Vedanta Opt. Maths Book 10 Final (2078)

Approved by the Government of Nepal, Ministry of Education, Science and Technology,
Curriculum Development Centre, Sanothimi, Bhaktapur as an Additional Learning Material

EvMexdcAaenlTtaiHn OEMptAioTnICaSl

10Book

Author
Piyush Raj Gosain

Hukum Pd. Dahal Editors P. L. Shah
Tara Bdr. Magar

vedanta

Vedanta Publication (P) Ltd.
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Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

vEMexdcAaenlTtaiHn OEMptAioTnICaSl

10Book

Author
Piyush Raj Gosain

All rights reserved. No part of this publication may
be reproduced, copied or transmitted in any way,
without the prior written permission of the publisher.

¤ Vedanta Publication (P) Ltd.

First Edition: B.S. 2077 (2020 A. D.)
Second Edition: B.S. 2078 (2021 A. D.)

Layout and Design
Pradeep Kandel

Printed in Nepal

Published by:

Vedanta Publication (P) Ltd.
j]bfGt klAns];g k|f= ln=

Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

Preface

The series of Vedanta Excel in Opt. Mathematics for class 9 and 10 is completely based on the

contemporary pedagogical teaching learning activities and methodologies. It is an innovative

and unique series in the sense that the contents of each textbooks of the series are written and

designed to ful ill the need of integrated teaching learning approaches.

Vedanta Excel in Opt. Mathematics has incorporated applied constructivism the latest trend of

learner centered teaching pedagogy. Every lesson of the series is written and designed in such a

manner that makes the classes automatically constructive and the learner actively participate in

the learning process to construct knowledge themselves, rather than just receiving ready made

information from their instructor. Even teachers will be able to get enough opportunities to play

the role of facilitators and guides shifting themselves from the traditional methods imposing

instructions. The idea of the presentation of every mathematical item is directly or indirectly

re lected from the writer's long experience, more than two decades, of teaching optional

mathematics.

Each unit of Vedanta Excel in Opt. Mathematics series is provided with many more worked out

examples, arranged in the hierarchy of the learning objectives and they are re lective to the

corresponding exercises.

Vedanta Excel in Opt. Mathematics class 10 covers the latest syllabus of CDC, the government of

Nepal, on the subject. My honest efforts have been to provide all the essential matter and practice

materials to the users. I believe that the book serves as a staircase for the students of class 10.

The book contains practice exercises in the form of simple to complex including the varieties of

problems. I have tried to establish relationship between the examples and the problems set for

practice to the maximum extent.

In the book, every chapter starts with review concepts of the same topic that the students have

studied in previous classes. Discussion questions in each topic are given to warm up the students

for the topic. Questions in each exercise are categorized into three groups - Very Short Questions,

Short Questions and Long Questions.

The project works are also given at the end of exercise as required. In my experience, the students

of class 10 require more practices on Trigonometry and Vector Geometry, the examples and the

exercise questions are given to ful ill it in the corresponding topics. The latest syllabus of the

subject speci ication grid and a model question issued by CDC are given at the end of the book.

My hearty thanks goes to Mr. Hukum Pd. Dahal, Tara Bahadur Magar and P.L. Shah, the series

editors, for their invaluable efforts in giving proper shape to the series. I am also thankful to my

colleage Mr. Gyanendra Shrestha who helped me a lot during the preparation of the book.

I am also thankful to my respected parents and my family members for their valuable support to

bring the book out in this form. I would also like to express my hearty gratitude to all my friends,

colleagues and beloved students who always encouraged me to express my knowledge, skill and

experience in the form of books. I am highly obliged to all my known and unknown teachers who

have laid the foundation of knowledge upon me to be such a person.

Last but not the least, I am hearty thankful to Mr. Pradeep Kandel, the computer and designing

senior of icer of the publication for his skill in designing the series in such an attractive form.

Efforts have been made to clear the subject matter included in the book. I do hope that teachers

and students will best utilize the series.

Valuable suggestions and comments for its further improvement from the concerned will be

highly appreciated. Piyush Raj Gosain

CONTENT

Unit 1 Functions 5
25
Unit 2 Polynomials 44
84
Unit 3 Sequence and Series 97
113
Unit 4 Linear Programming 128
154
Unit 5 Quadratic Equations and Graphs 199
269
Unit 6 Continuity 299
347
Unit 7 Matrices

Unit 8 Co-ordinate Geometry

Unit 9 Trigonometry

Unit 10 Vectors

Unit 11 Transformations

Unit 12 Statistics

Syllabus
Specification Grid
Model Questions

vedanta Excel In Opt. Mathematics - Book 10 1

Functions

1.0 Review

Study the following relations and state which are functions and which are not functions by
giving your reasons:

R1 = {(1,2), (2,3), (3,4)}
R2 = {1,2}, (1,4), (1,6)}
R3 = {(2,4), (3,6), (4,8)}
Discuss the following types of functions with examples related to our daily life:

(a) Onto Functions (b) Into Functions

(c) One to one Functions (d) Many to one Functions

1.1 Algebraic Functions

The functions which can be generated from a real variable x by a finite number of algebraic
operations (addition, subtraction, multiplication, division, and extraction of roots) are called
algebraic functions.

Quite often algebraic functions are algebraic equations.
Some examples of algebraic functions are
f(x) = 4x + 5, y = f(x) = x2 + 3x + 5

f(x) = 5 etc.

Note: function f(x) is also denoted by y. i.e., y = f(x)

Simple Algebraic Functions

The following are some simple algebraic functions: Y

(a) Linear Function (b) Constant Function 5 y=x+3
4
(c) Identity Function (d) Quadratic Function

(e) Cubic Function 3 (0, 3)
2
Let us define each of the above functions with their graphs.
1
X
(a) Linear Function X' (-3, 0) O 1 2 3 4 5

An algebraic function of the first degree expressed Y'
in the form of y = f(x) = mx + c is called linear
function. Here, m and c are constants. The linear
function always gives a straight line when plotted
in a graph.

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vedanta Excel In Opt. Mathematics - Book 10

In the above function, put m = 1 and c = 3,

we get y = x + 3.

To draw graph of y = x + 3, we take points as (1, 4) (-3, 0), (0, 3). The graph is shown

in above figure. Y

(b) Constant Function 4 y=4
An algebraic function that is expressed in the form

of y = f(x) = c i.e. y = c, (where c is a constant) is 3

called a constant function. It represents a straight 2

line parallel to X-axis. In constant function, the X' 1 X

value of y is constant for different values of x. O 1234

For example, y = f(x) = 4,

we get f(1) = 4, f(2) = 4, f(3) = 4.

The graph of y = 4 is shown in the figure, which Y'
is parallel to x-axis.

(c) Identity Function Y

An algebraic function expressed in the form of y=x is X' 2 y=x X
called identity function in which every x is associated 1
with itself. In this function, the function has same 45°
domain and range. -3 -2 -1 O 12 3
-1
The graph of identity function is straight line passing -2
through the origin and bisecting the angle between the
axes.

Note: An identity function is one to one onto function. Y'

(d) Quadratic Function

An algebraic function expressed in the form of y = f(x) = ax2 + bx + c, where a ≠ 0, a, b,
and c are constants, is called a quadratic function.

The graph of a quadratic function is a curve called parabola. The curve is symmetric
b b 4ac - b2
about the line x = - 2a with vertex = – 2a , 4a

The standard form of equation of parabola is y = a(x – h)2 + k,

For example : y = x2 – 6x + 8

Comparing it to y = ax2 + bx + c, Y

we get, a = 1, b = –6, c = 8

where vertex (h, k) = – b , 4ac - b2 y = x2 - 6x + 8
2a 4a

= – -6 , 4.1.8 - 36 = (3, -1)
2.1 4.1
from y = x2 - 6x + 8

x0123456 X' O X
y 8 3 0 -1 0 3 8

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vedanta Excel In Opt. Mathematics - Book 10

(e) Cubic Function Y

A function expressed in the form of

y = ax3 + bx3 + cx + d, where a ≠ 0, a, b, c and d are
constants, is called a cubic function.


Examples of cubic functions are
y = x3, y = x3 + 3x, y = 4x3 + 3x2 + 4x + 5 etc.


Let's as draw a graph of cubic function. y = x3 . X

x –2 –1 0 1 2 X'
y –8 –1 0 1 8




Its graph is given as shown in the figure.






1.2 Trigonometric Functions

Y'

A function involving trigonometric ratios like sine, cosine, tangent etc. is called trigonometric function.
Trigonometric functions are related to the angles and their measurements. Trigonometric
functions of a right angled triangle are defined in terms of ratios of sides of right angled triangle.
1. Discuss the values of trigonometric ratios of angles from -2Sc to 2Sc.
2. Note the maximum and minimum values of the trigonometric ratios of sine, cosine

and tangent.
3. Do the values of the trigonometric function repeat for different values of angles ?

Discuss it.
Trigonometric functions do not satisfy the algebraic operations like addition, subtraction,
multiplication, and division. They are different from algebraic functions.
The functions which are not algebraic are called transcendental functions. Trigonometric
functions are also transcendental functions.
Trigonometric functions are periodic functions. A function f which satisfies
f(x+k) = f(x) for all x belonging to its domain and smallest positive value of k.
Examples : sin(x + 2Sc) = sinx

cos(x +2Sc) = cosx
tan(x+Sc) = tanx
We can say that sinx, cosx, and tanx are the periodic function of periods 2Sc, 2Sc, and Sc
respectively.

Graphs of Trigonometric Functions

(a) Graph of y = sinx, (-2Sc ≤ x ≤ 2Sc)
For the sine function, y = sinx, take the values of x from -2Sc to 2Sc, we get the
corresponding values of y as shown in the table.

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vedanta Excel In Opt. Mathematics - Book 10

x -2Sc -3Sc -Sc -Sc 0 Sc Sc 3Sc 2Sc
2 2 2 2

y 0 1 0 -1 0 1 0 -1 0

What are maximum and minimum values of sinx ?

Plot the points (-2Sc,0), – 3Sc , 1 , Y
2
Sc Sc
(Sc, 0), – 2 , - 1 , (0, 0) 2 , 1 ,

(Sc, 0), 3Sc , –1 , and (-2Sc, 0) in a -1
2
X' X
graph paper and join the points. We -2cS 3Sc O Sc Sc
- 2 -Sc Sc -1 2 3Sc 2Sc
- 2 2

get a curve as shown in the figure

alongside.

(b) Grph of y = cosx, (-2Sc ≤ x≤ 2Sc) Y'

For y = cosx, take the values of x from -2Sc to 2Sc, we get corresponding values of y as
shown in the table.

x -2Sc -3Sc -Sc -Sc 0 Sc Sc 3Sc 2Sc
2 2 1 2 2

y 1 0 -1 0 0 -1 0 -1

Note the maximum and minimum values of cosx. Plot the above points in the graph
paper and join the points and we get the curve in the figure below :

Y

-1

X' -2Sc 3Sc 0 Sc X
2 -1 2
- -Sc - 3 Sc 3Sc 2Sc
Sc 2

Y'

(c) Graph of y = tanx, (-2Sc ≤ x ≤ 2Sc)

For y = tanx, the value of y is not defined for the values x = - 32Sc, - Sc , Sc , 3Sc .
2 2 2

Image of the function for these values of x are not defined.

x -2Sc -7Sc -3Sc -5Sc -Sc -3Sc -Sc -Sc 0 Sc Sc 3Sc Sc 5Sc 3Sc 7Sc 2Sc
4 2 4 4 2 4 4 2 4 4 2 4

y 0 1 ∞ -1 0 1 ∞ -1 0 1 ∞ -1 0 -1 ∞ -1 0

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vedanta Excel In Opt. Mathematics - Book 10

Y (1.249, 3)
4
3

2

1

X' 3Sc O X
- 2
-Sc - Sc Sc Sc 3Sc 2Sc
2 2 2
-1

-2

-3
Y'

The graph of tanx is shown in the above graph.

Worked out Examples

Example 1. Name the following algebraic functions.

Solution: (a) y = 5 (b) y = ax + b,
Example 2.
(c) y = x3 + 2x2 (d) y = 2x2 + x + 5

(a) Constant Function (b) Linear Function

(c) Cubic Function (d) Quadratic Function

Name the algebraic functions of the following graphs. Y
Y

X' O X X' O X

Y' Y'
(a) The given graph represents linear function.
Solution : (b) The given graph represents quadratic function.

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vedanta Excel In Opt. Mathematics - Book 10

Exercise 1.1

Very Short Questions
1. Classify the following functions:

(a) y = x (b) y = x2 (c) y = x3

(d) y = ax2 + bx + c (e) y = x3 - x2 (f) y = 7

(g) y = tanx

2. Define a function with an example.

3. Define an algebraic function.

4. What is a trigonometric function ?

5. What is a transcendental function ?

6. What is the minimum value of the cosine function y = cosx ?

7. Write the periods of the following functions:

(a) y = sinx (b) y = cosx (c) y = tanx

8. For what values of x, the image of tangent function y = tanx is not defined ?

9. Define each of following functions with an example:

(a) Linear Function (b) Constant Function (c) Identity Function

(d) Quadratic Function (e) Cubic Function (f) Periodic Function

10. Name the functions represented by the following graphs: Y
(a) Y (b)
4
2 3
1 2
1

45° X' X
X' X
-3 -2 -1 O 12 3 -3 -2 -1 O 12 3
Y
-1 -1
-2 -2
-3
Y' -4
(c) Y (d)
Y'

-1 -1

X' 0 X X' -2Sc 3Sc 0 Sc Sc X
2 -1 2
-2Sc - 3Sc -Sc - 3 Sc Sc 3Sc - -Sc - Sc 3Sc 2Sc
2 Sc -1 2 2 2 2

10 Y' Y'

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vedanta Excel In Opt. Mathematics - Book 10

Y

(e) Y (f)













X' 0 X

X' O X



y = -3
Y'








Short Questions



11. Draw graph for each of the following table: Y'

(a) x 2 3 4 5 6 (b) x 0 ±1 ±2 ±3
y 4 6 8 10 12 y0149

(c) x -3 -2 0 1 (d) x 1234
y -1 -2 -3 -4
y 6 4 0 -2

12. Draw graph of each of the following functions:

(a) y = 2x - 1 (b) y = x + 1 (c) y = x2

(d) y = -x (e) y = x3 (f) y = -x3

13. Draw the graphs of the following trigonometric functions:

(a) f(x) = sinx, (-Sc ≤ x ≤ Sc) (b) f(x) = cosx, (-Sc ≤ x ≤ Sc)

(c) f(S) = tanx, (-Sc ≤ x ≤ Sc)

Long Questions

14. Draw the graphs of the following trigonometric functions:

(a) f(x) = sinx, (-2Sc ≤ x ≤ 2Sc) (b) f(x) = cosx, (-2Sc ≤ x ≤ 2Sc)

(c) f(x) = tanx, (-2Sc ≤ x ≤ 2Sc)

Project Work

15. Record your body weight continuously for a week at the same time in the morning.
Taking days in X-axis and weight in Y-axis, draw the graph of it and discuss the type of
function represented by the graph drawn.

16. List the price of petrol per litre for a week. Taking days in X-axis and the price in Y-axis,
represent the information obtained in graph. Discuss the type of function represented
by the graph drawn.

17. Where are the functions f(x) = sinx and f(x) = cosx used in our daily life ? Investigate
and prepare a report and present in your class.

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vedanta Excel In Opt. Mathematics - Book 10

1. (a) Identity function (b) quadratic (c) cubic (d) quadratic
(g) trigonometric
(e) cubic (f) constant 6. –1
(c) Sc
7. (a) 2Sc (b) 2Sc (c) y = cosx 8. Sc
10. (a) identity (b) quadratic 2
(f) cubic
(e) constant (d) y = sinx

1.3 Composite Function

Let A = {1, 2, 3}, B = {2, 3, 4}, and C = {4, 6, 8} be three sets. Let us define two functions
f and g such that f : A → B and g : B → C such that

f = {(1, 2), (2, 3), (3, 4)} and g = {(2, 4), (3, 6), (4, 8)}

By representing above function in mapping diagram, we write it as below.

fg
A BC

1 24
2 36
3 48

gof

Here, f is a function defined from A to B. So write.
1  A o f(1) = 2  B
2  A o f(2) = 3  B

and 3  A o f(3) = 4  B
range of f = {2, 3, 4} which is domain for g.
Again, g is a function from B to C.

2  B o g(2) = 4  C
3  B o g (3) = 6  C
4  B o g (4) = 8  C
Now, let us define a new function from A to C. Such that,
1  A o 4 = g(2) = g(f(1))  C
2  A o 6 = g(3) = g (f(2))  C
4  B o 8 = g(4) = g (f(3))  C

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vedanta Excel In Opt. Mathematics - Book 10

The new function defined from A to C is called composite A gof
function of f and g, and it is denoted by gof. we can draw for gof
as below. 1 C
2
Writing the above composite function as a set of ordered pair, 3 4
6
we get, gof = {(1, 4), (2, 6), (3, 8)} 8

Definition

Let f : A → B and g: B → C be two functions. Then a new function defined from set A to

set C is called composite function of f and g. It is A B C

denoted by gof and defined by gof : A → C.

Then every element of A is associated with a x f(x) g(f(x))
unique element of set C.

Symbolically, we write.

If f : A → B and g: B → C, then the composite gof
function from A to C is defined by

gof : A → C, for every x  A, (gof) (x) = g (f(x))  C

Note :
(i) The composite function gof is read as "f followed by g".

(ii) In general gof ≠ fog

(iii) If (fog) (x) = x and (gof) (x) = x, then, the composite function is called identity
function.

Worked Out Examples

Example 1. If f = {(1, 1), (2, 4), (3, 9)} and g = {(1, 2), (4, 8), and (9, 18)}, then, show the
function gof in arrow diagram and find it in the set of ordered pairs.

Solution: Here, f = {(1, 1), (2, 4), (3, 9)}

g = {(1,2), (4,8), (9,18)}

Arrow diagram of composite function gof is given below,

f g
BC
A

11 2
24 8
3 9 18

gof 13

Writing above composite function in the ordered pairs, we get
gof = {(1,2), (2, 8), (3, 18)}

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vedanta Excel In Opt. Mathematics - Book 10

Example 2. Let f : R → R, f(x) = 4x + 3 and g : R → R, g(x) = x2 + 4, Find the following:
Solution:
(a) (fof) (x) (b) (gog) (x)

(c) (fog) (x) (d) (got) (x)

(e) (gof) (1) (f) (fog)(2)

Also, check is (gof) (x) = (fog)(x) ?

Here, f(x) = 4x + 3, g(x) = x2 + 4

(a) (fof)(x) = f(f(x)) = f(4x + 3) = 4(4x+3) + 3

= 16x + 12 + 3

= 16x + 15

(b) (gog)(x) = g(g(x)) = g(x2 + 4) = (x2 + 4)2 + 4

= x4 + 8x2 + 16 + 4

= x4 + 8x2 + 20

(c) (fog)(x) = f(g(x)) = f(x2 + 4) = 4(x2 + 4) + 3

= 4x2 + 16 + 3

= 4x2 + 19

(d) (gof) (x) = g(f(x))= g(4x + 3) =(4x + 3)2 + 4

= 16x2 + 24x + 9 + 4

= 16x2 + 24x + 13

(e) (gof) (1) = g(f(1)) = g(4.1 + 3) Alternative Method
= g(7) we have,

= 72 + 4 (gof)(x) = 16x2 + 24x + 23

= 49 + 4 (gof)(1) = 16.12 + 24.1 + 13

= 53 = 53

(f) (fog)(2) = f(g(2)) Alternative Method
= f(22 + 4) we have,

=f(8) (fog)(x) = 4x2 + 19

=4.8 + 3 ? (fog)(2) = 4.22 + 19

= 32 + 3 = 16 + 19

= 35 = 35

From above, we get,

(fog) (x) = 4x2 + 19

(gof) (x) = 16x2 + 24x + 13

? (gof)(x) ≠ (fog)(x)

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vedanta Excel In Opt. Mathematics - Book 10

Example 3. If f: R → R, and g: R → R are defined by f(x) = 2x and g(x) = 3x + 4 and
Solution: (gof)(x) = 22, find the value of x.

Example 4. Here, f(x) = 2x and g(x) = 3x + 4
Solution:
Now, (gof)(x) = g(f(x)) = g(2x) =3(2x) + 4 = 6x + 4
Example 5.
Solution: By question, (gof)(x) = 22

Example 6. ie. 6x + 4 = 22
Solution:
or, 6x = 22 - 4

or, x= 18
6
? x = 3

If f(x) = px + 4 and g(x) = 5x and (gof)(2) =4, find the value of p.
2
px + 4
Here, f(x) = 2 and g(x) = 5x

Now, (gof)(2) = 4

or, g(f(2)) = 4

or, g p.2+4 =4
2

or, g(p + 2) = 4

or, 5(p + 2) = 4

or, p+2= 4
5
4
or, p = 5 -2

or, p = 4 - 10 =- 6
5 5
6
? p =- 5

If (fog)(x) = 7x - 1 and f(x) = 3x + 5, find g(x) where g(x) is a linear function.
3
We have, f(x) = 3x + 5

Now, (fog) (x) = 7x - 1
3
7x - 1
or, f[g(x)] = 3

or, 3g(x) + 5 = 7x- 1
3
7x - 1
or, 3g(x) = 3 – 5

g(x) = 7x - 16
9

If f: R → R ; f(x) = 8 - x, Show that (fof)(x) = x.

Here, f(x) = 8 - x

(fof)(x) = f(8 - x) = 8 - (8 - x) = 8 - 8 + x = x

? (fof)(x) = x, proved

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vedanta Excel In Opt. Mathematics - Book 10

Exercise 1.2

Very Short Questions

1. Define a composite function of two functions.

2. Write the composite function of given functions f and g as shown in the mapping
diagram.

f g C

A B

x yz

3. From the adjoining figure, write the composite function gof of function f and g in the
ordered pair form.

f g C
B
A

ad x

be y

cf z

4. Write the composite function gof of the following functions in the ordered pair form:

(a) f = {(1, 2), (3, 4), (4, 5)} and g = {(2,6), (4,12), (5,15)}

(b) f = {(1,2), (3,4), (5,6)} and g = {(2,3), (4,6), (6, 8)}

(c) f = {(3,4), (4,5), (5,6)} and g = {(4,5), (5,6), (6.7)}

5. Write the composite function fog of the following functions in the ordered pair form:

(a) f = {(3,4), (4,5), (5,6)} and g = {(2,3), (3,4), (4,5)}

(b) f = {(4,2), (8,4), (16, 8)} and g = {(2,4), (4, 8), (8, 16)}

(c) f = {(2, 8), (4, 64), (8, 256)} and g = {(1, 2), (2, 4), (4, 8)}

Short Questions

6. (a) f = {(1,3), (2,1), (3,2)} and g = {(1,2), (2,3), and (3,1)}, find the domain and range
of the composite function gof.

(b) If f = {(1,2), (3,5), (4,1)} and g = {(2,3), (5,1), and (1,3)}, find the domain and
range of the composite function gof.

7. (a) If f = (1,5), (2,1), (3,3), (5,2)} and g= {(1,3), (2,1), (3,2), (5,5)}, find gof and fog.

(b) If h = {(5,2), (6,3), k = {(2,5), (3,6)} find hok and koh.

8. (a) From the function f = {(4,1), (5,3), (6,2)} and g = {(1,2), (3,6), (2,4)}, show gof in
arrow diagram and write it in ordered pair form.

(b) From the functions f = {(a,b), (b,c), (c,d)} and g = {(b,e), (c,f), (d,h)}, find gof in
ordered pair form and also show in mapping diagram.

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vedanta Excel In Opt. Mathematics - Book 10

9, If (fog) = {(2,1), (4,2), (6,3), and (8,4)} and g = {(2,4), (4,8), (6,12), and (8,16)}, find f in
ordered pair form. Also show fog in arrow diagram.

10. (a) If f(x+2) = 4x + 5, find f(x) and (fof) (x)

(b) If g(x+5) = x + 20, find g(x) and (gog) (x).

Long Questions

11. (a) Let f: R → R and g : R → R be defined by f(x) = x + 1 and g(x) = x3. Then (i) find
gof and fog. (ii) Is (gof) (x) = (fog)(x)?

(b) Let f: R → R and g: R→R be defined by f(x) = x3 - 1 and g(x) = x2

(i) Find (gof) (x) and (fog) (x) (ii) Is (gof)(x) = (fog)(x)

12. (a) If f: R → R and g: R → R are defined by f(x) = 2x + 3 and g(x) = 2x - 1,

find (gof) (3) and (gof)(-2).

(b) If f: R → R and g: R → R be defined by f(x) = x + 5 and g(x) = 5x - 5. Find (gof)(-2)
and (gog)(2).

13. (a) Find the value of x if f(x) = 4x + 5 and g(x) = 6x + 3, (gof)(x) = 75

(b) Find the value of x if f(x) = 3x + 4 and g(x) = 2x + 5, (fog)(x) = 25.

14. (a) If f(x) = px + 8 and g(x) = 3x + 5, (gof)(3) = 60, find the value of p.

(b) If f(x) = 7x + q and g(x) = x + 8, (fog) (4) = 24, find the value of q.
(c) If g(x) = 62 and h(x)= ax2 - 1 and (goh)(3)= 17, find the value of a.

x-2

(d) If f(x)= 3x + a and g(x) = 4x + 5 and (gof)(5) = 20, find the value of a.

(e) If f(x) = 3x , g(x)= Ex - 4, and (gof)(1) = 10, find the value of E.
x+2

15. (a) If f(x) = 4x - 5 and g(x) is a linear function and (gof) (x) = 5x + 1, find g(x).

(b) If g(x) = 5x + 3 and (gof)(x) = 2x + 5 and f(x) is a linear function, find the value
of f(x).

(c) If g(x) = 2x and (fog)(x) = 6x-2, find f(x), where f(x) is a linear function.

(d) If f(x) = 2x + 3 and (fog) (x) = 10x + 1, find the value of g(x).

(e) If f(x) = 2x + 8, (fog)(x) = 3x+4, find the function g(x).

(f) If f(x)= 2x -1, g(x) is a quadratic function and (fog)(x) = 6x2+4x + 1, find g(x).

16. (a) If g(x) = 10 - x, show that (gog)(x) = x.

(b) If h(x) = 20 -x, show that (hoh)(x) = x

17. (a) If p(x) = 4x+5 and q(x)= 9x-5 prove that (poq)(x) is an identity function.
9 4
3x+2
(b) If p(x) = 3 and q(x) = 3x-2 , prove that (poq) (x) is an identity function.
(a) If g(x3) = {(3,1), (5,1), (7,1)} prove that (gof) is constant
18. f = {(1,3), (4,5), (6,7)} and

function.

(b) If f = {(2,4), (4,5), (7,10)} and g = {(4,2), (5,2), (10,2)}. Show that gof is a constant
function.

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Project Work

19. Number of bacteria in food kept in a refrigerator is expressed by N(T)=20T2 - 80T +
500, (2 ≤ T ≤ 14), where T denotes temperature and
T(t) = 4t + 2, (0 ≤ t ≤ 3), where t represents time in hour.
(a) Find (NoT) (t)
(b) What is the number of bacteria in 2 hours?
(c) In how many hours does the number of bacteria in the food reach 3300 ?

2. gof 3. gof = {(a, x), (b, y), (c, z)}

4. (a) {(1, 6), (3, 12), (4, 15)} (b) {(1, 3), (3, 6), (5, 8)}

(c) {(3, 5), (4, 6), (5, 7)} 5. (a) {(2, 4), (3, 5), (4, 6)}

(b) {(2, 2), (4, 4), (8, 8) (c) {(1, 8), (2, 64), (4, 256)}

6. (a) D = {1, 2, 3), R = {1, 2, 3} (b) D = {1, 3, 4}, R = {3, 1}

7. (a) gof = {(1, 5), (2, 3), (3, 2), (5, 1)} fog = {(1, 3), (2, 5), (3, 1), (5, 2)}

(b) hok = {(2, 2), (3, 3)} koh = {(5, 5), (6, 6)}

8. (a) gof = {(4, 2), (5, 6), (6, 4)} (b) fog = {(a, e), (b, f), (c, h)}

9. F = {(4, 1), (8, 2), (12, 3), (16, 4)} 10.(a) 4x – 3, 16x – 15 (b) x + 15, x + 30

11. (a) gof = x3 + 3x2 + 3x + 1, fog = x3 + 1, No. (b) gof = x6 – 2x3 + 1, fog = x6 – 1, No.

12. (a) 17, –3 (b) 10, 20 13. (a) 7 (b) 1
4
31 19 45
14. (a) 9 (b) –36 (c) 51 (d) – 4 (e) 14

15. (a) 5x + 29 (b) 2(x + 1) (c) 3x – 2 (d) 5x – 1
4 5
3x
(e) 2 – 2 (f) 3x2 + 2x + 1 19. (a) 320t2 + 420 (b)1700 (c) 3 hours

1.4 Inverse Function

Let A = {1, 2, 3} and B = {1, 8, 27} be two sets and the function f is defined from A to B by
f = {(1,1), (2,8), (3,27)}

A fB

11
28
3 27

Here, domain of f = {1,2,3}
range of f = {1, 8, 27}

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co-domain of f = {1, 8, 27}
since range of f = co-domain of f, f is one to one onto function.
(or bijective function)
Now, let us define a new function g from B to A so that the domain and range of g are
obtained by interchanging the domain and range of f respectively.
Then g = {(1,1), (8,2), (27,3)}

g

11
82
27 3

domain of g = {1, 8, 27}
range of g = {1, 2, 3}
Here, g is also one onto function. g is called inverse of function of f.
This new function defined from B to A is called inverse of f. It is denoted by f -1.
Therefore the function must be one to one onto, to have its inverse function.
Let us take another function g as shown in the mapping diagram.

g

1x
2y
3z

Here, g is not a onto function. It is many to one into function.
Now, let us interchange the domain and range of g. Is it again a function ?

g -1

x1
y2
z3

This g-1 does not represent a function.

Hence, to have inverse of a function, the function must be one to one onto, i.e., bijective.

A function f is called bijective if it is one to and onto. Only the bijective functions have
their inverse functions.

Remember, f-1 ≠ 1 and f-1 (x) ≠ 1
f f(x)
Definition:

Let f: A → B be a one to one onto function. Then, the function f-1 : B → A is called inverse of
f for every x  B there corresponds f-1 (x) = y  A

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In functions f and f-1, the roles of x and y are interchanged.

Steps for finding inverse of an algebraic functions:
(i) Put y in place of f(x) i.e. y = f(x)
(ii) Interchange the roles of x and y.
(iii) Get new y in terms of x solving equation obtained in (ii).

Note :
(i) If the two functions f and g are inverse to each other, f-1 = g or g-1 = f and

(gof)(x) = x, (fog)(x) = x.
(ii) If f is a bijective function, (f-1)-1 = f.
(iii) If f–1 exists, then, domain of f is range of f–1 and range of f is domain of f–1.

Worked Out Examples

Example 1. Let f: A → B be a bijective function defined by f = {(1,2), (2,3), (3,4),(4,5)}.
Solution: Find f-1 in ordered pair form.
Example 2.
Solution : Here, f = {(1,2), (2,3), (3,4), (4,5)} Interchanging the elements of ordered
pairs, we get f-1 = {(2,1), (3,2), (4,3), (5,4)}

Let f: A → B be one to one and onto function, defined by f = {(2,4), (3,6), (4,8),
(5,10)}. Find f-1 and show f and f-1 in mapping diagrams.

Here, f = {(2,4), (3,6), (4,8), (5,10)}. Interchanging the components in each
ordered pair, we get f-1 = {(4,2), (6,3), (8,4), (10,5)}

In mapping diagrams, we show that

Af B B f A

2 44 2
3 66 3
4 88 4
5 10 10 5

Example 3. (a) If f(x) = 7x - 2, find f-1 (x)
Solution:
(b) If g : 5x - 3 , x ≠ - 1 , find g-1 (x) and g-1 (2)
2x + 1 2

(a) Here, f(x) = 7x - 2

To find f-1, y = 7x - 2

Interchanging the roles of x and y, we get

x = 7y - 2

or, x + 2 = 7y

y= x+2
7
x+2
? f-1 (x) = 7

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(b) Here, g (x) = 5x - 3
2x+ 1
To find g-1 (x), interchanging the roles of x and y, we get

x = 5y-3
2y+1
or, 2xy + x = 5y - 3

or, 2xy - 5y = - x - 3

or, y(2x - 5) = - (x + 3)

or, y = - (x+3)
2x - 5
x+3 5
y = 5 - 2x , x ≠ 2

g-1 (x) = x+3
5 - 2x
2+3 5
Also, g-1 (2) = 5-2×2 = 5 - 4 =5

? g–1 (2) = 5

Example 5. If f = {(x,y) : y = 7x - 4} find f-1 (x) and f-1 (4).
Solution:
Here, y = 7x - 4, to find f-1(x), interchanging the roles of x and y, we get
Example 6.
Solution: x = 7y - 4

or, 7y = x + 4

y = x+4
7
x+4
? f-1 (x) = 7

Now, f-1 (4) = 4 + 4 = 8
7 7
8
? f–1(4) = 7

If f = {(x, y) : y = 3x - k} and f-1(5) = 4, find the value of k.

Here, f = {(x,y) : y = 3x - k}

y = 3x - k

i.e. f(x) = 3x - k

To find f-1(x), interchanging the roles of x and y, we get,

x = 3y - k

or, y = x + k
3
x+k
f-1(x) = 3

and f-1 (5) = 5+k
3
But, f-1(5) = 4

or, 5+k =4
3

or, 5 + k = 12

? k=7

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Example 7. Given a function f(x+3) = 3x + 5, find f-1(x)
Solution:
Here, f(x+3) = 3x + 5
Example 8.
Solution: replacing x by x - 3, we get,

f(x - 3 + 3) = 3(x - 3) + 5

or, f(x) = 3x - 4 or, y = 3x - 4

To find f-1 (x) ,

Interchanging the roles of x and y, we get,

x = 3y - 4

or, y = x+4
3
x+4
? f-1(x) = 3

Alternate Method.

Here, f(x + 3) = 3x + 5

or, f(x + 3) = 3 (x + 3) - 4

f(D) = 3D - 4

where, D = x + 3

Since D is a dummy variable, it can be replaced by any other variable.

f(x) = 3x - 4 or, y = 3x - 4

To find f-1 (x),

interchanging the role of x and y, we get,

x = 3y - 4

or, y = x+4
3
x + 4
? f-1(x) = 3

If f(2x + 3) = 6x + 7, show that (fof-1) (x) is an identity function.

Here, f(2x + 3) = 6x + 7

x is replaced by x , we get,
2

( )f2. x =6× x 7
2 +3 2 +

or, f(x + 3) = 3x + 7

Again, x is replaced by x - 3, we get,

f(x) = 3(x - 3) + 7

or, f(x) = 3x - 2 or, y = 3x - 2

To find the f-1 (x), interchanging the role of x and y, we get,

x = 3y – 2

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y= x+2
3

( )Now, (fof-1) (x) = f (f-1 (x)) = f x
x+2 = 3. + 2 –2
3 3

= x + 2 - 2 = x.

Since (fof-1) (x) = x

Therefore, (fof-1)(x) is an identity function.

Exercise 1.3

Very Short Questions
1. Define one-to-one function with an example.
2. What is the required condition for existence of inverse of a function ?
3. Do all the functions have their inverses ?
4. Find the inverse functions of the following functions.

(a) f = {(a,x), (b,y), (c,z)}
(b) g = {(2,6), (3,9), (4, 12), (5, 15)}
(c) h = {(1,2), (2,3), (3,4), (4,5), (5,6)}
5. If f = {(2, 1), (4, 1), (5, 1)} Does f-1 exist ?
6. Let f : R →R be defined by f(x) = 2x - 3, find the formula that defines the inverse
function f-1.
7. Let f : R → R, then, find the inverse function in each of the following cases:
(a) f = {(x,y): y = 4x + 5}

{ }(b) g = 3x-1 3 ,
(x,y) : y = 3-2x , x≠ 2

{ }(c) h =
(x,y) : y = 4x+3 , x ≠ 1 ,
x-1
8. If f = {(a,x), (b,y), (c,z), (d,w)}, then, show f and f-1 in a mapping diagram.

Short Questions

9. (a) If g(x) = 3x - 5, the find g-1 (5), where g is a one to one onto function.

(b) If h(x) = 2x-3 , find h-1 (4), where h(x) is one to one.
4
2x+8
10. (a) If g(x) = 5 , find g-1(x) and g-1(10).

(b) If k(x) = 5x+7 , x ≠ 2, find k-1 (x) and k-1(4).
x-2
11. (a) If f(x) = 2x + k, f-1(4) = 20, find the value of k.

(b) If f(x) = 7x + k, f-1(3) = 12, find the value of k.

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12. (a) If f(x+2) = 4x + 5, find f-1(x)
(b) If f(2x + 5) = 10x + 7, find f-1(x)
(c) If f(4x + 5) = 12x + 20, find f-1 (x)

13. (a) If f(x + 2) = 7x + 13, find f-1(2)
(b) If f(x + 5) = 10x + 11, find f-1(4)
(c) If f(4x + 5) = 20x + 24, find f-1(x).

Long Questions

14. (a) If f and g are two one-to-one and onto functions defined by f(x) = 2x – 3 and
g(x) = 4x + 5, (fof)(x) = g-1 (x), find the value of x.

(b) If f(x) = 4x - 7 and g(x) = 3x - 5 are two one to one onto functions, and (fog-1)
(x) = 15, find the value of x.

(c) If f(x) = x -2 and g(x) = 1 and f-1(x) = (gof)(x), find the value of x.
2x +1 x
2x + 8
(d) If f(x) 4x-17 and g(x)= 5 and (fof)(x)= g-1(x), find the value of x.

15. (a) If f(x) = 2x - 4, then prove that, (fof-1)(x) is an identify function.

(b) If f(x) = 2x + 5 , x ≠ - 2, then, prove that (fof-1)(x) is an identity function.
x+2
2x + 1,
(c) If f(x) = 4 prove that (f-1of)(x) is an identify function.

16. If f(x) = x + 1 and g(x) = 3 - x , x ≠ 0, are two one-to-one onto function, then prove
that (f-1og-1)(2) = 0 x

17. If f = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}, g = {(4, 9), (5, 11), (6, 13), (7, 15), (8, 17)} and
3
(fog)–1(x) = x + 2, find the positive value of x.

Project Work

18. State the formula of conversion of temperature from degree Celsius to degree Fahrenheit
as a function. Write the inverse of the function. Convert 37°c into °F and vice versa.

2. One one and onto function 3. No.

4. (a) {(x, a), (y, b), (z, c)} (b) {(6, 2), (9, 3), (12, 4), (15, 5)}

(c) {(2, 1), (3, 2), (4, 3), (5, 4), (6, 5)} 5. does not exist

6. x+3 7.(a) x–5 (b) 3x + 1 , x ≠ – 3
2 4 2x + 3 2
x+ 3 10 19
(c) x– 4 , x ≠ 4 9.(a) 3 (b) 2

10. (a) 21 (b) –15 11.(a) –36 (b) –81
(d) 6
12. (a) x+3 (b) x + 18 (c) x–5
4 5 3
3 43 x+1
13. (a) 7 (b) 10 (c) 5

14. (a) 31 (b) 23 (c) ± 1 17. 4
15 2

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vedanta Excel In Opt. Mathematics - Book 10 7

Matrices

7.0 Review

Let us consider the following matrices:

P= 1 26 ,Q= 2 03 , R= 1 2 , S= 2 -2 4
3 21 4 12 3 5 32 1
0 32 2 1

Then answer the following questions.

(a) Write the orders of each of above matrices.

(b) Which of above matrices can be added ?

(c) Are PQ and QP defined ?

(d) Write QT and ST.

(e) Is Q2 = Q.Q ?

(f) Find 2P - 3S.
Discuss the answers of the above questions and conclude the conclusions.

7.1 Determinant of a Matrix

Let A = a b be a square matrix.
c d

Then write |A| = a b . It represents a number associated with given matrix A. It is
c d
called determinant of A. We also write det. A to represent |A|.

It is defined by |A| = a b =ad - bc.
c d

Example : Let A = 2 3
4 5

Then |A| = 2 3 = 2 × 5 - 4 × 3 = 10 - 12 = - 2
4 5

Definition: A determinant is a number associated with a square matrix. When the elements
of a square matrix are enclosed within vertical lines || without changing their positions, it
is called determinant of the given matrix. The determinant of a matrix A is denoted by |A|
or determinant of A or det. (A).

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Let A = a b then,
c d

|A| = a b = ad - bc
c d

Example : Let A = 1 2 , its determinant is given by ,
3 4

|A| = 1 2 = 1 × 4 - 3 × 2 = 4 - 6 = -2
3 4

Determinant of a square matrix of order 1.

Let A = [7], it is a square matrix of order 1.

Then |A| = |7| = 7

If B = [-7] then |B| = |-7| = - 7

Note : Absolute value of -7 = |-7| = 7.
det. [-7] = |-7| = - 7

Determinant of square matrix of order 2

Let A = 1 4 . It is a square matrix of order 2 × 2.
6 8
Then, its determinant is given by,

|A| = 1 4 = 1 × 8 - 6 × 4 = 8 - 24 = - 16

68

Example : Evaluate 2 5
7 3

Here, 2 5 = 2 × 3 - 7 × 5 = 6 - 35 = - 29
7 3

Worked out Examples

Example 1. Evaluate the following :

(a) |-10| (b) 2 3 (c) 1 -cos2x
4 5 1 sin2x
Solution: (a) |-10| = - 10, (determinant of (-10) = -10)

(b) 2 3 = 2 × 5 - 4 × 3 = 10 - 12 = -2
4 5

(c) 1 -cos2x = sin2x + cos2x = 1
1 sin2x

Example 2. Evaluate the determinant of the following matrices.

(a) [-20] (b) 4 5 (c) 2 4
6 7 -3 3

Solution: (a) Determinant of [-20] = |-20| = -20

(b) Determinant of 4 5 = 4 5 = 4 × 7 - 6 × 5 = 28 - 30 = -2
6 7 6 7

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(c) Determinant of 2 4 = 2 × 3 - (-3 × 4) = 6 + 12 = 18.
-3 3

Example 3. Find the value of x under the given conditions:

(a) 4 5 =x (b) x 3+2x =0
3 6 1 x

Solution: (a) Here, 4 5 =x
3 6
or, 4 × 6 - 3 × 5 = x

or, 24 - 15 = x

? x=9

(b) Here, x 3+2x =0
1 x
or, x2 - 3 - 2x = 0

or, x2 - 2x - 3 = 0

or, x2 - 3x + x - 3 = 0

or, x(x - 3) + 1 (x - 3) = 0

or, (x - 3) (x + 1) = 0

Either, x - 3 = 0 ? x=3

or, x + 1 = 0 ? x=-1

? x = -1,3

Example 4. If P = 2 3 and Q = 5 2 then find the determinant of,
Solution: 4 5 3 1
(a) 2P - 3Q (b) P2 - Q2

(a) Here, P = 2 3 ,Q= 5 2
4 5 3 1

Now, 2P - 3Q =2 2 3 -3 5 2
4 5 3 1

= 4 6 - 15 6
8 10 9 3

= 4-15 6-6 = -11 0
8-9 10-3 -1 7

Determinant of (2P - 3Q) = |2P - 3Q|

= -11 0 = -77 - 0 = - 77
-1 7
(b) For determinant of P2 - Q2

Here, P2 = P.P = 2 3 23
4 5 45

= 4+12 6 + 15 = 16 21
8+20 12+25 28 37

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Q2 = Q.Q = 5 2 5 2 = 25+6 10+2 = 31 12
3 1 3 1 15+3 6+1 18 7

Now, P2 - Q2 = 16 21 - 31 12
28 37 18 7

= 16-31 21-12 = -15 9
28-18 37-7 10 30

Determinant of (P2 - Q2 ) = |P2 - Q2| = -15 9
10 30

= –450 – 90 = –540

Example 5. If P = 2 1 , Q= 4 6 ,I= 1 0 , find the determinant of
Solution: -1 3 2 4 0 1
1 5
5P - 2 Q + 3I 15

Here, 5P = 5 2 1 = 10
-1 3 -5

1 Q = 1 4 6 = 2 3
2 2 2 4 1 2

3I =3 1 0 = 3 0
0 1 0 3

Now, 5P - 1 Q +3I = 10 5 - 2 3 +3 0

2 -5 15 12 03

10-2+3 5-3+0 = 11 2
= -5 -1 +0 15-2+3 -6 16

Determinant of 5P - 1 Q + 3I = 11 2
2 -6 16

= 176 + 12 = 188

Example 6. If A = 1 2 and B = 1 2
Solution: 3 4 4 5

Show that : |AB| = |A| |B|

Here, |A| = 1 2 = 4 - 6 = -2
3 4

|B| = 1 2 = 5 - 8 = -3
4 5

|A| |B| = (-2) × (-3) = 6

Also, AB = 1 2 12 = 1+8 2+10 = 9 12
3 4 45 3+16 6+20 19 26

|AB| = 9 12 = 234 - 228 =6
19 26

Therefore, |AB| = |A| |B| Proved.

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Exercise 7.1

Very Short Questions

1. (a) Write the condition to have determinant of a matrix.

(b) Evaluate (i) a b (ii) secɵ tanɵ
c d tanɵ secɵ

(c) Write the determinant of [-8]

(d) If A = [4] is a square matrix, find its determinant

(e) |-8| = -8 and |-8| = 8, write a difference between them.

2. Evaluate : (a) 1 0 (b) 0 1
3. Evaluate the following : 0 1 1 0

(a) 4 5 (b) 4 8 (c) x+y x-y
3 2 3 6 x-y x+y

4. Evaluate the determinant of the following matrices:

(a) [-8] (b) 23 (c) a b
Short Questions 4 -5 -b a

5. (a) If P = 1 0 ,Q= 3 1 , find the determinant of (i) P+Q+I (ii) P+Q - I
3 1 5 3

(b) If M = 2 4 ,N= 2 1 , find the determinant
5 2 3 4

(i) of 2M + 3N (ii) 3M - 2N (iii) M2 - N2

6. Evaluate (a) a+b a+b (b) a2+ab + b2 b2 + bc + c2
a+b a–b b–c a-b

7. Find the value of x in the following cases.

(a) 5 -4 =8 (b) x 4 = 0 (c) x x =9
x 4 x 2x 3x 4x

8. In each of the following cases, show that each |AB| = |A| |B|

(a) A= 2 3 , B= 0 -1
4 -1 5 2

(b) A= 2 3 , B = -3 4
4 5 2 1

Long Questions

9. (a) If P = 2 3 ,Q= 1 2 , find the determinants of PQ and QP.
4 5 4 5

(b) If A = 1 4 ,B= 2 3 , find the determinants of AB and BA.
4 5 3 4

10. If M = 1 2 and N = 2 3 , find the determinants of
4 5 3 5

(a) MT + NT (b) (M+N)T (c) (MN)T

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11. (a) If P = 0 -2 , find the determinant of 2P2 - 5P + 4I, where I is a unit matrix of
3 4

order 2 × 2.

(b) If P = 1 -2 and Q = -3 0 , find the determinant of 5P - Q2 + 5I, where I
3 4 1 -2
is a unit matrix of order 2 × 2.

12. If P = 1 2 and Q = 2 3 is |(P + Q)2| = |P2 + 2PQ + Q2|
4 5 4 5

1. (a) ad - bc (b) 1 (c) -8 (d) 4

(e) determinant and absolute value 2.(a) 1 (b) -1

3. (a) -7 (b) 0 (c) 4xy 4.(a) -8

(b) -22 (c) a2 + b2 5.(a) (i) 17 (ii) 1

(b) (i) -49 (ii) -94 (iii) 65

6. (a) -2ab - b2 (b) a3 - 2b3 + c3 7.(a) -3 (b) 0, 2

(c) ± 3 9.(a) 6, 6, |PQ| = |QP| (b) 11, 11, |AB| = |BA|

10. (a) -5 (b) -5 (c) -3

11. (a) 22 (b) 221 12. No

7.2 Inverse Matrix

Let us consider two matrices A = 2 1 and B = 3 -1 .
5 3 -5 2

Then, let us find AB and BA.

AB = 2 1 3 -1
and BA 5 3 -5 2

= 6-5 -2 + 2 = 1 0 =I
15 - 15 -5 + 6 0 1

= 3 -1 2 1
-5 2 5 3

= 6-5 3-3 = 1 0 =I
-10+10 -5 + 6 0 1

In both cases AB = I and BA = I where I is a unit matrix of order 2 × 2. Then B is called the
inverse matrix of A or vice versa. The inverse matrix of A is denoted by A-1 .

Before defining inverse matrix, let us define singular and non-singular matrix.

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Singular and Non-singular Matrices

Let A be a square matrix. Then, the matrix A is said to be singular if |A| = 0. Then the
matrix A is said to be non-singular if |A| ≠ 0.

Example 1. Let A = 4 5
8 10

Then |A| = 4 5 = 4 × 10 - 8 × 5 = 40 - 40 = 0 .
8 10
Since |A| = 0. the matrix A is a singular matrix.

Example 2. Let B = 5 2
3 6

Then |B| = 5 2 = 5 × 6 - 3 × 2 = 30 - 6 = 24 ≠ 0
3 6
Since |B| ≠ 0, B is not a singular matrix.

Definition of inverse Matrix

Let A be a square matrix whose determinant is not zero. If there exists a matrix B such that
AB = BA = I, where I is a unit matrix of order A or B, then the matrix B is called inverse of
matrix A. The inverse of matrix A is denoted by A-1.

Then, AA-1 = A-1A = I

Warning A-1 ≠ 1 , where A-1 is the inverse matrix of A.
A
Only non- singular matrices have their inverses.

Adjoint of a 2 × 2 Matrix

Let A = a b be a square matrix. then interchange the elements in the major diagonal
c d
(or principal diagonal) and change the signs of the element in the minor diagonal.

Then, the new matrix obtained is called adjoint of matrix A, i.e., d -b is called the
adjoint matrix of A. It is denoted by Adj. A -c a

Adj. A = d -b
-c a
2 3
Example : Find the adjoint of matrix A = 6 5 .

Here, adjoint of matrix A = Adj. A = 5 -3
-6 2
To find the inverse matrix of 2 × 2 order.

Let A = a b be a non-singular square matrix of order 2 × 2.
c d

Then |A| = a b = ad - bc.
c d
w x
Let the inverse of matrix of A be A-1 = y z .

Then by definition of inverse matrix, we write AA-1 = A-1 A = I, where I is a unit matrix of

order 2 × 2.

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Now, ab w x = 1 0
cd y z 0 1

or, aw + by ax + bz = 1 0
cw + dy cx + dz 0 1
Equating the corresponding elements of equal matrices:

aw + by = 1 .......... (i)

ax + bz = 0 .............. (ii)

cw + dy = 0 ............. (iii)

cx + dz = 1 ............... (iv)

Multiplying equation (i) by c and (iii) by a and subtracting (i) from (iii)
acw + ady = 0

acw + bcy = c

-- -

y(ad - bc) = - c

? y = -c
ad-bc
Putting the value of y in equation (iii)

cw + d × -c =0
ad-bc

or, cw = –d (– c )
? ad-bc
d
w = ad-bc

Again, multiplying equation (ii) by c and (iv) by a and subtracting (ii) from (iv), we get

acx + adz =a

acx + bcz = 0

-- -

z(ad - bc) = a

? z = a
ad-bc
Putting the value of z in equation (ii), we get,

a
ad-bc
( )ax = - b
b
? x = - ad-bc

Putting the values of w, x, y and z in A-1, we get,

A-1 = 1 d -b
ad-bc -c a

= 1 Adj. A, |A| ≠ 0.
|A|

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Working Rule to find inverse of a matrix.

Let A = a b be a non-singular square matrix.
c d
Its inverse can be found by using the following steps.

(i) Find |A| = ad - bc. If |A| = 0, the inverse of matrix A i.e. A-1 does not exists.

(ii) Find the adjoint matrix of A.

i.e. Adj. A = d -b
-c a

(iii) Find A-1 by using the formula,

A-1 = 1 Adj.A = 1 d -b
|A| (ad-bc) -c a

Note :
The following conditions are necessary for the existence of A-1.

(i) The matrix A must be square.

(ii) The matrix A must be non-singular. i.e. |A| ≠ 0.

(iii) AA-1 = A-1A = I must be satisfied.

Properties of Inverse of Matrices

Let A and B be two square matrices of same order. Then, the following properties are satisfied.

(i) The inverse of the product of two non-square matrices is equal to the product of inverses
taken in the reverse order.

i.e., (AB)-1 = B-1 A-1

(ii) For a non- singular matrix A. (A-1)-1 = A

(iii) For a non- singular matrix A (AT)-1 = (A-1)T

To prove above properties, here, we take 2 × 2 square matrices,

Let A= 4 5 ,B= 2 1
1 2 5 3

|A| = 4 5 =8-5=3
1 2

|B| = 2 1 =6-5=1
5 3

(i) To prove : (AB)-1 = B-1A-1

Here, AB = 4 5 . 2 1
1 2 5 3

= 8 + 25 4 + 15 = 33 19
2 + 10 1+6 12 7

|AB| = 33 19 = 231 - 228 = 3
12 7
1
LHS = (AB)-1 = |A| Adj. (AB)

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= 1 7 -19 7 - 19
3 -12 33 =3 3

-4 11 2 5
3
1 1 2 -5 3 -
|A| 3 -1 4
Also, A-1 = Adj. A = = 1 4
3 3
-

B-1 = 1 Adj.B = 1 3 1 = 3 1
|B| 1 -5 2 -5 2

2 - 5
3 -1 3 3
RHS = B-1A-1 = -5 2 ×
= 1 4
- 3 3
–139
2 + 1 –5 – 4 7
3 3 3
=
–130 – 2 25 + 8
3 3 3 –4 11

(ii) To show : (A–1)–1= A

2 - 5
3 3
|A-1| = = 8 - 5 = 3 = 1
- 1 4 9 9 9 3
3 3
1
Now, (A-1)-1 = |A-1| Adj. A–1.

4 5
3 3 4 5
= 1 1 2 = 1 2 =A
3 3
3
? (A-1)-1 = A

(iii) To show: (A-1)T = (AT)-1

From (i), we have,

2 - 5
3 3

A-1 = - 1 4
3 3

Again, AT = 4 1
5 2

|AT| = 4 1 =8-5=3
5 2
2 -13
LHS = (A–1)T = 3 4
-53 3
1 1 2 -1 2 -31
Now, RHS =(AT)-1 = |AT| Adj. AT = 3 -5 4 = 3 4
-35 3

? (A-1)T = (AT)-1 Proved.

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Worked out Examples

Example 1. Find the inverse matrices of the following matrices:
Solution :
(a) A= 2 3 (b) B = 3 6
4 5 2 4

(a) Here, A = 2 3
4 5

|A| = 2 3 = 10 - 12 = -2 ≠ 0
4 5
Since |A| ≠ 0, A-1 exists.

Adj. A = 5 -3
-4 2

Now, the inverse matrix of A is given by

A-1 = 1 Adj. A = 1 5 -3 = - 5 3
|A| -2 -4 2 2 2
2 -1
3 6
(b) Here, B = 2 4

|B| = 3 6 = 12 - 12 = 0
2 4

Since |B| = 0, B–1 does not have its inverse, i.e., B-1 does not exists.

Example 2. If P = 3 5 and Q = -3 5 , then, show that they are inverse to each
Solution: 2 3 2 -3
other.
Example 3.
Solution: Here, PQ = 3 5 -3 5
2 3 2 -3

= -9 + 10 15 - 15 = 1 0
-6 + 6 10 - 9 0 1

and QP = -3 5 35
2 -3 23

= -9 + 10 -15+15 = 1 0
6-6 10 - 9 0 1

Since PQ = QP = I, P, and Q are inverse matrix to each other.

If y 1 and 2 -1 are inverse matrices to each other,
5 x -5 3

find the values of x and y.

Since y 1 and 2 -1 are inverse to each other, we can write,
5 x -5 3

y 1 2 -1 = 1 0
x -5 3 0 1
5

or, 2y - 5 - y+3 = 1 0
10 - 5x -5 + 3x 0 1

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Equating the corresponding elements, we get,

2y - 5 = 1 or, 2y = 6 or, y = 3

-y + 3 = 0 or, y = 3

10 - 5x = 0 x=2 -5 + 3x = 1

or, 3x = 6 ? x = 2 ? x = 2, y = 3

Example 4. For what value of x the matrix P = x-3 0 does not have its
Solution: inverse? 0 3x+1

Here, P = x-3 0
0 3x+1
If P is a singular matrix, it does not have its inverse.

x-3 0
|P| = 0 3x+1 = 0

or, (x – 3). (3x + 1) = 0

Either x – 3 = 0, or, x = 3

or, 3x + 1 = 0 or, x = –1/3

? x = 3, – 1
3

Exercise 7.2

Short Questions

1. Find the adjoint matrices of the following matrices:

(a) A= 2 3 (b) B= 6 7 (c) C= 10 5
4 5 3 2 23

2. Which of the following matrices have their inverse?

(a) A= 4 6 (b) B= 4 10
2 3 2 5

(c) C= 2 4 2 24
6 7 (d) D = 2 8

3

3. Find the inverse matrices of the following matrices if possible:

(a) A= 2 1 (b B= 6 2 (c) C= 1 2
7 4 9 3 4 7
cosecT cotT
(d) D= 2 7 (e) E= cosT - sinT (f) F = cotT
4 14 sinT cosT cosecT

4. Show that inverse of each of the following matrix is the matrix itself:

(a) 1 0 (b) 01 (c) 0 -1 (d) -1 0
0 1 10 -1 0 0 -1

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5. In each of the following, prove that each matrix is an inverse of other:

(a) 3 1 and 2 -1 (b) 2 3 and 8 -3
5 2 -5 3 5 8 -5 2

(c) 5 3 and -4 -3
-7 -4 7 5

Long Questions

6. (a) If P = x 1 and Q = 2 -1 are inverse to each other find the value of x.
5 2 -5 x

(b) If M = x -1 and N = 2 -1 are inverse to each other, find the values of x
and y. 5 -2 y -3

(c) If -1 2 and q 2 are inverse to each other, find the value of o and q.
2p -7 4 1

7. (a) If P = 2 0 and Q = 4 1 ,
5 3 7 2

(i) Find P–1 and Q–1 (ii) Find (PQ)-1 (iii) Show that (PQ)-1 = Q-1P-1

(b) If A 5 3 and B = 3 2 , show that (AB)-1 = B-1A-1.
3 2 4 3

(c) If M = 2 3 and N = 4 5 , show that (MN)-1 = N-1M-1
4 5 6 7

53

8. (a) If A = 3 2 , then, show following:

(i) (A-1)-1 = A (ii) (A-1)T = (AT)-1 (iii) A-1 A = AA-1 I

(b) If B = 4 1 , then, show the following:
7 2

(i) (B-1)-1 = B (ii) (B-1)T = (BT)-1 (iii) BB-1 = B-1B = I

9. For what value of x the product matrix 3 2 2 -1 does not have its inverse
matrix. x 4 3 2

1. (a) 5 -3 (b) 2 -7 (c) 3 -5
-4 2 -3 6 -2 10

2. C 3.(a) 4 -1 (b) B-1 does not exist.
-7 2

(c) -7 2 (d) D-1 does not exist
4 -1

(e) cosT sinT (f) cosecT - cotT
-sinT cosT -cotT cosecT

6. (a) 3 (b) x = 3, y = 5 (c) p = 2, q = 7
9. x = 6
7. (a) (i) P-1 = 1/2 0 , Q-1 = 2 -1
-5/6 1/3 -7 4

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