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hehe kali ini saya akan beri kalian file lagi nih tentang matematika hehe lebih tepatnya ke materi matriks. disini diulas bagaimana ciri ciri matriks dan rumus rumus tentang matriks. file ini berbentuk ppt jadi
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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
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