CHAPTER 1
1
Matrix Algebra
INTRODUCTION
Matrix
• A rectangular array of elements or entries aij
involving m rows and n columns
• Matrix can be written as A = [aij ]
 a11
a
 21
 a31

A .
 ai1

 .
a
 m1
a12
a13
. a1 j
a22
a23
. a2 j
a32
a33
. a3 j
.
.
.
.
ai 2
ai 3
.
aij
.
.
.
.
am 2
am 3 . amj
n  coloumns
. a1n  

. a2 n  
. a3n  

. .   m  rows  [aij ]mn
. ain  

. . 
. amn  
2

Elements in matrix may be real or complex numbers,
or even functions.
1 3 2 
 2 9 1  real number


 2  3i 4i 
 complex number
 1

2i 

cos 
 sin 

sin  
 function

 cos  
3
If i = j, then the elements is called the leading diagonal
of matrix A.
 Example a11 , a22 , a33

4
EXAMPLE
5
SOLUTION
i.
ii.
iii.
Order of matrix : 3x3
Leading diagonal : 1,0,3
a12 = 2
a31= 5
a23= 2
6
TYPE OF MATRICES
7
Not
diagonal
Not
diagonal
Diagonal
8
Scalar
Scalar
Not scalar
9
Not
identity
matrix
Not
identity
matrix
10
 2 4 
A  


2

7


 3 0 
 B   9 3
 0 1
11
12
13
 1 3 7
AT   2 3 6 
 4 0 8 
9  1 
B 

3

2
4


T
14
5 1 3 
AT  1 2 2  since A T =A, so it is a symmetric matrix
3 2 7 
15
16
 0 1 5
 0 1 5
AT   1 0 3 , -A   1 0 3
 5 3 0 
 5 3 0 
since A T =-A, so it is a skew symmetric matrix
17
Row
echelon
form
Not Row
echelon
form
Row
echelon
form
Not Row
echelon
form
18
Reduced
Row
echelon
form
Not Reduced
Row echelon
form
Reduced
Row
echelon
form
Not Reduced
Row echelon
form
19
MATRIX OPERATION
i) Not possible
1 2
ii) A  B  
4 5
 8 10
A B  
14 16
3  7 8 9 


6  10 11 12 
12 
18 
20
i) Not possible
 1 5 0 
ii ) B  A  

10

1
2


21
MATRIX OPERATION
22
23
 2(3) 2(1) 2( 1) 
2A  

2(0)
2(

2)
2(4)


 6 2 2 
2A  

0

4
8


24
3( 1)  1(0)  4(4) 
 3(1)  1(2)  4( 6)
i ) AB  

5(1)

2(2)

(

2)(

6)
5(

1)

2(0)

(

2)(4)


 19 13 
AB  

21

13


25
26
Answer
3 2 3
AB  

3 0 1 
27
DETERMINANT
28
i) det  A  A
 3(1)  6(1)
9
ii ) det  AT   AT
 3 1
A 

6
1


T
AT  3(1)  (1)(6)
=9
29
-
-
- +
+
+
30
A  3(5)(7)  (1)(2)(0)  6(9)(4)   6(5)(0)  (2)(4)(3)  ( 1)(9)(7)
A  111  39
A  150
31
32
i ) M 11 
2
2
3
0
 2(0)  ( 2)(3)
 6
C11  (1) 2 (6)
C11  6
33
34
A  a11c11  a12 c12  a13c13

2

2

5 
2 5
3 9
4 9
A  (3) (1)
  (1) (1)
  (6) (1)

4
7
0
7
0
4






A  129  63  216
A  150
35
36
i) Find all the value of Cij
ii) Arrange all the value in matrix
form [Cij]
iii) Then transpose [Cij]
37
c11  2, c12  -6, c13  1
c21  3, c22  0, c23  3
c31  1, c32  3, c33  5
 2 6 1 
 2 3 1
T




Cij   3 0 3  , then adj Cij    6 0 3 
 1 3 5
 1 3 5
38
39
i ) A  3(6)  4(2)  10
A
1
1 6


10  2
4 
3 
40
41
c11  0, c12  9, c13  18
c21  2, c22  6, c23  11
c31  1, c32  3, c33  1
2 1
 0 9 18 
0
T


Cij   2 6 11 , then adj Cij    9 6 3 
 1 3
18 11 1 
1 
2 1
0
1
1
A   9 6 3 
9
18 11 1 
42
43
44
EXERCISE:
Find Inverse of matrix A using elementary row operation
1 0 1 
A  1 2 3
3 1 5 
Answer
 7
 2

A1   2
 5

 2
1
2
1
1

2

1

1

1

45
Find Inverse of matrix B using elementary
row operation
4 4 2


B  0 5 1 
 4 3 2 
Answer
1
7
 4 2

B 1   1
0
 5 1


3
 
2

1 
5 


46
SOLVING THE SYSTEMS OF
LINEAR EQUATION
47
1.6.1
Inversion Method
1.6.2
Gaussian Elimination and Gauss-Jordan
Elimination
1.6.3
Cramer’s Rule
1.6.1 INVERSION METHOD
48
AX = B
The left-hand side
can be simplified by
noting that
multiplying a matrix
by its inverse gives
the identity matrix
A1 AX  ( A1 A) X
 X
1
1
A AX  A B
X  A1B
Multiplying a matrix by
the identity matrix has no
effect and so
49
AX  B
 7 2   x  12 
3 1  y    5 

   
 1 2 
1
A 
7(1)  2(3)  3 7 
 1 2 


 3 7 
1
Then,
X  A1 B
2  12 
 x  1
 y    3 7   5 
  
 
1(12)  ( 2)(5) 


(

3)(12)

7(5)


2
 
 1
hence, x  2, y  1
50
x1  3x2  x3  6
8 x1  9 x2  4 x3  21
2 x1  x2  2 x3  3
51
52
1.6.2
GAUSSIAN ELIMINATION AND
GAUSS-JORDAN ELIMINATION
Gaussian elimination process.
• The procedure to reduce matrix A to
row echelon form (REF) by using
elementary row operation (ERO).
Gauss-Jordan elimination
• The procedure to reduce matrix A to
reduced row echelon form (RREF) by
using elementary row operation (ERO).
53
EXAMPLE 1.6.2
i.
Solve the system of linear equations by using
Gaussian elimination and Gauss-Jordan elimination.
2 x1  x2  2 x3  8
x1  3x2  3x3  4
4 x1  2 x2  x3  1
54
GAUSSIAN ELIMINATION
2 1 2 8 
1 3 3 4 
1 3 3 4  R  R  2 1 2 8  
R2  2 R1
2
1



 R3  4 R1 
 4 2 1 1 
 4 2 1 1 
1 3 3 4 
1 3 3 4  1


R2
4
16
R3 14 R2
0 7 4 16  
7
0 1 
 





7 7

0 14 13 17 
0 14 13 17 


4 
1 3 3
1 3 3 4 

 1 R3 

4
16
4
16
5
0 1 
 

 0 1 
7
7 
7 7


0 0 5 15
0 0
1
3 



55
From the Gaussian elimintion we have
x1  3 x2  3 x3  4
4
16
x2  x3 
7
7
x3  3
4
16
x2  (3) 
7
7
x2  4
x1  3(4)  3(3)  4
x1  1
56
From Gauss-Jordan Elimination,
2 1 2 8 
1 3 3 4 
1 3 3 4  R  R  2 1 2 8  
R2  2 R1
1

 2
 R3  4 R1 
 4 2 1 1 
 4 2 1 1 
1 3 3 4 
1 3 3 4  1


R2
4
16
R3 14 R2
0 7 4 16  
7
0 1 
 





7 7

0 14 13 17 
0 14 13 17 


4 
1 3 3
1 3 3 4 

 1 R3 

4
16
4
16
R1  3 R2
5
0 1 
 
 
 0 1 

7
7 
7 7


0 0 5 15
0 0
1
3 



57
9

1
0

7

0 1  4

7

0 0 1

20 
7
1 0 0 1

9
R

R
16 
1
3
0 1 0 4 
7

4

7  R2  7 R3 
0 0 1 3 

3

x1  1, x2  4, x3  3
58
59
1.6.3 CRAMER’S RULE

method of obtaining the solution of equations like
these as the ratio of two determinants.
a11 x  a12 y  b1
b1
a12
b2 a22
x
a11 a12
a21
a22
a21 x  a22 y  b2
 Ai
a11
b1
 A
a12
y
a11
b2
a12
a21
a22
 Ai
 A
60
61
62
1 3 1   x  8 
 2 1 1   y   7 

   
1 1 1  z   2 
then,
8 3
1
7 1
1
2 1 1 16
x

2
1 3 1
8
2 1 1
1 1 1
1 8
1
2 7
1
1 2 1 12 3
y


1 3 1
8 2
2 1 1
1 1 1
1 3 8
2 1 7
1 1 2 12 3
z
 
1 3 1
8 2
2 1 1
1 1 1
63
1.7 MATRIX APPLICATION
64
1.7.1
Application of Inversion Method
1.7.2
Application of Gaussian Elimination and
Gauss Jordan Elimination Method
1.7.3
Application of Cramer’s Rule
EXAMPLE 1.7.1.2 - APPLICATION OF INVERSION METHOD

Grand Canyon Tours offers air and ground scenic tours of the
Grand Canyon. Tickets for the 7.5 hours tour cost RM169 for an
adult and RM 129 for a child and each tour group is limited to 19
people. On three recent fully booked tours, total receipts were
RM2931 for the first tour, RM3011 for the second tour and RM
2771 for the third tour. Determine how many adults and how
many children were in each tour.
65
Let A= Adult
C= Children
From information above, the equation are
Tour 1
A  C  19
169 A  129C  3011
1   A  19 
 1
169 129 C   3011

  

 A
 129 1  19 
1

C  1 129  1 169  169 1  3011
 
   


1  129 1  19 
 
40  169 1  3011
1 
 129

 40
40   19 


169
1
3011




 40
40 
66
14 
 
5
A  14,
C 5
67
EXAMPLE 1.7.2.1 – APPLICATION OF GAUSSIAN
ELIMINATION AND GAUSS JORDAN
ELIMINATION METHOD

Ahmad inherited RM25000 and invested part of it in a money
market account, part in municipal bonds and part in a mutual
fund. After one year, he received a total of RM1620 in simple
interest from three investments. The money market paid 6%
annually, the bonds paid 7% annually and the mutually fund
paid 8% annually. There was RM6000 more invested in the bonds
than the mutual funds. Find the amount Ahmad invested in each
category using Gauss Elimination Method.
68
Solution
Let A= Money Market Account
B=Municipal bond
C=Mutual fund
A  B  C  25000
0.06 A  0.07 B  0.08C  1620
B  C  6000
Arrange in matrix form
 1
1
1 25000 


0.06
0.07
0.08
1620


 0
1
1 6000 
69
 1
1
1
1 25000 
1
1 25000 




0.06
0.07
0.08
1620
R

0.06
R
0
0.01
0.02
120
1

 2
 100 R2
 0
0
1
1 6000 
1
1 6000 
1 1 1 25000 
1 1 1 25000 




0 1 2 12000 
0 1 2 12000 
0 1 1 6000  R  R 0 0 3 6000  1
3
2
 R3
3
1 1 1 25000 


0 1 2 12000 
0 0 1 2000 
70
A  B  C  25000
B  2C  12000
C  2000
B  12000  2C
 12000  2(2000)
 12000  4000
 8000
A  25000  B  C
 25000  8000  2000
 15000
71
EXAMPLE 1.7.3.1-APPLICATION OF CRAMER’S RULE

A salesman has the following record of sales during three
months for three items A, B and C which have different
rates of commission. Find out the rates of commission on
the items A, B and C. Solve by Cramer’s rule.
Months
January
February
March
Sales of Units
A
B
90
100
130
50
60
100
Total commission drawn
C
20
40
30
800
900
850
72
Solution
Change into matrix form
800  2500   100  7000    20  47500 


 90  2500   100 1500    20 10000 
 90 100 20  A 800 
130 50 40  B   900

  

 60 100 30  C  850 

800 100 20
900
A
40
850 100 30
90 100 20
130
50
60
100 30
800 

50
 90 
50
350000
175000
2
40
40
100 30
50 40
100 30
 100 
 100 
900 40
850 30
130 40
60
30
  20 
  20 
900
50
850 100
130 50
60
100
73
90
800 20
130 900 40
B
60 850 30
90 100 20
130 50 40
60
 90 

100 30
900 40
850 30
  800 
130 40
60 30
175000
  20 
130 900
60
850
90  7000    800 1500    20  56500 


175000

700000
175000
4
74
C
90
100 800
130
50
900
60 100 850
90 100 20
130 50 40
60
 90 

100 30
50
900
100 850
 100 
130 900
60 850
175000
  800 
130
50
60
100
90  47500   100  56500   800 10000 


175000

1925000
175000
 11
75
1.8
INPUT OUTPUT MODEL
agriculture (A),
manufacturing (M),
and services (S)
Column A. Production of 1 unit agricultural products requires
the consumption of 0.2 unit of agricultural products, 0.2 unit of
manufactured goods and 0.1 units of services.
76
77
Example 1.8.1
TKK Corporation, a large conglomerate, has three
subsidiaries engaged in producing raw rubber,
manufacturing tires and manufacturing other rubber based
goods. The production of 1 unit of raw rubber requires the
consumption of 0.08 unit of rubber, 0.04 units of tires and
0.02 unit of other rubber based goods. To produce 1 unit of
tires requires 0.6 unit of raw rubber, 0.02 unit of tires and
0 units of other rubber based goods. To produce 1 unit of
other rubber based goods requires 0.3 unit of raw rubber,
0.01 units of tires and 0.06 unit of other rubber based
goods. Markets research indicates that the demand for the
following year will be RM 200 million for raw rubber, RM
800 million for tires and RM 120 million for other rubber
based products. Find the level of production for each
subsidiary in order to satisfy this demand.
78

Solution
79
80
81
82