Optional Math 9

Alternative way: 12,

i. nth term of linear sequence.
Taking Example 1 :

2, 4, 6, 8, 10,

22222

Here, constant difference of the terms of sequence is 2. Where nth term is taken as,

Ttnh=eacnon+sbtant difference is 2

So, a = 2.
Then,

tn = an + b
t1 = 2 × 1 + b
or, 2=2+b

\ b = 0
\ The nth term of the sequence is,

tn = 2n

Taking example 2 : 9, 11, 13,
1, 3, 5, 7,

22222 2

Here, the nth term is taken as,

Ttnh=eacnon+sbtant difference is 2, so a = 2
Then,

t1 = a × 1 + b

or, 1 = 2 + b

\ b = –1

\ nth term of the sequence is,

tn = an + b

= 2n – 1

ii) Quadratic nth term of quadratic sequence
Taking an example 4,

0, 2, 6, 12, 20, 30,

2 4 6 8 10

2222 47
Here, the nth term of the sequence is taken as,
tn = an2 + bn + c

PRIME Opt. Maths Book - IX

The second difference is the constant difference 2. So, 2a = 2
\ a = 1
Then,
tn = an2 + bn + c
t1 = 1 × 12 + b × 1 + c
or, 0 = 1 + b + c
\ b = – 1 – C ...................................... (i)
Again,
t2 = 1 × 22 + b × 2 + c
or, 2 = 4 – 2 – 2c + c
or, c = 0

From equation (i)
b = – 1 – C = – 1
\ nth term of the sequence is,
tn = an2 + bn + c
\ tn = n2 – n

• Taking another example 6, 11, 18, 27, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
Here,

6, 11, 18, 27,

579

22
The nth term of the sequences is,

Ttnh=easne2co+nbdnd+ifcference is the constant difference ‘2’. So, 2a = 2

\ a = 1
Then,

tn = an2 + bn + c
t1 = 1 × 12 + b × 1 + c
6 = 1 + b + c

or, b = 5 – C ................................. (i)

Again,

t2 = 1 × 22 + b × 2 + c [\ using (i)]
or, 11 = 4 + 10 – 2c + c

or, c = 3

From equation (i),

b = 5 – 3 = 2

\ nth term of the sequence is,

tn = an2 + bn + c
\ tn = n2 + 2n + 3

48 PRIME Opt. Maths Book - IX

Note :
i) If the sequence is written in terms of opposite sign as – , +, –, + etc

for each terms respectively tn = (–1)n
ii) If the sequence is written in terms of opposite sign as +, –, +, –,

+, – etc, the nth term tn = (–1)n + 1

Sum of the terms of the sequence :

If a1, a2, a3, a4, a5 are the terms of the finite sequence, the sum of the terms is written as,
S5 = a1 + a2 + a3 + a4 + a5

If a1 + a2 + a3 + a4 + ... + ... + ... + ... + ... + ... + ... + an are the ‘n’ terms of the sequence, the sum
can be written as,
Sn = a1, a2, a3, a4, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... + an

nIfoat1b, ae2,dae3f,ina4e,d...., ..., ..., ..., ..., ..., ..., ..., ..., ... are the infinite terms of the sequence, the sum can

Sum of the terms of a sequence using sigma notation:
In the sequence having ‘n’ terms aas1,,
The sum of them can be written a2, a3, a4, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... an

Sn = a1 + a2 + a3 + ... + ... + ... + ... + ... + ... + ... + ... + an

/n

= ak
k =1

Where ∑ (sigma) is used for symbol of summation of the terms and ‘k’ is the number

of terms from 1 to n.
To write the sum of the terms in sigma notation, nth term of the sequence should be

know.

Let us take an example.
1, 5, 9, 11, 15, 19.
For nth term of the sequence,

1, 5, 9, 11, 15, 19,

44444

Here, the nth term of this linear sequence is,
Ttnh=eacnon+sbtant difference is 4. So a = 4.

PRIME Opt. Maths Book - IX 49

Then,

tn = an + b
t1 = 4 + b
or, 1=4+b

\ b = – 3

\ nth term of the sequence, tn = 4n – 3.

The sum of the terms is

Sn = 1+ 5+ 9 + 11 + 15 + 19
= (4n – 3)
/6
n=1

Worked out Examples

1. If nth term of a sequence is tn = 3n – 2, find the first five terms of the sequence.
Solution :
The nth term of a sequence is,

Ttna=ki3nng – 2

n = 1, t1 = 3 × 1 – 2 = 1
n = 2, t2 = 3 × 2 – 2 = 4
n = 3, t3 = 3 × 3 – 2 = 7
n = 4, t4 = 3 × 4 – 2 = 10
n = 5, ftinr=st3fi×ve5t–er2m=s 13
Hence, the are 1, 4, 7, 10, 13.

2. Iffinad1, tah2,eav3a, .l.u.,e...o, .f..a, 1..,.,a.2..,, ... are the terms of a sequence where an + 1 = 2an – 3 and a0 = 4,
a3 and a4.
Solution :

In a sequence a1, a2, a3, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... are the terms.
an + 1 = 2an – 3 and
a0 = 4

Taking, n = 0
an + 1 = 2an – 3
or, a0 + 1 = 2 × a0 – 3
or, a1 = 2 × 4 – 3 = 5

Again, Taking n = 2, Taking n = 3,
Taking n = 1 a3 + 1 = 2 × a3 – 3
a4 = 2 × 11 – 3 = 19
a1 + 1 = 2a1 – 3 a2 + 1 = 2 × a2 – 3
\ a2 = 2 × 5 – 3 = 7 a3 = 2 × 7 – 3 = 11

50 PRIME Opt. Maths Book - IX

3. Write down in expanded form of /5 (–1)n (n2 +3n). Also find the sum.
Solution : n=1

The given sum of the terms of a sequence in sigma notation is,
/5
Sn
= (–1)n (n2 + 3n)

n=1

= t1 + t2 + t3 + t4 + t5 [Here, tn = (–1)n (n2 + 3n)]

= (–1)1 (12 + 3 × 1) + (–1)2 (22 + 3 × 2) + (–1)3(32 + 3 × 3) + (–1)4(42 +

3 × 4) + (–1)5(52 + 3 × 5)

= –4 + 10 – 18 + 28 – 40

= – 24

4. If sum of the ‘n’ terms of the sequence is Sn = n2 + n , find the 3rd and 4th terms of the
2
sequence.

Solution :
The sum of the ‘n’ terms is,
n2 + n
Sn = 2

Then,

3rd term (t3) = S3 – S2

= 32 + 3 – 22 + 2
2 2

=6–3

=3

4th term (t4) = S4 – S3

= 42 + 4 – 32 + 3
2 2

= 10 – 6

=4

5. Find the nth term of the sequence 3, 5, 8, 12, 17. Also write down in sigma notation.
Solution :
The given sequence is 3, 5, 8, 12, 17, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
Difference between the terms can be seen as follows,

3, 5, 8, 12, 17,

2 3 4 5 (is not equal)

111 (is equal)

Here, the constant difference is obtained in second step. So, the nth term is quadratic
form as,

PRIME Opt. Maths Book - IX 51

Ttnh=enan, 2 + bn + c

2a = 1 [constant difference]

\ a= 1
2

So, tn = an2 + bn + c
t1 =
1 × 12 + b × 1 + c
2
or,
b=3– 1 –c= 5 – 2c ....................... (i)
Again, 2 2

t2 = 1 × 22 + b × 2 + c
or, 2
or,
\ 5= 1 × 42 + 2c 5 – 2c m +c= 1
2 2 2

3=5–c

c=2

From equation (i)

b= b – 2×2 = 1
2 2

\ nth term is,

tn = 1 n2 + 1 n+2
2 2

= 1 (n2 + n + 4)
2

Again,
The sum of the terms of sequence in sigma notation is,

Sn = 3 + 5 + 8 + 12 + 17
/5
= 1 (n2 + n + 4)
2
n=1

= 1 /5 (n2 + n + 4)
2
n=1

6. Add two more patterns and find the nth term of the sequence so formed of the number

of dots. Also write down in sigma notation from the given patterns.

Solution : PRIME Opt. Maths Book - IX
Two more patterns in the given patterns.

52

The sequence according to number of dots from the diagrams as,
5, 10, 15, 20, 25, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...

For the general term (nth term)

5, 10, 15, 20, 25,

5 5 5 5 (is same)

The constant difference is 5 which is obtained in 1st step. So, the nth term becomes,
tn = an + b.

Here,

a = 5 (Constant difference)

t1 = 5 × 1 + b

or 5 = 5 + b

\ b = 0

Then,

the nth term becomes,

tn = 5n

Again, the sum of the terms of the sequence in sigma notation,

Sn = 5 + 10 + 15 + 20 + 25

= /5 (5n)
n=1

PRIME Opt. Maths Book - IX 53

Exercise 1.6

1. i) What is sequence? Write down one example.
ii) What do you mean by series? Write down with an example.
iii) Define the terms finite and infinite sequence with examples.
iv) What is progression? Write down with an example.
v) Write down the sum t1 + t2 + t3 + t4 + t5 + t6 in sigma notation.

2. Which of the followings are the finite or infinite sequence or series?
i) 2 + 5 + 8 + 11 + 14
ii) 3, 7, 11, 15, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... 15 terms
iii) 1, 4, 9, 16, 25, 36, ..., ..., ..., ..., ..., ..., ..., ..., ..., ...
iv) 2 + 6 + 18 + 54 + ... + ... + ... + ... + ... + ... +
v) a, a + d, a + 2d, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... , a + (n – 1)d

3. Find the followings.
i) Write down the corresponding series for the sequence 5, 12, 19, 26, 33, ..., ..., ...,

..., ..., ...
ii) Write down the corresponding sequence for the series 33 – 29 + 25 – 21 + 17 – 13
iii) Write down the next term of the sequence 1, 4, 9, 16, 25, ..., ..., ..., ..., ..., ... Is it finite or

infinite?
iv) What is the common difference of the sequence a, a + d, a + 2d, a + 3d, ..., ..., ..., ...,

..., ...
v) Is a sequence 2, 6, 11, 17, 25, ..., ..., ..., ..., ..., ... a progression? Why?

4. Find the next three terms of the sequences.

i) 1, 6, 11, 16, ..., ..., ..., ..., ..., ... ii) 1, 8, 27, 64, ..., ..., ..., ..., ..., ...

iii) 12 , 2 , 3 , 4 , ..., ..., ..., ..., ..., ... iv) 80, –75, 70, –65, ..., ..., ..., ..., ..., ...
3 4 5

v) – 3 , 7 , – 11 , 15 , ..., ..., ..., ..., ..., ...
7 12 17 22

5. Find the first five terms of the sequence whose nth term are given below.

i) tn = 2n – 1 ii) an = 4n + 3
iii) an = n2 + 4n – 5 iv) an = (–1)n n(n + 1)
v) tn = (–1)n + 1 a 2n
1 k
n2 +

6. Find first five terms of the followings where the terms are explained below.

i) an = an + 1 – 2 and a1 = 3 ii) un + 1 = un + 3 and u2 = 5
iii) tn – 1 = 2tn – 1 and t2 = 2 iv) an + 2 = an + 1 + an, a1 = 1 a2 = 2
tn – 2
v) tn = tn –1 , t1 = 1 and t2 = 2

7. Find the nth terms of the sequences.
i) 1, 4, 7, 10, 13, ..., ..., ..., ..., ..., ... ii) 5, 2, –1, –4, ..., ..., ..., ..., ..., ...

54 PRIME Opt. Maths Book - IX

iii) 10, 16, 22, 28, ..., ..., ..., ..., ..., ... iv) 1 , 3 , 5 , 7 , ..., ..., ..., ..., ..., ...
3 7 11 15
2 4 6 8
v) – 5 , 11 , – 17 , 23 , ..., ..., ..., ..., ..., ...

8. Find the general term of the sequences.

i) 1, 4, 9, 16, 25, ..., ..., ..., ..., ..., ... ii) 1, 3, 6, 10, 15, ..., ..., ..., ..., ..., ...

iii) 6, 11, 18, 27, 38, ..., ..., ..., ..., ..., ... iv) 0, – 1 , 2 , – 3 , 4 , – 5 , ..., ..., ..., ..., ..., ...
2 3 4 5 6
v) a, a + d, a + 2d, a + 3d, ..., ..., ..., ..., ..., ...

9. Write down in expanded form and evaluate.

i) /4 (2n + 3) ii) /5 (4n – 1)
n=1 n=1

iii) /5 n (n + 1) iv) /4 (–1)n [n (n + 2) + 3]
n=2 n=1

/5 (–1) n + 1 ` n n 1 j
+
v)
n=1

10. iiIvi)f)i )S n iSSSsnnnt===he22nnns(u22nm–++13o1,,f)f6f(i‘innn2’ddntetS+hr4m1,eS)5s5,toahfinftnetddrhmte45.t.s h eaqnudeiivi5n))t ch etewSSrnnhm==icsnh.n(in(sn2+g+i1v1)e),nf,ibnfiednldothwteh. e4t3h rtdearnmd. 6th terms.





11. Write down the following in sigma notation for the given series.

i) 2 + 5 + 8 + 11 + 14 + 17 ii) 1 + 3 + 6 + 10 + 15

iii) –4 + 7 –11 + 16 –22 + 29 iv) 1 – 2 + 3 – 4
v) a + ar + ar2 + ar3 + ar4 + ar5. 3 5 8 12

12. Write down the following sequence of number of dots used in diagrams in sigma
notation after calculating the general term by adding one more pattern for each.

i)

ii)

PRIME Opt. Maths Book - IX 55

iii)
iv)

v)

13. PRIME more creative questions

i) Find the nth term and write down in sigma notation of the series

1 × 3 + 2 × 7 + 3 × 11 + 4 × 15 + 5 × 19.
=a+ (n – 1)d where a = 3, d = 4, ftihnedstehreiets5.and t11.
ii) If nth term of the sequence is tn arn–1 where a = 3 and r = 2, find where a = 40 and
iii) If nth term of a sequence is tn = n
iv) If 2 = 5 and r = 2, find
d sum of first ‘n’ terms of a sequence is Sn = [2a + (n –1)d]
v) = – 4, find the value of S5.
If a (rn – 1)
sum of the first ‘n’ terms of a series is Sn = r –1 where a

the value of S4 + S5.

14. Project work
Prepare a chart of the way of finding the sigma notation for linear and quadratic

form of sequence and present in your classroom.

56 PRIME Opt. Maths Book - IX

Answer

1. Show to your teacher. ii) 125, 216, 243 iii) 5 , 6 , 7
2. Show to your teacher. 6 7 8
3. Show to your teacher. 19 23 27
v) – 27 , 32 ,– 37
4. i) 21, 26, 31

iv) 60, –55, 50

5. i) 1, 3, 5, 7, 9 ii) 7, 11, 15, 19, 23 iii) 0, 7, 16, 27, 40
iv) –2, 6, –12, 20, –30
v) 1, – 4 , 6 , – 8 , 10
5 10 17 26

6. i) 3, 5, 7, 9, 11 ii) 2, 5, 8, 11, 14 iii) 3, 2, 3 , 5 , 9
iv) 1, 2, 3 , 5, 8 2 4 8
1 1
v) 1, 2, 2 , 4, 8

7. i) 3n – 2 ii) 8 – 3n iii) 2(3n + 2)

iv) 24nn––11 v) (–1)n 62nn–1

8 . iiv) ) n(–2 1)n+1 ` n n– 1 j vii)) a2n+((nn+–11)) d iii) n2 + 2n + 3
iii) 68
9. i) 32 ii) 55 iii) 18
iv) 14
v) 37
60

10. i) 35, 53, 18 ii) 8
iv) 3, 6 v) 16, 25

6 5 n (n + 1)
(3n – 1) ii) 2
/ /11. i) /6 ( n2 + 3n + 4 )
2
n=1 n=1 iii) (–1) n
n=1

4 6

/ / iv) (–1)n+1 (arn–1)
n=1
a n2 2n 4 k v) n=1
+n+

12. i) /6 n (n + 1) ii) /5 (4n – 3) iii) /5 (2n + 1)
2
n=1 n=1 n=1

/6 5

iv) n2 v) Σ (3n + 2)
n=1 n=1

5 ii) 19, 43 iii) 3, 6, 12, 24, ..., ..., ...
v) 230
13. i) Σ n (4n –1) 57
n=1

iv) 160

PRIME Opt. Maths Book - IX

Algebra

Unit Test - 1 Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:
1. What is Cartesian product?

2. a. If (3x+y, y–2)and (10, 2y–6)are equal ordered pairs, find the value of ‘x’ and y;
b. What must be subtracted to 2x3 – 3x2 + 5x – 3 to get the polynomial x3 – 2x + 5?

c. Find the nth term of the sequence 3, 5, 8, 12, 17, 23, ................. .

3. a. Divide the polynomial x3 – 5x + 3x + 4 with x + 3.
b. If f(x) = 2x – 3, domain = {–2, –1, 0, 1, 2}, find range and show the function in

arrow diagram.

4. Add one more pattern in the given pattern and write down the sequence of dots in
sigma notation.



Unit Test - 2
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. If A × B = {(1, 3), (1, 4), (1, 5) (2, 3), (2, 4), (2, 5)}, find thee sets A and B.

2. a. If f(x + a) = f(x) + f(a), find f(0) and prove that f(–a) = –f(a).
b. If p(x) = 3x3 + 2x2 – 5x + 7 and q(x) = x3 –3x2 – 2x + 3, find p(x) + q(x). Also write
down its types according to degree.
c.
/5
3. a.
b. Write down in expanded form and find the sum of (–1)n+1 (n2 + 3n).

n=1

Find the nth term of the sequence 2 – 5 + 10 – 17 + 26 – 37. Also write down in
sigma notation.
Find the relation R = {(x, y) : x + y ≤ 10, x, y ∈ N}. Also show in arrow diagram.

4. If f(x) = 2sin x + 1, range = {1, 2, 3 + 1, 3}, find the domain. Which type of function is
it? Why? Also show in arrow diagram.

58 PRIME Opt. Maths Book - IX

Unit 2 Limit and Continuity

Specification Grid Table

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

No. of Questions 1 – 1 – 3 5 10
Weight 1 – 4 –

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

Objectives : At the end of the lesson
• Students are able to understand rational and irrational numbers and their

Sequence.
• Students are able to generalize the decimal numbers to the nearest whole

number.
• Students are able to find the limit value of a function in diagram.
• Students are able to know the limit value of infinite series and graph.
• Students are able to know the symbol of limit and indeterminate form.

Materials Required:
• Chart paper.
• Set of rational and irrational numbers.
• Graph paper.
• Model of graph of functions.
• Geometrical shapes.

PRIME Opt. Maths Book - IX 59

Limit

Enjoy the recall
• y = f(x) is a function f : A → B
where
Domain = { Set of element x} = A.
Co-domain = main set of range = B.
Range = {Set of element y}
Image = element of range = y
Pre - image = element of domain = x
• If f(x) = x2 – 2x + 3, we can find f(1), f(2), f(3), ...etc where f(1) = 12 – 2 × 1 + 3 = 2 and

so on ....................................... .
• If f = {(1, 5), (2, 6), (3, 7), (4, 8)} is a function where
Domain = {1, 2, 3, 4}, Range = {5, 6, 7, 8}
• Sequence in increasing order is 2, 5, 8, 11, ..., ..., ..., ..., ..., ...
• A sequence in decreasing order is, 20, 17, 14, 11, 8, 5, ..., ..., ..., ..., ..., ...
• When a football is dropped on the ground following result can be obtained.

• Round numbers of 2.467 is,

2. 467 = 2.47 (round up in hundred)

2. 467 = 2.5 (round up in tens)

2. 467 = 2 (round up in whole numbers.

• If value of ‘n’ goes on increasing, the value of 1 will decrease and be very close to
n
zero.

Diagramatic representation of a sequence and absolute numbers.
1. Let us consider a line segment AB which is going on dividing at mid-point

P again PA is divided by a mid-point Q and So on R, S, T can be obtained.

A Q R STP B

If AB is taken as 12cm. Then the parts of AB by dividing at mid point will be.
AP = 6cm
QP = 3cm
3
RP = 2 cm

SP = 3 cm
4

60 PRIME Opt. Maths Book - IX

TP = 3 cm
8
3 3 3 3
Where the sequence will be, 12, 6, 3, 2 , 4 , 8 , 16 , ..., ..., ..., ..., ..., ...

Here, The last mid - point of the line segment will be approaches to P and the length

becomes near to zero (0) but not equal to zero.

Note : If a real number is taken in a number line, it will approaches to + ∞ in right side

and approaches to – ∞ in left side.

2. Let us consider a triangle and the triangles obtained by joining the mid
- point of sides of previous triangle and so on.

AA

PQ PX Q

YZ

B R C BRC

According to the figures given in the above example, following conclusions can be

obtained.

• Area of triangle ABC is taken as 40 square units and the area of DPQR

becomes 10 square units. (i.e. Area goes on decreasing by 4 times.)

• The sequence so formed according to area becomes, 40, 10, 5 , 5 , 5 ,
2 8 32
............ and so on.

• The area of last triangle so formed will be very close to zero but not equal

to zero.

3. Let us consider a cylinder which is cutting down at the middle of the cylinder
and so on for each pieces.



According to the volume of cylinder goes on decreasing by half following conclusion

can be obtained where volume of 1st cylinder is taken as 20 cubic units.
• Volume of cylinder goes on decreasing by half and becomes 10 cubic units,
5
5 cubic units, 2 cubic and so on.

• The sequence so obtained of their volume is, 20, 10, 5, 5 , 5 , 5 , .......... The
2 4 8
volume of last cylinder so formed will be very close to zero but not equal

to zero.

PRIME Opt. Maths Book - IX 61

Sum of infinite series :
1. Let us consider a line segment PQ, having length 2cm.

P A C D EFQ

PQ is divided by A at mid - point
AQ is divided by C at mid -point
CQ is divided by D at mid - point,
DQ is divided by E at mid - point,
EQ is divided by F at mid - point
Then, The sequence of length of the line segment so formed becomes, 2, 1,
1,
1 , 1 8, , 1 , ..., ..., ..., ..., ..., ...
2 4, 16

After the last division the divided point will be nearest to Q where length becomes

very near to zero but not equal to zero.

Here, Sum of the length of the line segment becomes,
S1 = PA = 1cm = 1cm

S2 = PC = 1 + 1 cm = 1.5cm
2

S3 = PD = 1 + 1 + 1 = 1.75cm
2 4

S4 = PE = 1 + 1 + 1 + 1 = 1.875cm.
2 4 8

S5 = PF = 1 + 1 + 1 + 1 + 1 =
2 4 8 16

and So on S6, S8, ..., ..., ..., ..., ..., ..., Sn can be calculated which will be very close to 2 but
not equal to 2.

Here,

• For the infinite series, sum of the infinite terms can not be defined but
partial sum like S1, S2, S3, S4, .......... S∞ can be defined.

• 2 is the limit value of the sum of infinite series so formed.

• The common ratio of the series of the above sequence is w12 hwichhiicsh2i.s less
than 1. So, It has the limit value with fixed absolute value

• The limit value of the sum of infinite series can be defined using formula
a
also, which is S3 = 1– r .

• In the above example

The common ratio r = 1 , first term a = 1.
2
a 1 1
So, S3 = 1– = = 1 =2
r 1 2
1– 2

• This series is called convergent series where its limit exists.

62 PRIME Opt. Maths Book - IX

2. Let us consider a series 1 + 2 + 4 + 8 + 16 +... + ... + ... + ... + ... + ... +

Here,
The common ratio of the terms is 2 which is more than 1. So, the sum of the infinite

series will not be the fixed absolute value and limit does not exist.
• This series is called divergent series where its limit does not exist.



3. Let us consider a series,

–8+4–2+1– 1 + 1 – 1 + ... + ... + ... + ... + ... + ... + ...
2 4 8

Here,

The common ratio is – 1 which is less than ‘1’. Hence the series has the
2
limit value which will be a fixed absolute value.

4. Let us consider a series 0.25
The series 0.25 can be expressed as 0.25 + 0.0025 + 0.000025 + 0.00000025 + ... +

... + ... + ... + ... + ... + and so on and can be extended in terms of infinity where the last
term will be very close to zero but not equal to zero.

Hence,
The limit value of nth term will be the absolute value zero.

Exercise 2.1

1. i) What is the round number of 5.678 in two decimal system?
ii) Write down the round digit of 6.425.
iii) Define range and domain in a function y = f(x).
iv) Define image and pre-image in a function f(2) = 5.
v) Write down 4 terms of a sequence 0.5.

2. i) Draw a line segment of length 12cm and divide it at mid - point respectively 12
times. What conclusion do you get ? Explain.

ii) 0.1, 0.01, 0.001, ................ is a sequence. What will be its 10th term. Also write
down it’s limit value.

iii) Write down the limit value of the sequence 1.1, 1.01, 1.001, 1.0001, ......................

iv) Write down the limit value of 0.9, 0.99, 0.999, 0.9999, ...................
v) Draw a triangle having base of length 12cm and draw the line joining the mid

- points of other two sides and so on for the triangle obtained after joining the

mid - points continuously 5 times. What will be the length of the line segments

obtained by joining the mid - points respectively ? Also write down the limit

value of the length of such last line.

3. i) How many squares can be drawn in a square by joining the mid - point of sides
of a square continuously ? What is the limit value of area of square so formed ?
Show with diagram.

PRIME Opt. Maths Book - IX 63

ii) A cylindrical glass full of water is drank half/half in
each times after a interval. How many times the water Y

can drank ? What is the limit value of volume of water

in that activity ? Show in diagram.

iii) What will be the value of y on increasing the value of
‘x’ in the given graph? What is the limit value of y in

it? X
iv) The graph given in Q.No. (iv) can be used in our daily

life. Discuss with a suitable example for it in the O

classroom by preparing a graph in chart.

v) What is the limit value of the sum of the terms of the series, 3 + 3 + 3 + 3 +
2 4 8
3 3
16 + 32 ?

4. Which of the following series has the limit value will be the fixed absolute number?
1 1 1 1
i. 1+ 2 + 4 + 8 + 16 , ..., ..., ..., ..., ..., ...

ii. 4 – 2 + 1 – 1 + 1 ... + ... + ... + ... + ... + ...
2 4

iii. 8 + 40 + 200 + 1000 + ... + ... + ... + ... + ... + ...

iv. 0.6 + 0.06 + 0.006 + 0.0006 + ... + ... + ... + ... + ... + ...

v. 0.45

5. A square having side 1cm is dividing making half continuously to their parts as
shown in diagrams. Answer the followings.



i) Write down the sequence of the shaded area of region.
ii) What will be the area shaded region of figure.
iii) Write down the limit value of the shaded region.

6. i) If an = 1 + (–1)n 1 , find a1, a2, a3, a4 and a5
2n2

ii) If f(x) = 2x – 1, find f(2), f(1.99) and f(2.01).
iii) If f(x) = 3x + 1, find f(3), f(2.999) and f(3.001).
1
iv) tn = n + 1 , find the difference between t3 and t4.

v) Find the value and fill the given table.

x 0.1 0.01 0.001 0.0001 0.00001 0
....... .......
f(x) = 2x + 3 ....... ....... ....... ....... – 0.00001 0
....... .......
x – 0.1 – 0.01 – 0.001 – 0.0001

f(x) = 2x + 3 ....... ....... ....... .......



64 PRIME Opt. Maths Book - IX

7. Find the limiting value of sum of the following infinite series.
1
i) 16 + 4 + 1 + 4 + ... + ... + ... + ... + ... + ... + ...

ii) 0.5 + 0.05 + 0.005 + 0.0005 + ... + ... + ... + ... + ... + ... + ...

iii) 6+3+ 3 + 3 + 3 + ... + ... + ... + ... + ... + ...
2 4 8

iv) 8 – 4 + 2 – 1 + ... + ... + ... + ... + ... + ...

v) –20 + 10 – 5 + 5 – 5 + ... + ... + ... + ... + ... + ...
2 4
vi) 0.4

Answer

1. i) 5.68
ii) 6
iii) Domain = Set of the elements of n
Range = Set of the elements of y
iv) 5 is the image of 2 and 2 is the pre-image of 5
v) 0.5 + 0.05 + 0.005 + 0.0005

2. i) Limit value of the length will be zero.

ii) 0.0000000001, zero

iii) 1 iv) 1

v) 12, 6, 3, 3 3 ; zero
2, 4

3. i) Infinite, zero ii) Infinite, zero 6
iii) Decreasing zero iv) Show to your teacher v)

4. i, ii, iv, v

5. i) T12h,e34lim, 78it,in11g65v,a.l..u, e...,is...1, .... ii) Nearly equal to 1 cm2.
iii)

6. i) 1 , 9 , 17 , 33 , 49 ii) 3, 2.98, 3.02
2 8 18 32 50
1
iii) 10, 9.97, 10.03 iv) 20

v) Show to your subject teacher.

7. i) 21.3 a 64 k ii) 0.5 a 5 k iii) 12
3 9

iv) 5 1 v) – 430 vi) 0.4 a 4 k
3 9

PRIME Opt. Maths Book - IX 65

2.2 Limit using graph

i. Let us consider a function f(x) = 3x – 4 and range of the function can be taken
using domain as follows for x = 3.
From left side:

x 1 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9
f(x) –1 2 2.3 2.6 2.9 3.2 3.5 3.8 4.1 4.4 4.7

Here, we conclude that,
On increasing the value of x up to nearly equal to 3, the element of range f(x) also will

goes on increasing to nearly equal to 5.

Y

5

4

3

2

1 X
X’ O 123

Y’

Hence, the limit value of the function f(x) tends to 5 (very close to 5) from left side
which is called Left Hand Limit (LHL).

66 PRIME Opt. Maths Book - IX

Also from right side:
x 5 4 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1

f(x) 11 8 7.7 7.4 7.1 6.8 6.5 6.2 5.9 5.6 5.3

Here, we conclude that,
On decreasing the value of x up to nearly equal to 3, the value of f(x) also will goes on

decreasing to nearly equal to 5.

Y

7 X
6 12345
5
4
3
2
1
X’ O

Y’

Hence, the limit value of the function f(x) tends to 5 (very close to 5) from right side
which is called Right Hand Limit (RHL).

PRIME Opt. Maths Book - IX 67

ii. Let us consider a function f(x) = x2.
x 0 –1 1 –2 2 –3 3 –4 4
f(x) 0 1 1 4 4 9 9 16 16
Y

y = x2

X’ O X

Y’
Taking x = 1 and 4 in the graph.

The rectangle ABCD is formed where area A = AB × BC = 8 × 2 = 16cm2.
Y

AD
y = x2

X’ O BC

X

Y’

68 PRIME Opt. Maths Book - IX

Taking x = 1.5 and 2.5 Taking x = 1.9 and 2.1
Y Y
y = x2
y = x2
AD

X’ O BC X’ O

X X

Y’ Y’
The area of rectangle ABCD, The area of rectangle ABCD
A = AB × BC = 4 × 1 = 4cm2 A = AB × BC = 0.8 × 0.2 = 0.16 cm2

Conclusion:
When value of x approaches to 2 form left hand limit as well as from right hand limit

the rectangle ABCD will be fixed near 4 of f(x)

Limit of a function

Limit and Continuity can be defined by understanding the function y = f(x) and values
of f(x) in different situations. The term limit and Continuity are the fundamental
mathematical terms to introduce calculus can be defined only by studying the concept of
function, limit and Continuity one after another in a sequence.

If f is a function from non-empty set A to B which is written as y = f(x) where ‘y’ is the
image of x and ‘x’ is the pre- image of ‘y’.

Here, value of x = a is called the element of domain and y = f(x) = f(a) is called the range

of function.

• Let us take an example f(x) = 2x + 3 and x = 1 is element of domain.
Then, y = f(x) = 2x + 3
y = f(2) = 2 × 1 + 3 = 5 (a finite number)

• Taking a function f(x) = 1 and x = 1,
x –1
1
Then, y = f(x) = x –1

f(1) = 1 = 1 ( It is not a finite number)
1–1 0


PRIME Opt. Maths Book - IX 69

Taking a function f(x) = x2 – 1 and x = 1
x –1

Then, y = f(x) = x2 – 1
x –1

f(1) =be11d2e––t1e2rm=in00e(dI.t is indeterminant form)
i.e. can’t
It

Meaning of x → a in a limit

Let us consider an example by taking a variable x having values 1.9, 1.99, 1.999, 1.9999,
1.99999, ..........., approaching very close to 2 but which is not exactly equal to 2.
i.e. The difference between the numbers x and 2 is very small. (too much less)

Taking another example of the variable x having values 2.1, 2.01, 2.001, 2.0001, 2.000
...........2 approaching very close to 2 but which is not exactly equal to 2.
i.e., The difference between the numbers x and 2 is very small.

Here, two examples discussed above tell us the value of x is very close to 2 in both the
cases but not exactly equal to 2 which is read as,
x approaches to 2 or
x tends to 2 or
x → 2

Thus
For any constant ‘a’ of the variable ‘x’ we say, x → a. (x tends to a).
Limit can be understand on discussing the area bounded by the polygons which are
inscribed in a circle according to the number of sides of the polygons.
As the number of sides of the polygon increased the area bounded by them also goes on
increasing which is very close and equal to the area of circle for the large number of sides
taken as infinity (∞). It can be observed in diagrams.

The relation of number of sides of polygons and area is,

Wf(nh)e=reA‘nnf’oirs n→∞ of sides
number

f(n) is the area of polygons and An is the area of circle.
It can be written as,

n lim f(x) = lim
"3 n " 3An

From such examples limit of a function f(x) = ‘l’ can be defined

70 PRIME Opt. Maths Book - IX

Where, the limit for x → a can be written as

lim f(x) = l
x"a

A function f(x) is said to be have a limit ‘l’ when x → a, if the
numerical difference between the value of f(x) and ‘l’ can be
made very small as well as we please by making x sufficiently

close to a and we write it as,

lim f(x) = l
x"a

Meaning of infinity in a limit
If f(x) be a function of x which is very close to ‘a’ and the value of f(x) is obtained ultimately
very large (∞) we say that the limit of f(x) is at infinity as ‘x’ tends to a.

i.e. lim f(x) = ∞
x"a

Let us consider an example: f(x) = x2 1000 10000 100000
x 1 10 100

f(x) 1 100 10000 100000 1000000 10000000

The value of f(x) goes on increasing as increasing ‘x’ and will be closed to infinity for

x → ∞.

Limit at infinity:

Let us consider an example f(x) = 1 , as the value of ‘x’ increase here in the function, the
x2

value of f(x) will be decreased as the reciprocal value.

x1 10 100 1000

f(x) 1 0.01 0.0001 0.000001

On increasing the value of x and tends to ∞, the value of f(x) ultimately decreased and

tends to zero.

i.e. x lim f (x) = lim 1 =0
"3 x " 3 x2

Indeterminate form:
L Suetchuf(stxyc)po=ensoxixfd2–fe–o1r1rma=fuli00nkec(tI0itoc,na∞nf(, xn3)o=t+bxex32d––,e113tera–mn3idnteaedkt.c)inagrex
can not be defined for i0nd∞eterminate form. = 1. The function will be, and limit
called indeterminate form

PRIME Opt. Maths Book - IX 71

To find the limit of a function in indeterminate form should be changed by using algebraic

operation. Examples :

i) x2 – 1 = (x + 1)(x – 1) = (x + 1)
x –1 (x – 1)

ii) xx3 –8 = (x – 2)(x2 + 2x + 4) = x2 + 2x + 4
–2 (x – 2)

Properties of limit

i. xli"ma k = k (where k is any constant)

ii. xli"ma [f(x) + g(x)] = lim f(x) + lim g(x)
x"a x"a

iii. xli"ma [f(x) – g(x)] = lim f(x) – lim g(x)
x"a x"a

iv. xli"ma [f(x) . g(x)] = lim f(x) . lim g(x)
x"a x"a

v. xli"ma f (x) lim f (x)
g (x) = x"a
lim
x"a g (x)

Working rule to find limit

i. For polynomial function, lim f (x) = f(a)
x"a

ii. For rational function, lim f (x) = f (a) if it is not in indeterminate form.
x " a g (x) g (a)

iii. If the function is in indeterminate form, factorize and cancel the common factors
and find f(a).

Worked out Examples

1. Complete the given table.

x 0.9 0.99 0.999 1.01 1.001
.......... .......... ..........
f(x) = x3 – 1 .......... ..........
x–1
Solution : Taking,
x3 – 1
f(x) = x –1

= (x – 1)(x2 +x + 1)
(x – 1)

= x2 + x + 1

f(0.9) = 2.71

f(0.99) = 2.97

f(0.999) = 2.997

f(1.01) = 3.0301

f(1.001) = 3.003001

72 PRIME Opt. Maths Book - IX

It conclude that the whole number is near to 3 from both sides.

Hence, it can be written as,

lim f (x) = 3
x"1

x 0.9 0.99 0.999 1.01 1.001 x → 1

f(x) = x3 – 1 2.71 2.97 2.997 3.03 3.003 f(x) = 3
x –1

2. Show that the functional value does not exist but limit exist at x = 2 for the function

f(x) = x2 –4 .
x–2
Solution:
Again,
x2 – 4
Here, f(x) = x–2 For limit at x = 2

Functional value at x = 2 is, lim f(x) = lim x2 – 4
x"2 x"2 x–2

f(2) = 22 – 4 lim (x + 2)(x – 2)
2–2 x"2 (x – 2)
=
= 4–4
2–2
= (2 + 2)
= 0
0 =4
= It can not be determined. = It is finite number.

Hence, the functional value does not exist. Hence, limit exists.

3. Show that the limit lim x2 –4 exists.
x"2 x–2
Solution:

lim x2 – 4 = lim (x2 – 22)
x"2 x–2 x " 2 (x – 2

= lim (x + 2)(x – 2)
x"2 (x – 2)

= (2 + 2)

=4
= It is finite number.

= It exists limit.

4. Evaluate lim (2x–3)
Solution :x " 2

xl i"m3 (2 x – 3) =2×3–3
=6–3

=3

PRIME Opt. Maths Book - IX 73

5. Find the value of lim x2 + 2x – 3
Solution : x"1 x2 + x – 2

lim x2 + 2x – 3 = lim x2 + 3x – x – 3
x"1 x2 + x – 2 x"1 x2 + 2x + x – 2

= lim x(x + 3) –1(x + 3)
x"1 x(x + 2) –1(x – 2)

= lim (x + 3)(x – 1)
x"1 (x + 2)(x – 1)

= 1+3
1+2

= 4
3

6. Find the value of lim x3 – a3
Solution : x$a x–a

lim x3 – a3 = lim (x – a)(x² + ax + a²)
x"a x–a x$a (x – a)

= lim (x2 + ax + a2)
x"a

= a2 + a2 + a2

= 3a2

7. Calculate : x lim 0 3x2 + 2x
Solution : $ 2x

lim 3x2 + 2x = lim x (3x + 2)
x$0 2x x$0 2x

= lim 3x + 2
x$0 2

= 3×0+2
2
=1

8. Evaluate : lim 6x2 + x – 2
x$3 2x2 + 3x–1
Solution :

6x2 + x – 2

x2

lim 6x2 + x – 2 = x lim 2x2 + 3x – 1
x$3 2x2 + 3x–1 $3 x2

lim 6 + 1 – 2
$3 x x2
=
x + 3 1
2 x – x2

74 PRIME Opt. Maths Book - IX

= 6+ 1 – 2
3 3

2 + 3 – 1
3 3

= 6+0–0
2+0–0

= 3

Exercise 2.2

1. i) What is limit of a function? Write in symbol also.
ii) What is indeterminate form? Is limit define for it? What have to be done to find

limit value for it?
iii) Write down the round number of 3.002 and 4.999. Also present them as the

limit value with symbol for a function f(x).
iv) Complete the table given below.

x 0.9 0.99 0.999 1.01 1.001 x → ..........

f(x) = x –1 .......... .......... .......... .......... .......... f(x) ..........
x2 – 1 0.999 1.01 1.001 x → ..........

v) Complete the table given below.

x 0.9 0.99

f(x) = x2 – 1 .......... .......... .......... .......... .......... f(x) ..........
x –1

2. Answer the followings:
i) Write down in symbols of x approaches to 4.

ii) Write down in sentence of lim f(x).
x"3

iii) Write down in sentence of x → a.
iv) Taking a function f(x) = x2, discuss the value of function f(x) taking x nearly equal

to 3 using graph.
v) Taking a function f(x) = 2x – 1 and discuss left hand and right hand limit by using

graph for a element of domain 3.

3. Which of the following functions exists the limit value of x?

i) xli"m1 x2 –1 ii) xli"m0 xx2 2+–2xx++26
x–1

iii) x l"im3 x2 –2x iv) xli"m0 2x25–x 3x
3x

v) xli"m2 (x2 + x + 1)

PRIME Opt. Maths Book - IX 75

4. Find the limit value of the followings.
i) xli"m3 (3x – 2) ii) xli"m3 (2x + 3)
iii) xli"m2 (2x2 – 5x + 6) iv) xli"m4 (2x2 – 7x + 3)
v) xli"m0 (4x2 – 8x + 3)

5. Find the limiting value of the following functions.

i) xli"m2 3x – 2 ii) xli"m2 5xx+–23
2x + 1

iii) xli"m1 3x2 + 2x – 4 iv) xli"m3 62xx2 2++3xx – 12
x2 + 5x – 4 –4

v) xli"m2 x2 + 2x – 1
2x2 + 3x – 7

6. Compute the following limits.

i) xli"m1 x2 + 3x –4 ii) xli"m2 x2 –x5–x2+ 6
x–1

iii) xli"m3 2x2 – 5x – 3 iv) xli"m4 xx2 ––146
x–3

v) xli"ma x2 –a2
x–a

7. Compute the following limits.

i) xli"m0 4x3 –8x ii) xli"m0 3xx23––32xx
3x2 –2x

iii) xli"m2 x2 –5x + 6 iv) xli"m0 4x33x–2x+2 + 2x
x2 – x–2 4x

v) xli"m0 5x2 + 3x
x

8. Compute the following limits.

i) x lim 3x + 2 ii) x l"im3 34xx22++2xx–+51
"3 x +1

iii) x l"im3 2x2 iv) x l"im3 54xx22 ++ 34xx +2
3x2 + 2 –3

v) x lim 6x2 + 2x – 7
"3 3x2 + 2x + 1

9. PRIME more creative questions (Evaluation the limits)

i) xli"ma x23 – a23 ii) xli"ma xx –– a a
x –a

76 PRIME Opt. Maths Book - IX

iii) x l"im64 6 x –2 iv) xli"ma xx34 –– aa34
3 x –4

v) x lim 1
"3 x+a – x

10. Project work
Take a piece of paper having length 80 cm and breadth 1 cm. Divide the piece of paper

making two equal parts and one half again divide making half and go on continue
upto 5 times.
i) What will be the area of the smallest part?
ii) Write down the sequence formed in this condition.
iii) Write value with symbols.

Answer

1. i) Show to your teacher.

ii) Show to your teacher.

iii) 3 and 5, lim f(x), lim f(x)
x"3 x"5

iv) xli"m1 f(x) = 1 v) xli"m1 f(x) =2
2

2. Show to your teacher.

3. i) doesn’t exists ii) exists iii) doesn’t exists
iv) doesn’t exists v) exists

4. i) 7 ii) 9 iii) 4 iv) 7 v) 3
iv) 3 v) 1
5. i) 4 ii) 74 iii) 12
5

6. i) 5 ii) –1 iii) 7 iv) 8 v) 2a
7. i) 4
8. i) 3 ii) 2 iii) – 13 iv) 12 v) 3
3

ii) 3 iii) 32 iv) 54 v) 2
4

9. i) 2 ii) 1 iii) 14 iv) 43a v) 0
33 a 2a

PRIME Opt. Maths Book - IX 77

Limit and Continuity

Unit Test Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. Write down the 10th term of the sequence 5.1, 5.01, 5.001, 5.0001,------------

2. a. Round of the number 7.368 to the tenth digit, hundred digit and whole number.
b. What is the limiting value of the infinite series of 0.78?
x2 – 9
c. When does f(x) = x–3 give certain existing value? Does f(x) give limit value

for x → 3?

3. a. Complete the table for f(x) = 2x + 3.

x 0.1 0.01 0.001 0.0001 0.00001
.......... .......... ..........
f(x) .......... ..........

b. Evaluate: lim a 2x2 – 3x + 2 k
x"0 x+2

4. Taking an equilateral triangle of side 16cm. Take the mid-point of sides continuously
three times for the triangle so formed and show in diagram. What will be the limiting

value for area of triangles so formed.

78 PRIME Opt. Maths Book - IX

Unit 3 Matrices

Specification Grid Table

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

No. of Questions 1 2 1 – 4 9 20
Weight 1 4 4 –

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

Objectives : At the end of the lesson
• Student are able to know the rows and columns of a matrix and its order.
• Students are able to use matrix for the data used in daily life.
• Student are able to identify the types of matrices.
• Student are able to operate the matrices like addition, subtraction,

multiplication, transpose, etc.
• Students are able to use the difference laws for addition & multiplication

of matrix.

Materials Required:
• Cost chart, mark chart etc which are used in our daily life.
• Chart paper.
• Flash card.
• Sample of matrix of order m × n.
• Chart of laws of addition and multiplication of matrices.

PRIME Opt. Maths Book - IX 79

3.1 Matrices

The cost per kg of vegetables in Kalimati vegetable 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 reSSRSSSSTSS112p850rese322n054ted112u520sWWWWWWWWVXing square bracket as,

Here, rows represent the shops
column represents the type of vegetables.
Where the elements denote the cost.

The rectangular arrangement of the numbers in rows and
columns enclosed by ( ) or [ ] is called a matrix. (plural-

matrices) where horizontal arrangement are called rows

and vertical arrangements are called columns.

• The matrix is denoted by capital letter A, B, C, ...
• The members of the matrix are called elements which are denoted by

small letters a, b, c, d, e, f, ........................
• The horizontal arrangements of the matrix are called rows.
• The vertical arrangements of the 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.
Order of A = R × C
• The matrix is denoted by using the position of elements as the no. of
H erAreo,=w sTSSRSSSSSSaaaaa1321n1111d=caaaeo132222lleummaaaen132333nsWWWWWXWVWWtaisn, 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.

80 PRIME Opt. Maths Book - IX

General form of matrix of order m × n.

According to the number of rows and columns of a matrix, its elements can be denoted by
a11, a12, a13 .............. etc. Where aij is taken as the general element of ith row and jth column.

Here,
A matrix of order m × n can be written using general elements as,

a11 a12 a13 ............. a1j .............. a1n
a21 a22 a23 ............. a2j .............. a2n
a31 a32 a33 ............. a3j .............. a3n

A = ai1 ai2 ai3 ............. aij .............. ain

am1 am2 am3 ............. amj .............. amn

aij is the general element of matrix A of ith row and jth column.
If aij = 2i + 3j, then the matrix A of order 2 × 2 can be calculated as below.
a11 = 2 × 1 + 3 × 1 = 5
a12 = 2 × 1 + 3 × 2 = 8
a21 = 2 × 2 + 3 × 1 = 7
a22 = 2 × 2 + 3 × 2 = 10
\ Matrix of order 2 × 2 is, A = <5 8 F

7 10

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 C = [a]1×1

2. Column matrix:
The matrix having only one column is cSRSSSSSTSSSSS1532alWVWWWWWWXWWWWl4e×1d colu mn m atrix.
Ex : A = SSSSSRSTSSaaa132111WWVWWWWWWX3×1 B = C = [0]1×1

3. Null matrix (zero):
The matrix having all the elements zero is called null matrix. It is denoted by ‘0’.

Ex : 0 = <0 0 0F
0 0 0 2×3

PRIME Opt. Maths Book - IX 81

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

Ex : A = <2 3 4F
1 –3 –2 2×3

5. Square matrix:
The matrix having equal number of rows and coluCm=nsSSSSTRSSSS174is c528al369leWWVWWWXWWWd3×3square matrix.
Ex : A = [2]1×1
B = <1 2F
3 4 2×2

6. Diagonal matrix
The square matrix having the main diagonal elements (from left top to right bottom)

non-zero and remaining elements are zSSSSSTRSSS00pero00qis00rcWWWWWWWWVXa3l×l3ed diagonal matrix.
Ex: A = <a 0F B =

7. 0 b 2×2
Scalar matrix:
The diagonal matrix having all non-zero diagonal elements are equal is called scalar

matrix. SSSSRTSSSS00a 0 00aWWWVWWWXWW3×3 B = TRSSSSSSSS00k 0 00kXVWWWWWWWW 3×3
Ex : A = a k
0 0

8. Identity matrix (Unit matrix):
The square matrix having the main diagonal elements (from left top to right bottom)
all one and remaining elements are =zeSSSSSSSRTSr100o is c100aWWXWWWWVWWl3le×3d identity matrix. It is denoted by ‘I’.
0
Eg : I = <1 0F I
9. 0 1 2×2 1

0

Triangular matrix:
The square matrix having zero elements towards upper or lower parts of leading

diagonal of the matrix is called triaBng=uSSSSSSSSTRl100ar 024ma536tWXWWWWWWWVr3ix×.3

Eg : A = <1 2F L.T.M U.T.M
0 3
2 × 2

C = <1 0F D = SSRSSSSSTS124 0 006WWWWWWXVWW 3×3
2 3 2×2 3
5
Here,
Matrices A and B are called upper triangular matrices and C and D are called lower

triangular matrices.

82 PRIME Opt. Maths Book - IX

10. Equality of the matrices

Any two matrices having same order and same corresponding elements are called

equal matrices.

Ex : A = <2 3F , B= <2 3F
4 1 4 1

Here, A = B (equal matrices)

Worked out Examples

1. The cost of fruits in a fruits shop in different days is given below. Represent the

informations in a matrix with appropriate meaning.
Apple Orange Banana
Day 1 150 80 75
Day 2 120 100 70

Solution :

The cost of fruits given in table of two days are represented by matrix as follows

where,

rows → represents the days

columns → represents the type of fruits

A= <150 80 75F
120 100 70

2. If A = <–32 2 –51F
4

i) Find the order of matrix A.
ii) Find the elements a12, a21, & a23.
Solution,

i) No. of rows = 2

No. of columns = 3
\ Order of matrix A is 2 by 3 (2 × 3).

ii) a12 = element in 1st row 2nd column = 2
a21 = element in 2nd row 1st column = –2

a23 = element in 2nd row 3rd column = –1

3. If A = B, where A = =x + 2 13G and B = <6 3 13F , find the value of x & y.
y – 2y

Solution :

A = =x + 2 3G ,
y1

B= <3 3F
6 – 2y 1

Here,

Matrix A & B are equal.

PRIME Opt. Maths Book - IX 83

By equating the corresponding elements

x + 2 = 3 and y = 6 – 2y

or, x = 3 – 2 y + 2y = 6

or, x = 1 3y = 6

\ x = 1 y= 6
3

\ y=2

\ x = 1, y = 2

4. If aij = 3i – 2j is the general element of a matrix, find the matrix of order 2 × 3. Also
write down its type.

Solution :

Taihj =en3,i – 2j (general element)

2 × 3 matrix A = <a11 a12 a13F
where, a21 a22 a23

a11 =3×1–2×1 =1
a12 = –1
a13 =3×1–2×2 =–3
a21 =4
a22 =3×1–2×3 =2
a23 =0
\ A =3×2–2×1

=3×2–2×2

=3×2–3×2

= <1 –1 –3F
4 2 0

Here, Number of rows and column of the matrix are different.
So, it is rectangular matrix.

84 PRIME Opt. Maths Book - IX

Exercise 3.1

1. i) What is matrix? Write down one example.
ii) What do you mean by order of the matrix?
iii) What is scalar matrix? Write down its example.
iv) What is identify matrix? Write down its example.
v) Write down the standard form of the matrix.

2. Write down the following informations in matrix form with appropriate meaning.

Also write down the order of the matrices.
i) The number student in a class are given below.

1st column 2nd column

1st row 4 3

2nd row 5 2

3rd row 6 4

4th row 3 5

ii) The cost of clothes in three shops are given below.
Shirt Pants Vest

1st shop 500 700 300

2nd shop 450 600 350

3rd shop 400 550 250


iii) The production of crops in Jhapa district in different years in metric tonne is

given below. Rice Wheat Maize others


2072 BS 300 150 100 150

2073 BS 350 170 120 130

2074 BS 475 180 200 300

iv) What is matrix? Write short note on order of the matrix.
v) In what conditions equality of the matrices takes place? Write down with

example.

3. Write down the types of the matrices from the followings. Also write down the order

iiov)f )t heSTSSRSSSSS<13022ma02154tFXWVWWWWWWWr ices. ivi)) SSSSSSRTSS0000a 00b00 00c WWXWWWVWWW viiii)) SSSRTSSSSSSSSSSSSTSR100–141100502100132VXWWWWWWWW WWWWWWWVWX
PRIME Opt. Maths Book - IX 85

4. iiAvi)i)n i )s wET<e–lh2er2emthe32eleenfm–to12ilenlFno fwtirais2nt2g.roqwueasntidovnsiesicif)or onmdSSRSSSTSSS324cWWWWWWWXWVtoh leummant.rix STSSRSSSSS132 320 –151WXVWWWWWWW ix) 63 –2 1@
iivii)) Ea3l1em+ ae2n3t+aaij 3w3 here i = 3 & j = 2.


v) Write down the order of the matrix.

5. If the general element of a matrix is aij = 2i – 3j, what will be the matrices of following order.
i) 2 × 2 matrix ii) 2 × 3 matrix iii) 3 × 3 matrix

iv) 3 × 2 matrix v) 3 × 1 matrix

6. Find the value of ‘x’ and ‘y’ from the following equal matrices.

i) A = <3 x–1F & B = <2y3+ 1 7F
5 6 6

ii) A = =2x + 3 5 G , B = <7 – 2x 5F
2 + 1 2 3y – 5
y

iii) P= =2x + y 2 7G , B = <7 2 7F
3 5 y 3 5 2y – 3

iv) M = TSRSSSSSSS124 3 x–7+2yVWWXWWWWWW , N = SSSSSSTRSS124 3 –772WWWWWWWWVX
x–y 1
5
5

v) A= <2x5+ y 7F = <5 3x – 2yF
1 7 1

Answer

1. Show to your teacher. 2. Show to your teacher.
3. Show to your teacher.

4. i) 3 ii) 0 iii) 9 iv) 2ii i) SSSTSSSSRS113 ––042v) –––3573XWVWWWWWWW× 3

5. i) <–1 –4F ii) <–11 ––42 ––75F
1 –2 v) RSSSSSSSST–131WVXWWWWWWW
––024WWXWVWWWWW
iv) SSSSRSSSST–131

6. i) x = 8, y = 2 ii) x = 1, y = 3 iii) x = 2, y = 3
iv) x = 4, y = 3 v) x = 3, y = 1

86 PRIME Opt. Maths Book - IX

3.2 Operation on matrices

The simplification of two or more matrices in to a single matrix by using any kind of
mathematical operations indicates the operation on matrices.

3.2.1. Addition of matrices

Marks obtained by three students Sita, Pranav and Pranisha in two test in optional maths
in two successive months are as follows.



1st month 2nd month

1st 2nd 1st 2nd

Sita 60 75 Sita 70 75
Pranav 85 90 Pranav 95 80
Pranisha 95 85 Pranisha 80 95

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

1st 2nd

Sita 60 + 70 = 130 75 + 75 = 150
Pranav 85 + 95 = 180 90 + 80 = 170
Pranisha 95 + 80 = 175 85 + 95 = 180

Th is==inSRSSSSTSSSTSSRSSSSSSRSSSTSSSSS689f111689o055783055r005m+++789a550798111tWWWXWWWWWV050857io000+n789WXVWWWWWWWSSSRSSSTSS550c789a+++005n798b789550e505WVXWWWWVXWWWWexpressed in matrix form as,

The sum of any two matrices having same order is called a new
single matrix obtained by adding the corresponding elements of

the matrices where the single matrix also has the order same as the
given matrices.

PRIME Opt. Maths Book - IX 87

Points to remember:
• Order of the matrices should be same.
• Corresponding elements of the matrices should be added respectively.
• New matrix obtained after addition has the order same as the given matrices.

Eg. ITf hAe=n,<–21 53F , B = <–34 –21F


A + B = <2 3F + <3 2F
–1 5 –4 –1

= =2 + 3 3 + 2G
–1–4 5 – 1

= <5 5F
–5 4

3.2.2. Difference of the matrices (Subtraction)

Out of three persons in the above example, who has got more marks in which month can
be found out by subtraction.

Where,
Difference of marks of them

1st month 2nd month

1st 2nd 1st 2nd

Sita 60 75 Sita 70 75
Pranav 85 90 Pranav 95 80
Pranisha 95 85 Pranisha 80 95

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

1st 2nd

Sita 60 – 70 = –10 75 – 75 = 0
Pranav 85 – 95 = –10 90 – 80 = 10
Pranish 95 – 80 = 15 85 – 95 = –15

T his ==inSSSSSSSSSSRSSSSTRTSSfTSSSSSSSRSo689––689111r0550555m00–––7987a89000–t5501i10WWWWWWVWWXo05789n–WWWWWVXWWW550c–––SSSSRTSSSSa789798n005550bXWVWWWWWWW78e9505eWWWWWWWWVXxpressed in matrix form as, PRIME Opt. Maths Book - IX
88

The difference of any two matrices having same order is called
a new single matrix obtained by subtracting the corresponding
elements of the matrices where the single matrix also has the order

same as the given matrices.

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

Then,

P–Q = <2 3 1F – <–1 1 5F
5 –3 2 2 1 3

= =2 + 1 3–1 1 – 5G
5–2 –3 – 1 2–3

= <3 2 –4F
3 –4 –1

3.2.3. Multiplication of a matrix with a scalar.

The new matrix formed by multiplying the each
elements of a given matrix with a given scalar quantity

is called the multiplication of a matrix with a scalar.

Eg. If A = <13 ––12F , find 3A.

Solution : A = <1 –2F
3 –1

3A = 3 <1 –2F = <3 –6F
3 –1 9 –3

3.2.4. Transpose of the matrix.

In the example, cost of apple is given.

1st shop 2nd shop

Sunday 150 140

Monday 160 145
Tuesday 170 150

PRIME Opt. Maths Book - IX 89

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=chTSSSSSSSSR111e657x000am111p544l005esWWWWWWXVWW in matrix form. <150 160 170F
A = after changing row and columns = 140 145 150


It is called transpose of the matrix.

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`.

Order of transpose matrix will be different from matrix A from rectangular matrix but

same for the square matrix.

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

AT = SSSSSSRSST534 629XWWVWWWWWW

3.2.5. Symmetrical matrix:
The square matrix where same matrix is formed by interchanging the rows and
AT
columns is called =sySSSSTSRSSSm352m284et138riWWWXWWWWVWcal matrix. (i.e. A = = symmetric)
Eg : A = <a
b bF , B
c

3.2.6 Properties of matrix addition :

1. Closer property :

A + B is hold for matrices A and B of same order.

If A = <2 1F , B = <1 2F
Then, 3 1 4 3

A+B= <2 1F + <1 2F = =2 + 1 1 + 2G = <3 3F
3 1 4 3 3+1 1+3 7 4

It has same order of A and B.

90 PRIME Opt. Maths Book - IX

2. Commutative property

A+B=B+A

If A = <2 1F , B = <3 –1F
Then, 3 –1 2 1

A+B= <2 1F + <3 –1F = =2 + 3 1–1 G = <5 0F
3 –1 2 1 3+2 –1 + 1 5 0

B + A = <3 –1F + <2 1F = =3 + 2 –1 + 1G = <5 0F
2 1 3 –1 2 + 3 1–1 5 0

\ A + B = B + A(Hence proved)

3. Associative property :

(A + B) + C = A + (B + C)

If A = <1 2F , B = <3 1F , C = <1 3F
Then, 3 4 –1 2 2 –2

L.H.S. = (A + B) + C

= <1 2F + <3 1F + <1 3F
3 4 –1 2 2 –2

= =1 + 3 2 + 1G + <1 3F
3– 1 4 + 2 2 –2

= <4 3F + <1 3F
2 6 2 –2

= =4 + 1 3 + 3G
2 + 2 6–2

= <5 6F
4 4

R.H.S. = <1 2F + <3 1F + <1 3F
3 4 –1 2 2 –2

= =–31++12 1 + 3G
2–2

= <4 4F
1 0

= =1 + 4 2 + 4G
3 + 1 4 + 0

= <5 6F
4 4

\ L.H.S. = R.H.S. proved

PRIME Opt. Maths Book - IX 91

4. Additive inverse:

For any matrix A ∃ a matrix – A such that Here – A is called the additive in verse of A.

A + (–A) = (A) – A = 0

If A = <2 –1F
3 –2

\ –A= <–2 1F
–3 2

Then,

A + (–A) = <2 –1F + <–2 1F = =2 – 2 –1 + 1G = <00 00F
3 –2 –3 2 3–3 –2 + 2

(–A) + A = <–2 1F + <2 –1F = =–2 + 2 1 – 1G = <00 00F
–3 2 3 –2 –3 + 3 2–2

\ A + (–A) = (–A) + A = 0 proved

5. Additive identity :
For any matrix A ∃ a matrix ‘O’ of same order such that ‘O’ is called the additive

identity.

A+O=O+A=A

If A = <2 3F , O = <00 00F
Then, –2 4

A+O= <2 3F + <00 00F = =–22++00 3 + 0G = <2 3F =A
–2 4 4 + 0 –2 4

O+A= <00 00F + <2 3F = =0 + 2 0 + 3G = <2 3F =A
–2 4 0–2 0 + 4 –2 4

\ A + O = O + A = A proved.

6. Distributive property over scalar.

K(A + B) = KA + KB

where k is a scalar.

If A = <2 3F , B = <–1 1F
Then, 4 1 2 3

K(A + B) = K (<2 3F + <–1 1F2
4 12 3

= K =2 – 1 3 + 1G
4 + 2 1 + 3

= <K 4KF
6K 4K

92 PRIME Opt. Maths Book - IX

KA + KB = K <2 3F + K <–1 1F
4 1 2 3

= <2K 3KF + <–K KF
4K k 2K 3K

= = 2K – K 3K + KG
4K + 2K K + 3K

= <K 4KF
6K 4K

\ L.H.S. = R.H.S. is proved

7. Property of the matrices over transpose:

i) (AT)T = A ii) (A + B)T = AT + BT

If A = <2 3F If A = <1 –2F , B = <2 4F
4 1 Then, 3 4 1 –1

Then, (A + B)T = (<1 –2F + <2 4 T
AT = <2 3FT = <2 4F
3 4 1 –1 F2
41 31
=1 + 2 –2 + 4GT
(AT)T = <2 4FT = <2 3F = A = 3 + 1 4–1
31 41
<3 2FT
\ (AT)T = A = <2 3F = 4 3
4 1
<3 4F
= 2 3

AT + BT = <1 –2FT + <2 4 T
3 4 1
–1 F

= <1 3F + <2 1F
–2 4 4 –1

= =–12++24 3 + 1G
4–1

= <3 4F
2 3

\ (A + B)T = AT + BT proved

PRIME Opt. Maths Book - IX 93

Worked out Examples

1. If A = <–11 32F , B = <32 –12F , prove that (A + B)T = BT + AT.

Solution :

A + B = <1 2F + <3 –2F
–1 3 2 1

= <4 0F
1 4

\ L.H.S. = (A + B)T

= <4 0FT
1 4

= <4 1F
0 4

\ R.H.S. = BT + AT

= <3 2F + <1 –1F
–2 1 2 3

= <4 1F
0 4

\ L.H.S. = R.H.S. proved.

2. If A = <–11 32F , B = <12 –21F and C = <3 1F , prove that A + (B + C) = (A + B) + C.
2 –1

Solution :

L.H.S. = A + (B + C)= <1 2F + (<2 –1F + <3 1 F2
–1 3 1 22 –1

= <1 2F + <5 0F
–1 3 3 1

= <6 2F
2 4

R.H.S. = (A + B) + C = (< 1 2F + <2 –1F2 + <3 1F
–1 31 2 2 –1

= <3 1F + <3 1F
0 5 2 –1

= <6 2F
2 4

\ L.H.S. = R.H.S. proved.

94 PRIME Opt. Maths Book - IX

3. If A + B = <43 14F and A – B = <––52 32F , find the matrices A and B. Also prove the closer
property of A + B.

Solution :

A+B= <4 1F ................ (i)
3 4

A–B= <–2 3F ................ (ii)
–5 2

Adding (i) and (ii), we get,

A+B= <4 1F
3 4

A–B= <–2 3F
–5 2

2A = <2 4F
–2 6

\ A= <1 2F
–1 3


Putting the value of ‘A’ in equation (i)

B= <4 1F – <1 2F = <3 –1F
3 4 –1 3 4 1

\ A= <1 2F &B= <3 –1F
–1 3 4 1

Again,

B + A = <3 –1F + <1 2F
4 1 –1 3

= =3 + 1 –11++32G
4 – 1

= <4 1F
3 4

\ (A + B) = (B + A)
i.e. closer property is proved.

PRIME Opt. Maths Book - IX 95

Exercise 3.2

1. i) What do you mean by symmetrical matrix? Write down with example.
ii) Write down the conditions of matrix addition.
iii) Write down the condition of matrix multiplication with a scalar.
iv) Write down the associative property of matrix addition.
v) Write down the distributive property of matrix addition over scalar.

2. Operate the followings. SSSSSSSSTR120 –121WVXWWWWWWW + SSTRSSSSSS–322 102WWWWWWWXWV

i) <1 2 3F + <2 –1 –3F ii) 3 1
–1 4 0 0 1 2 1 –1
3 3

iii) <34 –2F + <2 –3F iv) <34 1 –2F – <5 0 –3F
1 2 1 –1 2 2 –2 –1

v) 2 <1 2F + 3 <1 0F
3 –1 –1 2

3. Find the following from the given matrices.

i) A= <1 2F and B = <3 –1F , A + B
–1 3 4 1

ii) M= <3 –2F , N = <2 7F , M + N
–1 4 6 1

iii) P = <–2 3F , Q = <2 –3F , 2P + Q
4 –1 –4 1

iv) A = <3 5F , B = <–2 –3F , 3A + 2B
–2 7 5 –3

v) M= <4 2 1F , N = <3 –1 2F , 3M – 2N
3 1 5 –2 1 0

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

i) 2A + B ii) 3A – B iii) A + 3B

iv) 3A – 2B v) 4A – 3B

5. i) If M = <1 –2F and N = <3 –6F prove that 3M – N is a null matrix.
2 3 6 9

ii) If A = <3 2F and B = <5 4F , prove that 2A – B is an identity matrix.
–1 –2 –2 –5

iii) If A = <1 2F , B = <2 1F and C = <–2 –4F , prove that 2A + B and B – C are equal matrices.
–1 3 3 –4 2 –6

96 PRIME Opt. Maths Book - IX