HASIL KALI ELEMENTER BERTANDA - Direktori File...
Transcript of HASIL KALI ELEMENTER BERTANDA - Direktori File...
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HASIL KALIELEMENTER
PERMUTASI:- GENAP- GANJIL
INVERSI/PEMBALIKAN
HASIL KALI ELEMENTER BERTANDA
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DETERMINAN MATRIKS
SYARATMATRIKS BUJUR SANGKAR(jumlah baris = jumlah kolom)
NILAI DETERMINAN SKALAR
NOTASI det (A) atau |A|
det (A) = 0 MATRIKS SINGULAR
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Matriks berordo 2 x 2
Jika A =
Maka, det (A) = |A| =
Contoh :
A =
dc
ba
dc
ba
76
54
bcad
76
54Maka |A| = - 6.54.7 = 28 - 30 = -2
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Matriks berordo 3 x 3
A =
Det (A)=
5/10/2010
f
a b c
d e
g h i
a b c
d e
g h i
f
a b
d e
g h
+ + - -= a e i b f g c d h c e g a f h b d i-
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Matriks berordo 3 x 3
A =
Det (A)=
5/10/2010
1
2 1 4
4 2
5 1 3
2 1 4
4 2
5 1 1
1
2 1
4 2
1 1
+ + - -= 2.2.1 1.1.1 4.4.1 5.2.1 1.1.2 1.4.1-
= 12 + 5 + 16 – 40 – 2 - 12= - 21
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MATRIKS ADJOINTMatriks Adjoint adalah transpose dari matriks kofaktornya.
Jika
dengan
adj
ijC
ijM
)( jiCAAdj
)( ijcA
ij
ji
ij MC .1
ija
ija
A11K 12K 13K
21K 22K 23K
31K 32K 33K
)(A
A11K 12K 13K
21K 22K 23K
31K 32K 33K
kofaktormatriks
kofaktormatriks
maka
maka
Jika
Dimana : = Kofaktor dari elemen
= Minor dari elemen
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Contoh :
Tentukan matriks adjoint dari
Jawab : 123
254
312
A
ij
ji
ij MC .1
11C
12C
13C
56112
31).1(.1 21
12M
12
25.111
11.1 M
13
24).1(12
21.1 M
13
31.1 M
23
54.1
21C
5 4 1
4 6 2
8 15 7
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79213
32.1.1 22
22
22 MC
13423
12).1(.1 23
32
23 MC
1315225
31.1.1 31
13
31 MC
812424
32).1(.1 32
23
32 MC
641054
12.1.1 33
33
33 MC
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adj
A
1 5 13
2 7 8
7 1 6
A
A
1 5 13
2 7 8
7 1 6
kofaktormatriks
didapat
111C
212C
713C
521C
722C
123C
1331C
832C
633C
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INVERS MATRIKS
NOTASI A-1
(A) adjointA
A11
RUMUS UMUM
11C
12C
dc
baA
INVERS MATRIKS 2X2
misal
1. Tentukan matriks kofaktornya dengan rumus
ij
ji
ij MC .1
11
11.1 M d.1 d
c12
21.1 M c).1(
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Jadi,
111
11
.)(
..
:
ABAB
IAAAA
Sifat
bbMC ).1(.1 21
12
21
aaMC .1.1 22
22
22
Matriks kofaktor ab
cdA
d
b
Maka adj A
)(adjoint11 AA
A
a
c
d b
bcad
11A
ac
bd
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contoh :
35
24A
1A
22
5
12
3
misalkan
bcad
11A
ac
bdmaka
5.23.4
1
45
23.
2
1
45
23
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det (A) = 0MATRIKS SINGULAR
tidak punya invers / balikan
Mengapa?
)adjoint(11 AA
A
tidak terdefinisi
Maka, matriks singular tidak mempunyai invers
Misalkan A matriks singular, maka det (A) = 0
)adjoint(0
1A
)int(0
1)int(
11 AadjoAadjoA
A
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