Post on 21-Jul-2016
description
1
MATRIKS
Matrix asalah susunan bilangan berbentuk segi-4 yang terdiri atas baris dan kolom yang ditulis dalam sepasang tanda kurung.
DEFINISI
NOTASI OF MATRIKS
mnmm
n
n
m aaa
aaaaaa
a
aa
A
...............
...
...
...
32
22322
11312
1
21
11
Nama Matriks
Amxn
ELEMENT MATRIKS
mnmm
n
n
m aaa
aaaaaa
a
aa
A
...............
...
...
...
32
22322
11312
1
21
11elementbaris
1
Letak elemenElemen kolom 1
ORDO
mnmm
n
n
m aaa
aaaaaa
a
aa
A
...............
...
...
...
32
22322
11312
1
21
11 Baris 1
Baris 2
Baris m
Kolom1 Kolom2
Kolom 3 Kolom n
Ordo m x n Notasi : A m x n
53
12
643970182
Ζ
1. Apakah nama matriks di atas?
Contoh:
53
12
643970182
Ζ
2. Sebutkan elemen baris 3 dan kolom 4!
53
12
643970182
Ζ
3. Sebutkan elemen baris ke-2!
53
12
643970182
Ζ
3. Sebutkan ordo matriks di atas dan notasinya!
Ordo 3 x 4
Notasi : Z 3 x 4
JENIS-JENIS MATRIKS
10
MATRIKS BARIS
4991N
MATRIKS KOLOM
603
S
MATRIKS DIAGONAL
100010004
M
MATRIKS IDENTITAS
000
000
W
100010001
W
Penjumlahan Perkalian
Matriks 0
1001
W
MATRIKS SEGITIGA
14002110341
W
Segitiga Atas
Segitiga bawah
28472030600290001
W
TRANSPOS MATRIKS
Transpos matriks A terjadi jika setiap baris pada matriks tersebut berubah menjadi kolom . Transpose matriks A ditulis A’ atau At. Sehingga A m x n menjadi A’ n x m.
Elemen baris 1 matriks A Kolom 1 matriks A’
Elemen baris 2 matriks A Kolom 2 matriks A’
dst
TRANSPOS MATRIKS
93
214
105
21'A
A 4 x 2
921102
3451
A
A’ 2 x
4
93
214
105
21tA
Tentukan transpose matriks berikut!
125
A
413221130
B
Tentukan transpose matriks berikut!
125
tA 125 A
Jawab
421123310
B
413221130
tB
413221130
B
hgc
fe
Bdb
aA
a = e
d = h
b = fc = g
PERSAMAAN MATRIKS
zx
2-1
B32-
1A
4
If A = B, tentukan nilai x dan z!
y4x2x52
Bz6yx2
A
Jika A = B, tentukan nilai x, y dan z!
y4x2x52
Bz6yx2
A
2 = 2
z = 4x - y
x + y = -56 = 2x
2 = 2
z = 4x – y
= 4.3 – (-8)
z = 12 + 8 = 20
x + y = -5
3 + y = -5
y = -5 - 3 = -8
6 = 2x
x = 6/2 = 3
205
652
B206
2A
2.36
20 4.3 – (-8)12 + 820
3 + (-8)-5
y4x2x52
Bz6yx2
A
1. PENJUMLAHAN DAN PENGURANGAN MATRIKS
Dua atau lebih matriks dapat dijumlahkan atau dikurang kan jika :
a. Matriks tersebut berordo sama
b. Yang dioperasikan elemen yang seletak
Contoh:
215
49
942
7010
8122
536
CBA
Dapatkah A dan C dijumlahkan?
Jika
Dapatkah A dan B dijumlahkan?
1164
12316
942
7010
8122
536
A + B = …
Untuk
942
7010
8122
536
BA
1780
234
8122
536
942
7010
B - A = …
2. PERKALIAN MATRIKS
a. Perkalian 2 buah matriks
=
tsrqpo
nm
lkj
ihg
fed
cba
CBA321
3 x 3 3 x 2 2 x 4
1. Dapatkah A dan C dikalikan?
2. Dapatkah A dan B dikalikan?
Contoh
05
20
12/1
43
802
536
CA
Dapatkah A dan C dikalikan?
Diberikan
A 3 x 2 C2 x 4
=
C2 x 4
Z3 x 4
05
20
12/1
43
802
536
CA
A x C = …
Untuk
A 3 x 2 C2 x 4
=
K1C K2C K3C K4C
B1A
B2A
B3A
K1C K2C K3C K4C
B1A
B2A
B3A
05
20
12/1
43
802
536
CAa = (6x3)+(2x4)
= 18 + 8
= 26
K1C K2C K3C K4C
B1A
B2A
B3A
05
20
12/1
43
802
536
CAa = (-3x3)+(0x4)
= -9 + 0
= -9
26
K1C K2C K3C K4C
B1A
B2A
B3A
05
20
12/1
43
802
536
CAa = (5x5)+(0x-8)
= 25 + 0
= 25
26
-9
K1C K2C K3C K4C
B1A
B2A
B3A
05
20
12/1
43
802
536
CA
26
-925
26
-925
05
20
12/1
43
802
536
CA
-17 10,5
-1,51 -4
016
30
-15A.C =
Kerjakan soal berikut!
232
140
421
42
53
21
ZX Y
Diberikan
Tentukanlah matriks :
1.X.Y
2.Z.X
1.
b. Perkalian Matriks dengan skala
Multiplication a real number with matrix A is multipilcation each elements of matrix A by that real number
k.A = [k.amn]
Example
802
536
A
Determine 2 x A if
Answer
2 x 82 x 02 x 2
5 x 23 x2
6 x 2 2.A =
1604
10612
=
DETERMINANT
Determinant of matrix
a.Only used in square
b.are substraction with elements 1st diagonal and 2nd diagonal, where each elements enclosed
a. DETERMINANT ORDO 2 X 2
If
dcba
A
than|A| = ad - bc
Example
Determine value of determinant matrix below
61-105
A
Answer:
|A| = 5.6 – 10.-1 = 30 + 10 = 40
DETERMINAN ORDO 3 x 3
If given
ihgfedcba
A
than |A| =
heb
gda
ihgfedcba
A
DETERMINAN ORDO 3 x 3
|A| =
heb
gda
ihgfedcba
A
= (a.e.i + b.f.g + c.d.h) –(c.e.g + a.f.h + b.d.i)
Example
Determine determinat of
531312740
A
Answer: = (0.1.5 + 4.-3.-1 + 7.2.3) –(-1.1.7 +3.-3.0 + 5.2.4)
= (0+12+42) – (-7+0+40)
= 54 – 33 = 21
314
120
531312740
A
4. ADJOINAdjoin matrix A is the result transpose from kofaktor matriks A.
Matrix A
Adjoin Matrix A
Minor Matrix A
Kofaktor Matrix A
Minor Jika maka minor
61-105
A
M11 = 6
61-105
AM12 = -1
61-105
A
M21 = 10
61-105
AM22 = 5
61-105
A
5101-6
A
a. Ordo 2 x 2
Kofactor
If than kofactor
61-105
A
M11 = 6 .-11+1 = 6M12 = -1. -11+2 = -1.-1 =1M21 = 10. -12+1 = 10. -1 = -10M22 = 5. -1 2+2 = 5.1 = 5
510-16
A
5101-6
A
-
-
Adjoin
If than Adjoin matrix A
Resulted from the its kofactor
61-105
A
510-16
Akofaktor
5101-6
AMinor
5110-6
A Adjoin
205321211
A
M11 = 2.-2 – (0.3)
= -4- 0
= -4
205321211
A
M12 = 1.-2 – (-5.3)
= -2 – (-15)
= 13
205321211
A
M13 = 1.0 – (-5.2)
= 0 – (-10)
= 10
If , minor matrix A showed
next
205321211
A
b. Ordo 3 x 3
205321211
A
M21 = 1.-2 – 0.2
= -2- 0
= -2
205321211
A
M22 = 1.-2 – (-5.2)
= -2 – (-10)
= 8
205321211
A
M23 = 1.0 – (-5.1)
= 0 – (-5)
= 5
205321211
A
M21 = 1.3 – (2.2)
= 3 - 4
= -1
205321211
A
M22 = 1.3 – (1.2)
= 3 – 2
= 1
205321211
A
M23 = 1.2 – (1.1)
= 2 – (1)
= 1
Kofactor
11158210134-
A
:Minor
A
11-15-82
1013-4-A
:Kofactor
Adjoin
111582111-
:AMinor
11-15-8211-1-
:AKofactor
15-11-81-1-21-
:A Adjoin
205321211
A
given If
Inverse matrix A AAdjoin.|A|
1A 1 0 A ,
5. INVERSE
a. Inverse ordo 2 X 2
,
dcba
AIf
acbd
|A|1Aor 1
AAdjoin.|A|
1A 1
Answer
dcba
Aif
acbd
|A|1Ao 1r
AAdjoin.|A|
1A , 1
Contoh:
61-105
A
Determine inverse from
Answer :
51106
|)10.1(6.5|11A
51106
|1030|11A
51106
4011A
.adjA|A|
1A 1
40/540/140/1040/61A
8/140/14/120/31A
II. MATRIX APPLICATIONUsing to determine variabel value of linear equation. If the equation have variabel x dan y, than ..
|A||X|x
|A||Y|y
Example Determine value of x dan y from the next
equations2x + 3y = 7x - 2y = 7
7
72132
yx
734(1.3)22.|A|21
32A
35211
4(7.3)27.|X|27
37X
774(1.7)2.|Y|1
72Y
17
7
5735
|A||X|x
17 -7|A|
|Y|y
Competence Check1. Given
(A.B)-1 = ….
455-6-
Ba4321
A nd
42-3-1
b.
2-1211
21
e.
21211-
21
c.
21-211-
21
d.
1234
a.
2. Determine solution set from the next l
are ….
54
yx
2132
)2,1( d.)2,1(b.
)1,2(e.)2,1( c.)2,1(a.
204321301
A
b. Find determinan and adjoint from the next matrix
3-1-12
B
3-1-12
B
Minor Jika maka minor
3-1-12
A
M11 = -3
3-1-12
AM12 = -1
3-1-12
A
M21 = 1
3-1-12
AM22 = 2
3-1-12
A
211-3-
A
Kofactor
If than kofactor
3-1-12
A
M11 = -3 .-11+1 = -3M12 = -1. -11+2 = -1.-1 =1M21 = 1. -12+1 = 1. -1 = -1M22 = 2. -1 2+2 = 2.1 = 2
2-1-13-
A
211-3-
A
-
-
SOAL
3-1-12
A
2-1-13-
A
211-3-
A
2
31
1-
MINOR
KOFACTOR
ADJOINT