ACCT-1205

Business Mathematics II
Matrix and Determinant

1
 Matrix:
Definition of Matrix: A System of 𝑚𝑛 numbers arranged in a rectangular array of m rows and
n columns is called a matrix of order 𝑚 × 𝑛 (read as 𝑚 by 𝑛 matrix) matrix. Where the
horizontal lines in a matrix are called rows and the vertical lines are called columns. The
numbers in the matrix are called entries. The plural form of matrix is matrices. A matrix is
expressed one of the following ways.

1 2 1 2 1 2
For example: 𝑖. (3 2) 𝑖𝑖. [3 2] 𝑖𝑖𝑖. ‖3 2‖
1 2 1 2 1 2

Types of Matrix
Rectangular Matrix: If the number of rows and columns of a matrix are not equal
𝑖. 𝑒. 𝑚 ≠ 𝑛, then the matrix is known as a rectangular matrix.
2 2 4
Such as: ( )
5 6 7
Classification of rectangular matrices are as follows: -

Square matrix: If 𝑚 = 𝑛 𝑖. 𝑒 the number of rows and columns of a matrix are equal, then
the matrix is of order 𝑛 × 𝑛 and is called a square matrix of order 𝑛.
2 3 1
2 3
For example: (1 5 2) this is 3 × 3 matrix and ( ) this is 2 × 2 matrix. Here all
4 5
7 6 9
matrices are square matrix.

Horizontal matrix: If the number of columns is more than the number of rows in a matrix
then it is called a horizontal matrix.
2 2 4 3
For example: ( ) is a horizontal matrix.
5 6 7 9

Row matrix: If there is only one row in a matrix then it is called a row matrix.
For example: (1 2 3) is called a row matrix.

Column matrix: If there is only one column in a matrix then it is called a column matrix.
1
For example: (2) is called a column.
3
Vertical matrix: If the number of rows is more than the number of columns in a matrix then
2 3
it is called a vertical matrix. For example: (3 5) is a vertical matrix.
4 6
Equal matrix: Two matrices are said to be equal if the satisfy the following conditions.

2
a) They are of the same type. i.e. if they have same number rows and columns.
b) The elements in the corresponding positions of the two matrices are equal.
2 1 3 2 1 3
For example: ( ) ( ) are equal matrices.
9 8 7 9 8 7

Null or Zero matrix: If all the elements of an 𝑚 × 𝑛 matrix are zero, then it is called a Null
or Zero matrix.
0 0 0
For example: (0 0 0) is a null matrix.
0 0 0

Unit or Identity matrix: A Square matrix having unity for its elements in the leading diagonal
and all other elements are zero is called unit or identity matrix.
1 0 0
For example: (0 1 0) is a 3 rowed unit matrix and we denote it by 𝐼3 .
0 0 1

Diagonal matrix: A square matrix in which all elements expect those in the main diagonal are
zero is known as a diagonal matrix.
2 0 0
For example: (0 4 0) is a diagonal matrix.
0 0 6

Sub matrix: A matrix which is obtained from a given matrix by deleting any number of rows
and columns is called a sub matrix of the given matrix.
8 2 6
1 2
For example: ( ) is a sub matrix of (5 1 2).
3 4
9 3 4

Operation of matrix

Addition of matrix:

2 3 5 4 6 3
i) 𝐴 = (1 9 7) , 𝐵 = ( 7 2 9)
3 4 5 −1 −3 −5

2+4 3+6 5+3 6 9 8

∴ 𝐴 + 𝐵 = (1 + 7 9 + 2 7 + 9) = (8 11 16)
3−1 4−3 5−5 2 1 0

4 9 8 5 4 8 2
ii) 𝐴 = (2 5 3 1), 𝐵 = (6 4 2)
2 8 7 5 2 1 9
These matrices are not suitable for addition. Because two matrices are suitable for addition
only, when they are of the same type.
3
 3 2 7 4 −2 −2 −7 −4
9 6 4 3 −9 −5 −4 −3
iii) 𝐴 = ( ) 𝐵=( )
2 9 0 1 −2 −9 1 −1
3 4 6 9 −3 −4 −6 −8

3−2 2−2 7−7 4−4 1 0 0 0

9−9 6−5 4−4 3−3 0 1 0 0
∴ 𝐴+𝐵 =( )=( ) = 𝐼4
2−2 9−9 0+1 1−1 0 0 1 0
3−3 4−4 6−6 9−8 0 0 0 1

# Subtraction of matrix:

3 2 7 4 −2 −2 −7 −4
9 6 4 3 −9 −5 −4 −3
𝐴=( ), 𝐵=( )
2 9 0 1 −2 −9 1 −1
3 4 6 9 −3 −4 −6 −8

3+2 2+2 7+7 4+4 5 4 14 8

9+9 6+5 4+4 3+3 18 11 8 6
∴𝐴−𝐵 =( )= ( )
2+2 9+9 0−1 1+1 4 18 −1 2
3+3 4+4 6+6 9+8 6 8 12 17
Two matrices are suitable for subtraction only when they are of same type.

Product of Matrices

2 3 4 2 3
i) 𝐴 = (2 1 5 ) , 𝐵 = (4 5) , here the matrix A is of type 3× 3, and the
6 2 5 6 9
matrix B of type 3 × 2 . And, BA= is not possible, since the number of columns of B
is 2 not equal to the number of rows of A is 3.

4 + 12 + 24 6 + 15 + 27 40 48
∴ 𝐴 × 𝐵 = ( 4 + 4 + 30 6 + 5 + 45 ) = (38 56)
12 + 8 + 30 18 + 10 + 45 50 73

2 3 4 2 0 1
ii) If, 𝐴 = (2 1 5) , 𝐵 = ( 3 5 −1) Show that 𝐴𝐵 ≠ 𝐵𝐴
6 2 8 −1 −2 −3

4+9−4 0 + 15 − 8 2 − 3 − 12 9 7 −13
∴𝐴×𝐵 = (4+3−5 0 + 5 − 10 2 − 1 − 15 ) = ( 2 −5 −14)
12 + 6 − 8 0 + 10 − 16 6 − 2 − 24 10 −6 −20

4
 4 + 0 + 16 6+0+2 8+0+8 20 8 16
∴ 𝐵 × 𝐴 = ( 6 + 10 − 6 9+5−2 12 + 25 − 8 ) = ( 10 12 29 )
−2 − 4 − 18 −3 − 2 − 6 −4 − 10 − 24 −24 −11 −38

∴ 𝐴𝐵 ≠ 𝐵𝐴

If A and B are two matrices, then the product AB can be calculated only if the number of
columns in A be equal to the number of rows in B. The two matrices A and B satisfying this
condition are called conformable to multiplication.

1 0 1 0
iii) Let 𝐴 = ( ), 𝐵 = ( )
0 −2 0 4
1 0 1 0 1+0 0+0 1 0
∴ 𝐴𝐵 = ( )×( ) =( )=( )
0 −2 0 4 0−0 0−8 0 −8

1 0 1 0 1+0 0−0 1 0
∴ 𝐵𝐴 = ( )×( ) =( )=( )
0 4 0 −2 0+0 0−8 0 −8
∴ 𝐴𝐵 = 𝐵𝐴

If A and B are two matrices and their products AB and BA are equal i.e, 𝐴𝐵 = 𝐵𝐴 then they
commute with each other or they satisfy the commutative law for Multiplication.

2 3 5
iv) If a matrix 𝐴 = ( 1 2 9) then find the matrix 3𝐴3 − 4𝐴2 + 2𝐴 − 4
1 0 2

2 3 5 2 3 5 4+3+5 6 + 6 + 0 10 + 27 + 10
∴ 𝐴 × 𝐴 = (1 2 9) × ( 1 2 9) = (2 + 2 + 9 3 + 4 + 0 5 + 18 + 18 )
1 0 2 1 0 2 2+0+2 3+0+0 5+0+4

12 12 47
= (13 7 41)
4 3 9

12 12 47 2 3 5
∴ 𝐴3 = 𝐴2 × 𝐴 = (13 7 41 ) × ( 1 2 9)
4 3 9 1 0 2

24 + 12 + 47 36 + 24 + 0 60 + 108 + 94 83 60 262
= ( 26 + 7 + 41 39 + 14 + 0 65 + 63 + 82 ) = (74 53 210)
8+3+9 12 + 6 + 0 20 + 27 + 18 20 18 65

83 60 262 249 180 786

3
∴ 3𝐴 = 3 × (74 53 210) = (222 159 630)
20 18 65 60 54 195

5
 12 12 47 48 48 188
∴ 4𝐴2 = 4 × (13 7 41 ) = ( 52 28 164)
4 3 9 16 12 36

2 3 5 4 6 10
∴ 2𝐴 = 2 × (1 2 9) = ( 2 4 18)
1 0 2 2 0 4

201 132 598

∴ 3𝐴3 − 4𝐴2 = (170 131 466)
44 42 159

205 138 608

3 2
∴ 3𝐴 − 4𝐴 + 2𝐴 = (172 135 484)
46 42 163

∴ 3𝐴3 − 4𝐴2 + 2𝐴 − 4𝐼3

205 138 608 4 0 0 201 138 608
= (172 135 484) − (0 4 0) = (172 131 484)
46 42 163 0 0 4 46 42 159

Some typical Solved Problems

Problem 1: A man buys 8 dozens of mangoes, 10 dozens of apples, and 4 dozens of
bananas. Mangoes cots Tk 18 per dozen, apples Tk 9 per dozen and bananas Tk 6 per
dozen. Represent the quantities bought by a row matrix and hence obtain the total cost.

Solution : A=[8 10 4] is 1× 3 row matrix represents the quantities purchased by the man.
18
B =[ 9 ] is 3×1 column matrix represents the prices of different fruits purchased.
6
The product of matrices A (1× 3) and B (3× 1) i.e. AB (1× 1) matrix will represent the total
cost.
18
AB=[8 10 4] [ 9 ] =[(8× 18) + (10 × 9) + (4 × 6)] =[144+90+24] =[258]
6
Hence the total cost is Tk 258.

Problem 2: A store has in stock 20 dozens shirts, 15 dozens trousers and 25 dozens pair of socks.
If the selling prices are respectively Tk 50, Tk 90 and Tk 12 per piece, then find the total price of
the entire stock of the store.

Solution: We know, 1 dozen = 12 pieces

6
Therefore, 20 dozens = 240 pieces
15 dozens = 180 pieces
25 dozens = 300 pieces
Converting the stock from dozens to single pieces and representing them by a row matrix we
get, 𝐴 = [240 180 300]
and representing the unit prices of each item by a column matrix we get
50
𝐵 = [90]
12
Now, the total price of the entire stock of the store will be the product of the matrices.

50
We have 𝐴𝐵 = [240 180 300] [90]
12
= [(240× 50) + (180 × 90) + (300 × 12)] =[12000+16200+3600]
=[31800]. Hence amount received for the entire stock is Tk.31800.

Problem 3: In a development plan of a city, a contractor has taken contract to construct

certain houses for which he needs building materials like stone, sand etc. There are three
firms A, B, C, that can supply him 40, 35 and 25 truck load of stones and 10, 5 and 8 truck
load of sand respectively. If the cost of one truck load of stone and sand are Tk 1200 and
Tk 500 respectively, then find the total amount is to be paid by the contractor to each of
the firms A, B, C respectively.

Solution: Representing the materials supplied by the three firms A, B, C in a matrix, we get
A B C
40 35 25
M=[ ]
10 5 8
Representing the cost of one truck of each material in a matrix, we get E = [1200 500]
Now, the product of matrices E and M will represent the amount paid to the firms.
40 35 25
EM = [1200 500] [ ]
10 5 8
= [(1200× 40 + 500 × 10) (1200 × 35 + 500 × 5) (1200 × 25 + 500 × 8)]
= [(48000+5000) (42000+2500) (30000+4000)]
A B C
= [ 53000 44500 34000 ]

7
Hence the total amount paid to the firms A, B and C for supplied materials are respectively
Tk.53000, Tk. 44500 and Tk. 34000.

Problem 4: A manufacture produces three products A, B, C, Which he sells in the two

markets of Dhaka and Rangpur. Annual sales volumes are indicated as follows:
Markets Products A B C
Dhaka 8,000 10,000 15,000
Rangpur 10,000 2,000 20,000

(a) If unit sale prices of A, B and C are Tk.2.25, Tk. 1.50 and Tk. 1.25 respectively,
find the total revenue collected from each of the markets of Dhaka and Rangpur
with the help of matrices.
(b) If the unit costs of the above three products are Tk. 1.60, Tk. 1.20 and Tk. 0.90
respectively, find the gross profit with the help of matrices.

Solution (a): Representing the unit selling prices of the products in a matrix, we get
A B C
𝑃 = [2.25 1.50 1.25]
Also, representing the sales volumes in a matrix, we get
Dhaka Rangpur
8000 10000
𝑉 = [10000 2000 ]
15000 20000
Now the total revenue collected from each of the markets of Dhaka and Rangpur is given by
the product of matrices 𝑃 and 𝑉.
8000 10000
𝑃𝑉 = [2.25 1.50 1.25] × [10000 2000 ]
15000 20000
= [2.25* 8000 + 1.50 * 10000 + 1.25 * 15000 2.25 * 10000 + 1.50 * 2000 + 1.25 * 20000]

=[18,000 + 15,000 + 18,750 22,500+3,000+25,000]

Dhaka Rangpur

=[51750 50500]

Thus the revenue collected from Dhaka = Tk. 51750

And the revenue collected from Rangpur = Tk. 50500.
Total revenue collected from both the markets is = Tk (51750 + 50500) = Tk. 102250

Solution (b):
We know, profit = selling price – cost price

8
Therefore, profit for unit sale of product A = Tk.(2.25 – 1.60) = Tk.0.65
Profit for unit sale of product B = Tk.(1.50 – 1.20) = Tk.0.30
Profit for unit sale of product C = Tk.(1.25 – 0.90) = Tk.0.35
Representing the profits for unit selling prices of the products in a matrix, we get
A B C
𝑃 = [0.65 0.30 0.35]
Also, representing the sales volumes in a matrix, we get
Dhaka Rangpur
8000 10000
𝑉 = [10000 2000 ]
15000 20000
Now the total profit collected from each of the markets of Dhaka and Rangpur is given by
the product of matrices 𝑃 and 𝑉.

8000 10000
𝑃𝑉 = [0.65 0.30 0.35] × [10000 2000 ]
15000 20000
= [0.65* 8000 + 0.30 * 10000 + 0.35 * 15000 0.65 * 10000 + 0.30 * 2000 + 0.35 * 20000]

=[5200 + 3000 + 5250 6500+600+7000]

Dhaka Rangpur

=[13450 14100]

Hence the profit from Dhaka market is Tk. 13450 and from Rangpur market is Tk.14100.
Thus total profit from the two markets is Tk. 27550.

Problem 5: A trust fund has Tk. 50,000 that is to be invested into two types of bonds.
The first bond pays 5% interest per year and the second bond pays 6% per year. Using
matrix multiplication, determine how to divide Tk. 50,000 among two types of bonds so
as to obtain an annual total interest of Tk. 2780.
Solution: Let Tk.50000 be divided into two parts Tk. X and Tk.(50,000 - X) out of which
first part is invested in first type of bonds and the second part is invested in second type of
bonds. The values of these bonds can be written in the form of a row matrix A given by

A = [X 50,000 – X] which is 1 × 2 matrix

And the amounts received as interest per taka annually from these two types of bonds can be
written in the form of a column matrix B given by a 2 × 1 matirx i.e.

5 / 100
B=   Here the interest has been calculated per taka annually.
6 / 100
9
Now the interest to be obtained annually is a single number i.e. a matrix of order 1 × 1 and the
same can be obtained by the product matrix AB, since this product matrix would be a 1
× 1 𝑚𝑎𝑡𝑟𝑖x.

5 / 100
Here AB = [X 50,000 - X] ×  
6 / 100
5 6
= [𝑋. 100 + (50,000 − 𝑋). 100]
𝑋
=[3000 − ]
100

Also we are given that the annual interest = Tk.2,780. Therefore, we must have
𝑋
[3000 − ]= [2780]
100

𝑋
or 3000 − = 2780
100

or 𝑋 = (3000 − 2780) × 100

or 𝑋 = (220 × 100 ) or 𝑋 = 22,000

Hence the required amounts are Tk. 22,000 and Tk. (50,000 – 22,000) = Tk. 28,000.

Problem 6: A finance company has offices located in every division, every district and
every thana in Bangladesh. Assume that there are five divisions, thirty districts and 200
thanas in the country. Each office has one head clerk, one cashier, one clerk and one peon.
A divisional office has in addition, one office superintendent, two clerks, one typist and
one peon. A district office has in addition one clerk and one peon. The basic monthly
salaries are as follows: office superintendent Tk 5000, head clerk Tk 2000, cashier Tk
1750, clerk and typist Tk 1500 and peon Tk 1000. Using matrix notation find-
(a) The total number of posts of each kind in all the offices taken together.
(b) The total basic monthly salary bill of each kind of office.
(c) The total basic monthly salary bill of all offices taken together.

Solution: Let us use the symbols Div, Dis, Tah for division, district, thana respectively and
O, H, C, Cl, T and P for office superintendent, Head clerk, cashier, clerk, typist and peon
respectively.
Then the number of offices can be arranged as elements of a row matrix A (say) given by
Div. Dis. Tah
A = [5 30 200]

The composition of staff in various offices can be arranged in a 3 × 6 matrix B (say) given
by

10
 O H C Cl T P

1 1 1 2 + 1 1 1 + 1
B = 0 1 1 2 + 1 0 1 + 1
0 1 1 1 0 1 

The basic of the monthly salaries of various types of employees of these offices correspond to
the elements of the column matrix C (say) given by

500 
200
 
175 
C=  
150 
150 
 
100 

(a) Total number of posts of each kind in all the offices are the elements of the product matrix
AB.

1 1 1 3 1 2
i.e. [5 30 200] × 0 1 1 3 0 2
0 1 1 1 0 1

i.e. [5+0+0, 5+30+200, 5+30+200, 15+60+200, 5+0+0, 10+60+200]

i.e. O H C Cl T P
[5 235 235 275 5 270]
i.e. Required number of posts in all the offices taken together are 5 offices superintendent,
235 Head clerks, 235 cashiers, 275 clerks, 5 typists and 270 peons.

(b) Total basic monthly salary bill of each kind of office are the elements of the products
matrix BC

500 
200
1 1 1 3 1 2  
0 1 1 3 0 2 × 175 
i.e.    
0 1 1 1 0 1 150 
150 
 
100 

11
  (1 * 500) + (1 * 200) + (1 *175) + (3 *150) + (1 *150) + (2 *100) 
= (0 * 500) + (1 * 200) + (1 *175) + (2 *150) + (0 *150) + (2 *100)
 (0 * 500) + (1 * 200) + (1 *175) + (1 *150) + (0 *150) + (1 *100) 

500 + 200 + 175 + 450 + 150 + 200 1675

=  0 + 200 + 175 + 300 + 0 + 200  =  875 
 0 + 200 + 175 + 150 + 0 + 100   625 

i.e. The total basic monthly salary bill of each divisional, district and thana offices are
Tk. 1675, Tk. 875 and Tk. 625 respectively

(c) Total basic monthly salary bill of all the officers is the element of the product matrix ABC

1675
i.e. [5 30 200] ×  875 
 625 

= [(5*1675)+(30*875)+(200*625)]
= [8375+2650+125000] = [159625]
i.e. total basic monthly salary bill of all the offices taken together is Tk. 159,625.

Uses of Matrix multiplication in computer science:

i. In terms of benchmarking the computing prowess of a processor or even sometimes a
General purpose GPU, a routine called DGEMM - Double precision General Matrix
Multiplication is performed. It is a common benchmarking routine and widely used.
ii. In a range of applications from image processing to genetic analysis, computers are often
called upon to solve systems of linear equations — usually with many more than two
variables. Even more frequently, they’re called upon to multiply matrices.
iii. One of the areas of computer science in which matrix multiplication is particularly useful
is graphics, since a digital image is basically a matrix to begin with: The rows and columns
of the matrix correspond to rows and columns of pixels, and the numerical entries
correspond to the pixels’ color values. Decoding digital video, for instance, requires matrix
multiplication.

iv. In the same way that matrix multiplication can help process digital video, it can help process
digital sound. A digital audio signal is basically a sequence of numbers, representing the
variation over time of the air pressure of an acoustic audio signal. Many techniques for

12
 filtering or compressing digital audio signals, such as the Fourier transform, rely on matrix
multiplication.
v. Matrix math intensive scientific software, calling some operations and matrix
multiplication is a demanding one. Matrix multiplier runs at light speed on our particular
computer.

However, few of us are likely to consciously apply multiplication of matrix in field of computer
science in our day to day lives. If all of us consciously apply matrix multiplication, then
applications of it will increase.

IDEMPOTENT MATRIX
Definition: -A square matrix A is called idempotent provided it satisfies the relation A2 = A.
1 0
Example: A=( ) is idempotent.
0 1
1 0 1 0
Solution: - A2 =A●A= ( )×( )
0 1 0 1
1●1 + 0●0 1●0 + 0●1
=( )
0●1 + 1●0 0●0 + 1●1
1 0
=( ). Therefore, A2 =A
0 1
Hence the matrix A is idempotent.
𝟐 −𝟐 −𝟒
Problem: -Show that the matrix 𝐀 = (−𝟏 𝟑 𝟒 ) is idempotent.
𝟏 −𝟐 −𝟑
2 −2 −4 2 −2 −4
Solution: - A2 = A●A =(−1 3 4 ) × (−1 3 4)
1 −2 −3 1 −2 −3
2●2 − 2●(−1) − 4●1 2●(−2) − 2●3 − 4●(−2) 2●(−4) − 2●4 − 4●(−3)
= (−1●2 + 3●(−1) + 4●1 −1●(−2) + 3●3 + 4●(−2) −1●(−4) + 3●4 + 4●(−3) )
1●2 − 2●(−1) − 3●1 1●(−2) − 2●3 − 3●(−2) 1●(−4) − 2●4 − 3●(−3)
2 −2 −4
= (−1 3 4) = 𝐴
1 −2 −3
Hence the matrix A is idempotent.

13
 INVOLUTORY MATRIX
Definition:-A square matrix A is called Involutory provided it satisfies the relation A2 = I,
where I is the identity matrix.
1 0
Example: - The matrix A = ( ) is involutory matrix, since
0 −1
1 0 1 0
A2 =A●A = ( )×( )
0 −1 0 −1
1●1 + 0●0 1●0 + 0●(−1)
=( )
0●1 − 1●0 0●0 + (−1)●(−1)
1 0
=( )=𝐼
0 1
Therefore, 𝐴2 = 𝐼.
Hence the given matrix A is involutory.
−𝟓 −𝟖 𝟎
Problem:- Show that the matrix 𝐀 = ( 𝟑 𝟓 𝟎 ) is involutory.
𝟏 𝟐 −𝟏
−5 −8 0
Solution:-Given that A = ( 3 5 0)
1 2 −1
−5 −8 0 −5 −8 0
Now, A2 = A●A = ( 3 5 0) × ( 3 5 0) =
1 2 −1 1 2 −1

−5●(−5) + (−8)●3 + 0●1 (−5)●(−8) + (−8)●5 + 0●2 (−5)●0 + (−8)●0 + 0●(−1)

( 3●(−5) + 5●3 + 0●1 3●(−8) + 5●5 + 0●2 3●0 + 5●0 + 0●(−1) )
1●(−5) + 2●3 + (−1)●1 1●(−8) + 2●5 + (−1)●2 1●0 + 2●0 + (−1)●(−1)
1 0 0
= (0 1 0) = 𝐼
0 0 1
Hence the given matrix A is involutory.

NILPOTENT MATRIX
Definition: - A square matrix A is called Nilpotent matrix of order (index) n, if it satisfies the
relation An = 0 and An−1 ≠ 0, where n is a positive integer and 0 is the null matrix.

14
 0 1
EXAMPLE: - The matrix A = ( ) is the nilpotent matrix,
0 0
0 1
Since, A=( ) ≠ 0, and
0 0
0 1 0 1 0●0 + 1●0 0●1 + 1●0 0 0
A2 = A●A = ( )×( )=( ) =( )=0
0 0 0 0 0●0 + 0●0 0●1 + 0●0 0 0
Therefore the given matrix is nilpotent matrix of order 2.
𝟏 𝟏 𝟑
PROBLEM: - Show that A=( 𝟓 𝟐 𝟔 )is a nilpotent matrix of order 3.
−𝟐 −𝟏 −𝟑
1 1 3
SOLUTION: - Given that A = ( 5 2 6)
−2 −1 −3
1 1 3 1 1 3
Now, A2 = A●A = ( 5 2 6 )×( 5 2 6)
−2 −1 −3 −2 −1 −3
1●1 + 1●5 + 3●(−2) 1●1 + 1●2 + 3●(−1) 1●3 + 1●6 + 3●(−3)
=( 5●1 + 2●5 + 6●(−2) 5●1 + 2●2 + 6(−1) 5●3 + 2●6 + 6●(−3) )
(−2)●1 + (−1)●5 + (−3)●(−2) (−2)●1 + (−1)●2 + (−3)●(−1) (−2)●3 + (−1)●6 + (−3)●(−3)

0 0 0
=( 3 3 9)
−1 −1 −3
0 0 0 1 1 3
A3 = 𝐴2 ●A = ( 3 3 9 )×( 5 2 6)
−1 −1 −3 −2 −1 −3
0●1 + 0●5 + 0●(−2) 0●1 + 0●2 + 0●(−1) 0●3 + 0●6 + 0●(−3)
=( 3●1 + 3●5 + 9●(−2) 3●1 + 3●2 + 9●(−1) 3●3 + 3●6 + 9●(−3) )
(−1)●1 + (−1)●5 + (−3)●(−2) (−1)●1 + (−1)●2 + (−3)●(−1) (−1)●3 + (−1)●6 + (−3)●(−3)

0 0 0
= (0 0 0 ) = 0
0 0 0
Therefore the given matrix is nilpotent matrix of order 3.

TRANSPOSED MATRIX
Transposed matrix: The matrix of order n × m obtained by interchanging the row and column
of a matrix A of order m× n is called transposed of matrix A or transposed matrix A and is
denoted by 𝐴′ and At . (Read as A transpose).

15
Another definition: If A = (aij ) be a matrix of order m × n then the matrix B = (bji ) of order
n × m such that bji = aij is known as transposed matrix of or the transpose of the matrix A.

2 3 4
Example: A = (9 −2 4)
7 2 0
2 9 7
At = (3 −2 2)
4 4 0
𝟏 𝟐
𝟏 𝟐 𝟑
Problem: If 𝐀 = ( ) 𝐁 = ( 𝟐 𝟎), show that 𝐁 𝐭 × 𝐀𝐭 = (𝐀𝐁)𝐭 .
𝟑 −𝟐 𝟏
−𝟏 𝟏
1+4−3 3−4−1
Solution: L.H.S = Bt × At = ( )
2+0+3 6+0+1
2 −2
=( )
5 7
1 2
1 2 3 1+4−3 2+0+3
Now, 𝐴𝐵 = ( ) × 2 0) = (
( )
3 −2 1 3−4−1 6+0+1
−1 1
2 5
=( )
−2 7
2 −2
R.H.S: 𝐴𝐵 𝑡 = ( )
5 7
L.H.S = R.H.S. (Showed)

COMPLEX CONJUGATE OF MATRIX

Complex conjugate of matrix: The matrix obtained from any given matrix A of order m × n
with complex element aij by replacing its element by the corresponding conjugate complex
number is called the complex conjugated or conjugate of A and is denoted by ̅A.
Example:
2 + 3i 4 − 3i 2 − 4i 2 − 3i 4 + 3i 2 + 4i
If A = (3 + 2i 4 − 2i 3 − 2i ) then ̅
A = ( 3 − 2i 4 + 2i 3 + 2i).
4 + 4i 6 − 5i 3−i 4 − 4i 6 + 5i 3+i

16