Materi kuliah ini berisi pokok bahasan invers matrik dengan meotde adjoin dan partisi

1. MODUL 2
INVERS
MATRIK
PRAYUDI

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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−
−
−−
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


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



111
230
132
653
432
321










−
−
−−
=
111
230
132





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


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=










100
010
001
653
432
321

3. Contoh
2 3 4 7 -3 -1
A = 3 5 7 B = -7 2 2
4 6 7 2 0 -1
14-21+8 -6+6+0 -2+6-4
A B = 21-35+14 -9+10+0 -3+10-7
28-42+14 -12+12+0 -4+12-7
1 0 0
A B = 0 1 0
0 0 1
14-9-4 21-15-6 28-21-7
B A = -14+6+8 -21+10+12 -28+14+14
4+0-4 6+0-6 8+0-7
1 0 0
BA = 0 1 0
0 0 1

4. 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

5. 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 :
A−1 =
1
det(A)
Adj(A)
dengan,
Adj A =
𝐶11 𝐶21 𝐶31 . . . 𝐶 𝑛1
𝐶12 𝐶22 𝐶32 . . . 𝐶 𝑛2
𝐶13 𝐶23 𝐶33 . . . 𝐶 𝑛3
. . . . . . . . . . . . . . .
𝐶1𝑛 𝐶2𝑛 𝐶3𝑛 . . . 𝐶 𝑛𝑛
dengan
𝐶𝑖𝑗 = (−1)𝑖+𝑗
𝑀𝑖𝑗

6. Kasus n=2
Jika A adalah matrik (2x2), yaitu :
𝐴 =
𝑎11 𝑎12
𝑎21 𝑎22
dan jika det(A)≠ 0, maka :
𝐴−1 =
1
𝑎11 𝑎22 − 𝑎12 𝑎21
𝑎22 −𝑎12
−𝑎21 𝑎11
Contoh :
𝐴 =
2 5
3 7
Karena, det(A) = 14 – 15 = –1, maka
𝐴−1 =
1
−1
7 −5
−3 2
=
−7 5
3 −2
Bukti :
𝐴𝐴−1 =
2 5
3 7
−7 5
3 −2
=
−14 + 15 10 − 10
−21 + 21 15 − 14
=
1 0
0 1

7. Kasus n=3
Jika A adalah matrik (3x3), yaitu :
A =
a11 a12 a13
a21 a22 a23
a31 a32 a33
dan jika det(A)≠ 0, maka :
A−1 =
1
det(A)
C11 C21 C31
C12 C22 C32
C13 C23 C33
Karena,
𝐶𝑖𝑗 = (−1)𝑖+𝑗 𝑀𝑖𝑗
Maka
A−1 =
1
det(A)
M11 −M21 M31
−M12 M22 −M32
M13 −M23 M33

8. Contoh :
Hitunglah invers matrik berikut ini jika ada.
A =
2 3 4
3 4 5
4 5 5
Jawab
Karena, det(A) = 40 + 60 + 60 – 64 – 45 – 50 = 1
maka,
A−1 =
1
1
4 5
5 5
−
3 4
5 5
3 4
4 5
−
3 5
4 5
2 4
4 5
−
2 4
3 5
3 4
4 5
−
2 3
4 5
2 3
3 4
=
(20−25) −(15−20) (15−16)
−(15−20) (10 − 16) −(10−12)
(15−16) −(10−12) (8−9)
=
−5 5 −1
5 −6 2
−1 2 −1

9. Kasus n=4
Jika A adalah matrik (4x4), yaitu :
A =
a11 a12 a13 a14
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 a44
dan jika det(A)≠ 0, maka,
A−1 =
1
det(A)
C11 C12 𝐶13 C14
𝐶21 C22 C23 C24
C31 C32 C33 C34
C41 C42 C43 C44
=
1
det(A)
M11 −M21 M31 −M41
−M12 M22 −M32 M42
M13 −M23 M33 −M43
−M14 M24 −M34 M44

10. 2 3 4 2
A = 3 5 7 6
4 6 7 5
4 7 12 7
Menghitung determinan ekspansi kofaktor baris 1
| A | =a11 M11 - a12 M12 + a13 M13 + a14 M14
= 2 5 7 6 - 3 3 7 6 +4 3 5 6 -2 3 5 7
6 7 5 4 7 5 4 6 5 4 6 7
7 12 7 4 12 7 4 7 7 4 7 12
= 2 (245+245+432 - 294 - 294 - 300) - 3(147+140+288 - 168 - 196 - 180)
+ 4 (126+100+168-144-140-105) - 2(432+140 + 196 - 168 - 240 - 147)
= 2 (34) - 3(31) + 4(5) - 2(-3) = 1
Menghitung determinan ekspansi kofaktor baris 2
| A | = -a21 M21 + a22 M22 - a23 M13 + a24 M24
= -3 3 4 2 +5 2 4 2 -7 2 3 2 +6 2 3 4
6 7 5 4 7 5 4 6 5 4 6 7
7 12 7 4 12 7 4 7 7 4 7 12
= -3 (147+140+144-98-168-180) + 5(98+80+96-56-112-120)
-7(84+60+56 -48-84-70) + 6(144+84+112-96-144-98)
= -3(-15) + 5(-14) - 7(-2) + 6(2) = 1
Contoh

11. Menghitung determinan ekspansi kofaktor baris 3
| A | = a31 M31 - a32 M32 + a33 M33 - a34 M34
= 4 3 4 2 -6 2 4 2 +7 2 3 2 -5 2 3 4
5 7 6 3 7 6 3 5 6 3 5 7
7 12 7 4 12 7 4 7 7 4 7 12
= 4(147+168+120-98-140-216) - 6(98+96+72-56-84-144)
+ 7(70+72+42-40-63-84) - 5(120+84+84-80-144-98)
= 4(-19) - 6(-18) + 7(-3) - 5(2) = 1
Menghitung determinan ekspansi kofaktor baris 4
| A | = -a41 M41 + a42 M42 - a43 M43 + a44 M44
= -4 3 4 2 +7 2 4 2 -12 2 3 2 +7 2 3 4
5 7 6 3 7 6 3 5 6 3 5 7
6 7 5 4 7 5 4 6 5 4 6 7
= -4(105+144+70-84-100-126) + 7 (70+96+42-56-60-84)
-12 (50+72+36-40-45-72) + 7 (70+ 84+72-80-63-74)
= -4(9) + 7(8) - 12(1) + 7(-1) = 1
Dengan demikian,
(34) -(-15) (-19) -(9) 34 15 -19 -9
A-1 = -(31) (-14) -(-18) (8) A-1 = -31 -14 18 8
(5) -(-2) (-3) -(1) 5 2 -3 -1
-(-3) (2) -(2) (-1) 3 2 -2 -1

12. 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


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


=





=





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=





=
68
47
A
63
53
A
55
43
A
32
21
A
:adalahAmatrikPartisi
2221
1211
















=
31554
13343
53632
23443
34532
A
CONTOH

13. INVERS : PARTISI MATRIK (2)
Andaikan A matrik bujur sangkar berordo (nxn) yang mempunyai
invers, yaitu : A–1 = B, dan partisinya masing-masing adalah :
A =
A11 A12
A21 A22
; B =
B11 B12
B21 B22
Karena, AB=BA=I maka diperoleh :
A11 A12
A21 A22
B11 B12
B21 B22
=
I 0
0 I
B11 B12
B21 B22
A11 A12
A21 A22
=
I 0
0 I
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

14. Dengan asumsi, A11
–1 ada, dan B22 = L–1 ada
Maka diperoleh rumus menghitung invers matrik B, yaitu :
(1). A11
–1(A11 B12 + A12 B22) = 0
B12 = –(A11
–1A12)L–1
(2). (B21 A11 + B22 A21) A11
–1 = 0
B21 = –L–1(A21 A11
–1)
(3). A11
–1(A11 B11 + A12 B21) = A11
–1 I
B11 = A11
–1 – (A11
–1A12) B21
= A11
–1 – (A11
–1A12) (–L–1)(A21 A11
–1)
= A11
–1 + (A11
–1A12) L–1 (A21 A11
–1)
(4). B21 A12 + B22 A22 = I
L{–L–1(A21 A11
–1)A12 + L–1 A22} = LI
L = A22 – (A21 A11
–1)A12

15. Contoh
2 3 4 2
A= 3 5 7 6
4 6 7 5
4 7 12 7
A21 A22
Menghitung A11-1 dan L-1
A11= 2 3 det(A11) = 10 - 9 = 1
3 5
A11-1 = 5 -3 = 5 -3
-3 2 -3 2
A11-1 A12 = 5 -3 4 2 = 20 - 21 10 - 18 = -1 -8
-3 2 7 6 -12 + 14 -6 + 12 2 6
A21 A11-1 = 4 6 5 -3 = 20 – 18 -12+12 = 2 0
4 7 -3 2 20 - 21 -12+14 -1 2

16. L= A22 - A21 (A11-1 A12)
= A22 - 4 6 -1 -8
4 7 2 6
= A22 - -4 + 12 -32+36 = 7 5 - 8 4
-4 +14 -32+42 12 7 10 10
= -1 1 del(L) = 3 - 2 = 1
2 -3
L-1 = 1 -3 -1 = -3 -1
-2 -1 -2 -1
Menghitung matrik B
B12 = -(A11-1 A12) L-1
= - -1 -8 -3 -1 = - 3+16 1+9 = - 19 9
2 6 -2 -1 -6-12 -2-6 -18 -8
= -19 -9
18 8

17. B21 = - L-1 (A21 A11-1)
= - -3 -1 2 0 = - -6 + 1 0 - 2 = - -5 -2
-2 -1 -1 2 -4 + 1 0 - 2 -3 -2
= 5 2
3 2
B11= A11-1 + [(A11-1 A12) L-1 ](A21 A11-1)
= A11-1 + 19 9 2 0
-18 -8 -1 2
= A11-1 + 38-9 0+18 = 5 -3 + 29 18
-36+8 0-16 -3 2 -28 -16
= 34 15
-31 -14
Jadi
A-1 = B11 B12 = 34 15 -19 -9
B21 B22 -31 -14 18 8
5 2 -3 -1
3 2 -2 -1

21. Contoh
2 3 7 9 11
A = 3 4 8 10 12
8 7 4 3 2
10 9 4 2 2
12 11 5 2 2
Menghitung A11-1 dan L-1
A11 = 2 3 det(A11) = 8 - 9 = -1
3 4
A11-1 = -1 4 -3 = -4 3
-3 2 3 -2
A11-1 A12 = -4 3 7 9 11 = -28+24 -36+30 -44+36 = -4 -6 -8
3 -2 8 10 12 21-16 27-20 33-24 5 7 9
A21 A11-1 = 8 7 -4 3 = -28+21 24-14 = -11 10
10 9 3 -2 -40+27 30-18 -13 12
12 11 -48+33 36-22 -15 14

22. L = A22 - A21 (A11-1 A12)
= A22 - 8 7 -4 -6 -8 = A22 - -32+35 -48+49 -64+63
10 9 5 7 9 -40+45 -60+63 -80+81
12 11 -48+55 -72+77 -96+99
= 4 3 2 - 3 1 -1 = 1 2 3
4 2 2 5 3 1 -1 -1 1
5 2 2 7 5 3 -2 -3 -1
del(L)=1 - 4 + 9 - 6 - 2 + 3 = 1
L-1 = (1+3) -(-2+9) (2+3) = 4 -7 5
-(1+2) (-1+6) -(1+3) -3 5 -4
(3-2) -(-3+4) (-1+2) 1 -1 1

23. Menghitung B
B12 = - (A11-1 A12) L-1
= - -4 -6 -8 4 -7 5
5 7 9 -3 5 -4
1 -1 1
= - -16+18-8 28-30+8 -20+24-8 = - -6 6 -4
20-21+9 -35+35-9 25-28+9 8 -9 6
= 6 -6 4
-8 9 -6
B21 = - L-1 (A21 A11-1)
= - 4 -7 5 -11 10
-3 5 -4 -13 12
1 -1 1 -15 14
= - 44+91-75 40-84+70 = - -28 26 = 28 -26
33-65+60 -30+60-56 28 -26 -28 26
-11+13-15 10-12+14 -13 12 13 -12

24. B11= A11-1 + [(A11-1 A12) L-1 ](A21 A11-1)
= A11-1 + -4 -6 -8 -28 26
5 7 9 28 -26
-13 12
= A11-1 + 112-168+104 104+156-96
-140+196-117 130-182+108
= -4 3 + 48 -44 = 44 -41
3 -2 -61 56 -58 54
Jadi
A-1 = B11 B12 = 44 -41 6 -6 4
B21 B22 -58 54 -8 9 -6
28 -26 4 -7 5
-28 26 -3 5 -4
13 -12 1 -1 1

28. Soal-soal Latihan
Hitung invers matrik A berikut ini dengan cara
1 . 𝐴 =
𝑏 𝑏 + 1 𝑎 − 1 𝑎 + 1
𝑏 + 1 𝑏 + 2 𝑎 𝑎 + 2
𝑎 + 2 𝑎 + 1 𝑏 + 1 𝑏 + 4
𝑎 + 1 𝑎 𝑏 + 1 𝑏 + 1
Hitung invers matrik A berikut ini dengan cara partisi
2 . 𝐴 =
𝑎 + 1 𝑎 + 2 𝑏 − 2 𝑏 𝑏 + 1
𝑎 𝑎 + 1 𝑏 − 3 𝑏 − 1 𝑏
𝑏 + 1 𝑏 𝑎 + 3 𝑎 𝑎
𝑏 − 1 𝑏 − 2 𝑎 + 1 𝑎 + 2 𝑎 + 1
𝑏 − 2 𝑏 − 3 𝑎 𝑎 − 4 𝑎 − 4
3 . 𝐴 =
𝑎 − 2 𝑎 − 1 𝑏 − 1 𝑏 − 2 𝑏 − 3
𝑎 − 1 𝑎 𝑏 − 2 𝑏 − 3 𝑏 − 4
𝑏 − 4 𝑏 − 3 𝑎 + 2 𝑎 − 1 𝑎 + 1
𝑏 − 3 𝑏 − 2 𝑎 − 1 𝑎 + 1 𝑎 − 3
𝑏 − 2 𝑏 − 1 𝑎 + 1 𝑎 − 3 𝑎 + 1

29. 
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