FIRST SEMESTER BACHELOR IN BUSINESS … Sem/Course 3... · know some of the difficult, unseen terms. ... Binomial Theorm Page : (131-146) Binomial Theorem for any positive Integer

BBA (S1) 03-02

KRISHNA KANTA HANDIQUI STATE OPEN UNIVERSITYPatgaon,Ranigate, Guwahati - 781 017

FIRST SEMESTERBACHELOR IN BUSINESS ADMINISTRATION

COURSE: 03Business Mathematics

BLOCK-2

CONTENTS

UNIT 6 : LOGARITHM

UNIT 7 : BINOMIAL THEOREM

UNIT 8 : MATRICES

UNIT 9 : LIMIT AND CONTINUITY

REFERENCES : FOR ALL UNITS

Subject Experts

1. Professor Nripendra Narayan Sarma, Maniram Dewan School of Mangement, KKHSOU

2. Professor Munindra Kakati, VC, ARGUCOM

3. Professor Rinalini Pathak Kakati,Dept.of Busniess Administration, GU

Course Coordinator : Harekrishna Deka, K.K.H.S.O.U

Dr, Chayanika Senapati,KKHSOU

Dr. Smritishikha Choudhury,KKHSOU

SLM Preparation T eamUnits Contributor

6 Dr. Rupam Barman, IIT Guwahati

7 Dr. Dilip Kumar Sarma, Cotton College

8 Dr. Dipu Nath sarma, Handique girls college

9 S. K. Jha, Lumding College, Lumding

Editorial T eamContent : Prof. N.R. Das, Gauhati University

Dr. Sanjay Dutta, PETC, Gauhati UniversityStructure Format & Graphics : Harekrishna Deka, KKHSOU

June 2017

This Self Learning Material (SLM) of the Krishna Kanta Handiqui State Open University is

made available under a Creative Commons Attribution-Non Commercial-Share Alike 4.0 License

(international): http://creativecommons.org/licenses/by-nc-sa/4.0/

Printed and published by Registrar on behalf of the Krishna Kanta Handiqui State Open University.

Headquarters : Patgaon, Rani Gate, Guwahati - 781017 Housefed Complex, Dispur , Guwahati-781006; W eb: www .kkhsou.in

The University acknowledges with thanks the financial support provided by the

Distance Education Bureau , UGC for the preparation of this study material.

BLOCK INTRODUCTION

This is the second block of the course ‘Business Mathematics’. The block consists of four units.

The sixth unit will focus on Logarithms.

The seven unit is an elementary introduction to Binomial Theory and its uses.

The eighth unit is about Matrix theory. In this unit,you will able to explores some of application of matrix

theory in this unit.

In nineth unit ,we discuss limits and continuity of a functions.

While going through a unit, you will notice some along-sode boxes, which have been included to help you

know some of the difficult, unseen terms. Some ‘ACTIVITY’ (s) has been included to help you apply your

own thoughts. Again, we have included some relevant concepts in ‘LET US KNOW’ along with the text.

And, at the end of each section, you will get ‘CHECK YOUR PROGRESS’ questions. These have been

designed to self-check your progress of study. It will be better if you solve the problems put in these

boxes immediately after you go through the sections of the units and then match your answers with

‘ANSWERS TO CHECK YOUR PROGRESS’ given at the end of each unit.

BACHELOR IN BUSINESS ADMINISTRATIONBUSINESS MATHEMATICS

Block 2

DETAILED SYLLABUS

UNIT 6: Logarithms Pages: (120-130)

Definition, Properties of Logarithms, Application of Logarithms.

UNIT 7: Binomial Theorm Page : (131-146)

Binomial Theorem for any positive Integer, General and Middle Terms

UNIT 8: Matrices Page : (147-215)

Definition of Matrix and Examples, Types of matrices,Transpose of a

matrix,symmetric and skew-symmetric matrix,algebra of matrices:Addition of

matrices,scalar multiplication,subtraction of matrices,multiplication of

matrices: Adjoint and inverse of a matrix and its existence, Rank of a matrix,

Solution of a system of linear equations by matrix method, Solution of a system

of linear equations by Crammer’s rule.

UNIT 9: Limits and continuity Page : (216-230)

Limit of a function Continuity of a function.

UNIT 6: LOGARITHM

UNIT STRUCTURE

6.1 Learning Objectives

6.2 Introduction

6.3 Logarithms

6.3.1 Properties of Logarithm

6.3.2 Common Logarithm

6.3.3 Antilogarithm

6.3.4 Application of Logarithm

6.4 Let Us Sum Up

6.5 Further Readings

6.6 Answer To check Your Progress

6.7 Model Questions

6.1 LEARNING OBJECTIVES

After going through this unit you will be able to:

l define logarithm

l explain the properties of logarithms

l explain common logarithm

6.2 INTRODUCTION

Logarithms play an important role in numerical calculations. In this

unit, we are going to discuss about logarithm and its properties. Here we

shall discuss about the laws of logarithms and their applications. In statistical

work, they are particularly useful while constructing ratio graphs, in computing

geometric mean, in finding powers and extracting roots and fitting trend

lines. Thus through this unit you will able to understand the importance of

logarithms and its application in different field.

6.3 LOGARITHMS

The logarithm of a given number to a certain base is the power to

125Business Mathematics (Block 2)

which the base must be raised in order to obtain that number. Thus, the

logarithms or simply 'log' of 15625 to base 5 is 6. It means that if 5 is the

base, raised to the power 6, we shall get the required number i.e.

5x5x5x5x5x5=15625.

Let a be a positive real number other than 1. Then for any real number

x, ax is a positive real number. Let

ax = N

Here we say that a is the base and x is the index of the power of a .

The index x is called the logarithm of N with respect to the base a, and we

write x = logaN.

Thus, if Nax = ( 1,0 ≠> aa ), then logaN = x,

or if logaN=x, then ax = N.

Examples:

(i) 823 = , hence 38log2 = . (ii) 9

13 2 =−

, hence 2)9

1(log3 −= .

(iii) 5)125( 3

1

= , hence 3

15log125 = etc.

Remark : Logarithm of any negative number is not defined. Because, for

any 1,0 ≠> aa , we can not find any rational number x such that ax is

negative.

Special Cases: Let 0>a and 1≠a .

1. We have a0 = 1, hence loga1 = 0.

2. Since ,1 aa = we have 1log =aa .

Exercise 9: Find Nalog , where 01.=a and 0001.=N .

Solution: Let xNa =log . Then Nax = .

Thus, 2)0001(.log 01. = .

6.3.1 Properties of Logarithms

1) )0,,1,0(,loglog)(log >≠>+= nmaanmmn aaa

Proof: Let xma =log and yna =log . Hence, nama yx == , .

yxmnaaamn ayxyx +=⇒==⇒ + )(log

nmmn aaa loglog)(log +=⇒

126 Business Mathematics (Block 2)

Unit 6 Logarithm

Hence, the logarithm of the product of two numbers is equal to the

sum of their logarithms.

Examples: (i) 7log5log)75(log35log 3333 +=×=

(ii) 19log3log)193(log57log2222

+=×=

Remark: We can extend this property to any product of finite

numbers. For example,

57(log3log))57(3(log105log 2222 ×+=××=

5log7log3log 222 ++=

2) )0,,1,0(,loglog)(log >≠>−= nmaanmn

maaa

Proof: Let xma =log and yna =log . Hence, ,max = na y = .

Now, yx

y

x

aa

a

n

m −== . Hence, nmyxn

maaa loglog)(log −=−= .

Examples:

(i) 5log7log)5

7(log 222 −=

(ii) 5log7log2log5log14log)5

14(log 222222 −+=−=

(iii) rqpr

qpaaaa logloglog)

.(log −+=

3) )0,1,0(,log)(log >≠>= maampm ap

a

Proof: Let xma =log . Hence, max = .

Now, mppxmaam ap

apxpxp log)(log)( ==⇒== .

Examples: (i) 2log32log 53

5 = (ii) 5log2

15log 2

2

1

2 = , etc.

4) (Change of base):

)0,1,1,0,(,logloglog >≠≠>×= mbababmm aba

Proof: Let , xma =log , ymb =log and zba =log .

Then, ,, mbma yx == and baz = . This implies yx ba = .

Now, )(,)( zzyyzxyx abzyxaaaba ==⇒==⇒= Q .

This implies that bmm aba logloglog ×= .

Remark: If we put a for m in both sides of


Logarithm Unit 6

bmm aba logloglog ×= , we get baa aba logloglog ×= .

1loglog =×⇒ ba ab ba

ab log

1log =⇒

Examples: (i) 2log5log5log 323 ×= (ii) 3log25log25log 737 ×= .

5) )0,1,0(,log >≠>= maama ma .

Proof: Let xa ma =log . Then,

mamxmx maa

a =⇒=⇒= logloglog .

Corollary: xmmx maax

aa == loglog .

If 1−=x , m

maa mm aa11loglog 1

=== −− −

.

Examples: (i) 32 3log2 = (ii) 353log

2

15 = .

Exercise 10: Prove that 1))256(log(loglog 223 = .

Solution: Since 821616256 =×= , we have

13log)2log3(log

)2(loglog)8(loglog))2log8((loglog))256(log(loglog

323

32323223223

======

Hence the result.

Exercise 1 1: Simplify 2.1log

1000log8log27log −+, where the base

of all the logarithms are same.

Solution: 2.1log

10log2log3log

2.1log

1000log8log27log 2

332

3

−+=−+

2.1log

)10log2log23(log2

3

2.1log

10log2

32log33log

2

3 −+=

−+=

2

3

2.1log

2.1log

2

3

2.1log10

43log

2

3 ==

×

=

Hence the required answer is 2

3.

Exercise 12: If cb

a

−log

= ac

b

−log

= ba

c

−log

, thus aa bb cc=1


Unit 6 Logarithm

Solution: Let each fraction = K

log a = K (b-a)

log b= K (c-a)

log c= K (a- b)

Let, A = aa bb cc

Log A = log (aa bb cc) = log aa +log bb+ log cc= a log a + b log b +c log c

= a K (b-c) +b K (c-a) + c K (a-b) = K {ab- ac+ bc- ab+ ac- bc}

= 0 = log 1

Hence, A = 1 = aa bb cc, Proved.

Exercise 13: Prove that

Solution:

L.H.S = logabca + logabcb + logabcC = logabc (abc) = 1, proved

CHECK YOUR PROGRESS

Q 1: State whether the following statements are True ( T ) or False ( F ):

(i) 1)3

1(log3 −= (T / F)

(ii) 1)2

1(log 5. = (T / F)

(iii) ya ya =log (T / F)

(iv) mama =)1

(log (T / F)

(v) 1log <xa if 10 << x . (T / F)

Q 2: If 1log >xa and 1>a , then which one of the following is true?

(i) xa > (ii) xa < (iii) xa = (iv) a

x1=

Q 3: Fill up the blanks:

(i) ........)27(loglog36 =

(ii) ........81logloglog 322 =

(iii) If ,9404.072.8log10 = then ........8720log10 =

(iv) If ),(log)(log nmmn aa += then


Logarithm Unit 6

(v) If ,722 xyyx =+ and ),log(log)3

(log yxmyx

aaa +=+ then m

= .............

6.3.2 Common Logarithms

We have learnt that any positive number other than 1 can be

used as base in logarithms. In particular, the logarithm in which 10

is used as base is known as common logarithm. Generally, the base

is not written in common logarithm . Of course, there is another

particular logarithm known as Natural or Napierian logarithm in

which the base is the irrational number. The value of e lies between

2 and 3. It should be noted that L+++=!2

1

!1

11e where for any

positive integer n, nnn ×−×××= )1(21! L . The approximate value

of e is 2.7182. Also in Unit-3, we shall discuss this in more detail.

(a). COMPUTATION WITH COMMON LOGARITHMS:

From the definition, we have

L,31000log100010

2100log10010

110log1010

01log110

103

102

101

100

=⇒=

=⇒=

=⇒=

=⇒=

As we have seen, the logarithm of a number lying between 1

and 10 lies between 0 and 1. Hence it is of the form:

0 + some positive proper decimal number.

For example, consider the number 7. Since 1071 << ,

17log0 << , and hence it is of the form

0 + some proper decimal number.

Similarly, the logarithm of a number lying between 10 and 100 will be

of the form: 1 + some positive proper decimal number, and so on.

Thus we have seen that common logarithm of a number greater

than or equal to 1 has two parts: the first part is the integral part and

the second part is the decimal part. The integral part is called the


Unit 6 Logarithm

Characteristic and the decimal part is called the Mantissa. If the

number is an integral power of 10, then the Mantissa is always

zero. Also, the characteristic of a number greater than or equal to 1 is

the number of digits in the integral part of the number reduced by 1.

Again,

L,30001.0log0001.010

201.0log01.010

11.0log1.010

01log110

103

102

101

100

−=⇒=

−=⇒=

−=⇒=

=⇒=

−

−

−

As we have seen, the logarithm of a number lying between 0.1 and

1 lies between -1 and 0. Hence it is of the form:

-1 + some positive proper decimal number.

For example, consider the number 0.7. Since 17.00 << ,

07.0log1 <<− , and hence it is of the form -1 + some proper

decimal number.

Similarly, the logarithm of a number lying between 0.01 and 0.1 will

be of the form: -2 + some positive proper decimal number, and so

on. Thus, the characteristic of the logarithm of a positive number

less than 1 is a negative integer and it is one more than the number

of zeros just after the decimal point. That is, if a positive number

less than one 1 has n zeros just after the decimal point, then the

characteristic of the logarithm of the number is –(n +1).

The mantissa of the logarithm of a number is obtained by using a

standard "Log -table".

(b) Important steps to find characteristic of a positive number N :

(1) If 1>N , count the number of digits in the integral part of N, say

it is m. Then the characteristic of N10log is )1( −m .

(2) If 10 << N , count the number of zeros just after the decimal

point, say it is . Then the characteristic of the logarithm of the

number is )1( +− n .

Thus, characteristic of 234, 23.4, 2.34, 0.000234 are 2, 1, 0, - 4,

respectively.


Logarithm Unit 6

(c) Important Properties of Mantissa: The mantissa does not depend

on the position of the decimal point in the number. That is mantissa

of the numbers 234, 23.4, 2.34, 0.000234 are same. Let N be any

positive number. If we multiply N by 10 or powers of 10, then the

significant digits shall remain same in the same order, only the position

of the decimal point will change. Similarly, when we divide N by 10

or by powers of 10, the position of the decimal point in will change.

Let, abcdmN .0log += , where abcd.0 represents a decimal

number. Now,

)1(.0)(

.010loglog)10log(

Labcdpm

pabcdmNN pp

++=++=+=×

Again,

)2(.0)(

.010loglog)10

log(

Labcdpm

pabcdmNN p

p

+−=

−+=−=

Thus in both the cases, mantissa remains same, only the

characteristic changes.

Exercise 12: If the mantissa of log 2 is 0.3010 and mantissa of

log23 is 0.3617, find the following logarithms. (You should find these

mantissa values using the standard Log-table.)

(i) log 2 (ii) log 200 (iii) log 0.0002.

Solution: (i) log 2 = 0 + 0.3010 = 0.3010.

(ii) log 200 = 2 + 0.3010 = 2.3010.

(iii) log 0.0002 = - 4 + 0.3010 and we denote it by 3010.4 .

CHECK YOUR PROGRESS

Q 4: Fill up the blanks.

(i) Characteristic of the logarithm of 0.00023

is .............

(ii) If log 32 = 1.5051, then log 32000 = ................

(iii) If log 0.000672 = 8274.4 , that is, -4 + 0.8274, then log

6720 = .........


Unit 6 Logarithm

Q 5: If the mantissa of log 2 is 0.3010 and mantissa of log23 is

0.3617, find the following logarithms: (i) log 230 (ii) log 46

6.3.3 Antilogarithm

In many cases we require to find the number of which

logarithm is given. If mx =log , then x is the antilogarithm of m and

we write .log xmanti = In the antilog-table, the number

corresponding to the mantissa of the logarithm of a number is given.

The process of finding the number from the antilog-table

corresponding to the given mantissa is similar to that of finding the

mantissa from the log-table.

6.3.4 Application of Logarithms

Exercise 13: Find the number of digits of 525, if log 2 = 0.3010.

Solution: First we find the characteristic of log 525. Now,

.475.17)3010.01(25

)2log10(log252

10log255log255log 25

=−×=

−×===

Hence the characteristic is 17 and this implies that the number of

digits of 525 is 17 + 1 = 18.

Exercise 14: Find the compound interest on Rs.5000.00 for 2 years

and 6 months at 6% compound interest per annum, interest being

compounded half-yearly. Given 0128.2103log,3010.02log == and

.7630.0794.5log =Solution: We know that the formula for finding amount (A) on a given

principal (P) at r% compound interest per annum for n years is given

by

nr

PA

+×=100

1 and for half-yearly calculation of interest the

formula is:

nr

PA

2

10021

+×= .

Here, P = Rs.5000.00, r = 6, n = 2.5 years and hence


Logarithm Unit 6

5.22

1002

615000

×

+×=A

.7630.30640.03)3010.01(

0128.051000log5log)03.1log(55000loglog

=++−=×++=×+=⇒ A

57947630.3log ==⇒ antiA (Find this value from antilog-table).

Thus the compound interest = Rs.5794.00- Rs.5000.00 = Rs. 794.00.

6.4 LET US SUM UP

We defined the real number x as the logarithm of ax, where a was a

positive real number other than 1. We then derived several properties of

logarithms. We have learnt about logarithms with base 10 known as common

logarithms. Common logarithm of a number greater than or equal to 1 has

two parts: the first part is the integral part and the second part is the decimal

part. The integral part is called the Characteristic and the decimal part is

called the Mantissa . Characteristic of a number can be found easily, but

the one requires log-table to find Mantissa.

6.5 FURTHER READING

1. Khanna V. K., Zameeruddin Qazi & Bhambri S.K.(1995). Business

Mathematics, New Delhi, Vikas Publishing House Pvt Ltd. .

2. Hazarika P.L. Business Mathematics, New Delhi. S.Chand & Co.

6.6 ANSWERS TO CHECK YOUR PROGRESS

Ans to Q No 1:

(i) T, (ii) F, (iii) T, (iv) F, (v) T.


Unit 6 Logarithm

Ans to Q No 2: (ii) xa < .

Ans to Q No 3: (i) 1, (ii) 2, (iii) 3.9404, (iv) 1−n

n, (v)

2

1.

Ans to Q No 4: (i) -4, (ii) 4.5051, (iii) 3.8274.

Ans to Q No 5: (i) =2.3617, (ii) 1.6627.

6.7 MODEL QUESTIONS

1. If zyx dcba === , prove that zyxabcda

1111)(log +++= .

2. Solve 2log

)1log( =+x

x.

3. Prove that log52 is irrational.

4. Find the values of )641(log

2 and 81logloglog323 .

5. If 9405.072.8log10 = , find the value of .87200log10

6. Solve

(i) 2)13(log)12(log 22 =−−+ xx

(ii) .1)1(log)23(log 1010 =−−+ xx

(iii) (log a)2 – (log b)2 = (l2og ab). Log

(iv) – 2log 0

(v) log2+ +7 log

7. If a2x-3. b2x = a6-x. b5x, prove that 3log a= x log b

a

8. Show that, x log y- logz. y logz-iogx. z log x- logy =1.

9. Find x if + +

10. Simplify .

*** ***** ***


Logarithm Unit 6

UNIT 7: BINOMIAL THEOREM

UNIT STRUCTURE


7.2 Introduction

7.3 Binomial Theorem

7.3.1 Pre-Requisites

7.3.2 Statement of Binomial Theorem

7.3.3 Some Particular Forms of the Binomial Theorem

7.3.4 Middle Term (Middle Terms)

7.4 Properties of Binomial Coefficients

7.4.1 Greatest Coefficients

7.4.2 Greatest Term

7.5 Let Us Sum Up


7.7 Answers To Check Your Progress

7.8 Model Questions



l discuss Binomial Theorem and Binomial coefficients.

l discuss the properties of Binomial coefficients.

l explain the middle terms and equidistant terms and how to find

them in a given binomial expansion.

l determine greatest coefficient and greatest term in a given

binomial expansion.

7.2 INTRODUCTION

We are already familiar with algebraic expressions which are very

common in mathematics. An algebraic expression having two terms is called

a binomial expression. One can easily find the square or cube of binomial

expressions like ba+ or . But it will be difficult to expand

(

by


repeated multiplication. Binomial theorem gives an easier way to

expand nba )( + , where n is an integer or a rational number. In this unit, we

shall consider

n

to be always a positive integer. In mathematics, the binomial

coefficient

knC

is the coefficient of the x k term in the polynomial expansion

of the binomial power (1 + x) n. In combinatorics, knC is interpreted as the

number of k-element subsets of an n-element set, that is the number of

ways that k things can be ‘chosen’ from a set of n things.

7.3 BINOMIAL THEOREM

We know that for two real numbers a and b ,

L,464

)33)(()(

33)(

2)(

)(

1)(

432234

32234

32233

222

111

0

babbabaa

babbaababa

babbaaba

bababa

baba

ba

++++=++++=+

+++=+++=+

+=+=+

We observe that the powers of a go on decreasing by 1 and the powers of

b

go on increasing by 1 in the successive terms. We arrange the co-

efficients in the above expressions as follows, and the diagram is called

the Pascal’s triangle.

Index Co-efficients

0 1

1 1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

Figure: 1


Binomial Theorem Unit 7

We have seen that for n= 0, 1, 2, 3, 4 there are terms in the expansion

of nba )( + . Also, the co-efficients of na and nb are equal to 1 for all n . The

intermediate number 2 in 3rd row in Fig-1 is the sum of the two numbers 1

and 1 of the preceding row which has been shown by the triangle. Thus

using Fig-1, we can find the co-efficients in the expansion of

(

as

follows:

1 4 6 4 1

1 5 10 10 5 1

Hence, 543223455 510105)( babbababaaba +++++=+

The expansion with the help of Pascal's triangle becomes lengthy. Now, we

discuss about Binomial theorem which makes the expansion easier.

7.3.1 Pre-requisites

(i)

.))(21)(21(

)1(21

)!(!

!

rnr

nn

rnr

nC r

n

−×××××××−×××=

−=

LL

L

Hence,

.1,,2

)1(,,1 210 =−=== n

nnnn Cnn

CnCC L

(ii) rnn

rn CC −= , (iii)

rn

rn

rn CCC 1

1+

− =+ .

7.3.2 Statement of Binomial Theorem

If n is a positive integer, then for all values of a and x,

nn

nnn

nrrnr

nnnnnn xCaxCxaCxaCaCxa ++++++=+ −−

−− 11

110)( LL

Proof : For 1=n , L.H.S. = xaxa +=+ 1)( and R.H.S. = xa + .

Hence the theorem is true for .

For 2=n , L.H.S. = 222 2)( xaxaxa ++=+ = R.H.S. Hence the

theorem is also true for 2=n . From this we can assume that the

theorem may be true for mn = . Then,


Unit 7 Binomial Theorem

.

)(1

1

110

mm

mmm

m

rrmr

mmmmmm

xCaxC

xaCxaCaCxa

++

++++=+−

−

−−LL

Multiplying both sides by (a+x), we get

]1,1[,

)()(

][

][

]

)[())(()(

11

01

01

11

1011

01

112110

11

10

11

110

1

====++

+++++=

++++++

+++++=

++++

+++=++=+

++++

++

−−

++

++−−

−++

−−

−

−+

mm

mmmmm

mm

rrmr

mr

mmmmmm

mm

mrrmr

mmmmm

mm

mrrmr

mmmmm

mm

mmm

mrrmr

m

mmmmmm

CCCCxC

xaCCxaCCaC

xCxaCxaCxaC

axCxaCxaCaC

xCaxCxaC

xaCaCxaxaxaxa

QL

L

LL

LL

L

L

].[, 11

11

1

)1(11)1(1

110

1

rn

rn

rnm

mm

rrmr

mmmmm

CCCxC

xaCxaCaC+

−+

++

−++−++++

=++

++++=

Q

LL

This shows that we can get the expansion of 1)( ++ mxa by writing

1+m for n in the Binomial theorem. Therefore the theorem is true

for 1+= mn

if we assume that the theorem is true for mn = . But

the theorem is true for 1== mn and 2== mn . Hence the theorem

is true for 312 =+=n . Since it is true for 3, it is also true for 4, and

so on. From this we conclude that the theorem is true for any positive

integer n . This completes the proof of the theorem.

Remark : In the expansion of

nxa )( +

, we see that

1. total number of terms is 1+n ,2. degree of each term is n , and

3. the coefficients

nnnn CCC ,,, 10 L

are called Binomial

coefficients and they are written in short as nCCC ,,, 10 L

General Term : In the expansion of nxa )( + , rrnr

n xaC − occurs as

)1( +r th term. This term is known as general term and it is denoted

by 1+rT . Hence,

=+1rT rrnr

n xaC −

Putting L,2,1,0=r we get T1, T2, T3, … etc.

Equidist ant Terms : In the expansion of nxa )( + , the )1( +r th term

from the beginning is rrnr

n xaC − . As the )1( +r th term from the



end is the )1( +− rn th term from the beginning and this term is

rnrnnrn

n xaC −−−−

)( which is nothing but rnrrn

n xaC −− .

But we know that rnn

rn CC −= . Hence the binomial coefficients of

the terms equidistant from the beginning and from the end are equal.

7.3.3 Some Particular Forms of the Binomial theorem

(1) Writing x− for in

(

and its expansion, we find

.)1(

)1(

)()(

)()()(

222

11

11

110

nn

rrnr

nrnnnnn

nn

nnn

n

rrnr

nnnnnn

x

xaCxaCxaCa

xCxaC

xaCxaCaCxa

−++

−+−+−=

−+−+

+−++−+=−

−−−

−−

−−

L

L

LL

Hence the general term: =+1rT rrnr

nr xaC −− )1( .

(2) nn

nnn

nrr

nnnn xCxCxCxCCx ++++++=+ −−

1110)1( LL

nxx

nnnx ++−++= L

2

!2

)1(1

Hence the general term: =+1rT rr

n xC .

(3) nn

nnrr

nrnnn xCxCxCCx )1()1()1( 10 −++−++−=− LL

nn xx

nnnx )1(

!2

)1(1 2 −++−+−= L

Hence the general term: =+1rT rr

nr xC)1(− .

Exercise 3.1: Expand 6

3

3

+ x

x.

Solution:

.72927

2

3

520

135486729

33

3

3

3

3

3

3

3

3

33

3

3

642

246

6

66

5

56

42

46

33

36

24

26

5

16

66

xxx

xxx

xC

x

xC

x

xC

x

xC

x

xC

x

xC

x

x

x

++++++=

+

+

+

+

+

+

=

+



Exercise 3.2: Find the 9th term in the expansion of 12)32( yx − .

Solution: Here n =12 and the 9th term is:

== +189 TT 8488444

1288128

12 5196312032)3()2( yxyxCyxC ==−−

Exercise 3.3: Find the co-efficient of 5x in the expansion of 8)32( x− .

Solution: Here 8=n and the )1( +r th term is:

=+1rT rrr

rrrr xCxC −− −=− 8888 2)3()3(2

If this term has 5x as a factor, then r must be 5. Thus, the co-

efficient of

5x

in the expansion of 8)32( x− is

][10886432

321

678)3(2 3

85

8535585

8 CCC =−=××××××−=−−

Q

Exercise 3.4: If n is a natural number, using binomial theorem prove

that

134 −− nn

is divisible by 9.

Solution : If 1=n , then 134 −− nn = 0 and hence divisible by 9. For

2≥n , using binomial theorem, we get

[ ]kn

CCn

CCCnnn

nnn

nnnnn

931

33331

33331)31(42

322

11

221

++=++×+×++=

+×++×+×+=+=−

−−

L

L

where k is an integer. This implies that 134 −− nn is divisible by 9.

7.3.4 Middle term (Middle terms)

If the number of terms in the expansion of nxa )( + is odd

then there is exactly one middle term, but the expansion will have

two middle terms if the number of terms is even.

Case 1: When n is even, say

mn 2=

, then the number of terms is

equal to

12 +m

which is odd. Hence the middle term is the )1( +m th

term, that is, )12

( +n

th term.

Case 2: If n is odd, say 12 += mn , then the number of terms is

equal to 2m +2 which is even. Hence there are two middle terms

and these are )1( +m th term and (m+2)th term. Putting

2

1−= nm ,



we get the middle terms as

2

1+nth term and

2

3+nth term.

Exercise 3.5: Find the middle term in the expansion of 8

2

1

−x

x .

Solution: Here n =8 and hence there is one middle term in the

expansion which is 5T . Hence

.8

35)

2

1( 44

48

5 =−=x

xCT

Exercise 3.6: Find the middle terms in the expansion of 7

2 12

+x

x .

Solution: Here n =7 and hence there are two middle terms which

are 4T and 5T . Now,

.2801

)2(,5601

)2( 24

324

75

53

423

74 x

xxCTx

xxCT =

==

=

CHECK YOUR PROGRESS

Q 1: Fill up the blanks.

i. The number of terms in the expansion ofnxa )( + = ................

ii. The 6th term in the expansion of 7)1( −x = ................

iii. The coefficient of 6x in 9)2( +x = ................

iv. The coefficient of middle term in the expansion of 10)1( x+ is

................

Q 2: Write True (T) or False (F):

i. If n is even, then there is one middle term in

(

. (T / F)

ii. If n is odd, then there are two middle terms in nxa )( + . (T / F)

iii. If there is a term in n

xx

− 1 that is independent of x, then n is

even. (T / F)



7.4 PROPERTIES OF BINOMIAL COEFFICIENTS

Following are some basic properties of binomial coefficients.

(1) Sum of the binomial coefficients is equal to n2 , that is,

.2110n

nnr CCCCC =++++++ −LL

Proof: Let us consider the expansion:

nn

nn

rr

n xCxCxCxCCx ++++++=+ −−

1110)1( LL .

Putting 1 for x in both sides, we get

nnr

n CCCCC ++++++= −1102 LL

Example: The sum of the binomial coefficients in the expansion of 6)1( x+

is .6426 =(2) The sum of the binomial coefficients of the odd terms is equal to the

sum of the binomial coefficients of the even terms and each is equal to

.2 1−n

Proof: We have,

nn

nn

rr

n xCxCxCxCCx ++++++=+ −−

1110)1( LL

Putting –1 for x in both sides, we get

)()(

)1(0

531420

210

LL

L

+++−+++=−+−+−=

CCCCCC

xCCCC nn

nnnnn

LL +++=+++⇒ 531420 CCCCCC

Suppose that kCCCCCC =+++=+++ LL 531420 . Then,

)()(2 531420 LL +++++++= CCCCCCk

1210 222 −=⇒=++++=⇒ nn

n kCCCCk L , [Using property (1)].

Hence, .2 1531420

−=+++=+++ nCCCCCC LL

Exercise 3.7: Find the value of 45

25

05 CCC ++ .

Solution: 45

25

05 CCC ++ is the sum of the binomial coefficients of the even

terms of the expansion of 5)1( x+ and hence equal to 1624 = .

Exercise 3.8: Prove that

(1) .232 1321

−×=×++×+×+ nn nCnCCC L

(2)

).12(1

1

13211210 −

+=

+++++ +nn

nn

CCCCL



Solution: (1) L.H.S. =

1321

)2)(1(3

21

)1(2 ×++

××−−×+

×−×+ n

nnnnnn L

{ } 11

12

11

10

1 2

1!2

)2)(1()1(1

121

)2)(1()1(

−−

−−−− ×=++++×=

++−−+−+×=

×++×

−−+−×+=

nn

nnnn nCCCCn

nnnn

nnnn

nnn

L

L

L

= R.H.S.

(2) L.H.S. =

1321210

+++++

n

CCCC nL

{ }

{ }

( ) ].1[,121

11

11

1

1!3

)1()1(

!2

)1()1(

1

11

1

32

)1(

21

011

01

11

31

21

11

01

11

31

21

11

=−+

=

−+++++×+

=

++++×+

=

++−+++++×

+=

+++

×−×++=

++

++

+++++

+++++

Cn

CCCCCCn

CCCCn

nnnnnn

n

n

nnn

nn

nn

nnnnn

nnnnn

Q

L

L

L

L

= R.H.S. Hence proved.

7.4.1 Greatest Coefficients

The binomial coefficients in the expansion of nxa )( + are r

n C where

.,,1,0 nr L=

Case 1: If n is even, then r

n C has the greatest value for

2

nr = .

Hence, if n is even, then the greatest coefficient is 2

nn C .

Case 2: If n is odd, then r

n C has the greatest value for

2

1−= nr or

for

2

1+= nr . Hence, if n is odd, then the greatest coefficient is



21̀−n

n C or 2

1̀+nn C , both being equal.

Exercise 3.9: Find the greatest coefficient in the expansion of

112 )43( yx + .

Solution: Here 11=n and this is odd. Hence the greatest coefficient

is 511

2111

11 CC =− or 611

2111

11 CC =+ and both are equal.

Now,

462237112345

7891011

!6!5

!115

11 =×××=×××

××××=×

=C , which

is the required greatest coefficient.

7.4.2 Greatest T erm

In the expansion of nxa )( + where 0, >xa , we have

)1.3(1

111

1LL

a

x

r

rn

xaC

xaC

T

Trrn

rn

rrnr

n

r

r ×+−== −+−−

−+

From equation (3.1), rr TT <=>+1 if

11 <=>×+−

a

x

r

rn,

that is, if

a

x

ra

xn +<=>+1

)1(,

that is, if )()1( xarxn +<=>+ ,

that is, if

xa

xnr

++>=< )1(

.

Thus,

rr TT >+1 when

xa

xnr

++< )1(

,

rr TT =+1 when

xa

xnr

++= )1(

and rr TT <+1 when

xa

xnr

++> )1(

.

This shows that the terms go on increasing till

xa

xnr

++< )1(

and

start decreasing when r exceeds

xa

xn

++ )1(

.



Case 1: If

mxa

xn =++ )1(

, an integer, then

rr TT >+1 for ;1,,2,1 −= mr L

rr TT =+1 for ;mr =

rr TT <+1 for .mr >

Hence the greatest terms are mT and 1+mT both being the same.

Case 2: If

fkxa

xn +=++ )1(

, where k is an integer and f is a positive

proper fraction, then

rr TT >+1 for ;,,2,1 kr L=

rr TT <+1 for .,,2,1 nkkr L++=

Hence the greatest term is 1+kT .

Remark: If 0=k , the 1T , that is, the 1st term is the greatest term.

(2) If

nxa

xn >++ )1(

, that is, nax > , then 1+nT is the greatest term which is

nothing but the last term.

(3) Since the corresponding terms in the expansions of nxa )( + and nxa )( −

are numerically equal, so the numerically greatest term in the expansion of

nxa )( + is the same as that of nxa )( − .

Exercise 3.10: Which term is the greatest in the expansion of 12)32( x+

when

6

5=x ?

Solution: We have, 12

1212

2

312)32(

+×=+ xx .

In the expansion of 12

2

31

+ x, r th term

1

112

2

3−

−

=r

rr

xCT and the r +1)th

term r

rr

xCT

=+ 2

3121

r

xrx

rr

rrx

C

C

T

T

r

r

r

r

2

3)13(

2

3

)!12(!!12

)!112()!1(!12

2

3

112

121 ×−

=×−××

+−×−×=×=∴

−

+



When

6

5=x , we get

r

r

T

T

r

r

4

)13(51 −×=+. This implies that

rr TT ≥+1 if

1

4

)13(5 ≥−×r

r, that is, if

9

27

9

65 +=≤r .

Thus, in this problem, k =7 and

9

2=f , which is a positive proper fraction.

Hence the greatest term in the expansion of 12

2

31

+ xis

7

712

7

712

8 4

5

2

3

×=

×= Cx

CT when

6

5=x . This implies that the greatest

term in the expansion of 12)32( x+ when

6

5=x is equal to

7

71212

4

52

×× C .

CHECK YOUR PROGRESS

Q 3: Write True (T) or False (F):

(i) If Nm∈ , then sum of the binomial coefficients

in the expansion of mx)1( + is m2 . (T / F)

(ii) .6477

57

37

17 =+++ CCCC (T / F)

(iii) .12078

58

38 =++ CCC (T / F)

(iv) If ,)1( 22

2210

2 nn

n xaxaxaaxx ++++=++ L then

nnaaaa 2

2210 3=++++ L . (T / F)

(v) If ,)32( 22

2210

2 nn

n xaxaxaaxx ++++=+− L then

02210 =+−+− naaaa L (T / F)

(vi) If n is even, the greatest coefficient is

12

+n

nC . (T / F)

(vii) If n is odd, the greatest coefficient is

2

1+nnC

. (T / F)

Q 4: Fill up the blanks:

(i) In the expansion of 50)1( x+ , the sum of the coefficients of



odd powers of x is ........................

(ii) In the expansion of 100)1( x+ , the sum of the coefficients of

even powers of x is .......................

(iii) The greatest coefficient in the expansion of 16)1( x+ is equal to

..........................

(iv) In the expansion of nxa )( + where 0, >xa , the terms rT go

on increasing till ......................

(v) In the expansion of nxa )( + where 0, >xa , the terms rT go

on decreasing when r exceeds .....................

7.5 LET US SUM UP

1. We have learnt about the Binomial theorem on the expansion of

nxa )( + . Without writing the complete expansion of nxa )( + , we know

how to get any term of the expansion using the general term:

=+1rT rrnr

n xaC − .

2. We have also learnt about middle terms. If n is even, then there is one

middle term and if n is odd, then there are two middle terms in the

expansion of nxa )( + . If n is even, then

+12

nth term is the middle

term. If n is odd, then

+−1

2

1nth and

+−2

2

1nth terms are the

two middle terms.

3. We have discussed about binomial coefficients and also we derived

certain nice properties of binomial coefficients.

4. If we consider the terms equidistant from the beginning and from the

end, then we have noticed that the corresponding binomial coefficients

are same.

5. If n is even, then

2n

n C is the greatest binomial coefficient in the



expansion of nxa )( + . Similarly, if n is odd, then the greatest coefficient

in the expansion of nxa )( + is 2

1̀−nn C or

21̀+n

n C , both being equal.

6. Finally we have learnt about the greatest term in the expansion of

nxa )( + where 0, >xa . If

mxa

xn =++ )1(

is an integer, then the

greatest terms are mT and

1+mT both being the same. Again if

fk

xa

xn +=++ )1(

, where k is an integer and f is a positive proper

fraction, then the greatest term is 1+kT .

7.6 FURTHER READINGS

1. Business Mathematics, Qazi Zameeruddin, V K Khanna, and S K

Bhambri, Vikas Publishing House Pvt Ltd, Second Reprint, 1995.

2. Business Mathematics, Dr. P.L. Hazarika, S.Chand.

7.7 ANSWERS TO CHECK YOURPROGRESS

Ans to Q No 1: (i) 1+n (ii) 221x− (iii) 672 (iv) 252

Ans to Q No 2: T, T, T.

Ans to Q No 3: T, T, T, F, F, F, T.

Ans to Q No 4: (i) 492 (ii) 992 (iii) 816C

(iv)

xa

xnr

++< )1(

(v)

xa

xn

++ )1(



7.8 MODEL QUESTIONS

1. Find the value of the term which is independent of x in the expansion

of 9

23

12

+x

x

2. Using Binomial theorem find the value of 5)01.1( .

3. Using Binomial theorem prove that 1723 −− nn is divisible by 49. [Hint:

Expand nnn )71(823 +== ]

4. Expand 6

2 3

+x

x by using binomial theorem and write down the

general term.

5. Show that the coefficient of the middle term in the expansion of nx 2)1( +

is equal to the sum of the coefficients of two middle terms in the

expansion of 12)1( −+ nx .

6. Find the coefficient of 9x in the expansion of 20

2 12

+x

x .

7. Find the coefficient of 8x in the expansion of 82 )1)(432( xxx −−+ .

8. Find the coefficient of 17−x in the expansion of 15

34 1

−x

x .

9. Find the middle term(s) in the expansion of 8

12

−

yx and

211

+x

x .

10. If three successive coefficients in the expansion of nx)1( + are 220,

495 and 792 respectively, find the value of n.

11. Using binomial theorem find the value of .89

69

49

29

09 CCCCC ++++

12. If nn

n xCxCxCCx ++++=+ L2

210)1( , prove that

(i) .2)2()1(32 1210

−×+=+++++ nn nCnCCC L

(ii)

.1

1

1)1(

3221

0 +=

+−+−+−

nn

CCCC nn

L



13. Prove that ( ) nnnn

n CCCCCCC 22

12

022

210 +++=++++ LL .

14. In the expansion of 10

2

−x

ax , the term free from x is 405. Find the

value of a.

15. Find the greatest term in the expansion of 6

32

+a

x

x

awhen

2

1=a

and

3

1=x .

*** ***** ***



UNIT 8 : MATRICES

UNIT STRUCTURE

8.1 Learning objectives

8.2 Introduction

8.3 Definition of Matrix

8.4 Types of Matrix

8.5 Equality of Matrices

8.6 Addition and Subtraction of Matrices

8.7 Scalar multiplication of Matrix

8.8 Multiplication of Matrices

8.9 Transpose of a Matrix

8.10 Symmetric and Skew-Symmetric Matrix

8.11 Adjoint and Inverse of Matrices and its existence

8.12 Rank of a Matrix

8.13 Solution of system of linear equations by matrix method

8.14 Solution of system of linear equations by Crammers rule.

8.15 Answers to check your progress


8.17 Model Questions


After going through this unit, you will be able to :

l understand the definition of a Matrix

l describe different types of Matrix

l learn matrix operations e.g. addition, subtraction and

multiplication

l learn about adjoint, inverse and rank of Matrices

l solve system of linear equations by matrix method and crammer’s

rule


8.2 INTRODUCTION

Matrix is one of the most powerful tools of modern mathematics.

Matrix notations and operations are used in Computer graphics,

programming and implementation of electronic spreadsheet programmes.

Matrix were introduced by the English Mathematician Arthur Cayley in

1858.Matrices are initially connected with linear transformation. Matrix theory

now occupies important position in Physics, Economics, Statistics,

Engineering etc.

8.3 DEFINITION OF MATRIX

A set of mn numbers (real or complex) arranged in the form of a

rectangular array having m rows and n columns is called an mxn matrix (to

be read as 'm' by 'n' matrix).

An m x n matrix is usually written as

=

mnm2m1

2n2221

1n1211

a...aa

............

a...aa

a...aa

A

The compact representation of the above matrix is

A =[aij], i = 1, 2, .... m, j = 1, 2, .... n.

or, simply, A =[aij]mxn. Sometimes, the braket [ ] can be replaced by ( ) .

The general element of the matrix A is ija which belongs to the i-th

row and j-th column. ija is sometimes denoted by (i,j)th element of the

matrix. Here i, the first suffix denotes the number of row and j, the second

suffix denotes the number of column in which the element ija occurs.


Matrices Unit 8

Note :

8.4 TYPES OF MATRIX

There are different types of matrices. We will discuss one by one.

(i) Row Matrix : Any 1x n matrix which has only one row and n column

is called a row matrix.

For example,

A = [2, 5,- 3, 0, 3 ] is a row matrix order 1 × 5.

In general, B = [ ija ] 1× n is a row matrix of order 1x n

(ii)Column Matrix : Any m x1 matrix which has m rows and only one

column is called a column matrix.

For example,

−=

2

3

0

6

B is a column matrix of order 4×1

1. A matrix having m-rows and n-

columns is called a matrix of order

mn (read as 'm' by 'n') or simply

m x n matrix.

2. In a matrix, the number of rows

need not be equal to the number

of columns.

3. A matrix is denoted by the capital

letters A, B, C… etc. whereas any

elements of a matrix is denoted by

small letters such as

⋅⋅⋅⋅⋅⋅,, , ijijij cba etc.

Illustrative example :

32325 3 4

0 2 1 or

53 4

02 1A

××

=

is a 2 x3 matrix i.e. it has 2 rows

and 3 columns. For A,

12a = (1,2)th element = 2



33

2i2 3

1 0 4 31

2 i 1

B

×

−

+

=

is a 3x3 matrix. For B,

11b = 1+ i, 31b = 3 , 22b =0 etc.


MatricesUnit 8

In general, is a column matrix of order m×1.

(iii) Square Matrix : A matrix whose numbers of rows is equal to the number

of columns is called a square matrix.

For the matrix A= [ ija ] m× n , if m = n, then the matrix A is said to be

square matrix of order m.

Example:

(a)

=

30

01A is a square matrix of order 2.

(b)

=105

132

210

B is a square matrix of order 3

In a square matrix A = [ ija ] m×m , the elements a11, a22, a33 ...... are

called the Diagonal Elements and the line along which they lie is called the

Principal Diagonal of the matrix.

(iv)Diagonal Matrix : A square matrix A is said to be a diagonal matrix if all

its non-diagonal elements be zero.

Examples:

100

010

003

B,20

01A

=

=

-

are diagonal matrices of order 2 and 3 respectively.

(v)Scalar Matrix : A diagonal matrix (i.e. all non-diagonal elements

being zero) where all the diagonal elements are equal is called a

Scalar Matrix.

Examples :

300

030

003-

B,20

02A

−−=

=

are two scalar matrices of order 2 and 3 respectively.


Matrices Unit 8

Thus, the square matrix is a scalar matrix if

ija = 0, when i ≠ j

ija = k(say) when i = j

(vi) Unit (or Identity) Matrix : A square matrix in which the diagonal

elements are unity and non-diagonal elements are all zero is

called a unit (or Identity)matrix.

Examples :

100

010

001

I,10

01I 32

=

=

Unit matrices are denoted by I.

I2 and I3 are Unit matrices of order 2 and 3 respectively.

Thus, the square matrix is a unit matrix if

ija = 0 when i ≠ j

ija = 1 when i = j

(vii) Null Matrix or Zero Matrix : A matrix of order m x n in which all the

elements are zero is called a null matrix (a zero matrix). It is denoted by O.

Thus,

000

000

000

000

,00

00

are all zero matrices of order 2 x2 and 4x3

respectively.

(viii) Upper Triangular Matrix and Lower T riangular Matrix: Let A be a

square matrix of order n.

=

nnn3n2n1

3n333231

2n232221

1n131211

a ... a aa

...........................

a ... a aa

a ... a aa

a ... a aa

A


MatricesUnit 8

The diagonal elements of A are a11, a12 ,….., ann i.e. ija for i= j.

All these elements of A above the principal diagonal are a12

a13……. a1n, a23……. a2n............ etc. i.e. aij for i < j

All these elements of A below the principal diagonal are a21

, a31

,

a32, …….an1, an2, ……. i.e. aij for i > j.

If all aij for i > j be zeros in a square matrix A, then A is called Upper Triangular

matrix.

Example :

5000

2100

1420

6213

A

−=

is Upper Triangular Matrix

If all aij for i < j be zeros is a square matrix A, the A is called Lower T riangular

Matrix.

Example :

=

5 3 1 2-

0 4 1- 2

0 0 2 1

0 0 0 3

Ais Lower Triangular Matrix

Remarks : A diagonal matrix is both upper and lower triangular .

8.5 EQUALITY OF MATRICES

Two matrices A = [aij] and B = [bij] are said to be equal if

(i) they are of the same size or order

(ii) each element of one matrix is equal to the corresponding element

of the other matrix i.e. aij= bij for all i and j.

If A and B are equal, then we write A = B.If A and B are not equal,

then we write A ≠ B.

Example :

1.2222

42

31B,

4 2

3 1 A If

××

=

= then A = B .


Matrices Unit 8

2. 2222

22

22

164

91B,

4 2

3 1 A If

××

=

= then A = B.

3. 3222

642

531B,

4 2

3 1 A If

××

=

= then A ≠ B.

4.

2323 34

21

03-

c z

by

ax

If

××

=

then, x = - 3, y = 1, z = 4, a = 0, b = 2, c = √3

Remarks: The inequality of two matrices arised due to the following

reasons :

(i) This orders may not be equal.

(ii) Elements in the corresponding places may not be equal.

CHECK YOUR PROGRESS 1

Q 1: If a matrix has 10 elements what are the

possible orders it can have?

Q 2: Find the values of a, b, c, d when

−=

++

253

114

d4c2b-a

d-5c b2a

Q 3: Are the following matrices equal where

=

=000

501

342

B ,

600

501

342

A

8.6 ADDITION AND SUBTRACTION OF MATRICES

Let A = [aij]mxn and B = [bij]mxn be two matrices of same order mxn.

Then their sum(or difference) denoted by A+B (or A-B) is defined as

another matrix C= [cij] of the order m xn such that any element of C is

the sum (or difference) of the corresponding element of A and B.

i.e. cij = aij+bij for all i, j

Thus, C =[cij] = [a

ij+b

ij]mxn


MatricesUnit 8

Example : If

−−

=

=

12

52B,

03

15A

Then, C = A+B =

−=

+−+−+

11

47

102)(3

5125

Similarly, D=A – B =

−=

−−−−−−

15

63

102)(3

5)(125

Remark : Two matrices A and B are said to be conformable for addition

or subtraction if they are of the same order.

Negative of a Matrix : If A be a given matrix, then the negative of A denoted

by – A is the matrix whose elements are the negative of the corresponding

elements of A.

Example:

=−

−

−=

6- 5

4- 0

1 2-

A then

65

40

12

A If

Properties of Matrix Addition :

The addition of matrices satisfies the following properties :

(i) Commut ative Law : If A = [aij] and B = [bij] are matrices of the same

order, say m xn, then

A + B = B + A

Proof :

A+B = [aij] + [bij]

= [aij + bij]

= [bij + aij] (Q addition of numbers is commutative)

= [bij] + [aij]

= B + A

(ii)Associative Law : If A = [aij], B=[bij] and C = [cij] are matrices of the same

order mxn, then (A+B) + C = A + (B+C)

Proof: (A+B) + C = ([aij] + [bij]) + [cij]

= [aij + bij] + [cij]

= [(aij + bij) + cij]


Matrices Unit 8

= [aij] + [(bij + cij)]

= [aij] + ([bij] + [cij])

= A + (B+C)

i.e. (i,j)th element of (A+B)+C = (i,j)th element of A+(B+C)

(iii)Existence of Additive Identity : If O be the zero matrix of order m×n

and A=[aij] be an m×n matrix, then A+ O = O + A = A

Proof :

A + O = [aij ] + [0]

= [ ]0+ija = [aij] = A

and O + A = [0] + [aij]

= [0+aij]

= [aij]

= A

∴ A + O = O + A = A

Hence, the zero matrix of the type m×n plays the role of additive

identity.

(iv)Existence of Additive inverse : If A = [aij] be a matrix of order m×n,

then we have another matrix - A = [- aij] such that

A+(–A) = (–A) + A = 0

Proof :

A + (-A) = [aij] + [-aij]

= [aij- a

ij]

= [0] = 0

(–A) + A = [-aij] + [aij]

= [- aij + a

ij]

= [0] = 0

Hence, A + (–A) = (–A) + A = 0

Here, –A acts as the additive inverse of A.

(v)Subtraction of two Matrices : If A and B be two matrices of order m×n,

then we define A – B = A + (–B).

In other words, the difference A – B is obtained by subtracting from

each element of A the corresponding element of B.

(vi)Cancellation Laws for addition of Matrices : If A, B, C are three160 Business Mathematics (Block 2)

MatricesUnit 8

matrices of the same order say m×n, then

A + B = A + C ⇒ B = C [Left Cancellation law]

B + A = C + A ⇒ B = C (Right Cancellation Law)

Proof : We have

A + B = A + C

⇒ –A + (A+B) = –A + (A+C) [Adding – A to both sides ]

⇒ (–A+A) + B = (–A+A) + C [Qmatrix addition is associative]

⇒ O + B = O + C

⇒ B = C

Similarly :

B + A = C + A

⇒ (B+A) + (–A) = (C+A) + (–A)

⇒ B + (A – A) =C + (A – A)

⇒ B + O = C + O

⇒ B = C

Example:(1) If

=

−=

26

13Band

13

05 A

, find A + B and A – B

Solution :

+

=+

26

13

13

05BA

=

39

18

=

39

18

−

=

26

13

13

05 B-A

=

−−−−2163

1035

=

−−−13

12


Matrices Unit 8

(2) If A =

=

−=

02

53C,

25

30B,

34

01

Show that (A+B) + C = A + (B+C)Solution :

A+B =

−+++

=

−+

2354

3001

25

30

34

01

=

19

31

( )

+

=++

02

53

19

31 CBA

=

++++

0129

5331

=

111

84

Again, B+C =

++

− 02

53

25

30

=

+−+++

0225

5330

=

− 27

83

−+

=++∴

27

83

34

01C)(BA

=

=

−+++

111

84

2374

8031

( ) C)(BACBA ++=++∴

(3) Find the additive inverse of

3251 0

132A

×

−=

Solution:A is a matrix of order 2×3.– A will also be a matrix of order 2×3.


MatricesUnit 8

=−

5- (-1)- 0

1-3-2-A

325- 1 0

1-3-2-

×

=


Q 1: Is it possible to define A+B when–

(i) A has 3 rows ,2 columns and B has 2 rows,

2 columns.

(ii) A has 3 rows ,3 columns and B has 3 rows, 3 columns.

(iii) A and B are square matrices of same order.

Q 2: What is the additive identity of 2x3 matrices ?

8.7 MULTIPLICATION OF A MATRIX BY A SCALAR

Multiplication of a matrix by a scalar is defined as follows:If A = [aij]m×n is a matrix of order m×n and k is a scalar, then

kA = k[aij]m×n = [kaij]m×n

i.e., (i, j)th element of kA is k times the (i,j) th element of A .Example:

If k = 3 and A =

321 0

3 2-

1-2

×

,then

×××−×

−××=

=13 03

332)(3

1)(323

1 0

3 2-

1-2

33A

=

233 0

9 6-

3-6

×

and,


Matrices Unit 8

2321

0

23

1

21

1

121

021

321

2)(21

1)(21

221

1 0

3 2-

1-2

21

A21

×

−

−

=

××

×−×

−××

=

=

PROPERTIES OF MULTIPLICATION OF A MATRIX BY A SCALAR :

If A = [aij] and B = [bij] are two matrices of the same order m x n ; k and

l are two scalars, then

(i) k (A+B) = kA + kB

(ii) (k+l) A = kA + lA

(iii) k (lA) = (kl) A

(iv) (–k) A= – (kA) = k(–A)

Proof of (i) :

Let A = [aij]m×n and B= [bij]m×n

Then k(A+B)=k([aij]+[b

ij])

= k([aij+bij)

=([kaij+kbij])

=[kaij+kb

ij]

= [kaij]+[kbij]

= k[aij]+k[bij]

= kA + kB.

Example :

If A = 2222

27

04B,

81

93

××

=

Then verify that 2(A+B) = 2A + 2BSolution :We have

A + B =

+

27

04

81

93

=

++++2871

0943


MatricesUnit 8

=

108

97

∴2 (A+B) = 2

××××

=

10282

9272

108

97

=

2016

1814

Again, 2A+2B =

+

27

042

81

932

=

××××

+

××××

2272

0242

8212

9232

=

+

4 14

08

162

186

=

++++

416 142

0 18 86

=

20 16

18 14

∴2(A+B) = 2A + 2B verified.

Proof of (ii)

Let A = [ ]ija , then

(k+l) A = (k+l) [ ]ija

=

+

ijla

ijka

= [ ] [ ]ijlaijka +

= [ ] [ ]ijalijak +

= kA + lA

Example :

If k = 3, l = –2 and A =

12

0 1

2 3

Then verify that (k+l) A = kA + lA


Matrices Unit 8

(k+l) A = (3–2)

12

0 1

2 3

= (1)

12

0 1

2 3

Again,

kA + lA = 3A – 2A = 3

−

12

01

23

2

12

0 1

2 3

=

−

24

02

46

36

0 3

6 9

=

−−−

−

2-3 46

002 3

4-66 9

=

12

0 1

2 3

∴(k+l) A = kA + lA. Verified

Proof of (iii) :

Let A = [ ]ija , then

k(lA) = k(l [ ]ija )

= k[l ija ]

= [k(l ija )]

= [(kl) ija ]

= (kl) [ ija ] = (kl) A.

Example : If k=4, l=2 and A =

12

01

23

, then

Verify that k(lA) = (kl)A.


MatricesUnit 8

Solution :

k(lA) = 4(2A) = 4

××××××

12 22

02 12

22 32

=

××××××××××××

124 224

024 124

224 324

=

××××××

18 28

08 18

28 38

=

12

01

23

8

= 8A

= (4×2) A

= (kl)A. verified

Proof of (iv) :

If A = [ ija ] then

(–k) A = [(–k) ija ]

= [–(k ija )]

= –[k ija ] = –(kA)

Also k(–A) = k[– ija ]

= [k.–( ija )]

= –[k ija ]

= –(kA)

∴(–k) A =–(kA) = k(–A)

Example :

==12

01

23

A and 5k If


Matrices Unit 8

Verify that (–k)A = –(kA)= k (–A)Solution :

(–k) A = (–5)

12

01

23

=

×−×−×−×−×−×−

1)(52)(5

0)(51)5)(

2(535)(

= -

××××

××

1)(52)(5

0)(51(5)

2(5)3(5)

= –(kA)

Also, k(–A) = 5

−−−

−−

12

01

23

=

−×−××−×

−×−×

1)(52)(5

051)(5

2)(53)(5

= -

××××××

1525

0515

2535

−=12

01

23

5 = –(kA) Verified.

Example : If A =

=

3 1 0

3- 0 4 Band

241

230, find 3A – 2B.

Solution :

3A - 2B = 3

2 1 0

3- 0 4 2-

241

230

−−

=

420

608

6123

690


MatricesUnit 8

−−−−−−−

=4621203

6)(60980

−=

2103

1298

8.8 MULTIPLICATION OF TWO MATRICES

Two matrices A and B are conformable for the product AB if and only if the

number of columns of A is equal to the number of rows of B.

Let A = [ ija ]m×n and B = [ jkb ]n×p,

Then the product A,B,C is the matrix of order m×p

i.e. AB = C = [ jkc ] m×p where

∑=

+++==n

1jnkin2ki21ki1jkijjk b a........ ......... b a b a b a C

i.e. the (j, k) th element of the matrix C= AB is found by multiplyingthe corresponding element of the j-th row of A and the k-th columnof B and then, adding the product.The rule of multiplication is row by column multiplication.

Example:

If A = 222221

1211

233231

2221

1211

b b

b b Band

a a

a a

a a

××

=

Then AB =

23 22321231 21321131

2222122121221121

2212121121121111

b a b ab a b a

b a b ab a b a

b a b a b a b a

×

++

++++

Note : 1) If the product AB exists, then it is not necessary that the productBA will also exist.

Example : A= [2 3]1×2 and B = 32

321

504

×

Here,

[ ]

=

321

50432AB

= [ ] [ ]196119106038 =+++


Matrices Unit 8

But BA = [ ]32321

504

which does not exist

∴ AB≠BAMatrix multiplication is not commutative

2) In the case when both A and B are square matrices of the sameorder then also both AB and BA are defined but still AB≠BA.

Example : If A =

=

14

53 B and

20

12

Then AB =

14

53

20

12

=

++++

2x10x5 2x40x3

1x12x5 1x42x3

=

28

11

1 0

BA =

20

12

14

53

=

++++

1x24x1 1x04x2

5x23x1 5x03x3

=

68

19

3

It is seen that AB ≠ BAExample :

If A =

=

3 1 2-

0 3 2 Band

01

12

Can you find AB and BA ?

Solution :

Here, A is of order 2×2

B is of order 2×3

∴AB will be order 2×3

Again, B is of order 2×3

A is of order 2×2

∴BA does not exist because the number of column of B is 3 and


MatricesUnit 8

the number of rows of A is 2.

So, they are not conformable for the product BA.

PROPERTIES OF MATRIX MULTIPLICATION :

The multiplication of matrices possesses the following properties:

(1) The multiplication of matrices is not always commutative.

(a) Whenever AB exists, it is not necessary that BA should exist.

(b) Whenever AB and BA both exist, it is not always necessary that

they should be matrices of same order.

(c) Whenever AB and BA exists and are of the same order, it is not

necessary that AB = BA.

(d) In some cases, AB = BA.

(e) Matrix multiplication is associative i.e. A(BC) = (AB)C

where A, B, C are matrices of order m×n, n×p, p×q respectively.

(f) Matrix multiplication is distribution w. r. t. addition of matrix .i.e,

A(B+C) = AB+AC

Where, A, B, C are matrices of order m×n, n×p and n×p

respectively.

(g) For every square matrix A, there exists the identity matrix I of

same order of A such that AI = IA = A.

MATRIX MULTIPLICATION IS ASSOCIATIVE

If A, B, C are matrices of order m×n, n×p, p×q, thenA(BC) = (AB) C

Proof :

Let A= [ ija ] m×n, B = [ jkb ] n×p, C = [ klc ]p×q

Then AB = U = [ iku ] m×p where

∑=

=n

1jjkijik ba u

Also, [ ] qnjlvVBC ×== where ∑=

=n

kkljkjl cbv

1

∴A(BC) = W = [wil]m×q when (i, I)th element of A(BC)


Matrices Unit 8

jl

n

1jijil va W ∑

=

==

∑ ∑= =

=

n

1j

n

1Kkljkij cba

∑ ∑= =

=n

1Kkl

n

1jklij Cba

klik

n

kcu∑=

=1

= (i, j)th element of (AB) C

∴ A(BC) = (AB) C

MULTIPLICATION OF MATRICES IS DISTRIBUTIVE WITH RESPECTTO ADDITION OF MATRICES :

If A, B and C are matrices of order m×n, n×p, n×p respectively,then A(B+C) = AB + AC

Proof :

Let A=[ ija ]m×n , B =[ jkb ]n×p , C=[ jkc ]nxp

We have B+C = [ jkb + jkc ]n×p

∴(i, k)th element of A(B+C) = ∑=

+n

1jjkjkij )c(ba

jk

n

1j

n

1jijjkij cab a∑ ∑

= =

+=

= (i, k)th element of AB + (i, k)th element of AC

= (i, k)th element of AB +AC

Here, A(B+C) = AB + AC

Similarly, we can show that,

(B+C) D = BD + CD

where B, C and D are matrices of suitable types.

Example :

If A = 32120

213B ,

2243

21

×

=

×


MatricesUnit 8

C = 32064

105

×

then show that

A(B+C) = AB + ACSolution :

B+C =

+

064

105

120

213

32

184

318

×

=

A(B+C)

32143384134483

123182114281

32184

318

2243

21

×

×+××+××+××+××+××+×

=×

×

=

=

133540

51716

Again, AB =

120

213

43

21

=

×+××+××+××+××+××+×

142324130433

122122110231

=

10119

453

=

064

105

43

21AC

×+××+××+××+××+××+×

=041364034453

021162014251

=

32431

11213

+

=+

32431

11213

10119

453ACAB

=

133540

51716

AC AB C)A(B +=+∴


Matrices Unit 8

Example :

If A = 2322

41

20

31

B,

213

302

111

××

−=

−

−

22

02

21C

×

=

Find A(BC) and (AB)C and show that A(BC) ≠ (AB) C.Solution :

2223

02

21

41

20

31

BC×

×

−=

=

×+×−×+××+××+×

×+××+×

04212411-

02202210

03212311

=2 7

0 4

27

A(BC) =

−

−

−

27

04

27

21 3

30 2

111

−×+×−+××+×−+××+×+××+×+××+×+××−+×+×

=2)(201)(237241)(73

(-2)30022734072

(-2)(-1)012171)(4171

=

−231

235

44

and AB =

−

−

−

41

20

31

213

302

111

×+×−+×−×+×−+××+×+×−+×+×

−+×+×−−+×+×=

4221)(331)(201)(13

4320321)3(0012

1)4(21311)1)((0111


MatricesUnit 8

=

−151

181

12

−=02

21

151

181

12

(AB)C

=

−=

×+××+××+×−×+××+××+×

231

235

48

0152121511

01821)(2181 (-1)

01222112

CHECK YOUR PROGRESS 3Q 1: Given

B =

=

=

47

15D,

24

13C,

43

21

B(C+D) = BC + BD

(B+C) D = BD + CD

Q 2: For the matrices A,B,C put the numerical values of their

orders, so that (A+B) C = AC + BCQ 3: Find A and B if

A + B =

=−

30

03BA,

52

07

Q 4: Given A =

−=

0i

i0B,

021

10, Show that

A2 = B2 = I (unit matrix)

Q 5: If

−=

21

13A , show that 07I5AA 2 =+−

Q 6: If

−=

αααα

α cossin

sincosA , then prove that

( )

−=

αααα

α cos3sin3

sin3cos3A 3

8.9 TRANSPOSE OF A MATRIX

Let A = [aij]m×n. Then the n×m matrix obtained from A by changing the rows ofA into columns and columns into rows is called the Transpose of matrix A


Matrices Unit 8

and is denoted by AI or AT.Therefore, Al or AT will have n-rows and m-columns.

Example :

If A = 32

42 1

502

×

Then,

23

/

45

20

12

A

×

=

SOME PROPERTIES OF TRANSPOSE OF MATRICES :

(i) The transpose of transpose of a matrix is the matrix itself i.e. if A is a

matrix, then // )(A = A

(ii) If A is a matrix f order mxn and k is a scalar, then // kA(kA) =

(iii) If A and B are matrices of same order, then /// BAB)(A +=+

(iv) If A and B are two matrices conformable to multiplication, the/// AB(AB) = .

This law is known as "reversal Law for transpose".Example:

If A =

35

24

43

then show that

( ) AA / =/

Solution :

A =

=

321

543Athen,

35

24

13/

=35

24

13

)(A //

Hence, A )(A // =Note : The reversal law for transpose can be extended for more than twomatrices.

etc.ABCD(ABCD)

ABC (ABC) /////

////

==

Note :

BA ABif AB(AB) iii)

BAB)(A ii)

kA(kA) i)

/

//'/

//

==+=+

=


MatricesUnit 8

Note :

i) If A is a symmetric

matrix, then AA / =

ii) A is a skew-sym-

metric matrix then

AA/ −=

Example :

If

=

=

43

21 B,

14

32A

Show that (i) // 2A (2A) =

(ii) /// B A B)(A +=+

(iii) /// AB (AB) =Solution :

(i)

=

14

32 A

=

=

13

42A,

28

64 2A /

=

=

=∴

26

84

13

422A,

26

84 (2A) // 2

( ) // A2A2 =∴

(ii)

+

=+

43

21

14

32 BA

=

++++

=57

53

4134

2312

=+

5 5

73 B)(A /

Again,

+

=+

42

31

1 3

42 BA //

=

++++

=55

73

4123

3412

Hence, /// BAB)(A +=+

(ii)

=

43

21

14

32AB

×+××+××+××+×

=41243114

43223312

=

=

1216

711 (AB)

127

1611 /and


Matrices Unit 8

Again,

=

13

42

42

31AB //

×+××+××+××+×

=14423422

13413321

=

1216

711

Hence, /// AB(AB) =


Q 1: If A= 32

23

212

431B and

20

11

34

××

=

−

Show that /// AB (AB) =

Q 2: With suitable example show that A)(A // =

Q 3: With suitable example, show that AA/ −=

8.10 SYMMETRIC AND SKEW-SYMMETRIC MATRIX

SYMMETRIC MATRIX :

A square matrix A= [ ]ija is said to be symmetric if for all values of i and

j, i.e, jiij aa =i.e. (i,j)th element is same as the (j,i)th element of A

For example

653

542

321

,

cfg

fbh

gha

are symmetric matrix.Theorem : The necessary and sufficient condition for a matrix A to besymmetric is that /AA =

Proof : Let A=[ ija ]nxn is a n-rowed square matrix.

Necessary : jiij aa =

Now, A/ which is the transpose of A is also a square matrix.

∴ (i,j)th element of A/ =(j, i)th element of A


MatricesUnit 8

= jia =(i j)th element of A

Hence A is symmetric.Sufficient : Let AA / =

∴ (i, j)th elements of A = (j, i) th element of A/

= (j, i)th element of A

Hence A is symmetric.

Skew symmetric Metrix : A square matrix A=[ ija ] is said to be skew-

synmetric if for all value of i and j, jiij aa −=

i.e. (i,j)th element is the negative of (j,i)th element of A

For example

−−−

−−−

043

402

320

,

0fg

f0h

gh0

are skew symmetric matrix.

If A is skew-symmetric then –

jiij aa −= ∀ i, j

∴ iiii aa −= ∀ i

⇒2 iia = 0

⇒ iia = 0

Thus the diagonal elements a11

, a22

, a33

...... of a skew-symmetric matrix

are all zero.

Theorem : Necessary and sufficient condition for a matrix A to be skew-

symmetric is that. A/= – A.

Prof : Necessary :

Let A= [ ija ]nxn be an n rowed skew symmetric matrix. Then jiij aa −=

∴(i,j) th element of /A =(j, i) th element of A

= jia

= ija−

= – (i,j)th element of A

∴ AA / −=

Sufficiency part :


Matrices Unit 8

Let A/= – A

∴(i, j)th element of /A =–(i, j)th element of A

⇒ (j, i)th element of A = –(i, j)th element of A

⇒ ijji aa −=

⇒ jiij aa −=

Hence A is skew-symmetric.

Example 1:

If A is symmetric (skew symmetric) matrix then kA is also symmetrix (skew

symmetric).

Solution :

i) Let A be symmetric. Then A/ = A

Now (kA)/ = kA/ =kA

⇒ kA is also symmetric.

ii) Let A be skew symmetric. Then A/ = – A

Now, (kA)/ =kA/ = k(–A) = –(kA)

⇒kA is also symmetric.

Example 2: Show the matrix A+B is symmetric or skew-symmetricaccording as A and B are symmetric or skew-symmetric matrix.Solution :i) A and B be two symmetric matrics

∴A/=A, B/ =B

Now, /B)(A + =A/ +B/ =A+B ∴A+B is symmetricii) A and B be to skew-symmetric matrices

∴A/=–A, B/= – B

Now, A/+B/=A/ +B/ = – A–B= – (A+B)

∴A+B is skew-symmetric.

Example 3:If A and B are symmetric matices, then AB is symmetic if and only if A andB commute i.e. AB=BA

Solution : A and B are symmetric. To prove that (AB)/ =AB

∴A/=A, B/=BAlso, AB = BA

Now, (AB)/ =B/A/

=BA=AB


MatricesUnit 8

∴AB is symmetric.Conversely , Let AB be symmetric. To prove that AB=BA

∴(AB)/ =AB – (1)

Also (AB)/=B/A/

= BA – (2)From (1) and (2), is equel

AB = BA Proved

Example 4:

If A is a n-squared matrix, then AA/and A/Aare both symmetric matrices.Solution :

we have

symmetricisAA

A)(AAA

A)(A)(AA

/

///

/////

∴==

=

Q

Also, ( ) ( ) AAAAAA ////// ==

symmetric isA A / .

Example 5:

Show that every square matrix is uniquely expressible as the sum of a

symmetric matrix and a skew-symmetric metrix.

Solution : Let A be any square matrix.

We can write

QP

)A(A21

)A(A21

A //

+=

−++=

)A(A21

Qand)A(A21

PWhere ′−=′+=

Now,

///// kA(kA))A(A21

)A(A21

P =+=

+=′ Q

/

{ }/// )(AA21 +=

P)A(A

21

A)(A21

/

/

=+=

+=


Matrices Unit 8

∴P is symmetric matrics.and

{ }///

//

/

)(AA21

)A(A21

)A(A21

Q

−=

−=

−=′

/

Q

)A(A21

A)(A21

/

/

−=

−−=

−=

∴Q is skew-symmetric matrix.

Uniqueness : To prove that the representation is unique,let us take another

representation. A=R+S. where R is symmetric and S is skew-symmetric.

To prove that R=P, S=Q.

SR

SRS)(RA ////

−=+=+=

)A(A21

R

2RSRSRAA

/

/

+=⇒

=−++=+∴

and

)A(A21

S

2SSR-SRA-A

/

/

−=⇒

=++=

Thus R=P and S=Q.

∴ The representation is unique.

CHECK YOUR PROGRESS – 5

Q 1: AAandAAfind,123

231A //

=


MatricesUnit 8

Q 2: Let and)A(A21

findthen,

714

582

361

A /+

=

symmetricis)A(A21

thatshow).A(A21 // +−

symmetric.skewis)A(A21 / −−

Q 3: Show that

08-4

803i

4-3i- 0

is skew-symmetric matrix.

8.10 ADJOINT OF A SQUARE MATRIX

Let A=

mmm2.m1

2m22.21

1m12.11

a ...aa

.............................

a ...aa

a ...aa

=[aij]mxn

The determinant of the matrix A,

|A| =

mmm2.m1

2m22.21

1m12.11

a ...aa

.............................

a ...aa

a ...aa

Let the co-factors of the element ija be ijA considering the matrix whose

elements are the co-factors of the corresponding elements of A, we get

∆ =

mmm2.m1

2m22.21

1m12.11

A ...AA

............................ .

A ... AA

A ...AA

B(say)

A .. . AA

............................ .

A ... AA

A .. . AA

mm2m.1m

m222.12

m121.11

=

=∆/

/B ∆= is called the adjoint of the matrix A and is denoted by adj A.


Matrices Unit 8

i.e. B = adj. ATheorem:

If A be any n-rowed square matrix, then(adj A) A= A(adj A) = |A| In where In is the n-rowed unit matrix.

Proof :

Let A= [ ] nnija × be any n-rowed square matrix and

let adj A = B = [bij]n×n

Then bij = co-factor of ija in |A|

= Aji

Since A and adj A are both n-rowed square matrices, therefore they areconformable for multiplication

i.e. Both A (adj A) and (adj A) A exist and are of the type n×n.Now, (i, j)th element of A (adj A)

=∑=

n

kkjikba

1

∑=

=n

1Kjkik Aa

=≠

=ji if|A|

jiif0

In other words, all the elements of A(adj A) along the principal diagonal areequal to |A| and the non-diagonal elements are all zero.

=∴

|A| ........ 0 0 0

.................................

0 ......... |A| 0 0

00A 0

0 ...0......... 0 |A|

A)A(adj

=

1 .......... 0 0 0

..............................

0 1........ 0 0

0........ 0 1 0

0 0........ 0 1

|A|

= |A| In

Similarly,(i, j)th element of (adj A) A

kj

n

1Kikab∑=

=


MatricesUnit 8

∑==

n

1kkjkiaA

nI

j

|A| A A)(adj Therefore,

ji if |A|

; iif0

=

=≠

=

Hence,

A(adj A) = (adj A) A = |A|In and A–1 = ( )adj.A.|A|

1 .

Example :Find the adjoint of the matrix A

where A =

−102

413

021

Solution :We have

=102

41-3

021

|A|

Now, A11 = cofactor of (1,1)th element of |A|

10110

41−=−−=

−=

12A = cofactor of (1,2)th element of |A|

58)(312

43=−−==

Similarly,

A13 = 221)(002

13=−−−=

−

A21 = – 210

02−=

A22

= 402

21A1,

10

0123 =−==

A31

= 761 13

21A4,

43

01A8,

41-

023332 −=−−=

−==−==


Matrices Unit 8

=∴7- 4 2

4- 1 5

8 2 -1-

A adj

Note :

(1) To find the Adj A, write down the determinant |A| and find the co-

factor of various element of |A|. Replace each element in A by its

co-factor and then take the transpose to get adj A.

(2) If A is a n-rowed square matrix, then A is said to be singular matrix

if its determinant is zero i.e |A| = 0.

∴A(adj A) = (Adj A)A= |A| In= 0 (null matrix)

Hence, in this case all the elements of A(adj A) and (adj A)A will be

zero’s.

(3) If |A| ≠ 0, then

==

|A|.... 0 0 0

.......................

...0|A| 0 0

0.... 0|A| 0

0 0....0|A|

I|A|AAdj| n| A

n

A= .

1-n|A| | AAdj| =⇒

8.11 INVERSE OF A MATRIX

Let A and B any n-squared matrices such that

AB = BA = In

Then B is called inverse of A and is denoted by A–1 i.e. B=A–1

Note : It is to be noted that every square matrix does not possess

its inverse.

PROPERTIES OF THE INVERSE :

(a) The inverse of a matrix is unique.

Let A be n-squared matrix whose inverse exists.

Let B and C be two inverses of A.

Then AB = BA = In

––– (1)


MatricesUnit 8

AC = CA = In

––– (2)For (1) C(AB) = C I

n= C ––– (3)

For (2) (CA) B = InB = B ––– (4)

Q C(AB) = (CA) BQ C = B

which shows that an invertible matrix possesses a unique inverse.

(b) The necessary and sufficient condition for a square matrix A to possess

an inverse is that A is non-singular i.e. |A| ≠ 0

Proof: Necessary :

Let A be a square matrix of order n and B be its inverseThen AB = In

∴ |AB| = |In|

0 |A|

1 |B||A|

≠⇒

=⇒

Sufficient :

Let |A| ≠0, we define a matrix B so that B = ( )adj.A.|A|

1

Then AB = A Aadj|A|

1

A) A(adj|A|

1=

nn II|A| |A|

1 ==

Similarly, BA = AAadj|A|

1

nn II|A|.|A|

1AA)(adj

|A|1 ===

Thus, AB = BA =In .

Hence, A is invertible and B, is the inverse of A.

Aadj|A|

1AB 1 ==∴ −

(c) Reversal law for the inverse of a product.

If A, B be two n-rowed non-singular, then AB is also non-singular and

(AB)–1= B–1 A–1.i.e. the inverse of the product is the product of the

inverses in the reverse order.

Proof : Let A and B be two non-singular matrices i.e. |A|≠0, |B|≠0,

So their inverses exist.


Matrices Unit 8

We have |AB| = |A||B|≠0

Hence, the matrix AB is also invertible .

Let C be the inverse of AB.

Let A–1and B–1 be the inverses of A and B respectively andAA–1 = A–1A = In and B B–1= B–1B = In

∴C (AB) = (AB) C = In

Choosing C = B–1 A–1 we getC(AB) = (B–1 A–1)(AB)

= B–1 (A–1A)B= B–1 In B= B–1 B= In ... (1)

Again, (AB) C = (AB) (B–1 A–1)= A(BB–1) A–1

= AInA–1

= AA–1

= In ... (2)By (1) and (2)

C(AB) = (AB) C = In

Hence, C is the inverse of ABor B–1 A–1 is the inverse of ABThus, (AB)–1 = B–1 A–1

(d) The operation of transposing and inverting are commutative

i.e. /-1-1/ )(A )(A =Proof :Let A be a non-singular sequare matrix i.e. |A|≠0

∴ A–1 existsWe get AA–1 = A–1A = In∴ n

//-1/-1 IA)(A )(AA ==

n/-1///-1 I)(A A A)(A ==⇒

which shows that /-1)(A is the inverse of A–1.

Hence, /-11/ )(A)(A =−

Note : If A, B are two n-rowed non-singular matrices. Then

(AB)–1 = B–1A–1.

Example :

Find the adjoint and inverse of the matrix

021

113

310


MatricesUnit 8

Solution :

Let A =

021

113

310

=∴021

113

310

|A|

21

133

01

131)(0 +−+=

1)3(61)(1)( −+−−=

016151 ≠=+=

exists A 1−∴Now, Co-factors of the elements of 1st row are –

5 1, 2,-i.e.21

13,

01

13,

02

11−

Co-factors of the elements of 2nd row are –

1 3,- 6,i.e.21

10,

01

30,

02

31−−

Co-factors of the elements of 3rd row are

3-9, 2,-i.e.13

10,

13

30,

11

31−

Matrix formed by the corresponding co-factors of the elements of |A| is

=3- 9 2-

1 3- 6

5 12-

C

∴ adj A = transpose of C =

3- 1 5

9 3- 1

2- 62-

( )Aadj|A|

1A 1- =∴

−−

−−=

315

931

262

161

⋅


Matrices Unit 8

−

−

−

=

163

161

165

169

163

161

81

83

84

Example : Find the inverse of the matrix

=431

341

331

A

Solution :

=431

3 41

331

|A|

31

413

41

313

43

34+−=

0167337

4)(333)(439)(16

≠=−=−−=−+−−−=

Now, Co-factors of the elements of the 1st row are –

1- 1,- 7, i.e.31

41

4 1

31

43

34

,

−

Co-factors of the elements of the 2nd row are –

0 1, 3, i.e.31

31

4 1

31

43

33

,

−− ,

Co-factors of the elements of the 3rd row are –

1 0, 3,- i.e.41

31,

3 1

31

34

33

,

−

−=∴

1 01-

0 11-

3-37

Aadj

−=

=∴

1 01-

0 11-

3-37

11

Aadj|A|

1A 1-


MatricesUnit 8

−=

1 01-

0 11-

037


Q 1: Find adj A if A =

1 103

3 21

210

Q 2: Find A–1 if A =

−−

αααα

cos sin

sin cos

Q 3: Show that if |A|≠0, then |A–1| = |A|–1

Q 4: If AB = BA, then prove that A–1 B–1 = B–1A–1

Q 5: If |A|≠0 and A is symmetric, show that A–1 is also symmetric.

Q 6: If A be a square matrix, then prove that |adj A| = |A|n–1.

Q 7: If |A|≠0, then show that adj adj A = |A|n–2A.

8.12 RANK OF A MATRIX

Sub-m atrix of a matrix : Let A be any matrix of order m×n. Then a marix

obtained by leaving some rows and columns from A is called a sub-matrix

of A.

Note : A matrix A itself is a sub-matrix of A.

Minors of a matrix : Let A be any matrix of order m×n. by leaving (m–r)

rows and(n–r) columns of A, we get a square sub-matrix of order r. The

determinant of the square sub–matrix of order r is called a minor of A of

order r.

Example :

A = 32

231

213

×

−−

is of order 2×3.So we get minors of order 1 and 2


Matrices Unit 8

Minor of order 2 are 23

21,

21

23,

31

13 −−−−

Now,

101931

13=+=

−, 826

21

23=+=

−, 462

23

21=+−=

−−

RANK OF A MATRIX :

The rank of a matrix A is r if

(a) Every minor of A of order r+1 is zero.

(b) There is atleast one minor of A of order r which does not vanish.

The rank of a matrix A is denoted by ρ (A)

Note : (i) The rank of a null matrix is zero.

(ii) Rank of a sub-matrix ≤ rank of the matrix.

WORKING RULE :

(1) Calculate the minors of highest possible order of the matrix.

(2) If it is not zero, then the order of the minor is the rank of the matrix.

(3) If it is zero and all other minors of same order are also zero, then

calculate the minor of next lower order.

(4) If it is not zero, then follow (2)

(5) If it is zero, then follow (3) and so on.Example : Find the rank of the matrices

(i) A =

=

−

1 0 0

0 1 0

0 0 1

A(ii)2- 4 1

362

(iii) A =

0 0

00

= 4 4 3

3 2 4

6 5 4

A(iv)

Solution :(i) As the matrix A is of order 2×3, the highest possible order of

minors will be 2

Minors of order 2 : 3 4

6 6

−−

= 6(-2) - 4 (-3)


MatricesUnit 8

= -12 + 12 = 0

Another minor of order 2 is 61424 1

6 2×−×=

02

68

≠=−=

Hence, there is atleast one minor of order 2 is non–zero. ∴ρ (A) = 2.(ii) As the matrix A is of order 3×3, the highest possible order of minor

will be 3.Minor of order 3 is

01

100

010

001

≠=

( ) 3A =∴ ρ(iii) As the matrix A is of order 2×2, the highest possible order of minor

will be 2.

Minor of order 2 is 000

00=

Also minor of order 1 is zero

0 ρ(A) =∴

(iv) As the matrix A is of order 3×3, the highest possible order of minor

will be 3.

Matrix of order 3 is

2)344(13345(43)444(2

443

321

654

×−×+×−×−×−×=

= 4(8–12) – 5(4–9) + 6(4–6)

= 4(–4) – 5 (–5) + 6(–2)

= -16 + 25 –12

= 25 – 28

= –3 ≠ 0

3ρ(A) =∴


Matrices Unit 8

CHECK YOUR PROGRESS 6Q 1: Find the rank of the matrices

(i) A =

− 412

310 (ii)

−=

3 16 0

1 2- 0

5-40

111

A

=4- 3- 4

1374

312

A (iii)

=

10

01A (iv)

8.13 LINEAR EQUATIONS

Linear equation are divided into two types

(i) Homogeneous linear equations

(ii) Non-homogeneous linear equations.

The nature of solutions of a system of linear equations can be studied with

the help of matrices. We shall first consider a system of homogeneous

linear equations. There are two types of equations

(i) Consistent equation

(ii) Inconsistent equation

(i) Consistent equation : A system of equations is said to be consistent

if they have one or more solutions.

Example :

(i) x + 2y = 0 (ii) x+2y = 4

3x+ y = 2 4x + 8y = 16

(ii)Inconsistent equations : A system of equation is said to be inconsistent

if they have no solution.

Example :

x + 2y = 4

3x + 6y = 6

Homogeneous linear equations :

A system of n-linear equations in n-unknown say, nx,x,x ,21 ⋅⋅⋅⋅ is

given as follows :


MatricesUnit 8

(1)

0 = xa + .………… + xa + xa

- -- - - - - - - - - - - - - - - - - - - - - - - - - -

0 = xa + .………… + xa + xa

0 = xa + .………… + xa +xa

nnm2m21m1

n2n2221 21

n1n212 1 11

1m1nn

2

1

nmmnm2m1

2n2221

1n1211

0

.

.

0

0

0,

x

.

.

x

x

X

a ........ aa

........aaa

........aaa

××

×

=

=

= ,ALet

Then the above system of equations can be written in the matrix form as

=

0

.

.

0

0

,

x

.

.

x

x

a ......... aa

.a..........aa

.a..........aa

n

2

1

mnm2m1

2n2221

1n1211

i.e. AX = 0 (2)

The matrix A is called the co-efficient matrix of the system of equation (1)

Obviously, x1= x2 = …………… = xn = 0

i.e. X = 0 is a solution of (1) called a trivial solution.

Again, let X1 and X2 be two solutions of (2). Then their linear combination

K1X

1 + K

2X

2 where K

1 and K

2 are any arbitrary numbers is also a solution of

(2).

Therefore, AX1= 0 and AX2 = 0 and hence

A(K1X1 + K2 X2) = K1(AX1) + K2 (AX2)

= K1 0 + K20

= 0 + 0

= 0Hence, K1 X1 + K2 X2 is also a solution of (2)

8.14 SOLUTION OF NON-HOMOGENEOUSSIMULTANEOUS EQUATIONS

Let a system of n simultaneous equation in n-unknown x1, x2, ………. xn be

a11 x1 + a12 x2 + ………………… + a1n xn = b1


Matrices Unit 8

a21 x1 + a22 x2 + ………………… + a2n xn = b2

…………………………………………....…

an1 x1 + an2 x2 + ………………… + ann xn = bn

This system can be written in the matrix form as

AX = B ... (1)

When A =

nx1n

2

1

1nn

.

.2

1

nnnnn2n1

2n2221

1n1211

b

.

.

b

b

B

x

x

x

X

a ............. aa

a ............. aa

a ............. aa

=

=

×

×

If |A|≠0, then A–1 exists.

Pre-multiplying both sides of (1) by A–1

A–1(AX) = A–1 B

⇒ (A–1A) X = A–1B

⇒ IX = A–1 B

⇒ X = A–1B which is the solution of the given system of equations.

Example : Solve :

x+y+z =1, x+2y+z= 2, x+y+2z = 0

Solution :

Given system of equation is –

x+y+z = 1

x+2y + z = 2

x + y + 2z = 0

It can be written as –

=

3

2

1

z

y

x

211

121

111

i.e. A X = B where A =

=

=

0

2

1

B

z

y

x

X,

211

121

111

,

Now,


MatricesUnit 8

2)1(11)1(21)1(4

211

121

111

A −+−−−==

= 3 – 1 – 1 = 1≠0Co-factors of the elements of 1st row are –

1,-1i.e.3,11

21,

21

11,

21

12−−


0 1,1,-i.e.11

21,

21

11,

21

11−

Co-factors of the elements of 3rd row are –

1 0,1,-i.e.21

11,

11

11,

12

11−

1 0 1-

011

113

Aadj −−−

=

1|A|Aadj|A|Aadj

A 1 ===∴ −Q

==

=∴0

2

1

1 0 1-

0 11-

1-1-3

BA

z

y

x

X 1-

−=

+=1

1

1

1-

2 1-

2-3

i.e. x = 1, y = 1, z = –1

CHECK YOUR PROGRESS 7Q 1: Solve x+y+z = 6, x–y+z = 2, 2x+y+z = 1

Q 2: Solve x–3y+z = -1, 2x+y – 4z = –1, 6x–7y+8z = 7


Matrices Unit 8

2.15 SOLUTION OF A SYSTEM OF N NON-HOMOGENOUS LINEAR EQUA TIONS IN N-UNKNOWNS BY CRAMMAR'S RULE

Let the system of n non-homogenous linear equations in n-unknowns linear

(1)bxa ............ xa xa 1n1n212111 →=+++

(2)bxa ............ xa xa 2n2n222121 →=+++………………………………………………..

(n)b xa . xa xa nnnn2n21n1 →=+…++The system can be written as –

b

.

.b

b

x

.

.x

x

a ..... aa

......................

a .... aa

a .... aa

1nn

1

1nn

1

nnnnn2n1

2n2221

1n121122

××

×

=

i.e. AX = BDeterminant of the co-efficient matrix A

=|A| = D (say)Multiplying the equations (1), (2), ……., (n) respectively by the co-factors

of ⋅⋅⋅⋅,, 2111 aa i.e. 12111 A,,A,A n⋅⋅⋅⋅ and adding we get

⋅⋅⋅++⋅⋅⋅++++⋅⋅⋅++ )xaxax(aA)xaxax(aA n2n22212121n1n21211111

n1n212111nnn2n21n1n1 AbAbAb)xaxax(aA +⋅⋅⋅++=+⋅⋅⋅+++

n1n2121111n1n12121111 AbAbAb)xAaAaA(a +⋅⋅⋅++=+⋅⋅⋅++⇒

11 DDx =⇒

Where

nnn2n

2n222

1n121

1

nnn2n1

2n2221

1n1211

a ........ a b

..........................

a ........ a b

a ........ a b

D and

a ........ a a

..........................

a ........ a a

a ........ a a

==D

D1 is the determinant obtained from D by replacing the elements of 1st

column by corresponding b's

0D ProvidedD1

Dx

1

1 ≠=⇒

Similarly multiplying the equations (1), (2), ………… (n) by co-factors of the


MatricesUnit 8

elements of 2nd column of |A| and adding, we get –

nnnn1

2n221

1n111

222

.....aba

.....aba

a.....ba

DwhereDDx

==

0D,D1

Dx

2

2 ≠=⇒

As above we will get

0D,D1

Dx

..........Dx

Dx

n

n

2

2

1

1 ≠====

The unique solution of the given system of equation provided D≠0 in the

coefficient matrix is non-singular.

i.e. the rank of the co-efficient matrix is n = number of variables.

Note : For a system of n non-homogeneous linear equations with n-

unknowns

0D,D1

Dx

..........Dx

Dx

n

n

2

2

1

1 ≠==

Example : Solve the equations3x + y +2z = 32x - 3y – z = –3x + 2y + z = 4Using Cramer's rule.

Solution : We have

1 2 1

1- 3- 2

2 1 3

|A|D

−

==

= 3(–3 + 2) – 1(2+1) + 2(4+3)= –3 –3 + 14= 8≠ 0

1D =

1 2 4

1- 3- 3

2 1 3

−

= 3(–3+2) –1 (–3+4) + 2(–6+12)= –3 – 1 + 12


Matrices Unit 8

= 8

2D =

1 4 1

1- 3- 2

2 3 3

162293

3)2(81)3(24)33(

=+−=+++−+−=

3D = 421

332

313

−−

= 3 (–12+6) –1(8+3) + 3 (4+3)= -18 – 11 + 21= –8

18

8

D

Dx 1 ===∴

28

16D

Dy 2 ===

188-

D

Dz 3 −===

Example : Solve completely the equations by Cramer's rule –x+2y+3z = 6, 3x–2y+z=2, 4x+2y+z=7

Solution :

124

123

321

D −= 0404224

8)3(64)2(32)21(

≠=++−=++−−−−=

1 2 7

1 2- 2

3 2 6

D1 = 40 54 10 24-

14)3(4 7)-2(2- 2)-6(-2

=++=++=

174

123

361

D2 =

= 1(2–7)–6(3–4)+3(21–8)= –5+6+39= 40


MatricesUnit 8

724

223

621

D3 −=

= 1(–14–4)–2(21–8)+6(6+8)= –18–26+84= 84 – 44= 40

14040

D

D z

14040

DD

y

14040

DD

x

3

2

1

===

===

===∴


Solve completely the system of equations by using

Cramer's rule :

Q 1: 3x + 5y = 8, –x+2y–z =0, 3x–6y+4z=1

Q 2: x1+x2+x3=7, x1– x2+ x3 = 2, 2x1 – x2+ 3x3 = 9

8.16 LET US SUM UP

l A rectangular array of mn numbers arranged in m-rows and n-columns

and enclosed in square [ ] or a round bracket ( ), is called a matrix of

order m×n (m by n).

l The numbers which constitute the matrix are called the elements of a

matrix.

l The element which occurs in the i-th row and j-th column is called the

(i,j) the element of the matrix and is denoted by ija .

l A matrix with exactly one row is called a row matrix. The order of a row

matrix is of the type 1×n.

l A matrix with exactly one column is called a column matrix. The order

of a column matrix is of the type m×1.


Matrices Unit 8

l A matrix A=[ ija ]m×n is a square matrix if m=n.

l A square matrix a=[ ija ] is a scalar matrix if ija =

≠=

j wherei0

ji wherek

l A square matrix A = [ ija ] is a diagonal matrix if ija = 0 where i≠ j.

l A square matrix A=[ ija ] is a unit matrix if ija = 1 where i = j

= 0 where i≠ j ;

A unit matrix of order n is denoted by l or In.

l A matrix in which all elements are zero is called a null matrix or zero

matrix. A null matrix is denoted by O.

l A matrix is said to be upper triangular if ija = 0 when i>j.

l A matrix is said to be lower triangular if ija = 0 when i < j.

l Two matrix are said to be equal if and only of they are of same order

and the corresponding elements are equal.

l If A and B are two matrix of same order. Then their sum or difference

denoted by A + B ,A–B or B–A can be obtained.

l Matrix addition in commutative i.e. A+B = B+A.

l Matrix addition in associative i.e. (A+B) + C = A+(B+C)

l Existence of addition identity i.e. A+O = A = O +A

l Existence of additive inverse i.e. A+(–A) = (–A) + A

l Scalar multiplication of matrix i..e if (a) A= [ ija ] then kA= [k ija ]

(b) k(A+B) = kA+ kB (c) (k+l) A = kA + lA

(d) k(lA) = (kl) A (e) (–k)A= k(–A)

l Matrix multiplication is not commutative i.e. AB ≠ BA (in general)

l Matrix multiplication is associative i.e. (AB)C = A(BC)

l Matrix multiplication is distributive over addition i.e.

(i) A(B+C) = AB + AC

(ii) (A+B) C = AC + BC

l If A = [ ija ]m×n then AT=[ ijb ]n×m when ijb = ija

l For matrices A and B of the same order (A+B)/ = A/ +B/

l For matrix A and B conformable to multiplication

( ) /'// ABAB =

l The matrix A is called symmetric if A/=A202 Business Mathematics (Block 2)

MatricesUnit 8

l The matrix A is called skew symmetric if A/= – A

l Every square matrix A can uniquely be expressed as A = P+Q where

P is a symmetric matrix and Q is a skew-symmetric matrix.

l For a square matrix A (adj A) A = A (adj A) = |A|I

l For a square non-singular matrix A, A–1 = |A| Aadj

l For two non-singular matrices A and B , ( ) -1-11 ABAB =− .

l Rank of a sub matrix ≤ Rank of the matrix.

l If x1 and x2 be two solutions of AX= 0, the k1x1+ k2x2 is also a solution of

AX = 0

l For system of n simultaneous non-homogenous equations with n-

unknown written as AX=B, the solution matrix is X = A–1B

provided |A| ≠ 0.

l For a system of n-simultaneous linear non-homogenous equation with

n-unknowns

0D.,D1

Dx

.............Dx

Dx

n

n

2

2

2

1 ≠====

8.17 ANSWER TO CHECK YOURPROGRESS


Answer Q No 1: Find all possible orders of a matrix with 10 elements.

110 0 1 52 10

25 10 10110

×=×=×=×=

Possible orders of matrices are 1×10, 2×5, 5×2 and 10×1.

Answer Q No 2: By equality of two matrices

=

+−++

253

114

42

52

dcba

dcba

2a+b=4 5c–d=11

a–2b=–3 4c+d=25

Solving these we get

a = 1, b = 2, c = 4, d = 9

Answer Q No 3: For equality of two matrices, their orders and the co-


Matrices Unit 8

rresponding elements must be same. Here, the order of the matrices

are same. But (3,3)th element of ( )th3,3A ≠ element of B BA ≠∴


Answer Q No 1:

(i) It is not possible for define A + B, since A is of order 3×2 and B is

of order 2×2.

(ii) It is possible to define A+B, since A is of order 3×3 and B is of

order 3×3.

(iii) It is possible to define A+B, since A and B are square matrices of

the same order.Answer Q No 2:

ofidentity additive the is000

0000

= 2 x3 matrices.

CHECK YOUR PROGRESS 3Answer Q No 1 :

+

=+

47

15

24

13

43

21D)B(C

=

61

28

43

21

1

++

++=

4x63x24x113x8

2x61x22x111x8

=

08

40

36

13

BC+BD

+

=

47

15

43

21

24

13

43

21

++++

+

++++

=4x43x14x73x5

2x41x12x71x5

4x23x14x43x3

2x21x12x41x3

+

=

1943

919

1125

511

=

++++

=3068

1430

19114325

951911


MatricesUnit 8

CDBCD)B(C +=+∴

Again (B+C)D =

+

47

15

24

13

43

21

=

++++

47

15

2443

1231

=

47

15

67

34

=

++++

6x47x16x77x5

3x44x13x74x5

=

3177

1641

BD+CD=

+

47

15

24

13

47

15

43

21

=

++++

+

++++

2x44x12x74x5

1x43x11x73x5

4x43x14x73x5

2x41x12x71x5

=

+

1234

722

1943

919

=

=

++++

3177

1641

12193443

792219

CDBDC)D(B +=+∴Answer Q No 2:

If, A is of order 4×3Then, B is of order 4×3and C is of order 3×2

Answer Q No 3:

Given, A + B =

=−

30

03BA,

52

07

+

==−++∴

30

03

52

072ABABA

=

=

++++

82

010

3502

0037

=

=

41

05

82

010

21

A


Matrices Unit 8

Again,A+B – (A–B) = A+B–A+B=2B

=

−

30

03

5 2

07

=

=

−−22

04

3-5 0-2

0037

=

=∴

11

02

22

04

21

B

Answer Q No 4:

==

01

10

01

10AAA 2

=

++++

0x01x10x11x0

1x00x11x10x0

= I10

01=

−

−==

0i

i0

0i

i0BBB2

=

+++−+

0x0ix(-i)0xiix0

(-i)x00x(-i)i)i(0x0

= I10

01

i0

0i2

2

=

=

−

−

Answer Q No 5:A2–5A+7I=A.A–5A+7I

=

+

−

10

01

1-

13

21-

13

1-

13 7

25

2

=

+

−

+−−+−++

70

07

105-

515

2x21)x1(1)2(1)x3(

1X23x1 1(-1)3x3

=

+

−

− 70

07

105-

515

35

58

= 000

00

7103055

05-5 7158=

=

+−++−++−


MatricesUnit 8

Answer Q No 6:

−=

αααα

α cossin

Sin cosA

−

−=

αααα

αααα

α cossin

Sin cos

cossin

Sin cosA 2

+−−+

=αααααα

αααααα22

22

cosSin- sincoscossin

cossinsincos Sin- cos

−=

αααα

cos2 sin2

Sin2 cos2

−

−==

αααα

αααα

ααα cos sin

Sin cos

cos2 sin2

Sin2 cos2AAA 23

+−+

=αααααααααααααααα

coscos2 .SinSin2- Sincos2 - .cossin2

.cosSin2 .Sincos2 .SinSin2- .coscos2

++−++

=))

)2)

αααααααα

cos(2 sin(2

Sin( cos(2

=

αααα

cos2 sin3-

Sin3 cos3


Answer Q No 6:

3x2

/

2x3

/

2 4

1 3

21

B ,2 1 3

0 1 4 A

=

−=

−=

2 1 2

4 31

2 0

1 1

3 4

AB

++++++

+++−=

40 20 4 0

24 13 2 1

616- 312- 64

−=

4 2 4

6 4 3

10 9- 2


Matrices Unit 8

=4 6 10-

2 4 9 -

4 3 2

(AB) /

=2 1 3

0 1 4-

2 4

1 3

2 1

AB //

−−=

+++++++++

=4 6 10

2 4 9

4 3 2

40 24 616-

20 13 3 12-

40 21 6 4-

.AB(AB) /// =∴ Answer Q No 2:

Let A =

c f g

f b h

g ha

=c f g

f b h

gha

A /

A

c f g

f b h

gha

)(A // =

=

Answer Q No. 3 :

matrixsymmetric skew a is

o f g-

f o h-

gho

A

−=

-A

o f g

f o h-

gho

-

o f g

f o h

gho

A / =

−−=

−−−

=


Answer Q No. 1 :

1 2

23

31

A 1 2 3

2 3 1 A /

=∴

=


MatricesUnit 8

1 2

23

31

1 2 3

2 3 1 AA /

=

1x12x23x31x22x33x1

2x13x21x3 2x23x31x1

++++++++

=

1411

11 14

=

1 2 3

2 3 1

1 2

23

31

AA /

=

1x12x21x22x31x32x1

2x13x22x23x32x33x1

3x11x2 3x21x3 3x31x1

+++++++++

=

= 58 5

8139

5910

AAAA // =∴

Answer Q No 2:

=′∴

= 75 3

186

421

A

71 4

582

361

A

+

=+ 75 3

186

421

71 4

582

361

21

)A(A21 /

+++++++++

= 7751 34

158862

432611

21

=

=

7327

38427

41

146 7

6168

782

21


Matrices Unit 8

−

=− 75 3

186

421

71 4

582

361

21

)A(A21 /

−−−−−−−−−

= 7751 34

158862

432611

21

−

=

=

02-21

202-21

20

04- 1

404-

1-40

21

Answer Q No 3: Let

= 08- 4

803i

4-3i-0

A

−−= 08 4-

803i

43i0

A /

A

08- 4

803i

4-3i-0

−=

−=

SymmetricSkewisA∴CHECK YOUR PROGRESS 6Answer Q No 2:

113

321

210

A =

Co-factors of the elements of 1st row are –

58,1,i.e.,13

21,

13

31,

11

32−−−


36,-i.e.1,,13

10,

13

20,

11

21−−



MatricesUnit 8

12,1,i.e.,21

10,

31

20,

32

21−−−

135

268

111

Aadj

−−−

−−=∴

Answer Q No 2:

01sincosAcossin

sincosA 22 ≠=+=∴

−= αα

αααα

∴Co-factor of the elements of 1st row are αα sincos − ,

Co-factor of the element of 2nd row are –

αααα cos,i.e.sincos),sin(−−

αααα

cossin

sincosAadj

−=∴

αααα

cossin

sincos

AadjA

A 1

−==∴ −

Answer Q No 3:

existsA0A 1−∴≠Now,

11

1

1

1

AA1

A

1AA

IAA

IAA

−−

−

−

−

==⇒

=⇒

=⇒

=

Answer Q No 4:Given that AB=BA

1111

11

BAAB

(BA)(AB)−−−−

−−

=⇒

=∴

Answer Q No 5:

existsA0A 1−∴≠also A/=A

Now, 11//1 A)(A)(A −−− ==symmetric.isA -1∴


Matrices Unit 8

Answer Q No 6:

nIAadjAA =

nIAadjAA =∴

AkkAIAAadjA nn

n ==⇒ Q

1An == nIQ

0AifAAadj1n ≠=⇒

−

1nA0Aadjthen0,AIf

−===Hence the result.Answer Q No 7:

nIAA)A(adjhaveWe

0A

=

≠

Replacing A by adj A ,we get

nIAadjA)adjA)(adj(adj =

(i).IA n

1n −= −

Premultiplying both side of (i) by A

}IAA{A)adjA)(adj(adjA n

1n−=

)(AIA n

1n−=

AA1n−=

AAA)adj(adjA)}{A(adj1n−=⇒

AAA)adj(adjIA1n

n

−=⇒

AAA)adj(adjA1n−=⇒

AAAadjadj2n−=⇒ 0]AAby sidesboth[Dividy ≠Q

CHECK YOUR PROGRESS 7(i) As the matix A is of order 2 x 3, the highest possible order of minor is 2.

minor of order 022x10x112

102 ≠−=−==

2(A)=∴ρ(ii) As the matix A is of order 4x3, the highest possible order of minor is 3.

minor of order

180

530

111

3

−−−−=


MatricesUnit 8

03740318

53≠−=−=

−−−−

=

3(A)=∴ρ(ii) As the matix A is of order 3x3, the highest possible order of minor is 3.

minor of order

134

134

312

3

−−= 3

( ) ( )

3(A)

056

725672

24)3(562x36

12)123(5243932x

=∴≠=

−+=−++=

−−+−−−+−=

ρ

(iv) As the matix A is of order 2x2, the highest possible order of minor is 2.minor of order 2

011

≠==0

01

2(A)=∴ρ

CHECK YOUR PROGRESS 8Answer Q No 1:

x + y + z = 6x – y + z = 22x + y – z =1

This system of equations can be written in matix form as –Ax=B

where

=

=

−−=

1

2

6

B,

z

y

x

x,

112

111

111

A

12

11

12

11

11

11

112

111

111

A−

+−

−−

−=

−−=

06330

2)(12)1(11

≠=++=++−−−−=

Now, Co-factors of the elements of 1st row are –

33,0,i.e.,12

11,

12

11,

11

11 −−

−−

−


Matrices Unit 8


13,2,i.e.,12

11,

12

11,

11

11−−

−−−


20,2,i.e.,11

11,

11

11,

11

11 −

−−

−

−−=∴

213

033

220

Aadj

−

−=

−−==∴ −

31

61

21

021

21

31

31

0

213

033

220

61

.adj.AA1

A 1

−

−==∴1

2

6

31

61

21

021

21

31

31

0

BAX 1-

=

−+

−

+

=

⇒

3

2

1

31

31

3

1331

32

z

y

x

3.z2,y1,x ===∴Answer Q No 2:

x – 3y + z = –12x + y – 4z = –16x – 7y + 8z = 7

The above system of equations can be written in matrix from as Ax=B.

When

−−

=

=

−−

−=

7

1

1

,

z

y

x

X,

876

412

131

A


MatricesUnit 8

Now,

876

412

131

−−

−=A

76

12

86

423

87

41

−+

−+

−−

=

= 8–28+3 (16+24)+(–14–6)= 8–28+120–20

080 ≠=Now,Co-factor of the elements of 1st row are –

2040,20,i.e.,76

12,

86

42,

87

41−−−

−−

−−

−

Co-factor of the elements of 2nd row are –

112,17,i.e.,76

3-1,

86

11,

87

13−

−−

−−

−

Co-factor of the elements of 3rd row are –

76,11,i.e.,12

3-1,

4-2

11,

4-1

13−

−−

−−=

−−−−

=∴7111

622

11171

20

71120

6240

111720

Aadj

==∴ −

71120

6240

111720

801

AadjA1

A 1

−−

−−−−

==∴ −

7

1

1

71120

6240

111720

801

BAX 1

=

=

+++−+−

=

⇒

1

1

1

80

80

80

801

491120

42240

771720

801

z

y

x

1z1,y1,x ===∴CHECK YOUR PROGRESS 9Answer Q No 1:Solve : 3x + 5y + 0z = 8

–x + 2y – z = 0 3x – 6y + 4z = 1


Matrices Unit 8

463

121

053

D

−−−=

43

115

46

123

−−−

−−

=

= 3 (8-6) – 5(–4+3) = 6+5 = 011≠

41

105

46

128

461

120

058

D1

−−

−−

=−

−=

= 8(8–6) –5x1=16–5=11

413

101

083

D2 −−=

43

118

41

103

−−−

−=

= 3(0+1)–8(–4+3) = 3+8 = 11

1-3

21

853

D3

6

0−=

63

218

13

015

16-

023

−−

+−

−=

= 3(2–0) –5(–1–0) +8(6–6) = 6+5 =11

11111

DD

x 1 ===∴

11111

DD

y 2 ===

11111

DD

y 3 ===

i.e. x =1, y =1, z =1.Answer Q No 2:

x1 + x2 + x3= 6x1 – x2 + x3= 22x1 – x2 + 3x3= 9


MatricesUnit 8

12

11xa

32

111x

31

111x

312

111

111

D−−

+−−−

=−−=∴

= (–3+1)–(3–2)+(–1+2)= –2–1+1= 02 ≠−

19

12x1

39

121x

31

116

319

112

116

D1 −−

+−−−

=−−=∴

= 6(–3+1)–(6–9)+(–2+9)= –12+3+7= – 2

392

121

161

D2 =

92

21

32

11

3

1+−= x6

9

2

= (6–9) –6(3–2) +(9–4)= –3 –6 +5= –9 + 5= – 4

9-2

2-1

611

D3

1

1=

12

116

92

21

91

21

−−

+−−−

=

= (–9+2) –(9–4) + 6 x(–1+2)= –7–5+6= –12+6= –6

122

DD

x 11 =

−−==

224

DD

x 22 =

−−==

326

DD

x 33 =

−−==


Matrices Unit 8


l G . Cullen, Charles. Matrices and Linear Transformation

l Eves, Howard. Elementary Matrix Theory

l N. Franklin, Joel. Matrix theory

l W. Lewis, David. Matrix Theory

l Bodeing, Ewald, Matrix Calculus

l Vatssa, B.S, Theory of Matrices, New Age International Publishers

l Kala, V.N. & Rana, Rajeshri. Matrices

l Narayan, Shanti & Mittal, A Text Book of Matrices

8.20 MODEL QUESTIONS

Q 1: In a Matrix

−−−

1031

572411

21934

Find (i) Order of the matrix(ii) The number of elements(iii) Write the elements a

23, a

12, a

32, a

34,,a

44.

Q 2: If A = B- Aand B A find,4- 3

0 2 Band

2 6

5 4+

=

Q 3: 5A-and3A find,

0 1

2 0

2 3

AIf

−=

Q 4: Find AB and BA if A = and

1 5

1 0

3 2

− B =

− 351

21 3

Q 5: What is the additive identity of

01

1 0 ?


MatricesUnit 8

Q 6: If A =

−=

20

31

21

B,

5 3 1-

3 2 0

1- 1 1

,

C = C (AB) and (BC) A find,0 2 2-

2 1 3

and show that A(BC) = (AB) CQ 7: Show that A(B+C) = AB + AC

where A = and301

542B ,

3 0

1 2

−=

−=

406

310C

Q 8: If A = A)(A thatshow 652

734 // =

Q 9: If

−=

41

12A and

=

321

654B , then show that /// AB(AB) =

Q 10: If A =

×−×−×+

=

−−

4214

44421 Athatshow then,

11

43 4

Q 11: Find A–1 where A =

−−

312

321

111

Q 12: Find the inverse of the matrix

431

341

334

Q 13: If A = Aadj find,

4100

264

242

−

−−

Q 14: If A = 1- Afind,

653

542

321

Q 15: Solve by matrix method :

(a) 3x + y + 2z = 3, 2x - 3y - z = -3, x + 2y + z = 4

(b) x + y + z = 0, 2x + 3y - 5z = 7, 3x-4y - 2z = -1

(c) x + y + z = 6, x + 2y - 3z = -4, -x - 4y + 9z = 18


Matrices Unit 8

Q 16: Solve completely the system of equations by Cramer's rule.

(a) 2x + 3y + z = 9, x+2y + 3z = 6, 3x + y + 2x = 8

(b) 4x + y + z = 4, x+ 4y - 2z = 4, -x + 2y - 4z = 2

(c) x + y + z + t = 4 , x + y + z - t = 2 , x - y + z - t = 0

*** ***** ***


MatricesUnit 8

UNIT 9: LIMIT AND CONTINUITY

UNIT STRUCTURE


9.2 Introduction

9.3 Limits

9.3.1 Finding Limits Analytically

9.3.2 Examples of Evaluations of Limits Using Various Rules9

9.3.3 Limits by the Method of Substitution

9.3.4 A Special Limit

9.3.5 Some Other Special Limits

9.3.6 One Sided Limits

9.4 Continuity

9.4.1 Basic Definitions and Example

9.4.2 Algebra of Continuous Functions

9.5 Differentiation

9.5.1 Basic Definitions and Examples

9.5.2 Differentiation of Some Elementary Functions

9.5.3 Some Rules of Differentiations

9.5.4 Product and Quotient Rules of Differentiation

9.5.5 Differentiation of Composite Functions

9.6 Applications of Derivatives to Simple Optimization Problems

9.6.1 Basic Definitions

9.6.2 Test for Points of Maximum and Minimum

9.7 Integration

9.8 Let Us Sum Up


9.10 Answer To Check Your Progress

9.11 Model Questions




l define the concept of limit.

l evaluate limits by algebraic rules.

l explain the concept of continuity of a function.

l show that some well known functions are continuous.

9.2 INTRODUCTION

In the earlier unit, we have learnt about Set theory and functions.

In this unit we are going to discuss about limit, continuity,

differentiation and Integration, which are again very important and

fundamental in Mathematics.

The concept of limit is an abstract one. We will learn various algebraic

rules to find limits at various points of well-known functions. We will also

learn about left and right hand limits. We will also discuss about the continuity

and differentiation of functions. Finally, we will discuss about the concept of

integration. We will also discuss about two powerful methods of integration,

namely, methods of substitution and the method of integration by parts.

9.3 LIMITS

We are already familiar with the concept of a real function and a real

variable. The concept of limit is an abstract one. Consider the function

3

9)(

2

−−=

x

xxf , which is defined at all points except x = 3. In the first column

of following table, some values of x close to 3 are written and in the second

column the corresponding values of f (x) are written.

The value of the variable The corresponding value of 3

9)(

2

−−=

x

xxf

2.97 5.97

2.98 5.98

2.99 5.99


Unit 9 Limit and Continuity

3.01 6.01

3.02 6.02

3.03 6.03

From the above table, we observe that if x is very close to 3, then 3

92

−−

x

x is

very close to 6. It is also true that for all values of x close to 3, 3

92

−−

x

x is

close to 6. We then say:

As x tends to 3, the function 3

92

−−

x

x tends to 6 or equivalently, we say: The

function 3

92

−−

x

xhas the limit 6 at the point 3.

In symbols, we write

.63

9lim

2

3=

−−

→ x

xx

In general, by lxfax

=→

)(lim we mean that as gets close to a then

gets close to a limit of l. This can be put in a more precise way

as: |)(| lxf − can be made arbitrarily small, by taking || ax − to be sufficiently

small.

The value of the function f at a is immaterial for finding the limit of f

at a. Only the values that are close to a but not at a matter.

The following two results are simple consequences of the definition.

1. Constant Rule: ccax

=→

lim , that is, the limit of the constant function

cxf =)( as x tends to a is the value c.

2. Identity Rule: axax

=→

lim , that is, the limit of the identity function xxf =)(

as x tends to a is the value a.

Uniqueness of Limit

If )(lim xfax→ exists, it is unique. There cannot be two distinct numbers

l1 and l

2 such that when x tends to a the function f (x) tends to both

l1 and l2.


Limit and Continuity Unit 9

9.3.1 Finding Limit s Analytically

It is possible to find limits by using algebraic techniques. The

limit behaves well with respect to the operations of addition,

subtraction, multiplication by a constant, multiplication, division, etc.

The most basic theorem governing analysis of limits is the Principal

Limit Theorem.

Principal Limit Theorem:

Let a and k be real numbers, ,0≥n and let f and g be functions with

limits at a such that lxfax

=→

)(lim and mxgax

=→

)(lim . Then

1. Sum Rule: .)(lim)(lim)]()([lim mlxgxfxgxfaxaxax

+=+=+→→→

2. Difference Rule: .)(lim)(lim)]()([lim mlxgxfxgxfaxaxax

−=−=−→→→

3. Product Rule: .)(lim)(lim)]()([lim mlxgxfxgxfaxaxax

•=•=•→→→

4. Quotient Rule: m

l

xg

xf

xg

xf

ax

ax

ax==

→

→

→ )(lim

)(lim

)(

)(lim provided

.0)(lim ≠=→

mxgax

5. Coefficient Rule: .)(lim)(lim klxfkxfkaxax

==→→

6. Power Rule: [ ] .)(lim)]([lim nn

ax

n

axlxfxf ==

→→

We can state the above rules in language. For example, Sum Rule

could be expressed as "The limit of a sum is the sum of the

limits."

9.3.2 Examples of Evaluation of Limit s Using V arious

Rules

Example 1. Find .lim 2

2x

x→

Solution: ( )22

2

2limlim xxxx →→

= (by Power Rule)

= 22 (by Identity Rule)

= 4



Example 2. Find .5432lim 23 +++→

xxxax

Solution:

5432lim 23 +++→

xxxax

= 5limlim4lim3lim2 23

axaxaxaxxxx

→→→→+++ (by Sum and Coefficient Rules)

= ( ) ( ) 5limlim4lim3lim2 23

axaxaxaxxxx

→→→→+++ (by Power Rule)

= .5432 23 +++ aaa (by Identity and Constant Rules)

Remark: It is clear from the above example that for all polynomial

functions, the limit at a point can be obtained by substituting that

value for the variable x. But, for non-polynomial functions, this may

not be true always. The next two examples show this fact.

Example 3. Find .2

12lim

31 −−

→ x

xx

Solution:

2

12lim

31 −−

→ x

xx

= 2(lim

)12(lim3

1

1

−

−

→

→

x

x

x

x

(by Quotient Rule)

= 2limlim

1lim2lim

1

3

1

11

→→

→→

−

−

xx

xx

x

x

(by Difference Rule)

= 2lim)lim(

1limlim2

1

3

1

11

→→

→→

−

−

xx

xx

x

x

(by Coefficient and Power Rules)

= 2)1(

1)1(23 −

− (by Constant and Identity Rules)

= –1.

Example 4. Find .23

4lim

2

2

2 +−−

→ xx

xx

Solution: Since the denominator function 232 +− xx has the limit 0

at the point 1, that is, ,023lim 2

2=+−

→xx

xso Quotient Rule cannot be

applied here. Therefore, we have to resort to another method.

Factorizing both the numerator and the denominator, we find that



.)2)(1(

)2)(2(

23

42

2

−−+−=

+−−

xx

xx

xx

x

When ,2≠x we can cancel the common factor x – 2.

Thus,

.1

2

23

42

2

−+=

+−−

x

x

xx

x

Therefore,

23

4lim

2

2

2 +−−

→ xx

xx

= 1

2lim

2 −+

→ x

xx

= 1limlim

2limlim

22

22

→→

→→

−

+

xx

xx

x

x

= 1222

−+

= 3

CHECK YOUR PROGRESS

Q 1: Write true (T) or false (F).

(a) .55)1(lim 2

1=+−

→x

x(T / F)

(b) .0)4)(2(lim2

=+−→

xxx

(T/F)

(c) .03

9lim

2

3=

−−

→ x

xx

(T / F)

(d) .21

1lim

2

1=

+−

−→ x

xx

(T/F)

9.3.3 Limitis by the Method of Substitution

Consider the example 1

1lim

2

1 −−

→ x

xx

. We find that

2)1(lim1

)1)(1(lim

1

1lim

11

2

1=+=

−+−=

−−

→→→x

x

xx

x

xxxx

The same limit can be evaluated with the help of substitution. If we

substitute ,1−= xy i.e. ,1+= yx in the expression ,1

12

−−

x

xthen

.21)1(

11 22

+=−+=−−

yy

y

x

x Also, note that when x tends 1, then y

tends 0. Therefore,



.2)2(lim1

1lim

0

2

1=+=

−−

→→y

x

xyx

In general,

).(lim)(lim0

hafxfhax

+=→→

Equivalently,

.),(lim)(lim hxywhereyfxfhayax

+==+→→

Example 5. Find .1

)1(lim

4

4

1 −−

→ x

xx

Solution: We have

.464

lim1)1(

lim1

)1(lim

234

4

04

4

04

4

1 yyyy

y

y

y

x

xyyx +++

=−+

=−

−→→→

.4

1

464

1lim

230=

+++=

→ yyyy

9.3.4 A Special Limit

Let us evaluate ,limax

ax nn

ax −−

→ where n is a positive integer.

By the method of substitution, we find that

aha

aha

ax

ax nn

h

nn

ax −+−+=

−−

→→

)(limlim

0

−++

+

+= −−

→

nnnnn

hahha

nha

na

h...

211

lim 221

0

[By Binomial Theorem]

.1

...21

lim 11121

0

−−−−−

→=

=

++

+

= nnnnn

hnaa

nhha

na

n

Thus,

,lim 1−

→=

−− n

nn

axna

ax

ax where n is a positive integer.

The above formula remains true, when n is a rational number,

provided a is positive.



Example 6. Find .9

27lim

2

3

3 −−

→ x

xx

Solution:

.3

9lim

3

27lim

3

9

3

27lim

9

27lim

2

3

3

3

23

32

3

3

−−÷

−−=

−−÷

−−=

−−

→→→→ x

x

x

x

x

x

x

x

x

xxxxx

.2

9627)3(2)3(3

3

3lim

3

3lim 2

22

3

33

3=÷=÷=

−−÷

−−=

→→ x

x

x

xxx

Example 7. If 122

2lim

2=

−−

→ x

x nn

xand if n is a positive integer, find n.

Solution: We have 1

22

2

2lim −

→=

−− n

nn

xn

x

x. According to the given

question, .2.3122 21 ==−nn Therefore, n = 3.

9.3.5 Some Other Special Limits

The following limit rules of trigonometric and exponential

functions are very useful.

1. Sine Rule: ,1sin

lim0

=→ x

xx

where x is measured in radian.

2. Cosine Rule: ,1coslim0

=→

xx

where x is measured in radian.

3. Exponential Rule: .11

lim0

=−→ x

ex

x

Example 8. Find .4sin

3sinlim

0 θθ

θ →

Solution: We have

θθ

θθ

θθ

θθ

θθ

θθ

θθ

θθθθθ 4sin

4lim

4

3lim

3

3sinlim

4sin

4

4

3

3

3sinlim

4sin

3sinlim

00000 →→→→→⋅⋅=

⋅⋅=

.4

31

4

31

4

4sinlim

1

4

3

3

3sinlim

04

03=÷

=⋅⋅=

→

→

θθθ

θ

θ

θ

Example 9. Find .tan

lim0 θ

θθ →



Solution: We have

.1coslim

1sinlim

cos

1lim

sinlim

cos

sinlim

00000

=÷=÷=→

→→→→ θθθ

θθθ

θθθ

θθθθθ

Example 10. Find .1

lim4

0 x

e x

x

−→

Solution: We have

.41

lim444

1lim

1lim

0

4

0

4

0=−=⋅−=−

→→→ y

e

x

e

x

e y

y

x

x

x

x

9.3.6 One Sided Limits

In this section we will discuss one sided limits.

The notation lxfax

=+→

)(lim means that when is close x to a and

greater than a then f (x) is close to l.. Similarly, lxfax

=−→

)(lim means

that when is close x to a and less than a, then f (x) is close to l.

)(lim xfax +→ is called the right limit of f at a and )(lim xf

ax −→ is called the

left limit of f at a. Both these are called one-sided limits of f at a

Example1 1. Find ||lim

0 x

xx +→ and ||

lim0 x

xx −→ .

Solutions: When x takes positive values, then .1||

==x

x

x

x

Therefore, we have .1||

lim0

=+→ x

xx

On the other hand, when takes negative values, then

.1||

−=−

=x

x

x

x So, .1

||lim

0−=

−→ x

xx

If )(lim)(lim xfxfaxax −→+→

≠ then )(lim xfax→

does not exist.

If lxfxfaxax

==−→+→

)(lim)(lim then )(lim xfax→ exists and .)(lim lxf

ax=

→

Example 12. Decide whether ||lim

0 x

xx→ exists or not.

Solution: From the solutions of Example 11, we note that



1||

lim0

=+→ x

xx

and .1||

lim0

−=−→ x

xx

Since the right limit and left limit are

not equal, therefore ||lim

0 x

xx→ does not exist.

CHECK YOUR PROGRESS

Q 2: Write true (T) or false (F).

(a) If f is a real function such that f (x) = f (–x) for all

x , then ).(lim)(lim00

xfxfxx −→+→

= (T /F)

(b) .1][lim1

=−→

xx

(T / F)

(c) .1]lim[1+→

=x

x (T / F)

Q 3: Find the following limits:

(i) 2

2lim

2 +−

→ x

xx

(ii) 1)1(

1)1(lim

2

6

0 −+−+

→ x

xx

(iii) 3

lim3

3 −−

→ x

eex

x

9.4 CONTINUITY

In this section, we will learn about the concept of continuity of a

function. We will show that a large class of well known functions is

continuous.

9.4.1 Basic Definitions and Examples

Let f be a real function and let a be in the domain of f If the

limit of f at a is same as the value of f at a, then f is said to be

continuous at a.

Thus, if f is to be continuous at a the following three conditions must

be satisfied.

a) f (a) is defined,

b) )(lim xfax→ exists,

c) ).()(lim afxfax

=→

Example 13. Show that every constant function is continuous at all



points.

Solution: Consider a constant function cxf =)( and an arbitrary

point a.

Then caf =)( and ,lim)(lim ccxfaxax

==→→

and so, ).()(lim afxfax

=→

Therefore, f is continuous at a

Example 14. Show that the identity function is continuous at all points.

Solution: Consider the identity function xxf =)( and an arbitrary

point a.

Then, we have aaf =)( and .lim)(lim axxfaxax

==→→

Thus, ),()(lim afxfax

=→

which shows that f is continuous at ‘a’.

Example 15. Show that the greatest integer function [x] is not

continuous at 0.

Solution: Let ].[)( xxf =

Now, 0][lim)(lim00

==+→+→

xxfxx

and 1][lim)(lim00

−==−→−→

xxfxx

.

Thus, the right hand and left hand limits of f at 0 are not equal. Which

means that ][lim)(lim00

xxfxx →→

= does not exist. Therefore,

though ,0]0[)0( ==f f is not continuous at 0.

Some more results are listed in the following table.

(c) The function 2x is continuous at 3.

(d) The function 14

−−

x

x is continuous at 2.

(e) The function xsin is continuous at .

(f) The function xsin is continuous at 0.

(g) The function xcos is continuous at 0.

(h) The function xe is continuous at 0.

(i) The function || x

x is not continuous at 0.

Definition:

A real function is said to be continuous in an open or closed interval

if it is continuous at every point of the interval.

Remark: When a function f is considered on a closed interval ],,[ ba



then f is said to be continuous at the end point a if ).()(lim afxfax

=+→

Similarly, f is said to be continuous at the end point b if

).()(lim bfxfbx

=−→

9.4.2 Algebra of Continuous Functions

Theorem: Let f and g be two functions, continuous at a and let c be

a real number. Then

a) gf + is continuous at a

b) gf − is continuous at a

c) cf is continuous at a

d) fg is continuous at a

e) g

f is continuous at a provided .0)( ≠ag

Proofs. Follow from the Principal Limit Theorem and the definition

of continuity.

If f and g are continuous functions, that is, they are continuous at

every point of their domains, then the above theorem can be

generalized in the following way.

Theorem: If f and g are continuous functions, then

(a) f + g is continuous,

(b) f – g is continuous,

(c) cf is continuous at,

(d) fg is continuous at,

(e) g

f is continuous, at those points where g does not take the

value zero.

The statement (a) above means that the sum of two (or even more)

continuous functions is continuous. Similarly, statement (b) means

that the difference of two continuous functions is continuous, and

so on.

Example 16. Show that the function 432 2 ++ xx is continuous at

all points.

Solution: We have already proved that the identity function xis



continuous at all points. Thus, 2x is continuous at all points being

the product of x and x. Again, the constant functions 2 and 4 are

continuous at all points. Therefore, 2x2, 3x and 4 are continuous at

all points. Finally, 432 2 ++ xx is continuous at all points being the

sum of three continuous functions.

Example 17. At What points is the function )5)(3(

2

−−−

xx

x

continuous?

Solution: The given function is not defined at the points 3 and 5.

The functions in the numerator, i.e., x – 2 and in the denominator,

i.e., )5)(3( −− xx are continuous at all points. Therefore, the function

)5)(3(2

−−−

xx

xis continuous at all those points where the denominator

does not take the value zero. Thus, it is continuous at all points

except at 3 and 5.

9.8 LET US SUM UP

l In this unit, we have learnt about the concept of limits and various

algebraic rules to find limits at various points of well-known functions.

We have also learnt about the left and right hand limits, and found that

they need not be equal always.

l We have learnt about the continuity of functions. In summary, we can

say that if the limit exists at a point and it is equal to the value of the

function at that point then the function is continuous at that point.




1. Khanna V. K., Zameeruddin Qazi & Bhambri S.K.(1995). Business

Mathematics, New Delhi, Vikas Publishing House Pvt Ltd. .

2. Hazarika P.L. Business Mathematics, New Delhi. S.Chand & Co.

9.10 ANSWERS TO CHECK YOURPROGRESS

Ans to Q No 1: (a) T, (b) T, (c) F, (d) F

Ans to Q No 2: (a) T, (b) F, (c) T.

Ans to Q No 3: (i) 0 (ii) 3, (iii) .3e

Ans to Q No 4: (a) 44, (b) No, (c) ,logaax (d) xe

Ans to Q No 5: (a) T, (b) T, (c) T, (d) T, (e) T, (f) F, (g) T.

Ans to Q No 6:

Ans to Q No 7:

Ans to Q No 8:

(2) None,

Ans to Q No 9: (a) F, (b) T, (c) T.

Ans to Q No 10: (a) F, (b) F, (c)T,

Ans to Q No 1 1: xxx −log

Ans to Q No 12: 4

2sin

2

xx + and .4

2sin

2

xx −

9.11 MODEL QUESTIONS

1. Evaluate the following limits:

(a) 34

127lim

2

2

3 +−+−

→ xx

xxx

(b ) x

xx

11lim

0

−+→



(c) x

xx 11sin

10sinlim

0→ (d) 55

1010

limkx

kxkx −

−→

2. Suppose a function f is defined as follows.

>−≥

=.0,

0,)(

xx

xxxf

By evaluating the left and right hand limits, show that .0)(lim0

=→

xfx

3. If f is an odd function (i.e., .),()( xxfxf ∀−=− ) and )(lim0

xfx→

exists, then show that the value of this limit is zero.

4. Decide whether the function given in Question 2 is continuous at

0=x or not.

5. A real function f is defined as follows.

=

≠=

.0,4

3

0,tan

)(x

xx

x

xf

Show that f is continuous at 0.

*** ***** ***



REFERENCES

l Sharma A.K. (2011), ‘Text book of differential calculus’. New Delhi,

Discovery Publishing House.

l Rao G.S.P. (2011), ‘An introductory mathematics to business &

economics’. New Delhi, Akansha Publishing House.

l. Mohanty R.K. (2004). Integral calculus. New Delhi, Anmol Publica-

tions Pvt. Ltd.

l Akilesh K.B. & Balasubrahmanyam S. (2009). Mathematics and

statistics for management. New Delhi, Vikash Publishiong House

Pvt. Ltd.

l Tamuli, B. K. : Contemporary Algebra (1977), New Book Stall,

Guwahati-781001.

l Pandey, R.K.,Differential Calculus, Jaipur, RBSA Publishers, 2009,1st

edition.

l Sharma, A.K., Advanced Differential Calculus,New Delhi, Discovery

publishing House, 2010,1st edition.

l Integral Calculus, B.C. Das and B.N. Mukherjee, U.N. Dhur & Sons

Pvt. Ltd., Kolkata.

l Integral Calculus, K.C. Maity & Ram Krishna Ghosh, New Central

Book Agency (P) Ltd., Kolkata.

l Integral Calculus, P.N. Chatterji, Rajhans Prakashan Mandir, Meerut.


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