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SMKN 2 PROBOLINGGO
Konsep Matriks
SMKN 2 PROBOLINGGO
Concept of Matrix
SMKN 2 PROBOLINGGO
Macam-macam Matriks
Kompetensi Dasar :
Mendeskripsikan macam-macam matriks
Indikator :
1. Matriks ditentukan unsur dan notasinya
2. Matriks dibedakan menurut jenis dan relasinya
Hal.: 3
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
Basic Competences :
Describing the kinds of matrix
Indicators :
1. Matrix is determined by its elements and
notations
2. Matriks matrix is distinguished by its kinds and
relations
Hal.: 4
Matriks
Adaptif
Macam – macam Matriks
SMKN 2 PROBOLINGGO
Pengertian Matriks
 Matriks adalah susunan bilangan-bilangan yang terdiri atas barisbaris dan kolom-kolom.
 Masing-masing bilangan dalam matriks disebut entri atau elemen.
Ordo (ukuran) matriks adalah jumlah baris kali jumlah kolom.
Notasi:
Matriks: A = [aij]
Elemen: (A)ij = aij
Ordo A: m x n
Hal.: 5
A=
a11
a21
:
ai1
:
am1
a12…….a1j ……a1n
a22 ……a2j…….a2n
:
:
:
ai2 ……aij…….. ain
:
:
:
am2……amj……. amn
baris
kolom
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
Definition of Matrix
 Matrix is the arrangement of numbers which
consists of rows and columns.
 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.
Notation:
Matrix: A = [aij]
Element: (A)ij = aij
Order A: m x n
Hal.: 6
A=
a11
a21
:
ai1
:
am1
a12…….a1j ……a1n
a22 ……a2j…….a2n
:
:
:
ai2 ……aij…….. ain
:
:
:
am2……amj……. amn
rows
column
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Macam-macam Matriks
1. Matriks Baris
Matriks baris adalah matriks yang hanya terdiri dari satu baris.
A12 
Hal.: 7
2
5
B1x 3 
1
-8
C1x 4 
-2
0
25
14
8
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
1. Row matrix
Row matrix is a matrix which consists of one row.
A12 
Hal.: 8
2
5
B1x 3 
1
-8
C1x 4 
-2
0
25
14
8
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Macam-macam Matriks
2. Matriks Kolom
Matriks Kolom adalah matriks yang hanya terdiri dari satu kolom
P21 
Q31 
2
-7
9
2
1
Hal.: 9
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
2. Column matrix
Column matrix is a matrix which consists of one column.
P21 
Q31 
2
-7
9
2
1
Hal.: 10
Matriks
Adaptif
Macam – macam Matriks
SMKN 2 PROBOLINGGO
3. Matriks Persegi
Matriks persegi (bujur sangkar) adalah matriks yang jumlah baris
dan jumlah kolom sama.
1
2
4
2
2
2
3
3
3
Trace(A) = 1 + 2 + 3
diagonal utama
Trace dari matriks adalah jumlahan elemen-elemen diagonal utama
Hal.: 11
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
3. Square matrix
Square matrix is a matrix which has the same numbers of rows and
columns.
1
2
4
2
2
2
3
3
3
Trace(A) = 1 + 2 + 3
Main diagonal
Trace from matrix is the total numbers from the main diagonal
elements.
Hal.: 12
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Macam- macam Matriks
4. Matriks Nol
Matriks nol adalah matriks yang semua elemennya nol
0
0 0
0
0
0
0
Matriks identitas adalah matriks persegi yang elemen diagonal
I3
I4
utamanya 1I2dan elemen lainnya
0
Hal.: 13
1
0
1
0
0
1
0
0
0
0
1
0
1
0
0
1
0
0
0
0
1
0
0
1
0
0
0
0
1
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
4. Zero matrix
zero matrix is a matrix which all of its elements are zero.
0
0 0
0
0
0
0
Matrix identity is a square matrix which its main diagonal
I4
3
element is I12 and the otherIelement
is 0.
Hal.: 14
1
0
1
0
0
1
0
0
0
0
1
0
1
0
0
1
0
0
0
0
1
0
0
1
0
0
0
0
1
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Macam-macam Matriks
5. Matriks ortogonal
Matriks A orthogonal jika dan hanya jika AT = A –1
0
A=
B=
1
0 1
-1
AT=
0
½√2 -½√2
BT=
½√2 ½√2
-1
0
½√2 ½√2
= A-1
= B-1
-½√2 ½√2
(A
A-1-1)T = (A
ATT)-1
Jika A adalah matriks orthogonal, maka (A-1)T = (AT)-1
Hal.: 15
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
5. Orthogonal Matrix
Matrix A is orthogonal if and only if AT = A –1
0
A=
B=
1
0 1
-1
AT=
0
½√2 -½√2
BT=
½√2 ½√2
-1
0
½√2 ½√2
= A-1
= B-1
-½√2 ½√2
(A
A-1-1)T = (A
ATT)-1
If A is orthogonal matrix, so (A-1)T = (AT)-1
Hal.: 16
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Macam – macam Matriks
A=
4
2
6
7
5
3
-9
7
AT = A’ =
4
5
2
3
6
-9
7
7
Definisi:
Transpose matriks A adalah matriks AT, kolom-kolomnya adalah
baris-baris dari A, baris-barisnya adalah kolom-kolom dari A.
[AT]ij = [A]ji
nxm
Jika A adalah matriks m x n, maka matriks transpose AT berukuran
………..
Hal.: 17
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
A=
4
2
6
7
5
3
-9
7
AT = A’ =
4
5
2
3
6
-9
7
7
Definisi:
Transpose matrix A is matrix AT, its columns are rows of A, its rows
is columns of A.
[AT]ij = [A]ji
nxm
if A is matrix m x n, so matrix transpose AT should be ………..
Hal.: 18
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Macam – macam Matriks
Kesamaan dua matriks
 Dua matriks sama jika ukuran sama dan setiap entri yang bersesuaian sama.
A=
C=
E=
G=
Hal.: 19
1
2
4
2
1
3
1
2 2
2
1
1
2
2
1
2
4
2
1
3
2
1
2
3
2
1
3
4
x
2
4
2
2
2
2
2
2
2
2
4
5
6
9
0
7
B=
D=
F=
H=
2
?
4
?
9
?
2
?
5
?
0
?
Matriks
A=B
C≠D
E = F jika x = 1
2
?
6
?
7
?
G=H
Adaptif
SMKN 2 PROBOLINGGO
Kind of Matrix
Similarity of two matrixes
 Two matrix are similar if its size is similar and each symmetrical entry is similar
A=
C=
E=
G=
Hal.: 20
1
2
4
2
1
3
1
2 2
2
1
1
2
2
1
2
4
2
1
3
2
1
2
3
2
1
3
4
x
2
4
2
2
2
2
2
2
2
2
4
5
6
9
0
7
B=
D=
F=
H=
2
?
4
?
9
?
2
?
5
?
0
?
Matriks
A=B
C≠D
E = F if x = 1
2
?
6
?
7
?
G=H
Adaptif
SMKN 2 PROBOLINGGO
Macam-macam Matriks
Matriks Simetri
Matriks A disebut simetris jika dan hanya jika A = AT
A=
A =
Hal.: 21
1
2
3
4
4
2
2
3
2
5
7
0
A’ =
3
7
8
2
4
0
2
9
4
2
2
3
A simetri
= AT
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
Symmetrical matrix
Matrix A is called symmetric if and only if A = AT
A=
A =
Hal.: 22
1
2
3
4
4
2
2
3
2
5
7
0
A’ =
3
7
8
2
4
0
2
9
4
2
2
3
A symmetric
= AT
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Macam-macam Matriks
Sifat-sifat transpose matriks
1. Transpose dari A transpose adalah A: (AT )T = A
A
(AT)T = A
AT
Contoh:
Hal.: 23
4
5
2
3
4
2
6
6
-9
5
3
-9
7
7
Matriks
4
5
7
2
3
7
6
-9
7
7
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
properties of transpose matrix
(AT )T = A
1. Transpose of A transpose is A:
A
(AT)T = A
AT
Example:
Hal.: 24
4
5
2
3
4
2
6
6
-9
5
3
-9
7
7
Matriks
4
5
7
2
3
7
6
-9
7
7
Adaptif
SMKN 2 PROBOLINGGO
Macam-macam Matriks
2. (A+B)T = AT + BT
T
T
A+B
T
(A+B)
Hal.: 25
=
A
T
A
=
Matriks
T
+
B
+
BT
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
2. (A+B)T = AT + BT
T
T
A+B
T
(A+B)
Hal.: 26
=
A
T
A
=
Matriks
T
+
B
+
BT
Adaptif
SMKN 2 PROBOLINGGO
Macam-macam Matriks
3. (kA)T = k(A) T untuk skalar k
T
T
kA
T
(kA)
Hal.: 27
k
=
A
T
k(A)
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
3. (kA)T = k(A) T for scalar k
T
T
kA
T
(kA)
Hal.: 28
k
=
A
T
k(A)
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Macam-macam Matriks
4. (AB)T = BT AT
Hal.: 29
B
AB
=
T
(AB)
= BTAT
AB
T
T
T
Matriks
A
Adaptif
SMKN 2 PROBOLINGGO
Kinds of Matrix
4. (AB)T = BT AT
Hal.: 30
B
AB
=
T
(AB)
= BTAT
AB
T
T
T
Matriks
A
Adaptif
SMKN 2 PROBOLINGGO
Macam-macam Matriks
Soal :
Isilah titik-titik di bawah ini
1. 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
Hal.: 31
Matriks
Adaptif
SMKN 2 PROBOLINGGO
Kind of Matrix
Quiz :
Fill in the blanks bellow
1. 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
Hal.: 32
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
Kompetesi Dasar
Menyelesaikan Operasi Matriks
Indikator
1.
2.
Dua matriks atau lebih ditentukan hasil
penjumlahan atau pengurangannya
Dua matriks atau lebih ditentukan hasil kalinya
Hal.: 33
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
Basic competence
Finishing operation matrix
Indicators
1.
2.
Two or more matrixes is defined by the result of
their addition or subtraction
Two or more matrixes is defined by the result of
their multiplication
Hal.: 34
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
Penjumlahan dan pengurangan dua matriks
Contoh :
A=
A+B=
A-B=
Hal.: 35
10
22
1
-1
B=
10+2
22+6
1+7
-1+5
10-2
22-6
1-7
-1-5
2
6
7
5
=
12
28
8
4
8
16
-6
-6
=
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
Addition and subtraction of two matrixes
Example:
A=
A+B=
A-B=
Hal.: 36
10
22
1
-1
B=
10+2
22+6
1+7
-1+5
10-2
22-6
1-7
-1-5
2
6
7
5
=
12
28
8
4
8
16
-6
-6
=
Matriks
Adaptif
SMKN 2 PROBOLINGGO
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.: 37
Matriks
Adaptif
SMKN 2 PROBOLINGGO
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.: 38
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
Jumlah dua matriks
K=
C=
1
3
5
5
7
C+D
K+L
=
=
4 -9
7 0
9 -13
6
2
7 3 1
-2 4 -5
9 -4 3
L =
1
25
30
5
35
10 15
D =
3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
D+C
=
L+K
=
Apa kesimpulanmu? Apakah jumlahan matriks bersifat komutatif?
Hal.: 39
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
The quantity of two matrixes
K=
C=
1
3
5
5
7
C+D
K+L
=
=
4 -9
7 0
9 -13
6
2
7 3 1
-2 4 -5
9 -4 3
L =
1
25
30
5
35
10 15
D =
3
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
D+C
=
L+K
=
What is your conclusion? Is the addition of matrixes commutative?
Hal.: 40
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
 Soal:
C=
A =
3
-8
0
4
7
2
-1
8
4
0
0
0
0
0
0
D =
3
7
2
5
2
6
-1
8
4
0
0
0
0
0
0
B =
E =
Feedback: C +D =



Hal.: 41
C + D =…
C+E=…
A+B=…
Matriks
2
7
2
5
2
6
6
-1
2
9
9
8
-2
16
8
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
 Exercise:
C=
A =
3
-8
0
4
7
2
-1
8
4
0
0
0
0
0
0
D =
3
7
2
5
2
6
-1
8
4
0
0
0
0
0
0
B =
E =
Feedback: C +D =



Hal.: 42
C + D =…
C+E=…
A+B=…
Matriks
2
7
2
5
2
6
6
-1
2
9
9
8
-2
16
8
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
Hasil kali skalar dengan matriks
A=
5
6
1
7
2
3
5x5
5A =
5x6
5x1
=
5x5
5x2
5x3
25
30
5
35
10
15
Apa hubungan H dengan A?
H=
250 300
50
350 100
150
H = 50A
Diberikan matriks A = [aij] dan skalar c, perkalian skalar cA
mempunyai entri-entri sebagai berikut:
(cA)ij = c.(A)ij = caij
Catatan: Pada himpunan Mmxn, perkalian matriks dengan skalar bersifat
tertutup (menghasilkan matriks dengan ordo yang sama)
Hal.: 43
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
The multiplication result of scalar matrix
A=
5
6
1
7
2
3
5x5
5A =
5x6
5x1
=
5x5
5x2
5x3
25
30
5
35
10
15
What is the relation between H and A?
H=
250 300
50
350 100
150
H = 50A
Given matrix A = [aij] and scalar c, the multiplication of scalar cA
have the following entries:
(cA)ij = c.(A)ij = caij
Note: In the set of Mmxn, the matrix multiplication with scalar have closed
properties (it will have matrix with the same ordo)
Hal.: 44
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
 K
3x3
K=
4K =
5K =
Hal.: 45
1
3
5
4 -9
7 0
9 -13
4 16 -36
12 28 0
20 36 -52
5 20 -45
15 35 0
25 45 -65
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
 K
3x3
K=
4K =
5K =
Hal.: 46
1
3
5
4 -9
7 0
9 -13
4 16 -36
12 28 0
20 36 -52
5 20 -45
15 35 0
25 45 -65
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
 Diketahui bahwa cA adalah matriks nol. Apa kesimpulan Anda
tentang A dan c?
Contoh:
AA ==
cc == 70
cA =
0*2 7*0
7*0
0*7 7*0
0*2
0*5 7*0
7*0
0*2 7*0
0*6
=
0
2
0
7
0
2
0
5
0
2
0
6
0
0
0
0
0
0
kesimpulan
Kasus 1: c = 0 dan A matriks sembarang.
Kasus 2: A matriks nol dan c bisa berapa saja.
Hal.: 47
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
 Known that cA is zero matrix. What is your conclusion about
A and c?
Example:
AA ==
cc == 70
cA =
0*2 7*0
7*0
0*7 7*0
0*2
=
0*5 7*0
7*0
0*2 7*0
0*6
0
2
0
7
0
2
0
5
0
2
0
6
0
0
0
0
0
0
Conclusion
Case 1: c = 0 and A is any matrix
Case 2: A is zero matrix and c can be any
number
Hal.: 48
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
Perkalian matriks dengan matriks
 Definisi:
 Jika A = [aij] berukuran m x r , dan B = [bij] berukuran r x n,
maka matriks hasil kalir A dan B, yaitu C = AB mempunyai elemenelemen yang didefinisikan sebagai berikut:
(C)ij = (AB)ij =
∑ aikbkj = ai1b1j +ai2b2j+………airbrj
k=1
• Syarat:
A=
Hal.: 49
A
mxr
B
rxn
2
3
4
5
8
-7
9
-4
1
-5
7
-8
AB
mxn
B=
Matriks
1
2
7
-6
4
-9
Tentukan AB dan BA
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
Multiplication between 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,
r
yaitu is C = AB has elements that defined as follows:
(C)ij = (AB)ij =
∑ aikbkj = ai1b1j +ai2b2j+………airbrj
k=1
• Condition:
A=
Hal.: 50
A
mxr
B
rxn
2
3
4
5
8
-7
9
-4
1
-5
7
-8
AB
mxn
B=
Matriks
1
2
7
-6
4
-9
Define AB and BA
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
Perkalian matriks dengan matriks
Contoh :
A=
2
3
4
5
8
-7
9
-4
1
-5
7
-8
B=
1
2
7
-6
4
-9
11
3
=
2.1 +3.7+4.4+5.11
AB =
-35
-49
-35
-94
-55
94 -35
=
-49 -35
-94 -55
BA tidak didefinisikan
Hal.: 51
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
The multiplication between matrixes
Example:
A=
2
3
4
5
8
-7
9
-4
1
-5
7
-8
B=
1
2
7
-6
4
-9
11
3
=
2.1 +3.7+4.4+5.11
AB =
-35
-49
-35
-94
-55
94 -35
=
-49 -35
-94 -55
BA is not define
Hal.: 52
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
1. Diberikan A dan B, AB dan BA terdefinisi. Apa kesimpulanmu?
A
B
B
A
mxn
nxk
nxk
mxn
m=k
ABmxm
AB dan BA
matriks persegi
ABnxn
2. AB = O matriks nol, apakah salah satu dari A atau B pasti matriks nol?
A=
2
2
3
3
B=
3 -3
-2 2
AB =
0
0
0
0
AB matriks nol, belum tentu A atau B matriks nol
Hal.: 53
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
1. Given A and B, AB and BA is defined. What is your conclusion?
A
B
B
A
mxn
nxk
nxk
mxn
m=k
ABmxm
AB and BA
square matrix
ABnxn
2. AB = O is zero matrix, is one of (A or B) is zero matrix?
A=
2
2
3
3
B=
3 -3
-2 2
AB =
0
0
0
0
AB is zero matrix. Matrix A and B is not certain
zero matrix
Hal.: 54
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
Contoh 1:
Tentukan hasil kalinya jika terdefinisi.
A=
C=
•
•
•
•
•
Hal.: 55
A B = ??
AC = ??
BD = ??
CD = ??
DB = ??
2
4
2
3
7
3
4
9
5
7 -11 4
3 5 -6
5
0
6
B=
D=
Matriks
1
-9
8
5
1
2
0
2
0
0
6
8 9 5 6
5 6 -9 0
-4 7 8 9
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
Example 1:
Define the multiplication result if it defined:
A=
C=
•
•
•
•
•
Hal.: 56
A B = ??
AC = ??
BD = ??
CD = ??
DB = ??
2
4
2
3
7
3
4
9
5
7 -11 4
3 5 -6
5
0
6
B=
D=
Matriks
1
-9
8
5
1
2
0
2
0
0
6
8 9 5 6
5 6 -9 0
-4 7 8 9
Adaptif
SMKN 2 PROBOLINGGO
OPERASI MATRIKS
Contoh 2:
A=
2
1
3
2
A2 =
2
1
3
2
2
1
3
2
2
1
3
2
2
1
3
2
A3 = A x A2 =
2
1
3
2
A0 = I
An = A A A …A
n faktor
An+m = An Am
Hal.: 57
Matriks
Adaptif
SMKN 2 PROBOLINGGO
OPERATION OF MATRIX
Example 2:
A=
2
1
3
2
A2 =
2
1
3
2
2
1
3
2
2
1
3
2
2
1
3
2
A3 = A x A2 =
2
1
3
2
A0 = I
An = A A A …A
n factor
An+m = An Am
Hal.: 58
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Kompetensi Dasar:
Menentukan determinan dan invers
Indikator :
1. Matriks ditentukan determinannya
2. Matriks ditentukan inversnya
Hal.: 59
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Basic Competence:
Define the determinant and inverse
Indicator :
1. Matrix is defined by its determinant
2. Matrix is defined by its inverse
Hal.: 60
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Determinan Matriks ordo 2 x 2
Nilai determinan suatu matriks ordo 2 x 2 adalah hasil kali elemenelemen diagonal utama dikurangi hasil kali elemen pada diagonal kedua.
Misalkan diketahui matriks A berordo 2 x 2, A =
a b
c d
Determinan A adalah
a b
det A =
c d
Hal.: 61
= ad - bc
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Determinant Matrix ordo 2 x 2
Determinant 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 =
a b
c d
Determinant A is
a b
det A =
c d
Hal.: 62
= ad - bc
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Contoh: Invers matriks 2x2
A=
A-1
Hal.: 63
3
2
4
1
1
 3.1-4.2
=
-4
 3.1-4.2
   15
= I 4
3
3.1-4.2 
 5
-2
3.1-4.2
Matriks

 53 
2
5
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Example: Matrix inverse 2x2
A=
A-1
Hal.: 64
3
2
4
1
1
 3.1-4.2
=
-4
 3.1-4.2
   15
= I 4
3
3.1-4.2 
 5
-2
3.1-4.2
Matriks

 53 
2
5
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT DAN INVERSE
Contoh :
a b 
1. Kapan matriks 
 TIDAK mempunyai invers?
c
d


2. Tentukan invers matriks berikut ini
a.
b.
c.
d.
Hal.: 65
5
1
1
2
0
1
0
2
0
0
4
1
1
0
0
1
a.
2/3
-1/5
-1/5
5/3
ad-bc = 0
b. tidak mempunyai invers
c. tidak mempunyai invers
d.
Matriks
1
0
0
1
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Example :
a b 
1. When matrix 
 Doesn’t have inverse?
c
d


2. Define the following matrix inverse
a.
b.
c.
d.
Hal.: 66
5
1
1
2
0
1
0
2
0
0
4
1
1
0
0
1
a.
ad-bc = 0
2/3
-1/5
-1/5
5/3
b. Doesn’t have inverse
c. Doesn’t have inverse
d.
Matriks
1
0
0
1
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
B adalah invers dari matriks A, jika AB = BA = I matriks
identitas, ditulis B = A-1
A
a
Jika A =
c
A-1
b
, maka
d
A-1
=
A
=
I
1  d  b


A 
ad  bc   c a 
1
dengan A  ad  bc  0
Hal.: 67
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
B is inverse of matrix A, if AB = BA = I matrix identities,
it is written B = A-1
A
If A =
a
c
A-1
b
, then
d
A-1
=
A
=
I
1  d  b


A 
ad  bc   c a 
1
with A  ad  bc  0
Hal.: 68
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Contoh 1 :
Tentukan invers dari matriks
2

A
 7
danB    5
17 
2
5
10 

 4
Jawab :
A
17  5 
1  d  b
1

 


A   c a   2.17   7 .5  7  2 
17  5 

 
 7  2
det B = (-5) . (-4) – (-2) . (-10) = 20 – 20 = 0 , sehingga matriks B
tidak memiliki invers
Hal.: 69
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Example 1 :
Defined the inverse of matrix
2

A
 7
andB    5
17 
2
5
10 

 4
Answer :
A
17  5 
1  d  b
1

 


A   c a   2.17   7 .5  7  2 
17  5 

 
 7  2
det B = (-5) . (-4) – (-2) . (-10) = 20 – 20 = 0 , So, matrix B doesn’t
have inverse
Hal.: 70
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Contoh 2 :
 4 2 1


 4 2
danB   2 2 1
A  
 2 2
 3 3 1


Diketahui matriks
Tunjukkan bahwa A.A-1 = A-1.A = I dan B.B-1 = B-1. B = I
4
2
2
½ -½
2
-½
=
1
2
-½
2
2
1
A-1
4
2
1
½
-½
1
2
2
1
-½
-½
1
3
3
1
0
3
-2
Hal.: 71
4
A-1
A
B
½ -½
B-1
=
=
A
2
1
½
-½
1
2
2
1
-½
-½
1
3
3
1
0
3
-2
Matriks
0
0
1
I
4
B-1
1
B
=
1
0
0
0
1
0
0
0
1
I
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Example 2 :
 4 2 1


 4 2
andB   2 2 1
A  
 2 2
 3 3 1


Known matrix
Show that A.A-1 = A-1.A = I and B.B-1 = B-1. B = I
4
2
2
½ -½
2
-½
=
1
2
-½
2
2
1
A-1
4
2
1
½
-½
1
2
2
1
-½
-½
1
3
3
1
0
3
-2
Hal.: 72
4
A-1
A
B
½ -½
B-1
=
=
A
2
1
½
-½
1
2
2
1
-½
-½
1
3
3
1
0
3
-2
Matriks
0
0
1
I
4
B-1
1
B
=
1
0
0
0
1
0
0
0
1
I
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Matriks ordo 3 x 3
Determinan Matriks Ordo 3 x 3
a b

MisalkanA   d e
g h

c

f .
i 
Dengan aturan Sarrus, determinan A adalah sebagai berikut.
a b c a
Ad e f d
_g _ h _ i+ g+
b
e
h
+
 aei  bfg  cdh  ceg  afh  bdi
 (aei  bfg  cdh)  (ceg  afh  bdi)
Hal.: 73
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Matrix ordo 3 x 3
Matrix Determinant Ordo 3 x 3
a b

exampleA   d e
g h

c

f .
i 
With Sarrus rule, determinant A is as follows
a b c a
Ad e f d
_g _ h _ i+ g+
b
e
h
+
 aei  bfg  cdh  ceg  afh  bdi
 (aei  bfg  cdh)  (ceg  afh  bdi)
Hal.: 74
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Sistem Persamaan Linear Dua Variabel dengan Menggunakan Matriks
a1x  b1 y  c1
Misal SPL
a2 x  b2 y  c2
Persamaan tersebut dapat di ubah menjadi bentuk matriks
berikut
 a1 b1  x   c1 

    
 a2 b2  y   c2 
Hal.: 75
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
The equation of linear with two variable using matrix
a1x  b1 y  c1
For example SPL
a2 x  b2 y  c2
The equation can be changed into the following matrix
 a1 b1  x   c1 

    
 a2 b2  y   c2 
Hal.: 76
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Misalkan
 a1 b1 
 C1 
 x
, P   danB   , maka dapat ditulis
A  
 y
 a2 b2 
 C2 
 a1 b1  x   c1 

    
 a2 b2  y   c2 
Hal.: 77

AP  B

1
PA B
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
a b
C
x
Example A   1 1 , P   andB   1 , Then can be written
 y
a b 
C 
 
 2 2
 2
as
 a1 b1  x   c1 

    
 a2 b2  y   c2 
Hal.: 78

AP  B

1
PA B
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Contoh :
Tentukan nilai x dan y yang memenuhi sistem persamaan linear
 2 x  3 y  16
x  4 y  13
Jawab :
Sistem persamaan
 2 x  3 y  16
x  4 y  13
Jika dibuat dalam bentuk matriks menjadi
  2 3  x    16 

   

 1  4  y   13 
Hal.: 79
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Example:
Define the value of x and y that fulfill the equation of linear system
 2 x  3 y  16
x  4 y  13
answer :
Equation system
 2 x  3 y  16
x  4 y  13
If in matrix
  2 3  x    16 

   

 1  4  y   13 
Hal.: 80
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Perkalian matriks berbentuk AP = B dengan
 2 3 
 x
  16 
, P   danB  

A  
 1  4
 y
 13 
  4  3 1   4  3
1

  

A 
 2.  4  1.3   1  2  5   1  2 
1
AP  B
 P  A1B
 x  1   4  3   16 
   


 y  5   1  2  13 
1  64  39  1  25   5 
  
   
 
5  16  26  5   10   2 
Jadi nilai x = 5 dan y = 2
Hal.: 81
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
The matrix multiplication in the form of AP =
B with
 2 3 
 x
  16 
, P   danB  

A  
 1  4
 y
 13 
  4  3 1   4  3
1

  

A 
 2.  4  1.3   1  2  5   1  2 
1
AP  B
 P  A1B
 x  1   4  3   16 
   


 y  5   1  2  13 
1  64  39  1  25   5 
  
   
 
5  16  26  5   10   2 
So, the value of x = 5 and y = 2
Hal.: 82
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Penyelesaian sistem persamaan linear dua variabel dengan
menggunakan determinan atau aturan Cramer.
Misal SPL
ax  by  c
px  qy  r
Maka dengan aturan Cramer, diperoleh
c
r
x
a
p
Hal.: 83
b
q
, dan
b
q
a
p
y
a
p
c
r
b
q
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
The solution of linear equation system with two variables
using determinant or Cramer rule
For example
SPL
ax  by  c
px  qy  r
Then, with Cramer rule, we get
c
r
x
a
p
Hal.: 84
b
q
, and
b
q
a
p
y
a
p
c
r
b
q
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Contoh :
Gunakan aturan Cramer untuk menentukan himpunan penyelesaian
sistem persamaan linear
3 x  4 y  5
2x  y  4
Jawab :
Dengan aturan Cramer diperoleh
5 4
4
1
(5).1  4.(4) 11
x

 1
3 4
3.1  2.(4)
11
2 1
3 5
2 4
3.4  2.(5) 22
y


2
3  4 3.1  2.(4) 11
2 1
Jadi, himpunan penyelesaiannya adalah {(1,2)}.
Hal.: 85
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Example :
Use the Cramer rule to define the solution set of linear equation
system
3 x  4 y  5
2x  y  4
answer :
With cramer Rule, we get
5 4
4
1
(5).1  4.(4) 11
x

 1
3 4
3.1  2.(4)
11
2 1
3 5
2 4
3.4  2.(5) 22
y


2
3  4 3.1  2.(4) 11
2 1
So, the solution set is {(1,2)}.
Hal.: 86
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
 Menyelesaikan Sistem Persamaan Linear Tiga Variabel dengan
menggunakan Matriks
 SPL dalam bentuk:
a11x1 + a12x2 + a13x3 +….. ..a1nxn
= b1
a21x1 + a22x2 + a23x3 +…….a2nxn
= b2
am1x1 + am2x2 + am3x3 + ……amnxn = bm
 Dapat disajikan dalam bentuk persamaan matriks:
=
a11 a12……...a1n
x1
b1
a21 a22 ……..a2n
x2
b2
:
:
:
:
:
am1 am2…… amn
xn
bn
x
A: matriks koefisien
b
Ax = b
Hal.: 87
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
 Finishing the equation of linear system with three variables
using matrix
 SPL in the form of:
a11x1 + a12x2 + a13x3 +….. ..a1nxn
= b1
a21x1 + a22x2 + a23x3 +…….a2nxn
= b2
am1x1 + am2x2 + am3x3 + ……amnxn = bm
 It can be written in the form of matrix equation:
=
a11 a12……...a1n
x1
b1
a21 a22 ……..a2n
x2
b2
:
:
:
:
:
am1 am2…… amn
xn
bn
x
A: matrix coefficient
b
Ax = b
Hal.: 88
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Contoh :
SPL
x1 + 2x2 + x3
= 6
-x2 + x3
= 1
4x1 + 2x2 + x3 = 4
Dapat disajikan dalam bentuk matriks sebagai berikut
1.x1 +2.x2 + 1.x3
0.x1 + -1.x2 + 1.x3
4.x1 +2.x2 + 1.x3
Hal.: 89
=
6
1
2
1
x1
1
0
-1
1
x2
4
4
2
1
x3
Matriks
6
=
1
4
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
Example :
SPL
x1 + 2x2 + x3
= 6
-x2 + x3
= 1
4x1 + 2x2 + x3 = 4
It can be written in the form of the following matrix
1.x1 +2.x2 + 1.x3
0.x1 + -1.x2 + 1.x3
4.x1 +2.x2 + 1.x3
Hal.: 90
=
6
1
2
1
x1
1
0
-1
1
x2
4
4
2
1
x3
Matriks
6
=
1
4
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
Perkalian dengan matriks identitas
A=
A.I =
I.A =
Hal.: 91
1
2
3
7
5
6
-9
3
-7
1
2
3
7
5
6
-9
3
1
1
0
0
0
1
0
-7
0
0
0
0
1
0
1
0
0
0
1
X
X
1
2
3
7
5
6
1
-9
3
-7
2
3
1
2
3
7
5
6
7
5
6
-9
3
-7
-9
3
-7
Matriks
=
=
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
The multiplication of identity matrix
A=
A.I =
I.A =
Hal.: 92
1
2
3
7
5
6
-9
3
-7
1
2
3
7
5
6
-9
3
1
1
0
0
0
1
0
-7
0
0
0
0
1
0
1
0
0
0
1
X
X
1
2
3
7
5
6
1
-9
3
-7
2
3
1
2
3
7
5
6
7
5
6
-9
3
-7
-9
3
-7
Matriks
=
=
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
AB = A dan BA = A, apa kesimpulanmu?
1
3
5
4 -9
7 0
9 -13
1
0
0
0
1
0
1
0
0
0
1
0
1
3
5
4 -9
7 0
9 -13
A
0
0
1
I
=
0
0
1
I
=
=
A
1
3
5
4 -9
7 0
9 -13
1
3
5
4 -9
7 0
9 -13
=
A
AB = A dan BA = A, maka B = I
(I matriks identitas)
Hal.: 93
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
AB = A and BA = A, what is your conclusion?
1
3
5
4 -9
7 0
9 -13
1
0
0
0
1
0
1
0
0
0
1
0
1
3
5
4 -9
7 0
9 -13
A
0
0
1
I
=
0
0
1
I
=
=
A
1
3
5
4 -9
7 0
9 -13
1
3
5
4 -9
7 0
9 -13
=
A
AB = A and BA = A, then B = I
(I identity matrix )
Hal.: 94
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
4
2
½ -½
2
2
-½
=
1
1
0
0
1
I
A-1
A
a
b
c
d
1
0
0
1
A-1
=
A-1
d
 ab-cd
=
 -c
 ab-cd

=
a 
ab-cd 
-b
ab-cd
1
d
-b
ad - bc
-c
a
Jika ad –bc = 0 maka A TIDAK mempunyai invers.
Hal.: 95
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
4
2
½ -½
2
2
-½
=
1
1
0
0
1
I
A-1
A
a
b
c
d
1
0
0
1
A-1
=
A-1
d
 ab-cd
=
 -c
 ab-cd

=
a 
ab-cd 
-b
ab-cd
1
d
-b
ad - bc
-c
a
If ad –bc = 0 then A doesn’t have inverse
Hal.: 96
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
1. Invers dari matriks jika ada adalah tunggal:
Jika B = A-1 dan C = A-1, maka B = C
(A-1)-1
2. (A-1)-1 = A
A=
A-1
=
4
2
2
2
½ -½
-½
1
½ -½
=
?
-½ 1
A-1
4
2
2
2
1
0
0
1
A
Hal.: 97
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
1. If there is inverse of matrix is only one:
If B = A-1 and C = A-1, then B = C
(A-1)-1
2. (A-1)-1 = A
A=
A-1
=
4
2
2
2
½ -½
-½
1
½ -½
=
?
-½ 1
A-1
4
2
2
2
1
0
0
1
A
Hal.: 98
Matriks
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
3. Jika A mempunyai invers maka An mempunyai invers dan
(An)-1 = (A-1)n, n = 0, 1, 2, 3,…
A=
A3
=
(A3)-1 =
½ -½
4
2
2
2
4
2
4
2
4
2
2
2
2
2
2
2
A-1 =
-½
0.625
-1
1
=
104
64
64
40
-1
1.625
sama
(A-1)3 =
Hal.: 99
½ -½
½ -½
½ -½
-½
-½
-½
1
1
Matriks
1
=
0.625
-1
-1
1.625
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
3. If A have inverse then An have inverse and
(An)-1 = (A-1)n, n = 0, 1, 2, 3,…
A=
A3
=
(A3)-1 =
½ -½
4
2
2
2
4
2
4
2
4
2
2
2
2
2
2
2
A-1 =
-½
0.625
-1
1
=
104
64
64
40
-1
1.625
The same
with
Hal.: 100
(A-1)3 =
½ -½
½ -½
½ -½
-½
-½
-½
1
1
Matriks
1
=
0.625
-1
-1
1.625
Adaptif
SMKN 2 PROBOLINGGO
DETERMINAN DAN INVERS
4. (AB)-1 = B-1 A-1
A=
4
2
2
2
(AB)-1 =
B-1 A-1 =
B=
16
24
10
14
½
2
2
-1
½ -½
1
B-1 =
½
5/4
½
-¾
-0.875 1.5
0.625 -1
=
-¾
-½
Hal.: 101
5
5/4
½
A-1 B-1 =
3
½ -½
-½
1
½
5/4
½
-¾
Matriks
=
=
-0.875 1.5
0.625 -1
-0.5
1
0.75 -1.375
Adaptif
SMKN 2 PROBOLINGGO
DETERMINANT AND INVERSE
4. (AB)-1 = B-1 A-1
A=
4
2
2
2
(AB)-1 =
B-1 A-1 =
B=
16
24
10
14
½
2
2
-1
½ -½
1
B-1 =
½
5/4
½
-¾
-0.875 1.5
0.625 -1
=
-¾
-½
Hal.: 102
5
5/4
½
A-1 B-1 =
3
½ -½
-½
1
½
5/4
½
-¾
Matriks
=
=
-0.875 1.5
0.625 -1
-0.5
1
0.75 -1.375
Adaptif