1. MODUL 5
INVERS MATRIK
PRAYUDI STT PLN

2. PENGERTIAN INVERS MATRIK
 Matrik bujur sangkar A dikatakan mempunyai invers, jika terdapat matrik
B sedemikian rupa sehingga :
AB = BA = I
dimana I matrik identitas
 B dikatakan invers matrik A ditulis A–1
, maka, AA–1
= A–1
A = I
 A dikatakan invers matrik B ditulis B–1
, maka, B–1
B= BB–1
= I
 Contoh ; AB = BA = I
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111
230
132
653
432
321
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111
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100
010
001
653
432
321

3. TEKNIK MENGHITUNG INVERS
 Metode Adjoint matrik
 Metode operasi elementer baris
 Metode Perkalian Invers Matrik Elementer
 Metode partisi matrik
 Program Komputer – MATCADS, MATLAB
 WS OFICE EXCELL

4. Metode Adjoint Matrik
 Andaikan A matrik bujur sangkar berordo (nxn), Cij=(-1)i+j Mij kofaktor
elemen matrik aij, dan andaikan pula det(A)≠0 maka A mempunyai invers
yaitu :
adj(A)
det(A)
1
A 1

dimana,
ij
ji
ij
nnnnn
n
n
n
MC
CCCC
CCCC
CCCC
CCCC
Aadj
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
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)1(
...
...............
...
...
...
)(
321
3332313
2322212
1312111
Kasus, n = 2 : maka
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2221
1211
aa
aa
A
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



1121
1222
21122211
1 1
aa
aa
aaaa
A

5. Kasus, n = 3
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333231
232221
131211
aaa
aaa
aaa
A
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332313
322212
312111
332313
322212
312111
1-
MM-M
M-MM-
MM-M
det(A)
1
CCC
CCC
CCC
det(A)
1
A
CONTOH :
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554
543
432
A det(A)= 1
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1-21-
26-5
1-55-
9)-(812)-(10-16)-(15
12)-(10-)16-10(20)-(15-
16)-(1520)-(15-25)-(20
43
32
54
32
-
54
43
53
42
-
54
42
54
53
-
54
43
55
43
-
55
54
(1)
1
A 1-

6. KASUS : n = 4
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44342414
43332313
42322212
41312111
1-
44434241
34333231
24232221
14131211
MM-MM-
M-MM-M
MM-MM-
M-MM-M
det(A)
1
A
aaaa
aaaa
aaaa
aaaa
A
CONTOH :
Hitunglah invers matrik berikut ini :
Ekspansi baris -1 :
det(a)=M11-2M12+3M13-4M14
=-10 – 2(5) + 3(9) – 4(2)= –1
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6863
4753
5532
4321
A
Ekspansi baris-2 :
det(A)=-2M21+3M22-5M23+5M24
=-2(-6) –3(4) + 5(-6) –5(-1)= –1
Ekspansi baris-3 :
det(A)=3M31-5M32+7M33-4M34
=3(-8) –5(3) + 7(6) –4(1)= –1
Ekspansi baris-4 :
det(A)=-3M41+6M42-8M43+6M44
=-3(-7) +6(2) - 8(5) + 6(1)= –1

7. INVERS : OPERASI ELEMENTER BARIS
Operasi Elementer baris yang
digunakan adalah :
(1). Hj  kHj
(2). Hj  Hi
(3). Hj  Hj + kHj
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1000
...1......
0010
0001
...
.........
...
...
21
22221
11211
nnnn
ii
n
n
aaa
a
aaa
aaa
Langkah-langkah sebagai berikut
(1). Bentuk matrik lengkap [A,I]
(2). Dengan serangkain operasi elelemter
baris reduksilah [A,I] menjadi matrik
berbentuk [I,B]
(3). A–1 = B
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nnnn
ii
n
n
bbb
b
bbb
bbb
...
.........
...
...
1000
...1......
0010
0001
21
22221
11211
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nnnn
ii
n
n
bbb
b
bbb
bbb
...
.........
...
...
AJadi,
21
22221
11211
1-
Operasi elementer baris
Gaouss-Jordan

8. CONTOH :
M.Asal
2 3 4 1 0 0
3 4 5 0 1 0
4 5 5 0 0 1
Iterasi-1
1 1.5 2 0.5 0 0 H1=(1/a11)H1
0 -0.5 -1 -1.5 1 0 H2=H2-(a21/a11)H1
0 -1 -3 -2 0 1 H3=H3-(a31/a11)H1
Iterasi-2
1 1.5 2 0.5 0 0
0 1 2 3 -2 0 H2=(1/a22)H2
0 0 -1 1 -2 1 H3=H3-(a32/a22)H2
Iterasi-3
1 1.5 2 0.5 0 0
0 1 2 3 -2 0
0 0 1 -1 2 -1 H3=(1/a33)H3

9. Iterasi-4
1 1.5 0 2.5 -4 2 H1=H1-(a13/a33)H3
0 1 0 5 -6 2 H2=H2-(a23/a33)H3
0 0 1 -1 2 -1
Iterasi-5
1 0 0 -5 5 -1 H1=H1-(a12/a22)H2
0 1 0 5 -6 2
0 0 1 -1 2 -1
Lanjutan :
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1-21-
26-5
1-55-
AJadi, 1-

10. Matrik Awal
1 2 3 4 1 0 0 0
2 3 5 5 0 1 0 0
3 5 7 4 0 0 1 0
3 6 8 6 0 0 0 1
Iterasi - 1
1 2 3 4 1 0 0 0 H1=(1/a11)H1
0 -1 -1 -3 -2 1 0 0 H2=H2-(a21/a11)H1
0 -1 -2 -8 -3 0 1 0 H3=H3-(a21/a11)H1
0 0 -1 -6 -3 0 0 1 H4=H4-(a41/a11)H1
Iterasi - 2
1 2 3 4 1 0 0 0
0 1 1 3 2 -1 0 0 H2=(1/a22)H2
0 0 -1 -5 -1 -1 1 0 H3=H3-(a32/a22)H2
0 0 -1 -6 -3 0 0 1 H4=H4-(a42/a22)H2
Iterasi-4
1 2 3 4 1 0 0 0
0 1 1 3 2 -1 0 0
0 0 1 5 1 1 -1 0 H3=(1/a33)H3
0 0 0 -1 -2 1 -1 1 H4=H4-(a43/a33)H3

11. Iterasi-5
1 2 3 4 1 0 0 0
0 1 1 3 2 -1 0 0
0 0 1 5 1 1 -1 0
0 0 0 1 2 -1 1 -1 H4=(1/a44)H4
Iterasi-6
1 2 3 0 -7 4 -4 4 H1=H1-a14*H4
0 1 1 0 -4 2 -3 3 H2=H2-a24*H4
0 0 1 0 -9 6 -6 5 H3=H3-a34*H4
0 0 0 1 2 -1 1 -1
Iterasi-7
1 2 0 0 20 -14 14 -11 H1=H1-a13*H3
0 1 0 0 5 -4 3 -2 H2=H2-a23*H3
0 0 1 0 -9 6 -6 5
0 0 0 1 2 -1 1 -1
Iterasi-8
1 0 0 0 10 -6 8 -7 H1=H1-a12*H2
0 1 0 0 5 -4 3 -2
0 0 1 0 -9 6 -6 5
0 0 0 1 2 -1 1 -1
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1-112
56-69-
2-34-5
7-86-10
AJadi, 1-

12. PERKALIAN MATRIK ELEMENTER
(1). Matrik elementer adalah matrik
yang diperoleh dari operasi
elementer yang dikenakan pada
matrik identitas.
(2). Setiap matrik elementer
mempunyai invers, dan setiap
matrik bujur sangkar berordo
(nxn) yang mempunyai invers
ekivalen baris terhadap matrik
identitas I.
(3). Akibatnya, jika :
EkEk–1Ek–2 …E2E1A = I, maka,
A–1 = EkEk–1Ek–2 …E2E1
Matrik elementer E diperoleh dari
transformasi matrik identitas dimana pada
kolom ke-I diganti dengan normalitas
vektor kolom :
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1......00
...................
0......00
...................
0......10
0......01
.
.
.
.
ik
ik
ik
ik
i
N
N
N
N
E
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iiin
aii
iii
ik
aa
aa
N
/
...
/1
...
/
,
,1
,
iiiik IAEEEN 121, ...
:dimana


13. CONTOH
Hitung invers matrik A
Jawab :
Menghitung E1
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554
543
432
A
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5
5
4
A;
5
4
3
A;
4
3
2
A 321
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102-
011.5-
000.5
10/aa-
01/aa-
001/a
E
1131
1121
11
1
Menghitung E2
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12-1
02-3
034-
102
011.5-
000.5
12-0
02-0
031
EE
12-0
02-0
031
1)(-1)/(-0.5-0
00.5-1/0
00.5-1.5/-1
E
1-
0.5-
1.5
5
4
3
102-
011.5-
000.5
AEN
12
2
212

14. Menghitung E3 dan Invers Matrik
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1-21-
26-5
1-55-
12-1
02-3
034-
1-00
210
1-01
EEE
1-00
210
1-01
1/(-1)00
(2)/(-1)-10
(-1)/(-1)-01
E
1-
2
1-
5
5
4
12-1
02-3
034-
AEEN
123
3
3123
Jadi Invers Matrik
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1-21-
26-5
1-55-
A 1-

15. CONTOH
Hitung invers matrik A
Jawab :
Menghitung E1
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
6863
4753
5532
4321
A
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
1003-
0103-
0012-
0001
E
3
3
2
1
AN
1
11
Menghitung E2
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
1003-
011-1-
001-2
0023-
1003-
0103-
0012-
0001
1000
011-0
001-0
0021
EE
1000
011-0
001-0
0021
100/(-1)-0
01(-1)/(-1)-0
001-1/0
002/(-1)-1
E
0
1-
1-
2
6
5
3
2
1003-
0103-
0012-
0001
AEN
12
2
212

16. Menghitung E3
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
11-12-
01-11
012-1
0113-
1003-
011-1-
001-2
0023-
1000
0100
0010
0001
EEE
11-00
01-00
0110
0101
1(-1)/(-1)-00
01/(-1)00
01/(-1)-10
01/(-1)-01
E
1-
1-
1
1
8
7
5
3
1003-
011-1-
001-2
0023-
AEEN
123
3
3123

17. Menghitung E4 dan Invers Matrik
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





























































1-11-2
56-69-
2-34-5
7-86-10
11-12-
01-11
012-1
0114-
1-000
5100
2-010
7-001
EEEEA
1-000
5100
2-010
7-001
1/(-1)000
5/(-1)-100
(-2)/(-1)-010
(-7)/(-1)-001
E
1-
5
2-
7-
6
4
5
4
11-12-
01-11
012-1
0113-
AEEEN
1234
1-
3
41234

18. INVERS : PARTISI MATRIK (1)
Partisi matrik A yang berordo (mxn)
adalah sub matrik-sub matrik yang
diperoleh dari A dengan cara
memberikan batasan-batasan garis
horisontal diantara dua baris dan atau
memberikan batasan-batasan garis
vertikal diantara dua kolom.
CONTOH













6863
4753
5532
4321
A


























68
47
A
63
53
A
55
43
A
32
21
A
:adalahAmatrikPartisi
2221
1211

















31554
13343
53632
23443
34532
A
CONTOH

19. INVERS : PARTISI MATRIK (2)
Andaikan A matrik bujur sangkar
berordo (nxn) yang mempunyai
invers, yaitu : A–1 = B, dan partisinya
masing-masing adalah :
Karena, AB=BA=I maka diperoleh :













2221
1211
2221
1211
BB
BB
B;
AA
AA
A




































I0
0I
AA
AA
BB
BB
I0
0I
BB
BB
AA
AA
2221
1211
2221
1211
2221
1211
2221
1211
Dari perkalian matrik diperoleh hasil :
(1). A11 B11 + A12 B21 = I
(2). A11 B12 + A12 B22 = 0
(3). B21 A11 + B22 A21 = 0
(4). B21 A12 + B22 A22 = I
Dengan asumsi, A11
–1 ada, dan
B22 = L–1 ada
Maka rumus untuk menghitung inver
matriknya adalah :
(1). B12 = –(A 11
–1 A12)L–1
(2). B21 = – L–1(A21 A11
–1)
(3). B11 = A11
–1+(A11
–1A12)L–1(A21 A11
–1)
(4). L = A22 – (A21A11
–1A12)

20. CONTOH :
Kasus n=4. Hitunglah invers matrik
berikut ini
Jawab :













6863
4753
5532
4321
A


























68
47
A
63
53
A
55
43
A
32
21
A
:adalahAmatrikPartisi
2221
1211
Menghitung L








































































































1-1
56-
1-(-1)-
(-5)-6-
5-6
1
LJadi,
6-1-
5-1-
129
98
-
68
47
31
2-1
63
53
-
68
47
AAAAL
03
11
1-2
23-
63
53
AA
31
2-1
55
43
1-2
23-
AA
1-2
23-
12-
2-3
)43(
1
A
1-
12
1-
112122
1-
1121
12
1-
11
1-
11

21. Menghitung Invers Matrik















































































4-5
6-13
3-3
8-13
1-2
23-
12-
6-9
31
2-1
1-2
23-
)A(AL)AA(AB
1-2
69-
03
11
1-1
56-
-
)A(A-LB
2-3
7-8
1-1
56-
31
2-1
-
L)A-(AB
1-
1121
1-
12
1-
11
1-
1111
1-
1121
1-
21
1-
12
1-
1112



















1-112
56-69-
2-34-5
7-86-10
BB
BB
A
2221
12111-
CONTOH :
Hitung invers matrik A berikut :
Jawab : Partisi matrik A

















31554
13343
53632
23443
34532
A



































315
133
536
A
54
43
32
A
234
345
A
43
32
A
2221
1211

22. Menghitung L
21-
10
01
2-3
34-
54
43
32
AA
567
6-7-8-
234
345
2-3
34-
AA
2-3
34-
23-
3-4
98
1
A
1-
1121
12
1-
11
1-
11



















































































































































1-1-1
1-2-0
101-
1)-(02)(-1-0)-(1
2)(-1-4)-(22)(-2-
0)-(12)(-2-1)-(0
1
1
L
Jadi,
21-2
1-01-
21-1
123
234
345
-
315
133
536
567
6-7-8-
54
43
32
-
315
133
536
AAAAL
1-
12
1-
112122

23. Menghitung Invers Matrik





































































































27-5
337-
25-2
303-
2-3
34-
3-2
4-1
22-
567
6-7-8-
2-3
34-
)A(AL)AA(AB
32-
41-
2-2
21-
10
01
1-1-1
1-2-0
101-
-
)A(A-LB
4172
5-20-2-
1-1-1
1-2-0
101-
567
6-7-8-
-
L)A-(AB
1-
1121
1-
12
1-
11
1-
1111
1-
1121
1-
21
1-
12
1-
1112
























1-1-132-
1-2-041-
101-2-2
417227-5
5-20-2-337-
BB
BB
AJadi,
2221
12111-

24. SOAL TUGAS IV
Hitung invers matrik A berikut ini dengan cara :

















111
4112
221
111
bbaa
bbaa
aabb
aabb
A
a. Metode Adjoint
b. Perkalian matrik elementer
c. Operasi elementer baris
d. Metode partisi matrik
Hitung invers matrik A berikut ini dengan 2 cara partisi yang berbeda:






















13112
31123
11234
4321
32112
aaabb
aaabb
aaabb
bbbaa
bbbaa
A






















4432
12121
31
131
1221
aaabb
aaabb
aaabb
bbbaa
bbbaa
A