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.
=
8 &inor dari baris 2 kolom 2 a22= 1
( 22 = $12!2
%152
= 1 2.% " 5.1
= $5
8 &inor dari baris 3 kolom 2 a32= 4
( 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 #
(olom 1 8 &inor dari baris 1 kolom 1 a11 = 1
( 11 = $11!1
43
23
= 1 3.4 " 3.2
=
8 &inor dari baris 2 kolom 1 a21 = 2
( 21 = $12!1
43
32
= $1 2.4 " 3.3
= 1
8 &inor dari baris 3 kolom 1 a31 = 3
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1#
( 31 = $13!1
23
32
= 1 2.2 " 3.3
= $ 5
(olom 2 8 &inor dari baris 1 kolom 2 a12 = 2
( 12 = $11!2
43
22
= $1 2.4 " 3.2
= $ 2
8 &inor dari baris 2 kolom 2 a22= 3
( 22 = $12!2
43
31
= 1 1.4 " 3.3
= $ 5
8 &inor dari baris 3 kolom 2 a32= 3
( 32 = $13!2
22
31
= $1 1.2 " 2.3
= 4
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1+
(olom 3 8 &inor dari baris 1 kolom 3 a13 = 3
( 13 = $11!3
33
32
= 1 2.3 " 3.3
= $ 3
8 &inor dari baris 2 kolom 3 a23= 2
( 32 = $12!3
33
21
= $1 1.3 " 2.3
= 3
8 &inor dari baris 3 kolom 3 a33= 4
( 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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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,$1
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9. Tuas
9.1. Penjumlahan / Penuranan Matrik
1. Jika A =
[2 1 1
3 4 1
] B =
[3 4 2
1 1 1
]a. A ! B b. A " B
2. Jika A ¿[2 7 5
4 3 0
5 2 4]B=[
1 0 4
3 2 1
6 5 1]
a. A ! B
b. A " B
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9.%. Perkalian Matrik
3. Jika A = [3 4 21 1 1] , B = [3 2
4 1
1 1]
A @ B
4. Jika A = [2 3 14 3 2] , B= [3 3
2 3
1 3]
A @ B
5. Jika A = [2 2 24 3 1] , B = [2 4
3 2
2 2]A @ B B @ A
. Jika A = [3 1 2
4 1 3
2 1 1] B=[
3 2 1
1 2 1
3 1 1]
A @ B B @ A
+.3. /rans0ose &atrik
-. Jika A = [2 2 7
1 4 4
8 6 1]
a. At
b. A ! At = &atrik 9imetris
. At
¿ ¿t = A
#. Jika A = [2 2 5
3 3 7
2 1 4]
a. At
b. A ! At = &atrik 9imetris
. A
t
¿ ¿t = A
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2
15. Jika A =
[4 2 5
2 2 2
4 1 3
]a. &enggunakan ara ko6aktor b. &enggunakan ara sarrus
1. Jika A = [4 2 1
2 2 1
2 3 2]
a. &enggunakan ara (o6aktor
b. &enggunakan ara 9arrus
9.2. In3ers Matrik
1-. Jika A = [2 2 4
1 2 1
4 2 4] A
−1
1#. A = [2 2 3
1 2 3
4 1 1]
A−1
+.-. Adjoint &atrik
1+. Jika A = [1 2 1
2 4 1
1 1 1]
Adj A
9.6. 7ien 8alue 0an 7ien 8ekt"r
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2-
2%. Jika diketaui A = [1 24 3] dan < = [1 00 1 ]/entukan >igen ?alue dan >igen ?ektor
21. Jika diketaui A = [1 −21 4 ] dan < = [1 00 1 ]/entukan >igen?alue dan >igen?ektor
1:. Rankuman
Definisi matrik
Yaitu suatu elemen yang disusun dengan adanya baris dan kolom aijimana : i= nomor baris dan j = nomor kolom.
1. Penjumlahan 0an Penuranan Matrik
)enjumlaan dan 0engurangan antara dua matrik dilakukan 0ada anggota
bariskolom yang sama.
%. Perkalian Matrik
&isal, a&% ;
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2#
/rans0ose suatu matrik adala matrik yang meruba kolom menjadi
baris dan baris menjadi kolom.
A + At Matrik Simetris= ukti A1% A%1
(At) t A
(A + ) t At + t
(A ' ) t t ' A t
-. Matrik Simetris
aij a ji
kunin$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$nilai λ dida0atkan arga
' dan y nya itula eigen ektornya.
35
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3%
11. S"al 0an Pem
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A @ B =
[3 4 2
1 1 1
] @
[3 2
4 1
1 1
] = [3.3+ 4.4+ 2.1 3.2+4.1+2.11.3+1.4+1.1 1.2+1.1+1.1]A @ B = [9+16+2 6+4+23+4+1 2+1+1 ] = [27 128 4 ]
4. Jika A = [2 3 14 3 2] , B= [3 3
2 3
1 3]
itanya: A @ B
Pen$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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¿
[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
b. A ! At = &atrik 9imetris
. A
t
¿ ¿t = A
Pen$elesaian #
a. At =
[
2 2 7
1 4 4
8 6 1
]
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At =
[2 1 8
2 4 6
7 4 1
] b. A ! A
t = [2 2 7
1 4 4
8 6 1] ! [
2 1 8
2 4 6
7 4 1]
= [2+2 2+ 1 7+ 81+2 4+ 4 4 +68+7 6+ 4 1+ 1]
=
[ 4 3 15
3 8 1015 10 2 ]Jadi, terbukti A ! A t = &atrik 9imetris
. A
t
¿¿t = A
A = [2 2 71 4 48 6 1
]t
= [2 1 82 4 67 4 1
]t
= [2 2 7
1 4 4
8 6 1]
Jadi, terbukti A
t
¿ ¿t = A
#. Jika A = [2 2 5
3 3 7
2 1 4]
itanya :
a. At
b. A ! At = &atrik 9imetris
. A
t
¿ ¿t = A
Pen$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 %
= %
8 &inor dari baris 3 kolom 1 → a31=4
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 %
= %
8 &inor dari baris 2 kolom 2 → a22=2
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 " #
= $
8 &inor dari baris 2 kolom 3 → a23=1
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-
8 &inor dari baris 3 kolom 3 → a33=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
¿¿ &inor dari baris 3 kolom 1 a31 = 4
( 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
¿¿ &inor dari baris 2 kolom 2 a22 = 2
( 22 = $12!2 |2 34 1|
= 1 2.1 " 4.3
= 1 2 $12
= $1%
¿¿ &inor dari baris 3 kolom 2 a32 = 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 $#
= $-
¿¿ &inor dari baris 2 kolom 3 a23 = 3
( 23 = $12!3 |2 24 1|
= $1 2.1 " 4.2
= 1 2 $ #
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5%
=
¿¿ &inor dari baris 3 kolom 3 a33 = 1
( 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$1 = [−1 1 011 10 −3−7 6 2 ]
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11.-. Adjoint &atrik
1+. Jika A = [1 2 1
2 4 1
1 1 1]
itanya : Adj A
Pen$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
8 &inor dari baris 2 kolom 1 → a21=2
K 21 = $1 ¿
2+1|2 11 1|= $1 2.1 " 1.1
= $1 " 1= $1
8 &inor dari baris 3 kolom 1 → a31=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
8 &inor dari baris 2 kolom 2 → a22=4
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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55
= [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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5
&arga,&.d
3.>rlangga. Jakarta.