konsep matriks
DESCRIPTION
MATRIKS. Konsep Matriks. MATRIX. Concept of Matrix. Macam-macam Matriks. Kompetensi Dasar : Mendeskripsikan macam-macam matriks Indikator : Matriks ditentukan unsur dan notasinya Matriks dibedakan menurut jenis dan relasinya. Kinds of Matrix. Basic Competences : - PowerPoint PPT PresentationTRANSCRIPT
Hal.: 3 Matriks Adaptif
Macam-macam Matriks
Kompetensi Dasar :Mendeskripsikan macam-macam matriks
Indikator :1. Matriks ditentukan unsur dan notasinya2. Matriks dibedakan menurut jenis dan
relasinya
Hal.: 4 Matriks Adaptif
Kinds of Matrix
Basic Competences :Describing the kinds of matrix
Indicators :1. Matrix is determined by its elements and
notations2. Matriks matrix is distinguished by its kinds
and relations
Hal.: 5 Matriks Adaptif
Pengertian Matriks Matriks adalah susunan bilangan-bilangan yang terdiri
atas baris-baris dan kolom-kolom.
a11 a12…….a1j ……a1n
a21 a22 ……a2j…….a2n : : : :ai1 ai2 ……aij…….. ain
: : : :am1 am2……amj……. amn
A = baris
kolom
Notasi: Matriks: A = [aij]Elemen: (A)ij = aij
Ordo A: m x n
Masing-masing bilangan dalam matriks disebut entri atau elemen. Ordo (ukuran) matriks adalah jumlah baris kali jumlah kolom.
Macam – macam Matriks
Hal.: 6 Matriks Adaptif
Definition of Matrix Matrix is the arrangement of numbers
which consists of rows and columns.
a11 a12…….a1j ……a1n
a21 a22 ……a2j…….a2n : : : :ai1 ai2 ……aij…….. ain
: : : :am1 am2……amj……. amn
A = rows
column
Notation: Matrix: A = [aij]Element: (A)ij = aij
Order A: m x n
Each of the numbers in matrix is called as entry or element. Order (size) of matrix is the value of the row number multiplied by the number of column.
Kinds of Matrix
Hal.: 7 Matriks Adaptif
Macam-macam Matriks
Matriks baris adalah matriks yang hanya terdiri dari satu baris.
21A 2 5
31xB
41xC
1 -8 25
-2 0 14 8
1. Matriks Baris
Hal.: 8 Matriks Adaptif
Kinds of Matrix
Row matrix is a matrix which consists of one row.
21A 2 5
31xB
41xC
1 -8 25
-2 0 14 8
1. Row matrix
Hal.: 9 Matriks Adaptif
Macam-macam Matriks
12P
2. Matriks Kolom
Matriks Kolom adalah matriks yang hanya terdiri dari satu kolom
2
-7
13Q9
2
1
Hal.: 10 Matriks Adaptif
Kinds of Matrix
12P
2. Column matrix
Column matrix is a matrix which consists of one column.
2
-7
13Q9
2
1
Hal.: 11 Matriks Adaptif
1 2 42 2 23 3 3
3. Matriks PersegiMatriks persegi (bujur sangkar) adalah matriks yang jumlah baris dan jumlah kolom sama.
Trace(A) = 1 + 2 + 3
Trace dari matriks adalah jumlahan elemen-elemen diagonal utama
diagonal utama
Macam – macam Matriks
Hal.: 12 Matriks Adaptif
1 2 42 2 23 3 3
3. Square matrixSquare matrix is a matrix which has the same numbers of rows and columns.
Trace(A) = 1 + 2 + 3
Trace from matrix is the total numbers from the main diagonal elements.
Main diagonal
Kinds of Matrix
Hal.: 13 Matriks Adaptif
4. Matriks Nol
Matriks nol adalah matriks yang semua elemennya nol
0 0 0 0 00 0
1 00 1
1 0 00 1 00 0 1
1 0 0 00 1 0 00 0 1 0 0 0 0 1
I2I3 I4
Matriks identitas adalah matriks persegi yang elemen diagonal utamanya 1 dan elemen lainnya 0
Macam- macam Matriks
Hal.: 14 Matriks Adaptif
4. Zero matrix
zero matrix is a matrix which all of its elements are zero.
0 0 0 0 00 0
1 00 1
1 0 00 1 00 0 1
1 0 0 00 1 0 00 0 1 0 0 0 0 1
I2I3 I4
Matrix identity is a square matrix which its main diagonal element is 1 and the other element is 0.
Kinds of Matrix
Hal.: 15 Matriks Adaptif
5. Matriks ortogonalMatriks A orthogonal jika dan hanya jika AT = A –1
0 -1
1 0A =
0 1
-1 0AT=
B = ½√2 -½√2
½√2 ½√2 BT= ½√2 ½√2
-½√2 ½√2
Jika A adalah matriks orthogonal, maka (A-1)T = (AT)-1
= A-1
= B-1
(A-1)T = (AT)-1 A-1 AT
Macam-macam Matriks
Hal.: 16 Matriks Adaptif
5. Orthogonal MatrixMatrix A is orthogonal if and only if AT = A –1
0 -1
1 0A =
0 1
-1 0AT=
B = ½√2 -½√2
½√2 ½√2 BT= ½√2 ½√2
-½√2 ½√2
If A is orthogonal matrix, so (A-1)T = (AT)-1
= A-1
= B-1
(A-1)T = (AT)-1 A-1 AT
Kinds of Matrix
Hal.: 17 Matriks Adaptif
Macam – macam Matriks
Definisi:Transpose matriks A adalah matriks AT, kolom-kolomnya adalah baris-baris dari A, baris-barisnya adalah kolom-kolom dari A.
4 2 6 7
5 3 -9 7A = AT = A’ =
4 5
2 3
6 -9
7 7
Jika A adalah matriks m x n, maka matriks transpose AT berukuran ………..
[AT]ij = [A]ji
n x m
Hal.: 18 Matriks Adaptif
Kinds of Matrix
Definisi:Transpose matrix A is matrix AT, its columns are rows of A, its rows is columns of A.
4 2 6 7
5 3 -9 7A = AT = A’ =
4 5
2 3
6 -9
7 7
if A is matrix m x n, so matrix transpose AT should be ………..
[AT]ij = [A]ji
n x m
Hal.: 19 Matriks Adaptif
Kesamaan dua matriks Dua matriks sama jika ukuran sama dan setiap entri yang bersesuaian sama.
1 2 4
2 1 3A =
1 2 4
2 1 3B =
1 2 2
2 1 3C =
2 1 2
2 1 3D =
1 2 4
2 2 2E =
x 2 4
2 2 2F =
2 2 2
4 5 6
9 0 7
G = H =? ? ?
? ? ?
? ? ?
A = B
C ≠ D
E = F jika x = 1
G = H
2 2 2
4 5 6
9 0 7
Macam – macam Matriks
Hal.: 20 Matriks Adaptif
Similarity of two matrixes Two matrix are similar if its size is similar and each symmetrical entry is similar
1 2 4
2 1 3A =
1 2 4
2 1 3B =
1 2 2
2 1 3C =
2 1 2
2 1 3D =
1 2 4
2 2 2E =
x 2 4
2 2 2F =
2 2 2
4 5 6
9 0 7
G = H =? ? ?
? ? ?
? ? ?
A = B
C ≠ D
E = F jika x = 1
G = H
2 2 2
4 5 6
9 0 7
Kind of Matrix
Hal.: 21 Matriks Adaptif
Matriks Simetri
Matriks A disebut simetris jika dan hanya jika A = AT
4 2
2 3A =
4 2
2 3A’ = A simetri
1 2 3 42 5 7 0
3 7 8 2 4 0 2 9
A = = AT
Macam-macam Matriks
Hal.: 22 Matriks Adaptif
Symmetrical matrix
Matrix A is called symmetric if and only if A = AT
4 2
2 3A =
4 2
2 3A’ = A symmetric
1 2 3 42 5 7 0
3 7 8 2 4 0 2 9
A = = AT
Kinds of Matrix
Hal.: 23 Matriks Adaptif
Sifat-sifat transpose matriks
A AT (AT)T
(AT )T = A1. Transpose dari A transpose adalah A:
4 2 6 7
5 3 -9 7
4 5
2 3
6 -9
7 7
4 5
2 3
6 -9
7 7
= A
Contoh:
Macam-macam Matriks
Hal.: 24 Matriks Adaptif
properties of transpose matrix
A AT (AT)T
(AT )T = A1. Transpose of A transpose is A:
4 2 6 7
5 3 -9 7
4 5
2 3
6 -9
7 7
4 5
2 3
6 -9
7 7
= A
Example:
Kinds of Matrix
Hal.: 27 Matriks Adaptif
Macam-macam Matriks
3. (kA)T = k(A) T untuk skalar k
kA
(kA)T = k(A)T
A
T T
k
Hal.: 31 Matriks Adaptif
Macam-macam Matriks
Isilah titik-titik di bawah ini1. A simetri maka A + AT= ……..2. ((AT)T)T = …….3. (ABC)T = …….4. ((k+a)A)T = ….....5. (A + B + C)T = ……….
Kunci:1. 2A 2. AT
3. CTBTAT 4. (k+a)AT 5. AT + BT + CT
Soal :
Hal.: 32 Matriks Adaptif
Kind of Matrix
Fill in the blanks bellow1. A symmetric then A + AT= ……..2. ((AT)T)T = …….3. (ABC)T = …….4. ((k+a)A)T = ….....5. (A + B + C)T = ……….
Answer keys:1. 2A 2. AT
3. CTBTAT 4. (k+a)AT 5. AT + BT + CT
Quiz :
Hal.: 33 Matriks Adaptif
OPERASI MATRIKS
Kompetesi DasarMenyelesaikan Operasi Matriks
Indikator1. Dua matriks atau lebih ditentukan hasil
penjumlahan atau pengurangannya2. Dua matriks atau lebih ditentukan hasil kalinya
Hal.: 34 Matriks Adaptif
OPERATION OF MATRIX
Basic competenceFinishing operation matrix
Indicator1. Two or more matrixes is defined by the result of
their addition or subtraction 2. Two or more matrixes is defined by the result of
their multiplication
Hal.: 35 Matriks Adaptif
Penjumlahan dan pengurangan dua matriksContoh :
10 22
1 -1A = 2 6
7 5B =
10+2 22+6
1+7 -1+5A + B =
12 28
8 4=
8 16
-6 -6= A - B = 10-2 22-6
1-7 -1-5
OPERASI MATRIKS
Hal.: 36 Matriks Adaptif
Addition and subtraction of two matixesExample:
10 22
1 -1A = 2 6
7 5B =
10+2 22+6
1+7 -1+5A + B =
12 28
8 4=
8 16
-6 -6= A - B = 10-2 22-6
1-7 -1-5
OPERATION OF MATRIX
Hal.: 37 Matriks Adaptif
OPERASI MATRIKS
Apa syarat agar dua matriks dapat dijumlahkan?
Jawab:Ordo dua matriks tersebut sama
A = [aij] dan B = [bij] berukuran sama,
A + B didefinisikan: (A + B)ij = (A)ij + (B)ij = aij + bij
Hal.: 38 Matriks Adaptif
OPERATION OF MATRIX
What is the condition so that two matrixes can be added?
Answer:The ordo of the two matrixes are the same
A = [aij] dan B = [bij] have the same size,
A + B is defined: (A + B)ij = (A)ij + (B)ij = aij + bij
Hal.: 39 Matriks Adaptif
Jumlah dua matriks
5 6 1
7 2 3C = 25 30 5
35 10 15D =
C + D = ? ? ?
? ? ?
1 4 -9 3 7 0 5 9 -13
K = 7 3 1-2 4 -5 9 -4 3
L =
K + L =
? ? ?
? ? ?
? ? ?
D + C =
L + K =
Apa kesimpulanmu? Apakah jumlahan matriks bersifat komutatif?
OPERASI MATRIKS
Hal.: 40 Matriks Adaptif
The quantity of two matrixes
5 6 1
7 2 3C = 25 30 5
35 10 15D =
C + D = ? ? ?
? ? ?
1 4 -9 3 7 0 5 9 -13
K = 7 3 1-2 4 -5 9 -4 3
L =
K + L =
? ? ?
? ? ?
? ? ?
D + C =
L + K =
What is your conclusion? Is the addition of matrixes commutative?
OPERATION OF MATRIX
Hal.: 41 Matriks Adaptif
Soal:
C + D =… C + E = … A + B = …
3 -8 0
4 7 2
-1 8 4C = D =
3 7 2
5 2 6
-1 8 4E =
2 7 2
5 2 6
0 0 0
0 0 0 A =
0 0 0
0 0 0B =
6 -1 2
9 9 8
-2 16 8C +D = Feedback:
OPERASI MATRIKS
Hal.: 42 Matriks Adaptif
Exercise:
C + D =… C + E = … A + B = …
3 -8 0
4 7 2
-1 8 4C = D =
3 7 2
5 2 6
-1 8 4E =
2 7 2
5 2 6
0 0 0
0 0 0 A =
0 0 0
0 0 0B =
6 -1 2
9 9 8
-2 16 8C +D = Feedback:
OPERATION OF MATRIX
Hal.: 43 Matriks Adaptif
Hasil kali skalar dengan matriks
5 6 1
7 2 3A = 5A = =
250 300 50
350 100 150H = H =
Diberikan matriks A = [aij] dan skalar c, perkalian skalar cA mempunyai entri-entri sebagai berikut:
(cA)ij = c.(A)ij = caij
Apa hubungan H dengan A?
5x5
5x5
5x6
5x2
5x1
5x3
25
35
30
10
5
15
Catatan: Pada himpunan Mmxn, perkalian matriks dengan skalar bersifat tertutup (menghasilkan matriks dengan ordo yang sama)
50A
OPERASI MATRIKS
Hal.: 44 Matriks Adaptif
The multiplication result of scalar matrix
5 6 1
7 2 3A = 5A = =
250 300 50
350 100 150H = H =
Given matrix A = [aij] aand scalar c, the multiplication of scalar cA have the following entries:
(cA)ij = c.(A)ij = caij
What is the relation between H and A?
5x5
5x5
5x6
5x2
5x1
5x3
25
35
30
10
5
15
Note: In the set of Mmxn, the matrix multiplication with scalar have closed properties (it will have matrix with the same orrdo)
50A
OPERATION OF MATRIX
Hal.: 45 Matriks Adaptif
OPERASI MATRIKS K 3 x 3
1 4 -9 3 7 0 5 9 -13
K =
5 20 -4515 35 0
25 45 -655K =
4 16 -36 12 28 0
20 36 -524K =
Hal.: 46 Matriks Adaptif
OPERATION OF MATRIX K 3 x 3
1 4 -9 3 7 0 5 9 -13
K =
5 20 -4515 35 0
25 45 -655K =
4 16 -36 12 28 0
20 36 -524K =
Hal.: 47 Matriks Adaptif
OPERASI MATRIKS
Diketahui bahwa cA adalah matriks nol. Apa kesimpulan Anda tentang A dan c?
0 0 0
0 0 0 A = A =
2 7 2
5 2 6c = 0c = 7
cA = 0*2 0*7 0*2
0*5 0*2 0*6
0 0 0
0 0 0 = cA =
7*0 7*0 7*0
7*0 7*0 7*0
Kasus 1: c = 0 dan A matriks sembarang. Kasus 2: A matriks nol dan c bisa berapa saja.
Contoh:
kesimpulan
Hal.: 48 Matriks Adaptif
OPERATION OF MATRIX
Known that cA is zero matrix. What is your conclusion about A and c?
0 0 0
0 0 0 A = A =
2 7 2
5 2 6c = 0c = 7
cA = 0*2 0*7 0*2
0*5 0*2 0*6
0 0 0
0 0 0 = cA =
7*0 7*0 7*0
7*0 7*0 7*0
Case 1: c = 0 and A is any matrixCase 2: A is zero matrix and c can be any number
Example:
Conclusion
Hal.: 49 Matriks Adaptif
OPERASI MATRIKS
Definisi: Jika A = [aij] berukuran m x r , dan B = [bij] berukuran r x n,
maka matriks hasil kali A dan B, yaitu C = AB mempunyai elemen-elemen yang didefinisikan sebagai berikut:
∑ aikbkj = ai1b1j +ai2b2j+………airbrj
k = 1
(C)ij = (AB)ij =
2 3 4 5
8 -7 9 -4
1 -5 7 -8
A = 1 2
7 -6
4 -9
B = Tentukan AB dan BA
A B ABm x r r x n m x n
• Syarat:
r
Perkalian matriks dengan matriks
Hal.: 50 Matriks Adaptif
OPERATION OF MATRIX
Definition: If A = [aij] have size m x r , and B = [bij] have size r x n,
then the matrix which is from the multiplication result between A and B, yaitu is C = AB has elements that defined as follows:
∑ aikbkj = ai1b1j +ai2b2j+………airbrj
k = 1
(C)ij = (AB)ij =
2 3 4 5
8 -7 9 -4
1 -5 7 -8
A = 1 2
7 -6
4 -9
B = Define AB and BA
A B ABm x r r x n m x n
• Condition:
r
Multiplication between matrix
Hal.: 51 Matriks Adaptif
Perkalian matriks dengan matriks
2 3 4 5
8 -7 9 -4
1 -5 7 -8
A =
1 2
7 -6
4 -9
11 3
B =
A B = 2.1 +3.7+4.4+5.11 -35
-49 -35
-94 -55
94 -35
-49 -35
-94 -55
=
OPERASI MATRIKS
=
Contoh :
BA tidak didefinisikan
Hal.: 52 Matriks Adaptif
The multiplication between matrixes
2 3 4 5
8 -7 9 -4
1 -5 7 -8
A =
1 2
7 -6
4 -9
11 3
B =
A B = 2.1 +3.7+4.4+5.11 -35
-49 -35
-94 -55
94 -35
-49 -35
-94 -55
=
OPERATION OF MATRIX
=
Example:
BA is not define
Hal.: 53 Matriks Adaptif
OPERASI MATRIKS1. Diberikan A dan B, AB dan BA terdefinisi. Apa kesimpulanmu?
2. AB = O matriks nol, apakah salah satu dari A atau B pasti matriks nol?
2 32 3A = 3 -3
-2 2B = 0 00 0AB =
B An x k m x n
m = k
ABmxm ABnxn
AB dan BA matriks persegi
AB matriks nol, belum tentu A atau B matriks nol
A Bn x km x n
Hal.: 54 Matriks Adaptif
OPERATION OF MATRIX1. Given A and B, AB and BA is defined. What is your conclusion?
2. AB = O is zero matrix, is one of (A or B) is zero matrix?
2 32 3A = 3 -3
-2 2B = 0 00 0AB =
B An x k m x n
m = k
ABmxm ABnxn
AB and BA square matricx
AB is zero matrix. Matrix A and B is not certain zero matrix
A Bn x km x n
Hal.: 55 Matriks Adaptif
Tentukan hasil kalinya jika terdefinisi.
• A B = ??• AC = ??• BD = ??• CD = ??• DB = ??
OPERASI MATRIKS
2 3 4 5 4 7 9 0 2 3 5 6
A = 1 2-9 0 8 0 5 6
B =
7 -11 43 5 -6
C = 1 8 9 5 6 2 5 6 -9 0 0 -4 7 8 9
D =
Contoh 1:
Hal.: 56 Matriks Adaptif
Define the multiplication result if it defined:
• A B = ??• AC = ??• BD = ??• CD = ??• DB = ??
OPERATION OF MATRIX
2 3 4 5 4 7 9 0 2 3 5 6
A = 1 2-9 0 8 0 5 6
B =
7 -11 43 5 -6
C = 1 8 9 5 6 2 5 6 -9 0 0 -4 7 8 9
D =
Example 1:
Hal.: 57 Matriks Adaptif
OPERASI MATRIKS
Contoh 2:2 31 2A =
A2 = 2 31 2
2 31 2
A3 = A x A2 = 2 31 2
2 31 2
2 31 2
A0 = IAn =
n faktor
An+m = An Am
A A A …A
Hal.: 58 Matriks Adaptif
OPERATION OF MATRIX
Example 2:2 31 2A =
A2 = 2 31 2
2 31 2
A3 = A x A2 = 2 31 2
2 31 2
2 31 2
A0 = IAn =
n factor
An+m = An Am
A A A …A
Hal.: 59 Matriks Adaptif
DETERMINAN DAN INVERS
Kompetensi Dasar:Menentukan determinan dan invers
Indikator :1. Matriks ditentukan determinannya2. Matriks ditentukan inversnya
Hal.: 60 Matriks Adaptif
DETERMINANT AND INVERSE
Basic Competence:Define the determinant and inverseIndicator :1. Matrix is defined by its determinant2. Matrix is defined by its inverse
Hal.: 61 Matriks Adaptif
DETERMINAN DAN INVERS
Determinan Matriks ordo 2 x 2
Nilai determinan suatu matriks ordo 2 x 2 adalah hasil kali elemen-elemen diagonal utama dikurangi hasil kali elemen pada diagonal kedua.
Misalkan diketahui matriks A berordo 2 x 2, A =
Determinan A adalah
det A =
dcba
dcba
= ad - bc
Hal.: 62 Matriks Adaptif
DETERMINANT AND INVERSE
Determinant Matrix ordo 2 x 2Determinant value of a matrix ordo 2 x 2 is the multiplication result of the main diagonal elements and subtract by the multiplication result of the second diagonal.
For example, known matrix A ordo 2 x 2, A =
Determinant A is
det A =
dcba
dcba
= ad - bc
Hal.: 63 Matriks Adaptif
Contoh: Invers matriks 2x2
3 2
4 1A =
I=
1 -23.1-4.2 3.1-4.2
3-43.1-4.2 3.1-4.2
=A-1
1 25 5
345 5
DETERMINAN DAN INVERS
Hal.: 64 Matriks Adaptif
Example: Matrix inverse 2x2
3 2
4 1A =
I=
1 -23.1-4.2 3.1-4.2
3-43.1-4.2 3.1-4.2
=A-1
1 25 5
345 5
DETERMINANT AND INVERSE
Hal.: 65 Matriks Adaptif
DETERMINANT DAN INVERSE
1. Kapan matriks TIDAK mempunyai invers? a bc d
2. Tentukan invers matriks berikut ini
1 0
0 1d.
5 1
1 2a.
0 1
0 2b.
0 0
4 1c.
1 0
0 1d.
2/3 -1/5
-1/5 5/3a.
ad-bc = 0
b. tidak mempunyai invers
c. tidak mempunyai invers
Contoh :
Hal.: 66 Matriks Adaptif
DETERMINANT AND INVERSE
1. When matrix Doesn’t have inverse? a bc d
2. Define the following matrix inverse
1 0
0 1d.
5 1
1 2a.
0 1
0 2b.
0 0
4 1c.
1 0
0 1d.
2/3 -1/5
-1/5 5/3a.
ad-bc = 0
b. Doesn’t have inverse
c. Doesn’t have inverse
Example :
Hal.: 67 Matriks Adaptif
DETERMINAN DAN INVERS
B adalah invers dari matriks A, jika AB = BA = I matriks identitas, ditulis B = A-1
dcba
A IA-1A-1 A= =
Jika A = , maka
acbd
bcadA 11
0 bcadAdengan
Hal.: 68 Matriks Adaptif
DETERMINANT AND INVERSE
B is inverse of matrix A, if AB = BA = I matrix identities, it is written B = A-1
dcba
A IA-1A-1 A= =
If A = , then
acbd
bcadA 11
0 bcadAwith
Hal.: 69 Matriks Adaptif
DETERMINAN DAN INVERS
Contoh 1 :Tentukan invers dari matriks
27517
5.717.211
acbd
AA
42
10517752
danBA
Jawab :
27517
det B = (-5) . (-4) – (-2) . (-10) = 20 – 20 = 0 , sehingga matriks B
tidak memiliki invers
Hal.: 70 Matriks Adaptif
DETERMINANT AND INVERSE
Example 1 :Defined the inverse of matrix
27517
5.717.211
acbd
AA
42
10517752
danBA
Answer :
27517
det B = (-5) . (-4) – (-2) . (-10) = 20 – 20 = 0 , So, matrix B doesn’t
have inverse
Hal.: 71 Matriks Adaptif
DETERMINAN DAN INVERS
1 0 0
0 1 0
0 0 1
4 2
2 2
½ -½
-½ 1
4 2 1
2 2 1
3 3 1
½ -½ 1
-½ -½ 1
0 3 -2
1 0
0 1
Contoh 2 :
4 2
2 2
½ -½
-½ 1= =
A A-1 A-1 A I
4 2 1
2 2 1
3 3 1
½ -½ 1
-½ -½ 1
0 3 -2
= =
B B-1 B-1 B I
Diketahui matriks
Tunjukkan bahwa A.A-1 = A-1.A = I dan B.B-1 = B-1. B = I
133122124
2224
danBA
Hal.: 72 Matriks Adaptif
DETERMINANT AND INVERSE
1 0 0
0 1 0
0 0 1
4 2
2 2
½ -½
-½ 1
4 2 1
2 2 1
3 3 1
½ -½ 1
-½ -½ 1
0 3 -2
1 0
0 1
Example 2 :
4 2
2 2
½ -½
-½ 1= =
A A-1 A-1 A I
4 2 1
2 2 1
3 3 1
½ -½ 1
-½ -½ 1
0 3 -2
= =
B B-1 B-1 B I
Known matrix
Show that A.A-1 = A-1.A = I and B.B-1 = B-1. B = I
133122124
2224
danBA
Hal.: 73 Matriks Adaptif
DETERMINAN DAN INVERS
Matriks ordo 3 x 3
.
ihgfedcba
MisalkanA
Determinan Matriks Ordo 3 x 3
Dengan aturan Sarrus, determinan A adalah sebagai berikut.
hgedba
ihgfedcba
A
_ _ _ + + +bdiafhcegcdhbfgaei
)()( bdiafhcegcdhbfgaei
Hal.: 74 Matriks Adaptif
DETERMINANT AND INVERSE
Matrix ordo 3 x 3
.
ihgfedcba
exampleA
Matrix Determinant Ordo 3 x 3
With Sarrus rule, determinant A is as follows
hgedba
ihgfedcba
A
_ _ _ + + +bdiafhcegcdhbfgaei
)()( bdiafhcegcdhbfgaei
Hal.: 75 Matriks Adaptif
DETERMINAN DAN INVERS
Sistem Persamaan Linear Dua Variabel dengan Menggunakan Matriks
Misal SPL 111 cybxa
222 cybxa
Persamaan tersebut dapat di ubah menjadi bentuk matriks
berikut
2
1
22
11
cc
yx
baba
Hal.: 76 Matriks Adaptif
DETERMINANT AND INVERSE
The equation of linear with two variable using matrix
For example SPL 111 cybxa
222 cybxa
The equation can be changed into the following matrix
2
1
22
11
cc
yx
baba
Hal.: 77 Matriks Adaptif
DETERMINAN DAN INVERS
Misalkan ,,2
1
22
11
CC
danByx
Pbaba
A maka dapat ditulis
2
1
22
11
cc
yx
baba
BAP
BAP 1
Hal.: 78 Matriks Adaptif
DETERMINANT AND INVERSE
Example ,,2
1
22
11
CC
andByx
Pbaba
A Then can be write
as
2
1
22
11
cc
yx
baba
BAP
BAP 1
Hal.: 79 Matriks Adaptif
DETERMINAN DAN INVERS
Contoh :
1632 yxTentukan nilai x dan y yang memenuhi sistem persamaan linear
134 yx
134 yxJawab :
Sistem persamaan 1632 yx
Jika dibuat dalam bentuk matriks menjadi
1316
4132
yx
Hal.: 80 Matriks Adaptif
DETERMINANT AND INVERSE
Example:
1632 yxDefine the value of x and y that fulfill the equation of linear system
134 yx
134 yxanswer :
Equation system1632 yx
If in matrix
1316
4132
yx
Hal.: 81 Matriks Adaptif
DETERMINAN DAN INVERS
Perkalian matriks berbentuk AP = B dengan
1316
,41
32danB
yx
PA
2134
51
2134
3.14.211A
BAP
BAP 1
1316
2134
51
yx
25
1025
51
26163964
51
Jadi nilai x = 5 dan y = 2
Hal.: 82 Matriks Adaptif
DETERMINANT AND INVERSE
The matrix multiplication in the form of AP = B with
1316
,41
32danB
yx
PA
2134
51
2134
3.14.211A
BAP
BAP 1
1316
2134
51
yx
25
1025
51
26163964
51
So, the value of x = 5 and y = 2
Hal.: 83 Matriks Adaptif
DETERMINAN DAN INVERS
Penyelesaian sistem persamaan linear dua variabel dengan menggunakan determinan atau aturan Cramer.
cbyax Misal SPL
rqypx
Maka dengan aturan Cramer, diperoleh
,
qpbaqrbc
x
qpbarpca
y dan
Hal.: 84 Matriks Adaptif
DETERMINANT AND INVERSE
The solution of linear equation system with two variables using determinant or Cramer rule
cbyax For example
SPLrqypx
Then, with Cramer rule, we get
,
qpbaqrbc
x
qpbarpca
y dan
Hal.: 85 Matriks Adaptif
DETERMINAN DAN INVERS
Contoh :Gunakan aturan Cramer untuk menentukan himpunan penyelesaian sistem persamaan linear
543 yx
42 yx
11111
)4.(21.3)4.(41).5(
12431445
x
Jawab :
Dengan aturan Cramer diperoleh
21122
)4.(21.3)5.(24.3
1243
4253
y
Jadi, himpunan penyelesaiannya adalah {(1,2)}.
Hal.: 86 Matriks Adaptif
DETERMINANT AND INVERSE
Example :Use the Cramer rule to define the solution set of linear equation system
543 yx
42 yx
11111
)4.(21.3)4.(41).5(
12431445
x
answer :
With cramer Rule, we get
21122
)4.(21.3)5.(24.3
1243
4253
y
So, the solution set is {(1,2)}.
Hal.: 87 Matriks Adaptif
DETERMINAN DAN INVERS
Menyelesaikan Sistem Persamaan Linear Tiga Variabel dengan menggunakan Matriks
SPL dalam bentuk:
Dapat disajikan dalam bentuk persamaan matriks:
a11x1 + a12x2 + a13x3 +….. ..a1nxn = b1
a21x1 + a22x2 + a23x3 +…….a2nxn = b2
am1x1 + am2x2 + am3x3 + ……amnxn = bm
a11 a12……...a1n
a21 a22 ……..a2n : : : am1 am2…… amn
x1
x2
:xn
= b1
b2
:bn
A: matriks koefisien
Ax = bx b
Hal.: 88 Matriks Adaptif
DETERMINANT AND INVERSE
Finishing the equation of linear system with three variables using matrix
SPL in the form of:
It can be written in the form of matrix equation:
a11x1 + a12x2 + a13x3 +….. ..a1nxn = b1
a21x1 + a22x2 + a23x3 +…….a2nxn = b2
am1x1 + am2x2 + am3x3 + ……amnxn = bm
a11 a12……...a1n
a21 a22 ……..a2n : : : am1 am2…… amn
x1
x2
:xn
= b1
b2
:bn
A: matrix coefficient Ax = b
x b
Hal.: 89 Matriks Adaptif
DETERMINAN DAN INVERS
x1 + 2x2 + x3 = 6
-x2 + x3 = 1
4x1 + 2x2 + x3 = 4
SPL
1 2 1
0 -1 1
4 2 1
x1
x2
x3
=6
1
4
1.x1 +2.x2 + 1.x3
0.x1 + -1.x2 + 1.x3
4.x1 +2.x2 + 1.x3
=6
1
4
Dapat disajikan dalam bentuk matriks sebagai berikut
Contoh :
Hal.: 90 Matriks Adaptif
DETERMINANT AND INVERSE
x1 + 2x2 + x3 = 6
-x2 + x3 = 1
4x1 + 2x2 + x3 = 4
SPL
1 2 1
0 -1 1
4 2 1
x1
x2
x3
=6
1
4
1.x1 +2.x2 + 1.x3
0.x1 + -1.x2 + 1.x3
4.x1 +2.x2 + 1.x3
=6
1
4
It can be written in the form of the following matrix
Example :
Hal.: 91 Matriks Adaptif
Perkalian dengan matriks identitas
1 0 00 1 00 0 1
A= 1 2 37 5 6-9 3 -7
A.I = 1 2 37 5 6-9 3 -7
=
1 0 00 1 00 0 1
I.A = =1 2 37 5 6-9 3 -7
1 2 37 5 6-9 3 -7
1 2 37 5 6-9 3 -7
X
DETERMINAN DAN INVERS
X
Hal.: 92 Matriks Adaptif
The multiplication of identity matrix
1 0 00 1 00 0 1
A= 1 2 37 5 6-9 3 -7
A.I = 1 2 37 5 6-9 3 -7
=
1 0 00 1 00 0 1
I.A = =1 2 37 5 6-9 3 -7
1 2 37 5 6-9 3 -7
1 2 37 5 6-9 3 -7
X
DETERMINANT AND INVERSE
X
Hal.: 93 Matriks Adaptif
DETERMINAN DAN INVERS
AB = A dan BA = A, apa kesimpulanmu?
1 4 -9 3 7 0 5 9 -13
1 4 -9 3 7 0 5 9 -13
AB = A dan BA = A, maka B = I
(I matriks identitas)
1 0 00 1 0 0 0 1
1 0 00 1 0 0 0 1
=
=
1 4 -9 3 7 0 5 9 -13
1 4 -9 3 7 0 5 9 -13
A AII A= =
Hal.: 94 Matriks Adaptif
DETERMINANT AND INVERSE
AB = A and BA = A, what is your conclusion?
1 4 -9 3 7 0 5 9 -13
1 4 -9 3 7 0 5 9 -13
AB = A and BA = A, then B = I
(I identity matrix )
1 0 00 1 0 0 0 1
1 0 00 1 0 0 0 1
=
=
1 4 -9 3 7 0 5 9 -13
1 4 -9 3 7 0 5 9 -13
A AII A= =
Hal.: 95 Matriks Adaptif
d -bab-cd ab-cd
-c aab-cd ab-cd
DETERMINAN DAN INVERS
4 2
2 2
½ -½
-½ 1
1 0
0 1
d -b
-c a
1
ad - bc
Jika ad –bc = 0 maka A TIDAK mempunyai invers.
=
A IA-1
a b
c d A-1
1 0
0 1=
A-1 = =
Hal.: 96 Matriks Adaptif
d -bab-cd ab-cd
-c aab-cd ab-cd
DETERMINANT AND INVERSE
4 2
2 2
½ -½
-½ 1
1 0
0 1
d -b
-c a
1
ad - bc
If ad –bc = 0 then A doesn’t have inverse
=
A IA-1
a b
c d A-1
1 0
0 1=
A-1 = =
Hal.: 97 Matriks Adaptif
DETERMINAN DAN INVERS
1. Invers dari matriks jika ada adalah tunggal: Jika B = A-1 dan C = A-1, maka B = C
4 2
2 2A =
½ -½
-½ 1A-1
4 2
2 2
1 0
0 1
2. (A-1)-1 = A
?
(A-1)-1
= ½ -½
-½ 1A-1 =
A
Hal.: 98 Matriks Adaptif
DETERMINANT AND INVERSE
1. If there is inverse of matrix is only one: If B = A-1 and C = A-1, then B = C
4 2
2 2A =
½ -½
-½ 1A-1
4 2
2 2
1 0
0 1
2. (A-1)-1 = A
?
(A-1)-1
= ½ -½
-½ 1A-1 =
A
Hal.: 99 Matriks Adaptif
DETERMINAN DAN INVERS
3. Jika A mempunyai invers maka An mempunyai invers dan (An)-1 = (A-1)n, n = 0, 1, 2, 3,…
4 2
2 2A =
4 2
2 2A3 =
4 2
2 2
4 2
2 2
½ -½
-½ 1A-1 =
=104 64
64 40
(A3)-1 = 0.625 -1
-1 1.625
(A-1)3 = 0.625 -1
-1 1.625
½ -½
-½ 1
½ -½
-½ 1
½ -½
-½ 1=
sama
Hal.: 100 Matriks Adaptif
DETERMINANT AND INVERSE
3. If A have inverse then An have inverse and (An)-1 = (A-1)n, n = 0, 1, 2, 3,…
4 2
2 2A =
4 2
2 2A3 =
4 2
2 2
4 2
2 2
½ -½
-½ 1A-1 =
=104 64
64 40
(A3)-1 = 0.625 -1
-1 1.625
(A-1)3 = 0.625 -1
-1 1.625
½ -½
-½ 1
½ -½
-½ 1
½ -½
-½ 1=
The same
with
Hal.: 101 Matriks Adaptif
DETERMINAN DAN INVERS
4. (AB)-1 = B-1 A-1
4 2
2 2A =
3 5
2 2 B = B-1 =
½ 5/4
½ - ¾
(AB)-1 = 16 24
10 14
-1= -0.875 1.5
0.625 -1
A-1 B-1 = ½ 5/4
½ - ¾
½ -½
-½ 1= -0.5 1
0.75 -1.375
B-1 A-1 = ½ 5/4
½ - ¾
½ -½
-½ 1= -0.875 1.5
0.625 -1