Prime Optional Mathematics 7

Grade
VII

PRIME Optional
Mathematics

Pragya Books &
Distributors Pvt. Ltd.

Author Editors

Dirgha Raj Mishra LN Upadhyaya
Rajkumar Mathema

DN Chaudhary
Narayan Shrestha

Khem Timsina
J.N. Aryal

Kadambaba Pradhan
Dinesh Silwal

Pragya Books & Distributors Pvt. Ltd.

Lalitpur, Nepal
Tel : 5200575
email : [email protected]

© Author

Author Dirgha Raj Mishra

Editors LN Upadhyaya
Rajkumar Mathema
DN Chaudhary
Narayan Shrestha
Khem Timsina
J.N. Aryal
Kadambaba Pradhan
Dinesh Silwal

First Edition 2076 B.S. (2019 A.D.)
Revised Edition 2077 B.S. (2020 A.D.)

ISBN 978-9937-9170-4-9

Typist Sachin Maharjan
Sujan Thapa

Layout and Design Desktop Team

Printed in Nepal

Preface

Prime Optional Mathematics series is a distinctly outstanding mathematics
series designed according to new curriculum in compliance with Curriculum
Development Centre (CDC) to meet international standard in the school level
additional mathematics. The innovative, lucid and logical arrangement of the
contents make each book in their series coherent. The representation of ideas in
each volume makes the series not only unique but also a pioneer in the evaluation
of activity based mathematics teaching.

The subject is set in an easy and child-friendly pattern so that students will
discover learning mathematics is a fun thing to do even for the harder problems.
A lot of research, experimentation and careful graduation have gone into the
making of the series to ensure that the selection and presentation is systematic,
innovative, and both horizontally and vertically integrated for the students of
different levels.

Prime Optional Mathematics series is based on child-centered teaching
and learning methodologies, so that the teachers can find teaching this series
equally enjoyable.

I am optimistic that, this series shall bridge the existing inconsistencies
between the cognitive capacity of children and the subject matter.

I owe an immense dept of gratitude to the publishers (Pragya Books team)
for their creative, thoughtful and inspirational support in bringing about the
series. Similarly, I would like to acknowledge the tremendous support of editors
team, teachers, educationists and well-wishers for their contribution, assistance
and encouragement in making this series a success. I would like to express my
special thanks to Sachin Maharjan (Wonjala Desktop) for his sincere support
of designing part of the book and also Mr. Gopal Krishna Bhattarai to their
memorable support to prepare this series.

I hope this series will be another milestone in the advancement of teaching
and learning Mathematics in Nepal. We solicit feedback and suggestions from
teachers, students and guardians alike so that I can refine and improvise the series
in the future editions.

– Author

Contents

S.N. Units Page

Algebra 1
1.1 Order Pairs 2
1.2 Cartesian Product 9
1.3 Polynomial 17
1.4 Surds 22

Matrices 31

2.1 Matrices 32

2.2 Operation on matrices 39

Co-ordinate Geometry 47

3.1 Rectangular co-ordinate axis 48
3.2 Mid- point of a line segment 58

Trigonometry 65

4.1 Measurement of angles 66

4.2 Introduction of right angled triangle 72

4.3 Trigonometric ratios 81

4.4 Trigonometric identities 88

4.5 Conversion of Trigonometric ratios 93

4.6 Trigonometric Ratios of some standard angles 98

4.7 Solution of right angled triangle 106

Vector 113

5.1 Vector quantities using co-ordinate 115
5.2 Vector operations 121

Transformation 129

6.1 Reflection 131

6.2 Translation 137

Statistics 143

7.1 Measure of Central Tendency 145

7.2 Median, Mode and Range 151

Model question 160

1 Algebra

1. Algebra

1.1 Ordered pairs
1.2 Cartesian Product
1.3 Polynomials
1.4 Surds

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods
7 16
No. of Questions 1 1 1 – 3
Weight 1 2 4 –

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : Materials Required:

At the end of the lesson • Chart paper
• Students are able to know the • Chart of number system.
• Chart of types of surds.
ordered pairs and Cartesian • Chart of types
products.
• Students are able to represent polynomials & properties
Cartesian product in arrow of addition and
diagram and graph. multiplication.
• Students are able to know the
number system including surds.
• Students are able to operate
the surds.
• Students are able to know the
polynomials, types and degree
of polynomials.

The word ‘algebra’ is derived from the arabic word al-jabr and this
come from the treatise written in the year 830 by the medieval Persian
mathematician, Muhammad ibnmusa al-khwarizmi.

Algebra can essentially be considered as doing computations similar
to those of arithmetic but with non-numerical mathematical objects.
However, until the 19th century, algebra consisted essentially of the
theory of equations. For example; the fundamental theorem of algebra
belongs to the theory of equations and is not now a days, considered as
belonging to algebra.

Here we are going to discuss the variables involving in different situation
like to make ordered pairs, relations, functions, elements of set, elements
of matrices, components of vector etc.

1.1 Order Pairs

Let us discuss the following examples to visualize the concept of ordered

pairs.

Temple Hinduism

Musjid Muslim

Church Christian

Gumba Buddhism

Gurudwara Shikh
In the examples given above, the first informations are of the name of holly

places and second informations are of the religions. The informations

can be written in different forms like ( Temple, Hinduism),(Musjid,

Muslim), (Church, Christian), (Gumba, Buddhism), (Gurudwara, Shikh).
Here name of the temples are written as first component and location

are written as second component but there is the vital role of ordered

between the components. This kind of way of writing the components is

called ordered pairs.

The pair of the components x and y Which is written
according to an order in the form of (x, y) is called
ordered pairs.

2 PRIME Opt. Maths Book - VII

• The First component is denoted by the variable ‘x’ which is
called antecedent.

• The Second Component is denoted by the variable ‘y’ which
is called consequence.

• Ordered pair of x and y is written as (x, y)

In a ordered pair (x, y)
• x is the first component (antecedent)
• y is the second component (consequence)

The first and second components are written in different sets A = {Temple,
Musjid, Church, Gumba, Gurudwara} which is called domain. It is the set
of temples. B = {Hinduism , Muslim, Christian, Buddhism, Shikh}. It is the
set of the locations which is called co-domain.

• Domain is the set of first components of
the ordered pairs.( set of antecedent)

• Co-domainisthesetofsecondcomponents
of the ordered pairs (set of consequence)

The domain and co-domain can be taken in balloon and the ordered pairs
can be shown using arrow which is called balloon diagram or arrow
diagram. Let us take ordered pairs of currency with respect to countries
in arrow diagram.

Japan Yen
USA $
UK £
Nepal Rs

PRIME Opt. Maths Book - VII 3

Equal ordered pairs:

Let us discuss the ordered pairs given below.

First ordered pair = (2, 3)

Second ordered pair = (6 – 4, 5 – 2)
4 6
Third ordered pair = ( 2 , 2 )

The Final result of the ordered pairs mention above is (2,3) where all the
ordered pairs are (2,3) which are called equal ordered pairs.

The two or more ordered pairs having equal
antecedent of all and equal consequence of all
are called equal ordered pairs.

• If (x, y) and (a, b) are the equal ordered pairs,
Antecedent should be equal
i.e. x = a
Consequence should be equal,
i.e. y = b

• If (x, 3) and (2, y) are equal ordered pairs,
Then,
(x, 3) = (2, y)
i.e x = 2 and y = 3

• If (x – 2, 4) = (2, y + 1)
Then,
(x – 2, 4) = (2, y + 1)

or, x – 2 = 2 and y + 1 = 4
or, x = 2 + 2 and y = 4 – 1

x = 4 and y = 3

4 PRIME Opt. Maths Book - VII

Worked out Examples

1. Write down any four ordered pairs of capitals with respect to
countries. Also show in arrow diagram.
Solution;
A = { Set of countries}
= {Nepal , India, UK, Bhutan}
B = {Set of capitals}
= {Kathmandu, New delhi, London, Thimpu}
Then,
Ordered pairs of capitals with respect to countries are (Nepal,
Kathmandu) (India, New Dhelhi),(UK , London), (Bhutan, Thimpu)

Arrow diagram: Kathmandu
New Delhi
Nepal
India London
UK Thimpu
Bhutan

2. If domain represents the name of Nepali poet and co-domain
represents the title of poet, write down the ordered pairs taking
four elements.
Solution.
Domain = {Bhanubhakta, Motiram, Lekhnath, Sambhu Pd. Dhungel}
Co-domain = {Aadikabi, Anshukabi, Siromani, Yubakabi}

Then,
Ordered pair
= (Bhanubhakta, Aadikabi), (Motiram, Yubakabi),

(Lekhnath, Siromani), (Sambhu Pd. Dhungel, Anshukabi)

3. If (x + 2, y + 5) & (4, 8) equal ordered pairs, find the value of x and y.
Solution:
The equal ordered pairs are :
i.e (x + 2, y + 5) = (4, 8)
Equating the first components and second components respectively.

PRIME Opt. Maths Book - VII 5

we get
or, x + 2 = 4 and y + 5 = 8
or, x = 4 – 2 and y = 8 – 5
or, x = 2 and y = 3
\ x = 2, y = 3

4. If (a – 2, a + b) and (1, 5) are equal ordered pairs, find the value of a

and b.

Solution:

The equal ordered pairs are:

(a – 2, a + b) = (1, 5)

Equating the corresponding components,

We get,

or, a – 2 = 1

or, a = 1 + 2

a=3

Also, a + b = 5

or b = 5 – a

or b = 5 – 3

\ b=2

\ a=3 b=2

5. If A = {1, 2, 3, 4} and B = {2, 3, 4, 5}, prepare the ordered pairs from

the set A to the set B where order is more than by one. Also shown

in arrow diagram

Solution:

The given sets are:
A = {1, 2, 3, 4}
B = {2, 3, 4, 5}

The ordered pairs of the element of set A to B where the order is

more then by 1 are : A B
(1, 2), (2, 3), (3, 4), (4, 5)

12

23

34

45

6 PRIME Opt. Maths Book - VII

Exercise : 1.1

1. Answer the following questions:
i. What is an ordered pair?
ii. Write down any three ordered pairs of capitals with respect to
countries.
iii. What do you mean by antecedent.
iv. What do you mean by consequence.
v. Define the term equal ordered pairs.

2. Write down the following ordered pairs. Also show in arrow
diagram.
i. Any four ordered pairs of district with respect to headquarters.
ii. Any four ordered pairs of holy places with respect to location.
iii. Any five ordered pairs of things with respect to location which
is famous for the things.
iv. Write down the ordered pairs of vowel alphabets with respect
to the first 5 odd natural numbers.
v. Write down any five ordered pairs of your teacher with respect
to subject related to them.

3. Find the value of x and y from the following equal ordered pairs

i. (x, y) and (3, 4) ii. (x, 2) and (1, y)

ii. (2, y) and (x, 3) iv. (x + 1, 4) and (3, y - 1)

v. (3, 2)and (x – 1, y + 1)

4. Find the value of the unknowns from the followings.

i. (a , 4) = (5, b - 2) ii. (a + 2, 5 + b) = (5, 8)

ii. (x – 3, 7) = (2 , y+10) iv. (m + n , 7) = (5, n + 4)

v. (p + q, q – 2) = (7 , 2)

5. Prime more creative questions.
i. If A = {2, 3, 4, 5}, B = {7, 8, 9, 10} find the ordered pairs from A
to B where element of B is more than the element of A by 5.
ii. If A = {Nepal, India, Bhutan, china}, B = {Beijing, New Delhi,

PRIME Opt. Maths Book - VII 7

Thimpu, Kathmandu}, find the ordered pairs from the set A to
B. Also show in arrow diagram.
iii. Find the set of antecedent and consequence from the given
arrow diagram. Also write down the ordered pairs.

a3
c5
e7
g1

iv. Find the value of x and y from the given equal ordered pairs
(2x + 2y , 3) = (2 , 6 + y)

v. If x = 2 and y = 4, prove that the ordered pairs (3x + y, 2) and
(x + 8, 3x – y) are equal.

Answer

1. Show to your teacher

2. Show to your teacher

3. i. x = 3, y = 4 ii. x = 1, y = 2

iii. x = 2 , y = 3 iv. x = 2, y = 5

v. x = 4 , y = 1

4. i. a = 5 , b = 6 ii. a = 3, b = 3

iii. x = 5 ,y = –3 iv. m = 2, n = 3

v. p = 3 , q = 4

5. i. (2, 7), (3, 8), (4, 9), (5, 10)

ii. Kathmandu
Nepal

India New Delhi

Bhutan Thimpu

China Beijing

iii. Set of antecedent ={a,c,e,g} set of consequence ={1, 3, 5, 7}
ordered pairs=(a, 1), (c, 3), (e, 5), (g, 7)

iv. x = 4 , y = –3

v. Both are (10, 2), Hence they are equal.

8 PRIME Opt. Maths Book - VII

1.2 Cartesian Product

Let us consider the two non-empty sets A = {a, b} and B = {x, y, z}. The
collection of all possible ordered pairs from the set A to B can be taken as
{(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)}. It is called the Cartesian product A
× B which is read as A cross B and is written as A × B = {(x, y) : x ∈ A and
y ∈ B} in set builder form.

The set of all possible ordered pairs from
the non- empty set A to the non-empty set
B is called Cartesian product A × B.
i.e. A × B = {(x, y): x ∈ A and y ∈ B}

• Cartesian Product A × B is from the set A to B
• Cartesian Product B × A is from the set B to A
• n (A × B) is the cardinality of the set A × B

i.e. [n(A × B) = n(A) × n(B)]
• n(B × A) is the cardinality of the set B × A.

i.e. [n(B × A) = n(B) × n(A)]
• A × B is not equal to B × A

i.e. [A × B ≠ B×A]
• Cardinality of the sets of A × B and B × A are always equal.

i.e. n(A × B) = n(B×A)

Representation of Cartesian product 9
i. Set of ordered pairs:

Let us consider the sets A = {2, 3} and B = {3, 4, 5}.
A × B = {2, 3} × {3, 4, 5}

= {(2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)}

B × A = {3, 4, 5} × {2, 3}
= {(3, 2),(3, 3), (4, 2), (4, 3), (5, 2), (5, 3)}

PRIME Opt. Maths Book - VII

ii. Arrow Diagram: B = {3, 4, 5} 3
A = {2, 3} B×A 9
A×B
23 2
4 4
35 3

iii. Tree diagram: B = {3, 4, 5}
A = {2, 3}
Set A Set B Ordered pairs of A × B
2 3 (2, 3)

3 4 (2, 4)

Set B 5 (2, 5)
3
3 (3, 3)
4
4 (3, 4)
5
5 (3, 5)
10
Set A Ordered pairs of B × A
2 (3, 2)

3 (3, 3)
2 (4, 2)

3 (4, 3)
2 (5, 2)

3 (5, 3)

set A PRIME Opt. Maths Book - VII

iv. Graphical from: B = {3, 4, 5}
A = {2, 3}
B×A
A×B
5
5 4
4 3
3 2
2 1
1
12345
12345
Set B
Set A
Set B
Set A

In the above examples, the cardinality of the sets
A × B and B × A can be calculated as,

n(A × B) = n(A) × n(B) = 2 × 3 = 6
n(B × A) = n(B) × n(A) = 3 × 2 = 6
\ n(A × B) = n(B × A)

Worked out Examples

1. If A = {1, 2, 3}, B = {4, 5}, find the Cartesian product A × B and B × A.
Solution:
A = {1, 2, 3}
B = {4, 5}
Then,
A × B = {(x, y) : x ∈ A and y ∈ B}
= {1, 2, 3} × {4, 5}
= {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

B × A = {(x, y) : x ∈ B and y ∈ A}
= {4, 5} × {1, 2, 3}
= {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}

PRIME Opt. Maths Book - VII 11

2. If A = {a, b}, B = {2, 3, 4, 5}, prove that A × B ≠ B × A and n (A × B) =
n(B × A)

Solution:
A = {a, b}
B = {2, 3, 4, 5}
Then,
A × B = {(x, y): x ∈ A & y ∈ B}

= {a, b} × {2, 3, 4, 5}
= {(a, 2), (a, 3), (a, 4), (a, 5), (b, 2), (b, 3), (b, 4), (b, 5)}
B × A = {(x, y) : x ∈ B and y ∈ A}
= {2, 3, 4, 5} × {a, b}
= {(2, a), (2, b), (3, a), (3, b), (4, a), (4, b), (5, a), (5, b)}
Here,
A×B≠B×A
Again,

n(A × B) = n(A) × n(B) = 2 × 4 = 8

n(B × A) = n(B) × n(A) = 4 × 2 = 8

\ n(A × B) = n(B × A) = 8

3. If P = {3, 4, 5}, Q = {6, 7, 8}, show the sets P × Q and Q × P in arrow

diagram.

Solution:
P = {3, 4, 5}
Q = {6, 7, 8}

Then, Q×P
P×Q

36 63
47 74
58 85

12 PRIME Opt. Maths Book - VII

4. From the given graph of A × B, find the sets A and B. Also find B × A.Set B

5
4
3
2

1

12345

Set A

Solution:
From the given graph of A × B the set of ordered pairs can be written as.
A × B = {(1,3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
Here, A = {1, 2, 3} and

B = {3, 4}
Also,
B × A = {(x, y) : x ∈ B and y ∈ A}

= {3, 4} × {1, 2, 3}
= {(3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3)}

Exercise : 1.2

1. Answer the following questions.

i. What is Cartesian product? Write down in set builder form.
ii. {(x, y): x ∈ B and x ∈ A} is the set builder form of Cartesian

product. Which Cartesian product is represented by it?
iii. {(x, y) : x ∈ P and x ∈ Q} is the set builder form of Cartesian

product. Which Cartesian product is represented by it?
iv. If A = {2, 4, 6} and B = {7, 8, 9}, find the value of n(A × B)
v. If M = {a, b}, find the cartesian product M × M.

2. Find the Cartesian product A × B from the followings
i. A = {1, 3}, B = {2, 4} ii. A = {x, y}, B = {a, b}
iii. A = {p, q, r}, B = {a, b} iv. A = {3, 4, 5}, B = {6, 7}
v. A = {2, 3}, B = {7, 8, 9}

PRIME Opt. Maths Book - VII 13

3. If A = {4, 5, 6}, B = {2, 3}, find the followings.

i. A × B ii. B × A iii. A×A

iv. B × B v. n(A × B) and n(B × A).

4. Prove that the followings from the sets P = {3, 4, 5}, Q{1, 7} and R =
{2, 6, 8}
i. P × Q ≠ Q × P ii. n(P × Q) = n(Q × P)
iii. n(P × R) = n(R × P) iv. Q × R ≠ R × Q

v. n(Q × P) = n(Q × R)

5. Prime more creative questions:
i. If A = {1, 2, 3} and B = {3, 4, 5}, find A × B and show in arrow

diagram.
ii. If M = {3, 4, 5} and N = {6, 7, 8}, find N × M and show in tree

diagram.
iii. If P = {2, 3, 4, 5} and Q = {3, 4, 5}, show P × Q and Q × P in graph.
iv. From the given arrow diagram, find the sets A and B. Also write

down B ×A in ordered pairs.
A×B

xaQ

yb

zc

d

v. Write down the sets P, Q, P × Q and Q × P from the given graph.

5
4
3
2
1

12345

P

14 PRIME Opt. Maths Book - VII

Answer

1. i. Show to your teacher. v. {(a, a), (a, b), (b, a), (b, b)}
ii. B × A iii. P × Q iv. 9

2. i. A × B = {(1, 2), (1, 4), (3, 2), (3, 4)}
ii. A × B = {(x, a), (x, b), (y, a) (y, b)}
iii. A × B = {(p, a), (p, b), (q, a), (q, b), (r, a), (r, b)}
iv. A × B = {(3, 6), (3, 7), (4, 6), (4, 7), (5, 6), (5, 7)}
v. A × B = {(2, 7), (2, 8), (2, 9), (3, 7), (3, 8), (3, 9)}

3. i. A × B = {(4, 2), (4, 3), (5, 2), (5, 3), (6, 2), (6, 3)}
ii. B × A = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
iii. A × A = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
iv. B × B = {(2, 2), (2, 3), (3, 2), (3, 3)}
v. n(A × B) = 6, n(B × A) = 6

4. Do yourself.

5. i. A × B = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)}

A×B

ii. 1 3
Set N 2 4
3 5
6
Set M Ordered pairs
3 (6, 3)
4 (6, 4)
5 (6, 5)

PRIME Opt. Maths Book - VII 15

3 (7, 3)
7 4 (7, 4)

5 (7, 5)
3 (8, 3)
8 4 (8, 4)
5 (8, 5)

iii. P × Q Q×P

55
44
33
22
11
Q
P

12345 12345

P Q

iv. A = {x, y, z}, B = {a, b, c, d}
B × A = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, y),(c, z),
(d, x), (d, y), (d, z)}

v. P = {2, 3, 4, 5}, Q = {3, 4}
P × Q = {(2, 3), (2, 4), (3, 3), (3, 4), (4, 3), (4, 4), (5, 3), (5, 4)}.

16 PRIME Opt. Maths Book - VII

1.3 Polynomial

The algebric expression is the combination of the algebraic terms by
using plus (+) or minus (–) sign. It can be defined as the polynomial
under the certain conditions only where power of the variables used in
the expression should be only positive integer.

Example

x3 + 2x2 – 3x + 2 " It is polynomial

x2 + 3x – 5 " It is not polynomial.
x

x2 – 5x + 1 " It is not polynomial.

2x 2

x3 + 2x2 – 3x – 5 – 2x–2 " It is not polynomial.

The algebraic expression which has positive
integers as the power (exponent) of the variables
is called polynomial.

Polynomial of variable ‘x’ : 2x3 – 3x2 + 5x – 1
Polynomial of variable ‘y’ : y2 – 3y + 2

The terms of the polynomial have to be written according to the
decreasing order of the power of the variables. The form of polynomial
which is written in such a way that mentioned above is called standard
form of polynomial.

Example
Polynomial : 2x2 – 3x2 + 5 – 4x + x4.
Standard form : x4 – 3x3 + 2x2 – 4x + 5.

Degree of the polynomials:

The sum of the powers of the variable used in an algebraic term of the
polynomial is called degree of the term. The degree of the polynomial is
the maximum degree among the degree of the terms.

PRIME Opt. Maths Book - VII 17

Polynomial = x2y3 + x3y – 3x2 = 2.
Degree of 1st term = 2 = 3 = 5
Degree of 2nd term = 3 + 1 = 4
Degree of 3rd term = 2
Degree of 4th term = 0
Here, the maximum degree is 5.

∴ 5 is the degree of the polynomial.

Coefficient:

A constant real number used as the multiple of the variable in an algebric
term of the polynomial is called coefficient
• Coefficient of the variables may be in letters also which is called

literal coefficient.
• Coefficient in the form of real number is called numerial coefficient.

Example

In the algebric term 2ax3.
2 = numeral coefficient
a = literal coefficient

Types of Polynomial:

• According to number of terms :

i) Monomail It has only one term.

p(x) = 3x

ii) Binomial It has two terms.

p(x) = ax + b

iii) Trinomial It has three terms.

p(x) = ax2 + bx +c

iv) Polynomial It has more then three terms.

P(x) = ax4 + bx3 + cx2 + dx +e.

• According to degree :

i) Constant polynomial zero degree .

p(x) = 5

ii) Lines polynomial First degree.

p(x) = ax + b

18 PRIME Opt. Maths Book - VII

iii) Quadratic polynomial Second degree.

p(x) = ax2 + bx + c

iv) Cubic polynomial third degree.

p(x) = ax3 + bx2 + cx + d

v) Biquadratic polynomial Fourth degree.

p(x) = ax4 + bx3 + cx2 + dx + e

Exercise : 1.3

1. Answer the following questions:

i. What do you mean by algebraic term ?

ii. What is algebraic expression ?
iii. Define the term polynomial ?
iv. Write short not on degree of the polynomial with an example.
v. Differentiate between numeral and literal coefficients with an

example.

2. Which of the followings are the polynomial ? Write down with

reason.

i. x4 +3x² + 2x – 5x + 2

ii. x³ + 2x² – 3x + 3 – x
2

iii. x4 + 2x² + 3x² – 7x – 2 – 5x–2

2

iv. x² – 5x + 7 – 3x3

v. x³ + 5x² –7x – 6

3. Write down the coefficients of the following algebraic terms.

i. 3x² ii. 2px³ iii. 5x4y

iv. –7ax5 v. 12xy³

4. Write down the degree of the following polynomials.
i. 3x² + 2x³ – 3x + 5 + x4
ii. 2x³y² + 3x²y² – xy² + y³

PRIME Opt. Maths Book - VII 19

iii. 3x²yz³ – 2xy²z² + 3xy²
iv. x4y² – 4x³y + 2xy²
v. x² – 5 + 3x³ – x4 + 2x + x5

5. Write down the polynomials in standard form. Also write down the
types of polynomials.
i. x² – x + 3x³ + 2x
ii. 12 – 5x² + 2x + 3x³ + x4
iii. 4x – 7 + 3x²
iv. 5 + 2x
v. 3x³ – 4x² + 3 – 2x4 + x5 – 7x

6. Add the following polynomials.
i. f(x) = x2 + 3x – 2, g(x) = x2 – x – 1
ii. p(x) = 3x – 2, q(x) = 4 – 2x + x2
iii. 2 + 3x2 – x and x2 – 5 + 3x
iv. p(x) = 2x3 – 2x2 + 3x – 2, q(x) = 2x3 – x2 + 2x – 5
v. f(x) = 2x3 – 3x2 + 5x –1, g(x) = 2 – 5x + 2x2 –x3

7. Subtract the following polynomials.
i. f(x) = 2x2 + 2x – 3, g(x) = x2 – x – 1
ii. p(x) = 5 –2x + 3x2, q(x) = 3x – 2 – x2
iii. 3 + 2x – x2 + 2x3 and 5x – 3 – 2x2
iv. p(x) = 2x3 –2x + x2 – 1 and q(x) = x3 – 3 – x + 2x2
v. f(x) = 2x2 + 1 – 3x – x3 and g(x) = 3x2 –2x3 – 2x –1

8. Multiply the following polynomials:
i. p(x) = 2x + y, q(x) = 4x² – 2xy + y²
ii. f(x) = x – 3y, g(x) = x² + 3xy + 9y²
iii. p(x) = x² – 3x + 5, q(x) = 2x² + x – 3
iv. f(x) = x² + 2x + 4, g(x) = x² – 2x +4
v. f(x) = x³ + 2x² – 3x – 2, g(x) = x² + 4x – 3

9. Prime more creative questions:
i. Subtract x3 + 3x2 – 3x + 1 from the sum of 2x3 – x2 + 5x – 3 and
3x2 – 6x – 2

20 PRIME Opt. Maths Book - VII

ii. What must be subtracted from the polynomial 5x³ – 3x² + 2x +
5 to get the polynomial 2x³ – x² + 3x – 2

iii. If p( x) = x² + 2x – 3 and q(x) = 2x² – x + 2, write down the types
of polynomial according to degree by calculating p(x).q(x).

iv. What must be subtracted from the sum of 2x³ – 3x² + 2x – 5
and x³ – x² – 3x + 2 to get x³ – 2x² + x + 3.

v. Subtract 2x³ – 2x² + 5x – 2 from the sum of 3x³ – x² – 2x + 5 and
2x³ – 3x² + 7x – 3.

Answer

1. Show to your teacher.

2. i. Polynomial ii. Polynomial v. Polynomial
v. 5
3. Do yourself

4. i. 4 ii. 5 iii. 6 iv. 6

5. Do yourself

6. i. 2x² + 2x – 3 ii. x² + x + 2
iii. 4x² + 2x – 3 iv. 4x³ – 3x² + 5x – 7
v. x³ – x² + 1

7. i. x² + 3x – 2 ii. 4x² – 5x + 7

iii. 2x³ + x² – 3x + 6 iv. x³ – x² – x + 2

v. x³ – x² – x + 2

8. i. 8x³ + y³ ii. x³ – 27y³

iii. 2x4 – 5x³ – 11x² – x + 6

iv. x4 + 4x² + 16 v. x5 + 8x4 + 2x³ – 11x² – x + 6

9. i. x³ – x² + 2x – 6 ii. 3x³ –2x² – x + 7
iii. 2x4 + 3x³ – 6x² + 7x – 6
iv. 2x³ – 2x² – 2x – 6
v. 3x³ – 2x² + 4

PRIME Opt. Maths Book - VII 21

1.4 Surds

Number system

The set of natural numbers is the set of all counting numbers. We denote
the set of natural numbers by N. The least natural number is 1 and the
highest is not defined.

N = {1, 2, 3, 4, 5 ...........................}

The set of counting numbers including zero is called whole numbers.
The set of whole numbers is denoted by W The least whole number is
zero and the highest is not defined.

W = {0, 1, 2, 3, 4, 5, ........................}

The set of all counting numbers including zero and their negatives is
called integers. The set of integers is denoted by Z.

Z = {..................., –4, –3, –2, –1, 0, 1, 2, 3, 4, .........................}

In the given examples set of natural numbers
is the sub–set of whole number and set of
whole number is the sub set of integers.

Rational numbers:

Let us consider the elements of integers taken in different mathematical

operations as follows,

4 – 6 = –2 (Integer) 4 + 6 = 10 (Integer)

4 × 6 = 24 (Integer) 4÷6= 2 (Not Integer)
3

But 4 ÷ 2 = 2 (Integer)

Thus, the division of any two integers may not be always an integer. To
define such numbers a new number system is introduced which is called
the rational number.

22 PRIME Opt. Maths Book - VII

The set of numbers in the form of p/q where p & q

are the integers and q ≠ 0 is called the set of rational
numbers.
i.e. Q = {...., –2, –3/2, –1, 0, 1/2, 1, 3/2, 3, ............} is the
set of rational numbers.

Note :

i. The set of rational numbers is denoted by Q.

ii. All integers can be expressed as a rational number.
p
e.g. 2 = 2 = 4 etc which is in the form q , q ≠ 0 and p, q ∉ z.
1 2

So, all integers are rational numbers. But all rational numbers are not integers.

i.e., Z ⊂ Q.

Also N ⊂ W ⊂ Z ⊂ Q.

Let us consider the rational number 19 .
4

We have

4 19 4.75

16

30

28

20

20

×
Here, the process of division is terminated (finished) with the decimal

quotient 4.75. This shows that a rational number some time posses a

terminating decimal number.

Again consider the rational number 19
3

We have

3 19 6.33

18

10

9

10

9

1

PRIME Opt. Maths Book - VII 23

Here, the process of division doesn’t terminate and a number is occurred
again and again after decimal point. Similarly, we can see sometimes a
block of numbers repeats after decimal point.

Thus a rational number posses an integer or a terminating decimal

number or a non terminating recurring decimal number.

Try to find the recurring decimal nouf mrabtieornfaolrn2u72mbers between any two
Note : There are infinite number

rational numbers.

Irrational numbers:

Some Examples :
2 = 1.414213562....................
3 = 1.732...............
5 = 2.236067977..
4=2
9=3

Here, 2, 3 , 5 have the non terminating and non recurring decimal
parts while 4 & 9 have the integer numbers as their roots 2, 3 , 5
are called the irrational numbers.

Note:

i. All the roots which are not rational are irrational

ii. The non terminating and non recurring decimal numbers are

irrational.

iii. p= length of circumference of a circle is an irrational number.
diameter

iv. Length of a body is either a rational number or an irrational number.

Introduction of Surds:
• The numbers 2, 3, 3 2 etc are the roots of rational numbers
where the result is irrational are called surds.

• 2, 3 are unlike surds of same order (same orders but
different radicands)

24 PRIME Opt. Maths Book - VII

• 2, 3 2 are unlike surds of different order (different orders
but same radicands)

• 3, 2 3 are the like surds (same orders and same radicands)
• 3 5 , 3 10 are unlike surds
• Any number in the from n a which can not be expressed in

p
the form of q , where q ≠ 0, p, q ∉ z, is called a surd. Here n
is called the order, ‘a’ is called the radicand and called the
sign of radical.

Like surds are the surds having
same orders and same radicands.

• 12 is the pure surd (having only irrational factors)
• 2 3 is the mixed surd (having both irrational factors and

rational factors rather than 1.)
• For the operation of the surds, pure surds should be changed

into mixed surds.
• Only like surds can be operated using plus and minus sign.

Worked out Examples

1. Insert any three rational numbers between 2 and 3 .
3 4
Solution :

The given rational numbers are 2 and 3
3 4

The other three rational numbers between 2 and 3 are as follows.
3 4

2 + 3
3 4
1st rational no. =
2

= 1 ` 17 j = 17
2 12 24

2nd rational no. = 1 a 2 + 17 k
2 3 24

= 1 ` 33 j = 11
2 24 16

PRIME Opt. Maths Book - VII 25

3rd rational no. = 1 ` 17 + 3 j
2 24 4

= 1 a 35 k = 35
2 24 48

So, on any other can be found out.

2. Express the surd 50 into mixed surd and 23 5 into pure surd.
Solution :
50 = 52 × 2 = 5 2 (mixed surd)
2 3 5 = 3 23 × 5 = 3 40 (pure surd)

3. Express the surds 2, 3 5, 4 6 into the surds of same order and
arranged in ascending order.
Solution :
The given surds are 2, 3 5, 4 6
L.C.M. of 2, 3, & 4 is 12.
Where,
2 = 2× 6 26 = 12 26 = 12 64

3 5 = 3× 4 54 = 12 54 = 12 625

4 6 = 4 × 3 63 = 12 63 = 12 216

The surds in ascending order
12 64 , 12 216 , 12 625
i.e. 2, 4 6 , 3 5

Note : If n a and n b are two surds and a > b then n a > n b

4. Simplify : 5 18 + 72 – 3 32
Solution :
5 18 + 72 – 3 32
= 5 32 × 2 + 62 × 2 – 3 42 × 2

= 15 2 + 6 2 – 12 2

= (15 + 6 – 12) 2

=9 2

26 PRIME Opt. Maths Book - VII

5. Multiply the surds 3 and 3 2 .
Solution
3 ×3 4

LCM of 2 and 3 is 6.
= 6 33 × 6 42
= 6 27 × 16
= 6 432

6. Rationalise the denominate of 1 2.
Solution : 3+

1 = 1 × 3– 2
3+ 2 3+ 2 3– 2

= [ a Multiply both numerator
and denominator by conjugate
= 3– 2 of 3 + 2 which is 3 – 2 ]
3–2

= 3– 2
1

= 3– 2

7. Simplify : 4 2– 3 3– 1
6– 6+ 3+ 2
Solution :

= 4 2 – 3 3 – 1
6– 6+ 3+ 2

= 4× 6+ 2– 3 × 6+ 3– 1 × 3+ 2
6– 2 6+ 2 6+ 3 6+ 3 3+ 2 3+ 2

=––

4( 6 + 2) 3( 6 – 3) 3 + 2
= 4 – 3 –1

= 6+ 2 – 6+ 3 – 3– 2
=0

PRIME Opt. Maths Book - VII 27

Exercise : 1.4

1. i) What is natural number? Write down the set of natural
numbers.

ii) How can you say that set of natural numbers is the sub–set of
whole number?

iii) What do you mean by irrational numbers?
iv) What is the set of rational numbers?
v) Differentiate between like surds and unlike surds with

examples.

2. Pick out the like surds from the following with reason.

i) 3 2, 2, 5 2 ii) 2 3 5 , 53 2, 3 3

iii) 3 24 , 33 3, 3 81 iv) 210, 75, 4 144

v) 20 , 2 45 , 3 250

3. Find the followings:

i. Insert 2 rational numbers between 1 and 7 .
2 8

ii. Insert 3 rational numbers between 2 and 5 .
3 6

iii. Convert 32 into mixed surd.

iv. Covert 3 54 into mixed surd.

v. Covert 2 5 into purse surd.

4. Express the followings as indicated in brackets.
i) 3 and 3 2 (into same order)
ii) 3 3, 4 4 (which is greater)
iii) 3, 3 4, 6 20 (in ascending order)
iv) 3 3, 2, 4 5 (in descending order)
v) 3 4, 4 5, 12 50 (in ascending order)

5. Simplify:

i) 18 + 50 – 3 8 ii) 2 108 – 6 75 + 5 48

iii) 3 40 + 2 3 320 – 3 3 135 iv) 3 700 – 2800 – 2 63

v) 4 3 432 – 5 3 128 + 7 3 2000

28 PRIME Opt. Maths Book - VII

6. Multiply the surds. ii) 3 4 × 6 8
i) 2 × 3 3 iv) 2 5 × 3 3 4

iii) 3 × 3 5 × 6 4

v) 3 2 × 2 4 8 × 5 6 4

7. Rationalise the denominator.

i. 1 ii. 1 iii. 4
5 3– 2 6– 2

1– 3 2+ 3
iv. 3 + 1 v.

2– 3

8. Simplify the following.

i) 2 3 + 1
5+ 3– 2

ii) 1 2 – 3+ 4
3+ 6– 3 6– 2

iii) 3 2 + 2 5 + 5
5+ 7+ 2– 7

x+ a x– a x+ 2 x– 2
iv) + v) –
x– a x+ a x– 2 x+ 2

1. Show to your teacher. Answer

2. Show to your teacher.

4. i) 6 27 and 6 4 ii) 3 3 is greater.
iii) 3 4 < 6 20 < 3 iv) 4 5 > 3 3 > 2
v) 12 50 < 4 5 < 3 4

5. i) 2 2 ii) 2 3 iii) 3 5 iv) 4 7 v) 74 3 2

6. i) 6 72 ii) 2 6 2 iii) 6 4050
iv) 6 6 2000 v) 20 12 32
iv) 2^x + ah v) 4 2x
8. i) 5 + 2 ii) 0 iii) –2 2 x–a x–2

PRIME Opt. Maths Book - VII 29

Alzebra

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. What do you mean by ordered pair ?
2. a. Which is greater between 2 and 3 3 ?

b. If f(x) = x³ + 2x² – 3x – 2 and g(x) = x³ – 3x² + x – 1, find f(x) + g(x).
c. If A = {1, 2}, B = {3, 4}, find A × B and n(A × B).

3. a. Simplify : 4– 3 – 1
6– 2 6+ 3 3– 2

b. Subtract the polynomials p(x) = (x² – 2x + 1) and q(x) = (2x²–x +
2 + x³). Also write down the type of polynomials so farmed ?

4. Multiply p(x) = x³ – 2x² + 3x – 2 by the polynomials q(x) = x³ – 2x + 3.

Unit Test - 2
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Convert 12 into mixed surd.
2. a. Write down any four ordered pairs of countries with respect to

capitals.
b. If Cartesian product A × B = {(3, 0), (4, 1), (5, 2)}, Find the sets

A and B.
c. Write down the polynomial 5 – x² + 2x³ – 3x in standard form.

Also write down its degree.
3. a. Write down in ascending order of 3 4, 3 and 6 15 in ascending

order by converting into like surds.
b. If A = {a, b, c} B = {2, 3}, find A × B. Also show in arrow diagram.
4. If p(x) = x² – 3x + 4 and q(x) = x² + x – 2, find p(x).q(x). Also write

down its types according to degree.

30 PRIME Opt. Maths Book - VII

2 Matrices

Specification Grid Table TM Periods

K(1) U(2) A(4) HA(5) TQ

No. of Questions – 1 1 – 2 6 10

Weight –24–

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : Materials Required:
At the end of the lesson • Chart paper.
• Students are able to • List of price chart of

represents the numbers in goods of a market.
matrix form. • List of properties of
• Students are able to know the
concept of order of matrix matrix addition
and types of matrices. • List of types of matrices
• Students are able to
operate the matrix addition, 31
subtraction, transpose and
multiplication by scalar.
• Students are able to know
the properties of addition
matrices.

PRIME Opt. Maths Book - VII

2.1 Matrices

The arrangement of the elements in different rows and columns can be
taken as follows by defining the rows and columns as given in example.

This type of arrangement of elements is called matrix. The cost per kg of

vegetables in Kalimati vegetables market in a particular day is as follows.
Potato Cauliflower Cabbage

Shop A 15 25 20

Shop B 20 30 15

Shop C 18 24 12

Above cost of the vegetables can be represented using square bracket as,

Here, Rows represents the shops and column represents the type of
vegetables. This type of presentation of numbers is called a matrix.

The rectangular array of the numbers in
different rows and columns which are taken
into a square bracket or round bracket is called

a matrix.

• A matrix is denoted by capital letters A, B, C, ................
• The members used in a matrix are called elements which are

denoted by small letters a, b, c, d, e, f, ........................
• The horizontal arrangement of the number in a matrix are

called rows.
• The vertical arrangement of the number in a matrix are called

columns.
• The no. of rows (m) and no. of columns (n) can be written as

m × n which is called the order of the matrix. We read m × n
or m by n.
• An element of a matrix is represented by the no. of row and
no. of columns. It lies as lower subscript e.g. the element lies

32 PRIME Opt. Maths Book - VII

in 2nd row and 3rd column is denoted by a23. The elements of a
matrix according to their position can be shown as,
A=

Here, a11 = element in first row and first column.
a23 = element in second row and third column.

• Order of the matrix A = 3 by 3 written as 3 × 3.
• The matrix A can be written as A3×3

Types of matrices
1. Row matrix

The matrix having only one row is called row matrix.

Ex : A = [a11 a12 a13]1 × 3 B = [2 4]1 × 2

2. Column matrix:

The matrix having only one column is cBall=edSSSRSSSSSSSTS1352cWWWWWWWWWXVWWo4×l1umn matrix.
Ex : A = SSSSSSSSTRaaa132111WWWWVXWWWW3×1

3. Null matrix:

The matrix having all the elements zero is called null matrix. It is

denoted by ‘O’.

000
Ex : O = < F
0 0 0
2×3

4. Rectangular matrix:
The matrix having unequal number of rows and columns is called

rectangular matrix.

Ex : A =

PRIME Opt. Maths Book - VII 33

5. Square matrix:

The matrix having equal number of rows and columns is called

square matrix. C = SSSSTRSSSS174 369WWXVWWWWWW3×3

Ex : A = [2]1×1 12 2
B = <3 4F 5
8
2×2

6. Equality of the matrices

Any two matrices having same order and same corresponding

elements are called equal matrices.

23 23
Ex : A = < F , B= < F
4 1 4 1

Here, A = B

7. General element of a matrix:

The element of the matrix is denoted by aij as the general element
where i is no. of rows and j is no of columns.

Ex : If aij = 2i – j
Then, a11 = 2 × 1 – 1 = 1
a12 = 2 × 1 – 2 = 0
a21 = 2 × 2 – 1 = 3
a22 = 2 × 2 – 2 = 2
<a11 a12F 1 0
\ 2 × 2 matrix is, A = a21 a22 = <3 2F

Worked out Examples

1. Represent the cost of articles given below in a matrix.

pen pencil

Shop A 15 10

Shop B 12 8

Solution:

The cost of pen and pencil of shop A and B respectively are as

follows where,

Rows " Represents the shops.

Columns " Represents the articles.

15 10
A = <12 8 F

34 PRIME Opt. Maths Book - VII

2. If a matrix A= 2 3 1 , find the order of the matrix. Also write
<4 –2 0F

down the elements a21 and a13.
Solution:

The given matrix is,

231
A = < F
4 –2 0

No. of rows = 2

No. of columns = 3

` Order of the matrix = 2×3

a21 = element in second row and 1st column = 4
a13 = element in 1st row and 3rd column = 1

x 22 2 are equal matrices, find the value of x and y.
3. If < F and < F
3 y + 2 3
5

Solution:

x 2 22
The equal matrices are: < F = < F
3 y + 2 3 5

By equating the corresponding element,

x = 2 and y + 2 = 5

x = 2 and y = 5 – 2

` x = 2 and y = 3

` x =2

y =3

4. If aij = 3i + j, find a11, a12, a21 and a22. Also write down 2 × 2 matrix.
Solution:

a11 = 3 × 1 + 1 = 4
a12 = 3 × 1 + 2 = 5
a21 = 3 × 2 + 1 = 7
a22 = 3 × 2 + 2 = 8
45
` 2 × 2 matrix = < F
7 8

PRIME Opt. Maths Book - VII 35

Exercise : 2.1

1. Write down the following informations in matrix form. Also write

down the order of the matrices.

i. The cost of fruits in different shops.

Orange Apple

Shop - 1 Rs. 6 Rs. 8

Shop - 2 Rs. 5 Rs. 9

ii. The numbers of boys and girls of different schools.

Boys Girls

School - A 260 320

School - B 420 300

School - C 510 420

iii. The expenditure of three students in different days of school
tiffin.

student - A student - B student - C

Day - 1 Rs. 40 Rs. 60 Rs. 50

Day - 2 Rs. 45 Rs. 40 Rs. 35

2. Write down the types of [email protected]. .SSSSTRSSSSRSSSSSSSST132–242WWWWWWWXWV
3. order ii.
1 24
i. <F v.
–3 1 2 of
ii.
23 3 115XVWWWWWWWW
iv. <F v. 0
–1 2 –3

Wiiv. .riteSSSTRSSSSTRSSSSSSSSS132–d242XVWWWWWWWWow–30n3th115eWWWWWVXWWW 63 2 1@ 2 3
iii. <–1 2F

00 1 24
00 vi. < F
–3
12

36 PRIME Opt. Maths Book - VII

4. If a mWartirteixdAow=RSSSSSSSTSn364or–12d2er––0o53fWXWWVWWWWWm, fiantdrixthAe. followings.
i.

ii. Write down the element in second row and third column.

iii. Write down the element a32.
iv. Find the sum of a11 + a21 + a32.
v. Find the element aij where i = 3, j = 1.

5. Find 2 × 2 matrix for the followings.

i. aij = 2i + j ii. aij = 2i + 3j iii. aij = 3i – 2j

6. Find x and y from the following equal matrices.

2x 21
i. < F and < F
y 4 3 4

ii. A = =3 x + 1G and B = 3 3
1 y + 2 <1 5F

iii. P = =2 x - 2G and Q = 2 1
1 y + 3 < F
1
4

iv. 3 5 and =3 y - 2G
<x - 1 2F 2 2

3 1 =x - 2 y + 4G
v. A = < F and B = 0 2
0
2

7. Prime more creative questions.

i. What do you mean by matrix.
ii. Define the term square matrix with an example.

iii. If =x + y 7G = 5 z 7 2F , find the value of x, y and z.
x–y 2 <1 +

iv. Find 3 × 3 matrix where the general element is aij = i + 2j.
v. Collects the marks obtained by your four best friends in two

unit tests and represent the informations in matrix form.

PRIME Opt. Maths Book - VII 37

1. i. <56 89F2×2 Answer iii. <4450 60 5305F2×3
40
ii. RTSSSSSSSS524162000 334022000VXWWWWWWWW3×2
2. i. Rectangular iii. Column
iv. Square ii. Null vi. Square
v. Row
3. i. 3 × 3 iii. 2 × 2
iv. 3 × 1 ii. 1 × 3 vi. 2 × 3
v. 2 × 2
4. i. 3 × 3 iii. 2
iv. 9 ii. – 5
v. 6
5. i. <53 64F
6. i. 1, 3 ii. <57 180F iii. <14 –21F
iii. 3, 1
iv. 3, 7 ii. 2, 3
v. 5, –3

7. iii. 3, 2, 0 iv. SSSSRSTSSS354 5 789WWWWWWWWXV
6
7

38 PRIME Opt. Maths Book - VII

2.2 Operation on matrices

The simplification of two or more matrices in a single matrix by using
any kind of mathematical operations indicates the operation on matrices.
Which are addition, subtraction, multiplication with scalar, transpose
etc. Here we are discussing some of them in grade VII.

Addition of matrices

Marks obtained by three students Sita, Pranav and Pranisha in two

monthly test examinations in optional maths are as follows of two

months.

Ashadh 1st 2nd Shrawan 1st 2nd

Madhab 60 75 Madhab 70 75

Pralika 85 90 Pralika 95 80

Pranisha 95 85 Pranisha 80 95

Total marks obtained by them in two tests in two months as,

1st 2nd

Madhab 60 + 70 = 130 75 + 75 = 150

Pralika 85 + 95 = 180 90 + 80 + 170

Pranisha 95 + 80 = 175 85 + 95 = 180

TSSTRSSSSSS689h055is 789in550fWWWWWWWXWVor+mSSSRSSSTSS789a005tio789n505cWWWWWWWVXWa=n be expressed in m=atSSSSSRTSSSr111i378x050for111m578000aWWWWWWWWXVs,

The sum of any two matrices having same order
is called a new single matrix obtained by adding
the corresponding elements of the matrices.
The single matrix so formed has the order same
as the given matrices.

PRIME Opt. Maths Book - VII 39

eg. If A = <–21 53F, B = <–34 –21F

Then,

23 3 2
A+B = < F + < F
–1 5 –4 –1

= =2 + 3 3 + 2G
–1–4 5–1

55
= <–5 4F

Difference of the matrices

The difference of any two matrices having same order is called a new

single matrix obtained by subtracting the corresponding elements of the

matrices. The single matrix so formed has the order same as the given

matrices.

eg. If P = <52 3 12F, Q = <–21 1 53F
–3 1

Then,
2 3 1 –1 1 5

P – Q = <5 –3 2F – < 2 1 3F

=

3 2 –4
= <3 –4 –1F

Transpose of the matrix.

Let us taking an example where cost of apple of a shop of three days are

given.

1st shop 2nd shop

Sunday 150 140

Monday 160 145

Tuesday 170 150

40 PRIME Opt. Maths Book - VII

It can be written in such a way by changing the information of rows and

column as,

Sunday Monday Tuesday

1st Shop 150 160 170

2nd Shop 140 145 150

SAu=chSSSSSSSRTS111e567x000am111p454l005eWWWWWXVWWWs can be written in matrix form as,
After changing row and columns =

It is called transpose of the matrix A and denoted by AT.

The new matrix obtained by interchanging the

rows and columns of a matrix is called transpose

of the matrix.
• Transpose of A is denoted by AT or A` or � .

Note : Order of the transpose matrix is different than matrix A for the

rectangular matrix but same for the square matrix.

eg. If A = <32 4 95F then.
6

MultAiTp=licSSSSTSRSSS534at629ioWWWVWWWWXW n of a matrix with a scalar.

Let us consider the cost of apple given above becomes double in the next
week which can be written as

(old cost) (new cost)

PRIME Opt. Maths Book - VII 41

The new matrix formed by multiplying each element
of a given matrix by a given scalar quantity is called
the multiplication of a matrix with the scalar.
i.e. If A = <ac dbF; then, KA = <kkac kkdbF

Eg. If A = <13 ––21F, �ind 3A.

Solution : A = <13 ––21F
3A = 3<13 ––21F = <39 ––36F

Worked out Examples

1. If A= :13 2 D and B = :12 1 D , �ind A + B.
0 3
Solution:
:13 2
A = 0 D

B = :12 1 D
3
then,
:13 2 :12 1
A + B = 0 D + 3 D

= :13 + 2 2 + 1 D
+ 1 0 + 3

= : 3 3 D
4 3

2. If P= 831 4 B and Q= :12 –1 D , �ind 2P – Q.
2 –3
Solution:
831 4
P = 2 B

Q = :12 –1 D
–3
then,
= 2831 4 :12 –1
2P – Q 2 B – –3 D

42 PRIME Opt. Maths Book - VII

= :26 - 2 8 + 1 D
- 1 4 + 3

= 8 4 9 B
1 7

3. ISfoAlu=tio:n23: –1 D and B = 8 3 1 B , �ind (A + B)T
1 –2 4

A = : 2 –1 D , B = 8 –23 1 B
3 1 4

then, :23 –1 3 1
1 –2 4
(A + B)T = D – 8 B

= : 4 +3 - 1+ 1 D
3 -2 1+ 4

= <17 50FT

= <70 15F

4. If A + B = <32 12F, B = =10 -32G, �ind the matrix A.
Solution:

A+B = <32 12F , B = =1 - 2G
0 3

now,
A + B = <32 12F – B

A = <32 12F – =1 - 2G
0 3

= =32 - 1 2 + 32G
- 0 1 -

= <22 -42F

PRIME Opt. Maths Book - VII 43

Exercise : 2.2

1. Add the following matrices.

2 1 –1 3
i. A = < F and B = < F
1 0 2 1

3 –2 1 –4
ii. P = < 4 F and Q = < F
–1 –2 –1

3 –2 43
iii. M = <1 1 F and N = <2 4F

1 3 –2 21 2
iv. A = <1 2 4 F and B = <3 1 –3F

3 2 –1 3
v. P = <1 –4F and Q = < 2 1F

2. Subtract the pair of matrices given below.

3 2 –1 3
i. P = < F and Q = < F
1 –4 2 1

1 3 –2 21 2
ii. A = < F and B = < F
1 2 4 3 1 –3

3 –2 43
iii. M = < F and N = < F
1 1 2 4

3 –2 1 –4
iv. P = < 4 F and Q = < F
–1 –2 –1

2 1 –1 3
v. A = <1 0F and B = < 2 1F

3 12 ––32F, find the following matrices.
3. If A = < 4F , B = <
2 1

i. 2A + B ii. 3A + 2B

iii. AT + BT iv. (A + B)T

v. (2A + B)T

4. Prime more creative questions:

1 42 1 , find the matrices B.
i. If A + B = < F and A = < F
3 1 1
0

3 21 1 , find the matrices A.
ii. If B = < F and A –B = < F
1 –1 1
5

44 PRIME Opt. Maths Book - VII

12 –2 –4
iii. If P = <–3 –2F and Q = < 6 4 F, prove that 2P + Q is null matrix.

–4 2 –2 1
iv. If A = < 4 –2F and B = < 2 –1F, prove that A – 2B is a null
matrix.

v. If A + B = 2 5 4 1 , find the matrices A and B.
<3 5F and A – B = <–1 3F

vi. If P + Q = 2 5 and P – Q = 4 –1 , find the matrices P and Q.
< F < F
3 –1
–3 –5

Answer

14 4 –6 71
1. i. <3 1F ii. <–3 3 F iii. <3 5F

340 25
iv. <4 3 1F v. <3 –3F

4 –1 –1 2 –4 –1 –5
2. i. <–1 –5F ii. <–2 1 7 F iii. <–1 –3F

22 3 –2
iv. <1 5F v. <–1 –1F

80 13 –1 53
3. i. <F ii. <F iii. <F
55 86 –1 1

53 85
iv. <F v. <F
–1 1 05

–1 3 43 3 3 –1 2
4. i. <F ii. <F v. < F , < F
21 24 1 4 2 1

vi. 3 2 and –1 3
< F < F
1
–4 2 1

PRIME Opt. Maths Book - VII 45