4

+1.1

3.4+2.5+2.6 3.3+ 2.2+ 2.1 3.2+2.4 +2.1] = [

4+10+24 3+4+4 2+8+48+25+6 6+10+1 4+20+1

12+10+12 9+4+2 6+8+2 ] = [

38 11 14

39 17 25

34 15 16]

( A × B)t =

[

38 39 34

11 17 15

14 25 16

]B

t × A

t = [4 5 6

3 2 1

2 4 1]×[

1 2 3

2 5 2

4 1 2]

= [4.1+5.2+6.4 4.2+5.5+6.1 4.3+5.2+6.23.1+2.2+1.4 3.2+2.5+1.1 3.3+2.2+1.22.1+4.2+1.4 2.2+4.5+1.1 2.3+4.2+1.2]


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41

=

[4+10+24 8+25+26 12+10+12

3+4+4 6+10+1 9+4+22+8+4 4+20+1 6+8+2

]B

t × A

t = [38 39 34

11 17 15

14 25 16]

At @ B

t = [1 2 3

2 5 2

4 1 2] @ [

4 5 6

3 2 1

2 4 1]

=

[1.4+2.3+3.2 1.5+2.2+3.4 1.6+2.1+3.1

2.4+5.3+2.2 2.5+5.2+2.4 2.6+5.1+2.14.4+1.3+2.2 4.5+1.2+2.4 4.6+1.1+2.1]

= [ 4+6 +6 5+ 4+12 6 +2+38+15+4 10+10+ 8 12+5+ 216 +3+4 20+ 2+ 8 24 +1+2]

= [16 21 11

27 28 19

23 30 27]

Jadi, terbukti ( A+B)t = B t × At A t @ B t

[38 39 34

11 17 15

14 25 16]=[

38 39 34

11 17 15

14 25 16]≠ [

16 21 11

27 28 19

23 30 27]

11.- Matrik Simetris

13. Jika A = [1 1 4

4 5 1

3 6 2]

itanya : &atrik simetris A

Pen$elesaian #

&atrik 9imetris = A ! At


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42

A=

[1 1 4

4 5 1

3 6 2

]At = [1 4 3

1 5 6

4 1 2]

A ! At = [

1 1 4

4 5 1

3 6 2]+[

1 4 3

1 5 6

4 1 2]

=

[1+1 1+4 4+3

4+1 5+5 1+63+4 6+1 2+2 ]

= [2 5 7

5 10 7

7 7 4] &atrik 9imetris

14. Jika A = [1 1 4

4 5 1

3 6 2]

itanya : &atrik 9imetris A

Pen$elesaian #

&atrik 9imetris = A ! At

A= [1 1 2

4 5 2

3 2 1]

At= [1 4 3

1 5 2

2 2 1]

A ! At = [1 1 2

4 5 2

3 2 1] ! [1 4 3

1 5 2

2 2 1]= [

1+1 1+4 2+34+1 5+5 2+23+2 2+2 1+1]

= [1 5 5

5 10 4

5 4 1]


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43

11.5. eterminan

15. Jika A = [4 2 5

2 2 2

4 1 3]itanya : a. &enggunakan ara ko6aktor

b. &enggunakan ara sarrus

Pen$elesaian #

a. ara (o6aktor

det A = | A| = a11 k 11+a12 k 12 +a13 k 13

A = [4 2 52 2 2

4 1 3]K"l"m 1 8 &inor dari baris 1 kolom 1

K 11 = $1 ¿

1+1|2 21 3|= 1 2.3 " 2.1

= " 2

= 4

K"l"m %8 &inor dari baris 1 kolom 2

K 12 = $1 ¿

1+2|2 21 3|= $1 2.3 " 1.2

= $1 " 2

= $1 4

= $ 4

K"l"m &8 &inor dari baris 1 kolom 3

K 13 = $1 ¿1+3

|2 2

1 1|= 1 2.1 " 1.2 = 1 %

= %

Jadi , | A| = 4 4 ! 2 $4 ! 5 % = 1 $#

= #

b. ara 9arrus


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44

| A| =

||4 2 5

2 2 2

1 1 3

|4 2

2 2

1 1

|= 4.2.3 ! 2.2.1 ! 5.2.1 " 2.2.3 ! 4.2.1 ! 5.2.1 = 24 ! 4 !1% " 12 !# ! 1% = 3# " 3%

= #

1. Jika A = [4 2 1

2 2 1

2 3 2]

itanya : a. &enggunakan ara (o6aktor

b. &enggunakan ara 9arrus

Pen$elesaian #

a. ara (o6aktor

det A = | A| = a11 k 11+a12 k 12+a13 k 13

A =

[

4 2 1

2 2 1

2 3 2

]et A = | A| = |4 2 12 2 12 3 2

|= 4 |2 13 2| " 2 |2 12 2| ! 1 |2 22 3|= 42.2 " 3.1 " 22.2 " 2.1 ! 12.3 " 2.2

= 44 " 3 " 24 " 2 ! 1 " 4

= 41 " 22 ! 12

= 4 " 4 ! 2

= 2

b. ara 9arrus

et A = | A| = |4 2 1

2 2 1

2 3 2||

4 2

2 2

2 3|

= 4.2.2 ! 2.1.2 ! 1.2.3 " 1.2.2 ! 4.1.3 ! 2.2.2

= 1 ! 4 ! ! 4 ! 12 ! 2

= 2 " 24

= 2


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45

11.2. In3ers Matrik

1-. Jika A = [2 2 4

1 2 1

4 2 4]

itanya : A−1

Pen$elesaian #

A−1

At kofaktor

| A|

A = [2 2 4

1 2 1

4 2 4](olom 1 8 &inor dari baris 1 kolom 1 → a11=2

K 11 = $1 ¿

1+1|2 12 4|= 1 2.4 " 2.1

= # " 2

=

8 &inor dari baris 2 kolom 1 → a21=1

K 21 = $1 ¿

2+1|2 42 4|= $1 2.4 " 2.4

= $1 %

= %


K 31 = $1 ¿

3+ 1|2 42 1|= 1 2.1 " 2.4

= 2 " #

= $


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4

K"l"m %8 &inor dari baris 1 kolom 2 1 → a12=2

K 12 = $1 ¿

1+2

|1 1

4 4

|= $1 1.4 " 4.1

= $1 %

= %


K 22 = $1 ¿

2+2|2 44 4|= 1 2.4 " 4.4

= # $1

= $#

8 &inor dari baris 3 kolom 2 →a32=2

K 32 = $1 ¿

3+ 2|2 41 1|= $1 2.1 " 1.4

= $1 2 " 4

= $1 $2

= 2

K"l"m & 8 &inor dari baris 1 kolom 3 → a13=4

K 13 = $1 ¿

1+3|1 24 2|= 1 1.2 " 4.2

= 2 " #

= $


K 23 = $1 ¿

2+3|2 24 2|= $1 2.2 " 2.4

= $1 4 " #

K 23 = $1 $ 4

= 4


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4-


K 33 = $1 ¿

3+ 3

|2 2

1 2

|= 1 2.2 " 2.1

= 4 " 2

= 2

A ko6aktor = [ K

11 K

12 K

13

K 21

K 22

K 23

K 31

K 32

K 33

], A t ko6aktor = [ K

11 K

21 K

31

K 12

K 22

K 32

K 13

K 23

K 31

]A ko6aktor = [

6 0 −60 −8 4−6 2 2 ] , A t ko6aktor = [

6 0 −60 −8 2−6 4 2 ]

| A| = a11 k 11+a12 k 12+a 13 k 13

= 2. !2.% !4. $

= 12 $24

= $ 12

A−1=

A t

| A|

A−1=

−112 |

6 0 −60 −8 2

−6 4 2 |= |

−6/12 0 6/120 8/12 −2/12

6/12 −4 /12 −2/12| A

−1 = |−1/2 0 1/2

0 2/3 −1/61/2 −1/3 −1/6|

1#. A = [2 2 3

1 2 3

4 1 1]

itanya : A−1

Pen$elesaian #


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4#

A =

[2 2 3

1 3 1

3 1 4

]K"l"m 1 ¿¿ &inor dari baris 1 kolom 1 a11 = 2

( 11 = $11!1 |2 31 1|

= 1 2.1 " 3.1

= 1 2 $3

= $1

¿¿ &inor dari baris 2 kolom 1 a21 = 1

( 21 = $12!1 |2 31 1|

= 1 2.1 " 3.1

= 1 2 $3

= $1


( 31 = $13!1 |2 32 3|

= 1 2.3 " 2.3

= 1 $ = %

K"l"m % ¿¿ &inor dari baris 1 kolom 2 a12 = 2

( 12 = $11!2 |1 34 1|

= $1 1.1 " 4.3


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4+

= $1 1 $12

= 11


( 22 = $12!2 |2 34 1|

= 1 2.1 " 4.3

= 1 2 $12

= $1%


( 32 = $13!2 |2 31 3|

= $1 2.3 " 3.1

= $1 $3

= $3

K"l"m & ¿¿ &inor dari baris 1 kolom 3 a13 = 3

( 13 = $11!3 |1 24 1|

= 1 1.1 " 4.2

= 1 1 $#

= $-


( 23 = $12!3 |2 24 1|

= $1 2.1 " 4.2

= 1 2 $ #


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5%

=


( 33 = $13!3 |2 21 2|

= 1 2.2 " 2.1

= 1 4 $2

= 2

A (o6aktor : [−1 11 −7

1 −10 60 −3 2 ]

At (o6aktor : [−1 1 011 −10 −3−7 6 2 ]

| A| = a11 k 11 ! a12 k 12 ! a13 k 13

= 2.$1 ! 2$11 ! 3$-

= $2 ! 22 ! $21 = $1

A−1=

At

| A|

A$1 =1

−1 [−1 1 011 10 −3−7 6 2 ]

A$1 = [−1 /−1 1 /−1 0 /−111/−1 10 /−1 −3/−1−7 /−1 6 /−1 2 /−1 ]

A$1 = [−1 1 011 10 −3−7 6 2 ]


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51

11.-. Adjoint &atrik

1+. Jika A = [1 2 1

2 4 1

1 1 1]

itanya : Adj A

Pen$elesaian #

K"l"m 18 &inor dari baris 1 kolom 1 → a11=1

K 11 = $1 ¿

1+1|4 11 1|= 1 4.1 " 1.1

= 4 " 1

= 3


K 21 = $1 ¿

2+1|2 11 1|= $1 2.1 " 1.1

= $1 " 1= $1


K 31 = $1 ¿

3+1|2 14 1|= 1 2.1 " 4.1

= 2 " 4

= $ 2

K"l"m %8 &inor dari baris 1 kolom 2 → a12=2

K 12 = $1 ¿

1+2|2 11 1|= $1 2.1 " 1.1

= $1 2 " 1

= $1 " 1

= $1


K 22 = $1 ¿

2+2|1 11 1|= 1 1.1 " 1.1


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53

Pen$elesain #

>igen ?alue

| λ I − A|=|[ λ 0

0 λ ]−[1 2

4 3]| = |[ λ−1 −2−4 λ−3]|

| λ I − A|=[ ( λ−1 ) ( λ−3 )−(−4 ) (−2 ) ]

= λ2−4 λ+ 3 ¿−8

= λ2

−4 λ " 5

= λ−5 ¿ ( λ+1 )→ λ=5 dan λ=−1

>igen alue adala akar " akar dari λ maka eigen aleunya adala λ=5 dan

λ=−1

7ien 8ekt"r

Untuk λ=5

| λ I − A|=[ λ−1 −2−4 λ−3]

= [5−1 −2−4 5−3] ¿[ 4 −2−4 2 ] [ 4 −2−4 2 ][ x y ]=[00 ]

4' $ 2y = % ' = %.5

$ 4' ! 2y = % y = 2

Jadi >igen ?ektornya 0adaλ = 5 adala %.5,2

Untuk λ=−1

| λ I − A|=[ λ−1 −2−4 λ−3]


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= [−1−1 −2−4 −1−3] = [−2 −2−4 −4]

[−2 −2−4 −4] [ x y ]=[00 ]$2' $ 2y = % ' = $1

$ 4' $ 4y = % y = 1

Jadi >igen ?ektornya 0adaλ = $1 adala $1,1

21. Jika diketaui A = [1 −21 4 ] dan < = [1 00 1 ]itanya : /entukan >igen?alue dan >igen?ektor

Pen$elesain #

7ien 8alue

| λI − A|=|[ λ 0

0 λ

]−[1 −21 4

]| = |[ λ−1 2−1 λ−4]|¿ [ ( λ−1 ) ( λ−4 )−(−1 ) (2 ) ]

= λ2−5 λ +4 ¿+2

= λ2−5 λ !

= λ−3 ¿ ( λ−2 ) → λ=3 dan λ=2

>igen ?alue adala akar " akar dari λ maka , >igen ?aluenya adala

λ=3 dan λ=2

7ien 8ekt"r

Untuk λ=3

| λI − A|=[ λ−1 2−1 λ−4]


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= [3−1 2−1 3−4 ]| λI − A|=[ 4 −2−4 −2]

[ 2 2−1 −1][ x y ]=[00] 2' ! 2y = % ' = 1

$ ' " y = % y = $1

Jadi >igen ?ektornya 0adaλ = 3 adala 1,$1

Untuk λ=2

| λI − A|=[ λ−1 2−1 λ−4] = [2−1 2−1 2−4 ]| λI − A|=

[

1 2

−1 −2][ 1 2−1 −2] [ x y ]=[00 ]' ! 2y= % ' = 1

$ ' $2y= % y = $%.5

Jadi >igen ?ektornya 0adaλ = 3 adala 1,%.5

SUM7R PUSTAKA

Ayres, rank, Jr. 1+#5. Matriks, >rlangga. Jakarta


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&arga,&.d

3.>rlangga. Jakarta.

Gabungan Bab II Matrik

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