Final PDF file of OPT Class 9 2077

In the figure below, the regular polygons are circumscribed in the circles of same area A. While increasing
the number of sides of the regular polygon, the area of the polygons gradually go on decreasing and the
difference between the area of the circle and the area of the polygon gradually decreases.

a3 a4 a6 a8 a10

A AA AA Limit

n=3 n=4 n=6 n=8 n = 10

When the number of sides of the polygon approaches to infinity (i.e. n→∞), then the area of the polygon

approaches the area of the circle. If a represents the area of the polygon, then we can write the information

of this example as no. of sildiems→∞ Area of polygon = Area of circle or lim an = A
n→∞

Let us discuss one more activity to have the concept of limit which we have done in previous classes. We
use the area of a circle with radius r is pr2. Here we recall the process of finding the area of a circle.

First, we divide the given circle into even number of equal parts (like 4, 8, 16, 32, ... ... etc.) fig.(a) and
arrange the parts as shown in fig.(b).

Number of parts Fig. (a) Fig. (b)

4

8

16

32
Concept of Limit by Sequence of Diagrams ~ 47

From the above figure, it is clear that when we divide the circle into very large number (n) of equal parts,
then the fig.(b) approaches to a rectangle. The limit of the shape of fig.(b) as n tends to infinity is a rectangle.
It can be written in limit notation as

no. of equal parltism(sector)→∞ The shape of fig.(b) = Rectangle shape

i.e. lim fig.(b) = Rectangle
n→∞

b=r

l = C = 2pr = pr
n 2

In each case, the area of the circle is equal to the area of the rectangle, the breadth of the rectangle is equal
to the radius of the circle (i.e. r) and the length of the rectangle is equal to half of circumference of the circle
(i.e. pr) [As the circumference of the circle = 2pr].

Hence, the area of circle = Area of rectangle = l × b = pr × r = pr2.

Example 1:

In an equilateral triangle, a new triangle is inscribed by joining the mid-points of the sides

successively. Again, another triangle is inscribed in the second triangle by joining the mid-points of

it's sides successively. This porcess is continuied infinitely many times.

a. What the length of sides of 3rd and 4th­ triangle.

a. What is the limit of length of sides of triangle. 64cm 64cm

c. What is the limit of perimeter of triangle.

 a. The length of the first triangle is 64cm.

The length of the second triangle is 1 × 64 = 32cm 64cm
2
1
The length of the third triangle is 2 × 32 = 16cm

The length of the fourth triangle is 1 × 16 = 8cm
2

b. Here, the length of the side of each new triangle is half of the side of the previous one.
The sequence of the length of the sides of triangle is 64, 32, 16, 8, ... ...
Hence, the sides of triangle approaches zero i.e. the limit of side of triangle is 0.

c. Here, the length of perimeter of each new triangle is half of the perimeter of the previous one.
The sequence of length of the perimeters of triangle is 3×64, 3×32, 3×16, 3×8, ... ...
i.e. 192, 96, 48, 24, ... ...
Hence, the perimeters of triangle approaches zero. i.e. the limit of perimeter of triangle is 0.

48 ~ Perfect Optional Mathematics Class 9

Example 2:

Each side of a square is 64cm. Second square is inscribed by joining the mid-points successively.

Again, another square is inscribed in the second square by joining its mid-points succesively. If this

process is continuied up to large number of times, then answer the following questions.

a. What is the length of sides of 3rd and 4th­ squares? 64cm
a. What is the limit of the length of the sides of the square? 32cm

c. What is the limit of the area of the square? 32cm

Limit

 a. The length of the 1st square is 64cm.

64 2 64 2
2 2
The length of the 2nd square is = + = (32)2 + (32)2 = 32 2cm

32 2 2 32 2 2 322×2 322×2
2 2 4 4
The length of the 3rd square is = + = +

322 + 322 = (32)2 = 32cm
2 2

32 2 32 2
2 2
The length of the 4th square is = + = (16)2 + (16)2 = 16 2cm

b. The sequence of the sides of the squares is 64, 32 2, 32, 16 2, ... ...
The length of the sides of the square approaches zero.
Hence, the limit of the length of the sides of the square is zero.
c. The sequence of the areas of the squares is 642, 32 2 2, 322, 16 2 2, ... ...
i.e. 4096, 2048, 1024, 512, ... ...
The areas of the square approaches zero. Hence, the limit of the area of the square is zero.

EXERCISE : 2.2

1. Each side of an equilateral triangle is 32cm, second triangle is inscribed by joining the mid-points of
the sides succesively. This process is continuied infinitely many times.

a. Find the length of the sides of 2nd and 3rd­ triangles.

b. Find the limit of the length of the side of the triangle.

c. Find the limit of the perimeter of the triangle.

d. Find the limit of the area of the triangle. 32cm

2. The side of a square is 10 2cm. A second square is inscribed by joining the mid-points of the sides
succesively. If this process is continuied up to large number of steps,

a. what is the length of the sides of 3rd and 4th­ squares?

b. What is the limit of the length of the sides of the square?

c. What is the limit of the perimeter of the square?

d. What the limit of the area of the square? 10 2cm

Concept of Limit by Sequence of Diagrams ~ 49

3. In an equilateral triangle, other equilateral triangles are drawn inside the given triangle as shown.
a. Add the next two more patterns.
b. Write the limit of the area of the triangle.
c. Write the limit of the perimeter of the

triangle.

4. In a circle, concentric circles are drawn inside the given circle as shown.
a. Add the next two more patterns.
b. Write the limit of the area of the circle.
c. Write the limit of the perimeter of the

circle.
d. How many concentric circles can be

drawn inside the circle?
5. In a regular hexagon, regular hexagons are drawn inside the given hexagon as shown.
a. Add the next two more patterns.
b. How many regular hexagons can be

drawn inside the hexagon?
c. What will be the limit of the area of the

hexagon?
d. What will be the limit of the perimeter

of the hexagon?

6. In the adjoining figure, diameters are drawn inside the circle as shown.
a. Add the next two more patterns.
b. How many sectors can be formed?
c. What is the limit of the area of sector of

the sector?

7. In the adjoining figure, circles are drawn on the curved surface area of a cylinder from top side.
a. Add the next two more patterns.
b. How many circles can be drawn from

the cylinder?
c. What will be the limit of the number of

circles?



50 ~ Perfect Optional Mathematics Class 9

2.3 Concept of Limit by Sum of Infinite Series

Introduction Limit

In general, the sum of infinite series is infinity. The detail study of infinite series is beyond the scope of the
secondary level school curriculum. Here, we study the limit of sum of some infinite series.

a. Limit of sum of infinite series tends to infinity

Here, we study two types of series having the limit of sum infinity.

i. Limit of sum of infinite series with common difference.

Take an example the infinite seires as 1 + 2 + 3 + 4 + 5 + 6 + ..... .

In this series, the common difference is (d) = t2 – t1 = 2 – 1 = 1 (= t3 – t2 = t4 – t3 etc.)
Now, S1 = 1
S2 = 1 + 2 = 3
S3 = 1 + 2 + 3 = 6
S4 = 1 + 2 + 3 + 4 = 10
... ... ...

S∞ = 1 + 2 + 3 + 4 + 5 + 6 + ... ... not possible.

SN Series Common difference Limit of sum Result

1. 5 + 10 + 15 + 20 + 25 + ... ... d = 10 – 5 = 5 ∞

2. 20 + 22 + 24 + 24 + ... ... d = 22 – 20 = 2 ∞ not a fixed
3. 30 + 28 + 26 + 24 + ... ... d = 28 – 30 = – 2 number

– ∞

Hence, the limit of sum of infinite series is not possible for the series with common difference.

ii. Limit of sum of infinite series with common ratio r (|r| > 1).

Let us take an example of infinite series 2 + 4 + 6 + 8 + 10 + 12 + ... ...

In this series the common ratio is t2 = 4 = 2 = t3 = t4 etc.
t1 2 t2 t3

Now, S1 = 2

S2 = 2 + 4 = 6

S3 = 2 + 4 + 8 = 14

S4 = 2 + 4 + 8 + 10 = 24

... ... ...

S∞ = 2 + 4 + 6 + 8 + 10 + 12 + ... ... not possible.

Concept of Limit by Sum of Infinite Series ~ 51

SN Series Common Condition Limit of sum

1. 1 + 3 + 9 + 27 + 81 + ... ... r = 3 = 3 |r| = |3| = 3 > 1
2. 3 – 9 + 27 – 81 + ... ... 1 |r| = |– 3| = 3 > 1

r = – 9 = – 3 not a fixed real
3 number

3. 6 – 12 + 24 – 48 + ... ... r = – 12 = – 2 |r| = |– 2| = 2 > 1
6

Hence, the limit of the sum of infinite series is not possible for the series with common ratio greater than 1.

b. Limit of sum of infinite series tends to a fixed number

Here we study two types of series having the limit of sum a fixed number.

i. Limit of sum of infinite series with common ratio r (|r| < 1).

Suppose we take any object with a unit length and divide it into two equal pieces. Now, repeat the process
on the second of the two pieces, and continue in this way as long as you like.
The lengths of the parts of the object generate the sequence 12, 41, 81, 116, ... ... as shown below.

1 1 11
2 4 8 16

1

Taking the parts of objects separately, we have

1
2n

1 = 1
24 16
1 1
23 = 8

1 = 1
22 4
1 1
21 = 2

Taking the sum of all these parts as sum of infinite series, we get the following figure.

1+ 1 + 1 + ... ... + 1 1
2 4 8 2n 2n

1 1
2n
After n steps, the sum of series reaches at a point which is at a distnace of from 1 as shown above.

This can be written as 1 + 1 + 1 + ... ... ... + 1 = 1 – 1 1 1
2 4 8 2n 2n 128

32 1
1 64
The value of the sum approaches 1 as n becomes larger and larger. We can 1
2 81
write this as 1 + 1 + 1 + ... ... ...+ 1 → 1
2 4 8 2n 16

The limit of the series is 1 as n approaches ∞. 1
4
We also write this as xl→im∞  1 + 1 + 1 + ... ... ... + 1 = xl→im∞∑ 1 = 1
2 4 8 2n 2n

52 ~ Perfect Optional Mathematics Class 9

Alternately,

This sum of infinite seris can be represented by calculation process as follows.

1 + 1 + 1 + 1 + 1 + ... ... ...
2 4 8 16 32

If S1, S2, S3, S4, .... .... represent the sum of the first one, two, three, four, ... terms respectively, then we can

have the values of these sums as

S1 = 1 = 0.5 Limit
2

S2 = 1 + 1 = 0.5 + 0.25 = 0.75
2 4

S3 = 1 + 1 + 1 = 0.5 + 0.25 + 0.125 = 0.875
2 4 8

S4 = 1 + 1 + 1 + 1 = 0.5 + 0.25 + 0.125 + 0.0625= 0.9375
2 4 8 16

.... .... ...

Sn = 1 + 1 + 1 + 1 + 1 + ... ... ...
2 4 8 16 32

= 0.5 + 0.25 + 0.125 + 0.0625 + 0.03125 + ... ... ... → 1 (approaches 1)

\ The sum of above infinite series approaches 1.

In this series the common ratio is t2 = 0.25 = 0.5 = t3 = t4 etc.
t1 0.5 t2 t3

Hence, the limit of sum of series is 1.

SN Series Common Condition Limit of sum

1. 1 + 1 + 1 + 1 + ... ... r = 1/25 = 1 = 0.2 |r| = |0.2| = 0.2 < 1
5 25 125 625 1/5 5

2. 1 + 1 + 1 + 1 + ... ... r = 1/9 = 1 = 0.33 |r| = |0.3| = 0.3 < 1 a fixed real
3 9 27 81 1/3 3 number

3. 0.3 – 0.03 + 0.003 – 0.0003 + ... r = – 00..303 = – 110 =  – 0.1 |r| = |– 0.1| = 0.1 < 1

Hence, the limit of sum of infinite series is not possible for the series with common ratio greater than 1.

Finding a Rectangle of Maximum Area

Suppose we have been given 32cm of wire and asked to form a rectangle whose area is as large as possible.
What dimensions should the rectangle have?

Let w represent the width of the rectangle and let l represent the length of the rectangle.

Concept of Limit by Sum of Infinite Series ~ 53

The perimeter of the rectangle is 32cm. 16 – w

i.e. 2( + w) = 32

or,  + w = 16 w
or,  = 16 – w

So the area of the rectangle is

A =  × w

or, A = (l6 – w) × w

or, A = 16w – w2

Using this model for area, we can complete the experiment with different values of w to see how
to obtain the maximum area. After trying several values, it appears that the maximum area occurs when
w = 8cm as shown in the table below.

Width, w 7 7.5 7.9 8 8.1 8.4 8.7 9
Length, l 9 8.5 8.1 8 7.9 7.6 7.3 7
Area, A 63 63.75 63.99 64 63.99 63.84 63.51 63

In limit terminology, we can say that "the limit of A as w approaches 8 is 64." This is written as

wl→im∞ A = wl→im∞ (16w – w2) = 64

Proof without calculation

Here is an example of the limiting sum of infinite 1
geometric series from figure (proof without words) is 64
given. The figure itself shows the limit of the sum of the
infinite series. 1
16
Here, a unit square is divided into four equal squares and
each square has area 41. The small square on the right top 1
corner is again divided into four equal squares and the 4

process is continued. If the process is repeated infinite

number of times, what happens? We can see from the figure

that the unit square is divided into three sets of congruent

squares. The different sets of squares are represented by

different colours.

The limiting sum of the areas of each set of squares is 1 as there are three equal sets of squares from a unit
3
square. The sum of the areas of each set of squares represents the sum of the series.

Hence, we can write it as no. of lim   1 + 1 + 1 + 1 + ... ... ... + 1 = 1
terms→∞ 4 16 64 256 1024 3

54 ~ Perfect Optional Mathematics Class 9

EXERCISE : 2.3

1. Give the reason that the sum following series do not have the limit as a fixed real number.

a. 5 + 10 + 15 + 20 + 25 + ... ...

b. 10 + 8 + 6 + 4 + ... ...

c. 3 + 6 + 12 + 24 + ... ...

d. – 3 + 6 – 12 + 24 – ... ... Limit

e. 1 + 3 + 9 + 27 + ... ...
2 4 8
2. Give reason that the following series have the limit as a fixed real numbers.

a. 1 + 1 + 1 + 1 + ... ...
3 9 27 81
1 1 1 1
b. 1 – 2 + 4 – 8 + 16 – ... ...

c. 0.3 + 0.03 + 0.003 + 0.0003 + ... ...

d. 0.45 + 0.0045 + 0.000045 + ... ...

3. A designer draws a right angled triangle with hypotenuse 8m whose area is as large as possible.

a. Draw and label a diagram that shows the base x and height y of P 8cm
the triangle.

b. Verify that the area of the triangle is given by y

A = 21xy = 1 x 8 – x2 Q x R
2
2.2 2.5 x
c. Complete the following table and use the table to find xl→im2 A x
12cm
x 1 1.5 1.9 1.99 2 2.01 2.1 12 – 2x
A xx

4. From a square piece of material 12cm on a side, cut equal squares x
from each of its corners and turn up the sides to create an open box
whose volume is as high as possible.

a. Draw and label a diagram that represents the box that shows the
removed squares each of side x.

b. Verify that the volume V of the box is given by V = 4x(6 – x)2.

x

c. Complete the following table and use the table to find xl→im2 V x x

x 1 1.5 1.9 1.99 2 2.01 2.1 2.2 2.5
V

Concept of Limit by Sum of Infinite Series ~ 55

5. Find the series represented by the areas of the shaded parts of
the shapes. Also find the sum of the series without using any
formula (i.e. proof without words).

a. Here, the given shape is an equilateral triangle with its
area one square unit. First, it is divided into four equal
equilateral triangles by joining the midpoints of the sides.
The left bottom part is shaded. Again, the top triangle is
divided into four equal triangles by the same process and
its left bottom part is shaded. This process in continued
for a large number of times to get the final figure given.

Hints for drawing


Step 1 Step 2 Step 3 Step 4

1 41  1 41  1 2
4 4 4
[Series : + + + ... + ... ... ... ... ... ... = ?]

b. Here, the given shape is a square with area one square unit.
First, it is divided into two parts horizontally in such a way
that the bottom part is one-third of the whole. Again the given
square is divided into two parts vertically in such a way that
the left part is one-third of the whole. Then the square is
divided into four parts of three different sizes. The square on
the left bottom corner is shaded and the process is repeated
on the square of right top corner. Continue this process for a
large number of times to get the final figure.



Hints for drawing

Step 1 Step 2 Step 3 Step 4 Step 5

1 91  2 91  2 2 91  2 3
9 9 9 9
[Series : + + + + ... ... ... ... ... ... = ?]



56 ~ Perfect Optional Mathematics Class 9

2.4 Limit of a Function

Meaning of x→3

Let x be a variate value which is greater than or less than 3. When the difference between x and 3 is very Limit
small, we say that x approaches 3 or x tends to 3. Symbolically, this statement is written as x→3. Thus
x→3 and x = 3 are different. When x approaches 3 from the values less than 3 i.e. x takes the values
2.5, ..., 2.9, ..., 2.99, ..., 2.999, ... ... ... , we say that x tends to 3 from the left and it is denoted by x→3– .
When x approaches 3 from the values greater than 3 i.e. x takes the values 3.5, ..., 3.1, ..., 3.01, ..., 3.001, ...,
3.0001 ... ... , we say that x tends to 3 from the right and it is denoted by x→3+. But x = 3 means, x takes
value 3 exactly.

3

2 2.5 2.9 3.1 3.5 4

x→3–  x→3+

A Limit That Exist

We introduce the concept of the limit of a function by the following example.

Consider f(x) = x + 1. We let the variable x approach 2 through the values that are less than 2 and greater

than 2, and find the corresponding values of f(x).

Y

Input (x) Output f(x) = x + 1 f(x) = x + 1

Input x → 2 1.7 1.7 + 1 = 2.7 These outputs → 3 These outputs → 3 3
from the left 1.8 1.8 + 1 = 2.8

1.9 1.9 + 1 = 2.9 2
1.99 1.99 + 1 = 2.99

1.999 1.999 + 1 = 2.999 1
23

Input x → 2 2.0001 2.0001 + 1 = 3.0001 X' O X1 x→2–  x→2+ 3
from the right 2.001 2.001 + 1 = 3.001

2.01 2.01 + 1 = 3.01

2.1 2.1 + 1 = 3.1 Y'
2.4 2.4 + 1 = 3.4

From the table, it is clear that f(x) approaches (or tends) to 3 as x approaches 2 from the left or from the

right of 2. The limit of function f(x) when x approaches 2 from the values less than 2 i.e. x→2– , is known

as left hand limit of f(x) at x = 2. It is denoted by xl→im2– f(x).

From the above table, we have, lim f(x) = 3.
x→2– 

Limit of a Function ~ 57

Similarly, when x approaches 2 from the values greater than 2 i.e. x→2+, the limit of function f(x) is known

as right hand limit of f(x) at x = 2. It is denoted by lim f(x).
x→2+

From the above table, we have, lim f(x) = 3.
x→2+

When the left hand limit and the right hand limit of f(x) at x = 2 are equal i.e. xl→im2– f(x) = lim f(x) = 3,
x→2+

the limit of f(x) at x = 2 is said to be exist. It is written as lim f(x) = 3.
x→2 

When the left hand limit and the right hand limit of the function f(x) at x = a are equal and each is

equal to L i.e. lim f(x) = lim f(x) = Limit value. Then the limit of f(x) at x = a is said to be exist. It
x→a–  x→a+

is written as lim f(x) = L.
x→a

Estimating a Limit Numerically

We can calculate the limit of the given function f(x) as x→a by finding the values of f(x) for two sets of
x-values. One set of x approaches a from the left and another set approaches a from the right of a. Study the
examples below to estimate a limit of a function numerically.

Example 1:

Use a table to estimate the limit numerically : lim (2x + 1)
x→1

 Let f(x) = 2x + 1. Then construct a table that shows the values of f(x) for two sets of x-values: one

set of x that approaches 1 from the left and another set that approaches 1 from the right.

Input (x) Output f(x) = 2x + 1

Input x → 1 0.8 2 × 0.8 + 1 = 1.6 + 1 = 2.6 These outputs → 3 These outputs → 3 Y
from the left
0.9 2 × 0.9 + 1 = 1.8 + 1 = 2.8 5

0.99 2 × 0.99 + 1 = 1.98 + 1 = 2.98 4 f(x) = 2x + 1

0.999 2 × 0.999 + 1 = 1.998 + 1 = 2.998 3

0.9999 2 × 0.9999 + 1 = 1.9998 + 1 = 2.9998 2

 1 3 1

Input x → 1 1.0001 2 × 1.0001 + 1 = 2.0002 + 1 = 3.0002 X' Ox→1–  x→1+ 3 4 5 X
from the right 1.001 2 × 1.001 + 1 = 2.002 + 1 = 3.002

1.01 2 × 1.01 + 1 = 2.02 + 1 = 3.02 Y'

1.1 2 × 1.1 + 1 = 2.2 + 1 = 3.2

1.2 2 × 1.2 + 1 = 2.4 + 1 = 3.4

From the above table, it appears that the closer x gets to 1, the closer f(x) gets to 3. So we can estimate the
limit of f(x) to be 3 as x→1. The graph of the function also supports this conclusion.

58 ~ Perfect Optional Mathematics Class 9

Example 2:

Use a table to estimate the limit numerically : lim x2 – 9
x→3 x–3
 x2 – 39.
Let f(x) = x – Then construct a table that shows the values of f(x) for two sets of x-values: one set

of x that approaches 3 from the left and another set that approaches 3 from the right.

Input (x) Output f(x) = x2 – 9 Y
x – 3
7 Limit
2.8 2.82 – 39 = 5.8 These outputs → 6 6
2.8 –
Input x → 3
from the left 2.9 2.92 – 39 = 5.9
2.9 –

2.99 2.992 – 39 = 5.99 5
2.99 –
4

2.999 2.9992 – 39 = 5.999 x2 – 9 3
2.999 – x–3 2
f(x) = 1
3 6

Input x → 3 3.0001 33..000000112 – 9 = 6.0001 These outputs → 6 X' O X1 x→3–  x→3+ 5
from the right – 3

3.001 3.0012 – 39 = 6.001
3.001 –

3.01 3.012 – 39 = 6.01 Y'
3.01 –

3.1 3.12 – 39 = 6.1
3.1 –

From the above table, it appears that the closer x gets to 3, the closer f(x) gets to 6. So we can estimate the
limit of f(x) to be 6 as x→3. The graph of the function also supports this conclusion.

Example 3:

Use a table to estimate the limit numerically : lim f(x) where f(x) = 2x + 1, x<2
x→2 7 – x, x≥2

 Then construct a table that shows the values of f(x) for two sets of x-values: one set of x that approaches

2 from the left and another set that approaches 2 from the right.

Input (x) f(x) = 2x + 1, x<2 These outputs → 5 These outputs → 5 Y
7 – x, x≥2
Input x → 2 6
from the left 1.5 2 × 1.5 + 1 = 4
4
1.8 2 × 1.9 + 1 = 4.8 y = f(x)
2
1.99 2 × 1.99 + 1 = 4.98

1.999 2 × 1.999 + 1 = 4.998

25

Input x → 2 2.0001 7 – 2.0001 = 4.9999 X' O x→2–  x→2+ 4 6 X
from the right 2.002 7 – 2.002 = 4.998 Y'
2.01 7 – 2.01 = 4.99
2.1 7 – 2.1 = 4.9
2.5 7 – 2.5 = 4.5

Limit of a Function ~ 59

From above table, it appears that the closer x gets to 2, the closer f(x) gets to 5. So we can estimate the limit
of f(x) to be 5 as x→2. The graph of the function also support this conclusion.

A Limit That Does Not Exist

Consider a function f(x) = 2x + 1, x < 2
7 – x, x ≥ 2

For this function, we see the limit of the function at x = 1. So, we let the variable x approaches 1 through
the values that are less than 1 and greater than 1. Then find the values of f(x) as shown in the following
table.

Input (x) f(x) = 2x + 3, x≤1 Y
5 – x, x>1
These outputs → 5 6
Input x → 1 0.5 2 × 0.5 + 3 = 4
from the left 4
0.6 2 × 0.6 + 3 = 4.2 y = f(x)

0.9 2 × 0.9 + 3 = 4.8 2

0.99 2 × 0.99 + 3 = 4.98 X' O x→2–  x→2+ 4 6 X

1 0.999 2 × 0.999 + 3 = 4.998 5 Y'
1.0001 5 – 1.0001 = 3.9999 4
Input x → 1 These outputs → 4
from the right 1.001 5 – 1.001 = 3.999

1.01 5 – 1.01 = 3.99

1.1 5 – 1.1 = 3.9

1.5 5 – 1.5 = 3.5

From the table above, it appears that the closer x gets to 1 from the left, the closer f(x) gets to 5. That is the

left hand limit of f(x) as x approaches 1. It is denoted by lim f(x) = 5
x→1–

Also, it appears from the table that the closer x gets to 1 from the right, the closer f(x) gets to 4. That is the

right hand limit of f(x) as x approaches 1. It is denoted by lim f(x) = 4.
x→1+

It is clear that lim f(x) ≠ lim f(x)
x→1 x→1+

i.e. Left hand limit ≠ Right hand limit

In this case, we conclude that the limit of f(x) does not exist as x approaches 1.

Hence, xl→im1f(x) does not exist.
When the left hand limit and right hand limit of the function f(x) at x = a are not equal

i.e. lim  f(x) ≠xl→ima+  f(x). Then the limit of f(x) at x = a does not exist.
x→a

60 ~ Perfect Optional Mathematics Class 9

Example 4:

Use a table to show the limit does not exist numerically:

lim f(x) where f(x) = x2, x < 2
x→a–  – x + 4, x ≥ 2

 Construct a table that shows the values of f(x) for two sets of x-values: one set of x that approaches 2

from the left and another set that approaches 2 from the right.

Input (x) f(x) = x2, x < 2 These outputs → 4 Y Limit
– x + 4, x ≥ 2
Input x → 2 6
from the left 1.5 (1.5)2 = 2.25
4
1.8 (1.8)2 = 3.24 y = f(x)

1.99 (1.99)2 = 3.9601 2

2 1.999 (1.999)2 = 3.9960 4 X' O x→2–  x→2+ 4 6 X
2.0001 – 2.0001 + 4 = 1.9999 2
Input x → 2 These outputs → 2 Y'
from the right 2.005 – 2.002 + 4 = 1.995

2.01 – 2.01 + 4 = 1.99

2.1 – 2.1 + 4 = 1.9

2.2 – 2.2 + 4 = 1.8

From the table, it appears that the closer x gets to 2 from the left, the closer f(x) gets to 4. So the left hand

limit of f(x) as x approaches 2 is 4. It is denoted by lim f(x) = 4.
x→2– 

Again, it appears from the table that the closer x gets to 2 from the right, the closer f(x) gets to 2. That is the

right hand limit of f(x) as x approaches 2 is 2. It is denoted by xl→im2+ f(x) = 2.

It is clear that lim f(x) ≠ lim f(x)
x→2–  x→2+

i.e. Left hand limit ≠ Right hand limit

Hence, we conclude that the limit of f(x) does not exist as x approaches 2.

Hence, lim f(x) does not exist. EXERCISE : 2.4
x→2

1. Write the mathematical sentences by notational representation.

a. x approaches to 3 b. x approaches to a

c. x approaches to – 2 b. x approaches to – a

2. Write the following notational representation into mathematical sentences.

a. x → a b. x → – a

c. x → 8 d. x → – 2

e. xl→ima– f. lim
x→a+
lim lim
g. x→a– f(x) h. x→a+ f(x)

Limit of a Function ~ 61

3. a. What do you mean by the value of function?

b. What do you mean by limit of a function?

4. Complete the table and use the result to find the limit numerically.

a. lim (3x + 4) b. lim (x2 + 2x + 4)
x→2 x→1

Input (x) f(x) = 3x + 4 Input (x) f(x) = x2 + 2x + 4

1.4 ... ... ... ... ... ... 0.4 ... ... ... ... ... ...
1.5 ... ... ... ... ... ... 0.6 ... ... ... ... ... ...
1.9 ... ... ... ... ... ... 0.9 ... ... ... ... ... ...
1.99 ... ... ... ... ... ... 0.99 ... ... ... ... ... ...
1.999 ... ... ... ... ... ... 0.999 ... ... ... ... ... ...

2 2.001 ... ... ... ... ... ... ?  1 ?
2.01 ... ... ... ... ... ... 1.001 ... ... ... ... ... ... ?
2.1 ... ... ... ... ... ... 1.01 ... ... ... ... ... ...
2.2 ... ... ... ... ... ... d. 1.1 ... ... ... ... ... ...
2.5 ... ... ... ... ... ... 1.3 ... ... ... ... ... ...
1.5 ... ... ... ... ... ...

c. lim x2 – 4 lim x – 1
x→2 x–2 x→1 x2 + x – 2

Input (x) f(x) = x2 – 4 Input (x) f(x) = x–1
x–2 x2 + x – 2

1.3 ... ... ... ... ... ... 0.4 ... ... ... ... ... ...

1.7 ... ... ... ... ... ... 0.6 ... ... ... ... ... ...

1.9 ... ... ... ... ... ... 0.9 ... ... ... ... ... ...

1.99 ... ... ... ... ... ... 0.99 ... ... ... ... ... ...

1.999 ... ... ... ... ... ... ? 0.999 ... ... ... ... ... ...
 2  1

2.001 ... ... ... ... ... ... 1.001 ... ... ... ... ... ...
2.01 ... ... ... ... ... ...
2.05 ... ... ... ... ... ... 1.01 ... ... ... ... ... ...
2.1 ... ... ... ... ... ...
2.4 ... ... ... ... ... ... 1.05 ... ... ... ... ... ...

1.1 ... ... ... ... ... ...

1.5 ... ... ... ... ... ...

5. Use the table to estimate the limits numerically.

a. lim (2x – 1) b. lim (5x + 1) c. lim (7x – 15)
x→2 x→1 x→3

d. lim (4x – 5) e. lim (2x – 7) f. lim (2x – 3)
x→0 x→5 x→4

6. Use the table to estimate the limits numerically.

a. lim x2 – 11 b. lim x2 – 146 c. lim x2 – 2x
x→1 x – x→4 x – x→0 x

d. lim x2 – 3x + 2 e. lim x2 – 4x + 3 f. lim x2 – 2x – 3
x→2 x –2 x→1 x –1 x→3 x2 – 5x + 6

62 ~ Perfect Optional Mathematics Class 9

7. Use the table to estimate the limits numerically.

a. lim f(x), for f(x) = x – 1, x <2 b. xl→im1f(x), for f(x) = 3x + 1, x ≤ 1
x→2 3 – x, x ≥2 6 – 2x, x < 1

c. lim f(x), for f(x) = x2, x <2 d. xl→im3f(x), for f(x) = x + 1, x > 3
x→2 6 – x, x ≥2 x2 – 5, x < 3

8. Use the table to show that the limit does not exist numerically.

a. lim f(x), for f(x) = x + 1, x <1 b. xl→im3f(x), for f(x) = – x + 3, x < 3 Limit
x→1 2x – 3, x ≥1 x + 7, x > 3

c. lim f(x), for f(x) = x2 + 1, x < 2 d. xl→im3f(x), for f(x) = x2 – 5, x ≤ 3
x→2 4 – x, x > 2 2x – 7, x > 3

9. Use the given graph to find the value of f(2), lim f(x) and lim f(x). Also find the limit of the
x→2–  x→2+
function as x→2 or state that it does not exist.

a. Y b. Y

6 6

44

22

X' O 2 4X X' O 2 4X
c. d.
Y' Y'

Y Y

6 6

44

22

X' O 2 4X X' O 2 4X
e. f.
Y' Y'

Y Y

6 6

44

22

X' O 2 4X X' O 2 4X

Y' Y'

Limit of a Function ~ 63

10. Use the given graph to each limit or state that it does not exist.

a. lim f(x) b. lim f(x) c. lim f(x)
x→5+ x→5–  x→5

d. lim f(x) e. lim f(x) f. x→lim – 3 f(x)
x→1 x→10– 

Y

2

X' – 6 – 4 – 2 O 2 4 6 8 10 X

– 6

– 4

Y'

11. Sketch the graph of the function to find the given limit or state that it does not exist.

a. lim (3x + 1) b. lim (2x – 3) c. lim x2 – 2x
x→1 x→2 x→0 x

d. lim x2 –9 e. lim x2 – 4 f. lim 1
x→3 x +3 x→3 x – 2 x→0 x

g. lim (x2 – 3) h. lim (x – 1)2 i. lim (x2 + 1)
x→2 x→1 x→0

12. Draw the graph of the following functions and find the limit (if it exists) as x approches 2.

a. f(x) = 2x, 3, x < 2 b. f(x) = 3, x ≤ 2
– x + x ≥ 2 2x – 1, x > 2

c. f(x) = 5, x < 2 d. f(x) = – x, x<2
2x + 1, x > 2 x – 3, x>2

e. f(x) = 5, x = 2 f. f(x) = x2, x ≤ 2
x + 1, x ≠ 2 x + 1, x > 2



64 ~ Perfect Optional Mathematics Class 9

2.5 Techniques of Evaluating Limit

Techniques of Evaluating Limits Limit

In the privious section, we calculate the limit of a function by calculating its numerial values from the both
sides of a point. Now, in this section, we calculate the value of a function by substituting x value in the
function.

Evaluating Limits by Direct Substitution

The limit of some functions at the given point can be calculated by substituting the value of x in the
given function directly.

Example 1:

Evaluate : xl→im3  f(x) = 2(3x + 5)

 Now, lim (3x + 5) = 3 × 3 + 5 = 9 + 5 = 14.
x→3

Example 2:

Evaluate : lim 2t2 + 1 4
t→0 3t2 + 4t –

 Now, lim 2t2 +1 4 = 2.0 + 1 4 = – 1 .
t→0 3t2 + 4t – 3.0 + 4.0 – 4

Evaluating Limits by Dividing Out

The limit of some functions can be evaluated by direct substitution. In this section, we are going to study

dividing out technique for calculating limit. This technique should be applied when direct substitution
0
produces 0 in both the numerator and the denominator. An expression such as 0 has no meaning as a real

number. For example, lim x2 – 2x – 3 = 32 – 2×x – 3 = 9 – 2 = 00. Here 0 is called indeterminate form.
t→3 x–3 3–3 2 – 2 0

When we try to evaluate a limit of a rational function by direct substitution method and encounter this

indeterminate form, we can conclude that the numerator and the denominator must have a common factor.

After factorizing and dividing out, we should try direct substitution again.

Example 3:

Evaluate : lim x2 +x– 6
x→2 x–2

 Here, making direct substitution of x = 2 for lim x2 +x– 6, we have
x→2 x–2

lim x2 + x – 6 = 22 + 2 – 6 = 6–6 = 0 , which is an indeterminate form. So, we factorize the
x→2 x–2 3–3 2–2 0

expression in the numerator as follows.

Techniques of Evaluating Limit ~ 65

Hence, lim x2 +x– 6 = lim x2 + 3x – 2x – 6
x→2 x–2 x→2 x–2

= lim x(x + 3) – 2(x – 3)
x→2 x – 2

= lim (x + 3)(x – 2)
x→2 (x – 2)

= lim (x + 3) = 2 + 3 = 5
x→2

Example 4:

Evaluate : lim 3y2 + 2y
y→0 y

 When y = 0, 3y3 + 2y = 3.02 + 2.0 = 0 form, which is an indeterminate form.
y 0 0

∴ lim 3y2 + 2y = lim y(3y3 + 2) = lim (3y2 + 2) = 3.0 + 2 = 2.
y→0 y y→0 y y→0

Example 5:

Evaluate : lim x2 – 5x + 4
x→1 x2 + 2x – 3

 At x = 1, x2 – 5x + 4 = 12 – 5.1 + 4 = 1 –5+ 4 = 0 form, which is an indeterminate form.
x2 + 3x – 3 12 + 2.1 – 3 1 +2– 3 0

∴ lim x2 – 5x + 4
x→1 x2 + 3x – 3

= lim x2 – 4x – x + 4
x→1 x2 + 3x – x – 3

= lim x(x – 4) – 1(x – 4)
x→1 x(x + 3) – 1(x + 3)

= lim (x – 4)(x – 1)
x→1 (x + 3)(x – 1)

= lim x – 4
x→1 x + 3

= 1 – 4 = – 3
1 + 3 4

Evaluating Limits by Rationalizing

In some of the rational functions containing radical expressions, the direct substitution of the limiting value
of x produces 0 on both the numerator and the denominator and becomes an indeterminate form. When we
encounter this form, we multiply the numerator and the denominator of the given function by the conjugate
of either the numerator or the denominator of both. This is because the product of two conjugates containing
radicals will, itself, contain no radical expression.

66 ~ Perfect Optional Mathematics Class 9

Example 6:

Evaluate : lim x+5 –3
x→4 x–4

 At x = 4, x+5– 3 = 4+5–3 = 9–3 = 0 form, which is an indeterminate form. In this case , the
x–4 4–4 4–4 0

numerator contains radical expression, so we can rewrite the function by rationalizing the numerator.

∴ lim x+5–3 Limit
x→4 x–4

= lim x+5–3 x+5+3
x→4 x–4 x+5+3

= lim (x ( x+ 5)2 – 32 3)
x→4 – 4)( x+5+

= lim (x – x+ 5 – 9 + 3)
x→4 4)( x + 5

= lim (x – (x – 4) + 3)
x→4 4)( x +5

= lim 1
x→4 x+5+3

= 1
4+5 +3

= 1
9+3

= 3 1 3
+

= 1
6

Example 7:

Evaluate : lim x–a
x→a x2 – a2

 At x = a, x–a = a–a = 0 = 0 form, which is an indeterminate form.
x2 – a2 a2 – a2 0 0

\ lim x–a
x→a x2 – a2

= xl→ima ( x – a)2
(x – a)(x + a)

= xl→ima x–a x–a
x–a x+a

= lim x – a
x→a x + a

= a–a
a + a

= 20a = 0

Techniques of Evaluating Limit ~ 67

Example 8:

Evaluate : lim 2x – 3x – a
x→a x– a

 At x = a, 2x – 3x – a = 2a – 3a – a = 2a – 2a = 0 form, which is an indeterminate form.
x– a a– a a– a 0

Therefore, lim 2x – 3x – a
x→a x– a

= lim 2x – 3x – a × 2x + 3x – a = lim ( – 1( x– a)( x + a)
x→a x –a 2x + 3x – a x→a x– a)( 2x + 3x – a)

= lim ( 2x)2 – ( 3x – a)2 = lim – 1 ( x + a)
x→a x – a)( 2x + 3x – x→a 2x + 3x – a
( a)
– 1 ( a + a)
= lim 2x – (3x + a) = 2a + 3a – a
x→a – a)( 2x + 3x
( x – a) – 1( a + a)
2a + 2a
= lim 2x – 3x + a =
x→a a)( 2x + 3x – a)
( x– – 2a
2 2a
= lim – x + a =
x→a a)( 2x +
( x– 3x – a) – 2 a
2 2 a
= lim –1(x – a) =
x→a a)( 2x + 3x – a)
( x– –1
= 2
= lim ( – {( x)2 – ( a)2}
x→a x – a)( 2x + 3x – a)

Example 9:

Evaluate: lim x2 – a2
x→a x– a

 At x = a, a2 – a2 = 00, which is an indeterminate form.
a– a

\ lim x2 – a2 = lim (x – a)(x + a)
x→a x – a x→a x– a

Multiplying the numerator and the dinominator by x + a, we have

= lim (x – a)(x + a) × x+ a
x→a x– a x+ a

= lim (x – a)(x + a)( x + a)
x→a ( x)2 – ( a)2

= lim (x – a)(x + a)( x + a)
x→a (x – a)

= lim (x + a)( x + a)
x→a

= (a + a)( a + a)

= 2a × 2 a
= 4a a = 4a23

68 ~ Perfect Optional Mathematics Class 9

EXERCISE : 2.5

1. Evaluate the limit by direct substitution.

a. lim (3x – 2) b. lim (x2 + 3x – 2)
x→2 x→ – 2

c. lim x2 – 3x +2 d. lim x2 + 3x – 4
x→ – 2 x2 – 5x + 1 x→ – 2 x2 – 5x + 1
Limit
e. lim 2x + 3 f. lim 3 x2 – 1
x→ – 1 x→ 3

b. lim x2 – aa2 h. lim 2x2 + a2
x→a x + x→ a x2 – 4ax + a2

2. Evaluate the limit by dividing out technique.

a. lim x2 – 11 b. lim y – 2
x→1 x – y→2 y2 – 4

c. lim x2 – 1x6 d. lim x3 – b3
x→4 4 – x→b x – b

e. lim x2 – aa23 f. lim 5x
x→a x3 – x→0 x2 – 10x

g. lim x2 – 27 6 h. lim x2 – 8x +15
x→3 x2 – 5x + x→3 x2 – 2x – 3

3. Evaluate the limit by rationalizing technique.

a. lim x +4 – 2 b. lim 4 – 18 – x
x→3 x x→3 x – 2

c. lim xx––93 d. lim x2 – 16
x→3 x→3 x – 2

e. lim x – bb f. lim x – a
x→3 x – x→3 x2 – a2

g. lim 5x x –b 3b h. lim 3a – x – x + a
x→3 – 8x – x→a 4(x – a)

i. lim xx2­––aa2 j. lim x2 – b2
x→a x→b x2 – b2



Techniques of Evaluating Limit ~ 69

3 MATRICES

Curriculum Contents Included Periods

a. Introduction of matrix, order and components 20
b. Types of matrix: row, column, zero, square, diagonal, scalar, unit, equal, symmetric and

triangular matrices

c. Operation of matrices (Addition and subtraction)
d. Properties of matrix addition
e. Transpose of a matrix
f. Multiplication matrix
- Scalar multiplication
- Multiplication of matrices
- Properties of matrix multiplication (closure, associative and distributive)

Objectives

On completion of this unit, students will be able to:
define matrix and state components of matrix.

find the matrix with given formula for its elements and order.

state the kinds of matrices and perform the addition and subtraction of matrices and also state the properties
of addition of two matrices.

find the transpose of the given matrix.

find the matrix multiplication and test the properties of matrix multiplication.

Materials

We can represent different information about the data in the matrix form. It is very clear by representing the given
information in the matrix form.

The following table shows the number of boy and girl students in three different classes in a school. The
information can be displayed in the form of matrices. The first row represents the number of boys, the second row
represents the number of girls. Similarly, the first column represents students of class VIII, the second column
represents the students of class IX and the third column represents the students of class X.

Class VIII IX X Class VIII IX X

Boys 15 18 16 Boys 15 18 16

Girls 18 15 13 Girls 18 15 13

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Matrices 1 mark each 5 marks each Questions Marks
2 marks each 4 marks each
1 – 4 9
21

70 ~ Perfect Optional Mathematics Class 9

3.1 Representation and Types of Matrices

Introduction

A set of numbers arranged in a rectangular form in rows and columns and enclosed by round (or square)
bracket is known as a matrix. The plural form of matrix is matrices. The theory of matrices has become one
of the most important and powerful tool in Mathematics. It has wide applications in various fields such as
Statistics, Engineering, Economics etc. The following table gives the number of students in three classes of
two schools.

Class X Class XI Class XII Matrices

School A 27 25 20

School B 34 32 31

The above information can be arranged in following rectangular form between round brackets.

School A X XI XII 1st row
School B 27 25 20 2nd row
34 32 31
1st col. 2nd col. 3rd col.

The numbers 27, 25 and 20 are in the first horizontal line and are said to be in the first row (R1) and the
numbers 34, 32 and 31 are said to be in the second row (R2). Again the numbers 27 and 34 are in the first
vertical line and are said to be in the first column (C1) and 25, 32 and 20, 31 are said to be in the second
column (C2) and the third column (C3) respectively.

The first row shows the number of students in three different classes of school A and the first column shows
the number of students in class X of school A and B and so on.

Order of matrix

Each number of a matrix is known as an element or entry or component of the matrix. The number of rows
and columns taken together gives an order or size or dimension of a matrix. The matrices are denoted by
capital letters and the elements are denoted by small letters.

For example, A = a b c
d e f

A is a matrix and a, b, c, d, e and f are its elements. The number of rows of matrix A = 2 and its number of

columns = 3, so the order of matrix A is 2×3. A matrix ‘A’ of order 2×3 is denoted by A2×3.

The elements of a matrix are generally denoted by the corresponding small letters with double suffix where

the first suffix indicates the number of rows and the second suffix indicates the number of columns of that

element. b 11 b12 b13
b21 b22 f23
For example, B =

Here, b12 is an element belongs to the first row and the second column. Similarly, b13 is the element of the
first row and third column.

Representation and Types of Matrices ~ 71

Example 1:

2 3
If M = 11 – 4 , find

– 7 2

a. The order of M and

b. The values of m21, m12 and m32.
2 3

 Here, M = 11 – 4

– 7 2

a. Number of rows = 3 and number of columns = 2, the order of M is 3×2.

b. m21 = the element belongs to the 2nd row and the 1st column = 11.
m12 = the element belongs to the 1st row and the 2nd column = 3.
m32 = the element belongs to the 3rd row and the 2nd column = 2.

Types of Matrices

There are different types of matrices according to the dimensions of matrices and the nature of the elements.

a. Row Matrix or Row Vector

A matrix having only one row is known as a row matrix or a row vector.
For example, A = (2 5), B = (x y z).
A and B are row matrices of orders 1×2 and 1×3 respectively.

b. Column Matrix or Column Vector

A matrix having only one column is called a column matrix or column vector.

a 3
b
For example: P = , Q = – 2 , P and Q are column matrices of orders 2×1 and 3×1 respectively.
4

c. Zero or Null Matrix

A matrix in which each of its element zero is called a zero matrix or null matrix.

For example, M = 0 0 , N = 0 0 0 , M and N are null matrices of orders 2×2 and 2×3
0 0 0 0 0

respectively.

d. Rectangular Matrix

A matrix in which the number of rows and columns are different is called a rectangular matrix.

2 4 – 2 2 3
1 – 5 – 1
For example: C = , D = 4 1 , C and D are rectangular matrices of orders 2×3 and
– 1 0

3×2 respectively.

72 ~ Perfect Optional Mathematics Class 9

e. Square Matrix

A matrix having equal number of rows and columns is called a square matrix. For example:

R= a b ,Q= 1 – 1 7
c d 2 2 8 R and Q are square matrices of orders 2×2 and 3×3 respectively. A
3 – 4 0

square matrix of order n×n is often said to be a square matrix of order n.

f. Diagonal Matrix

A square matrix having all non-diagonal elements zero is called a diagonal matrix. The elements

belonging to same number of rows and columns are called the diagonal elements. a11, a22, a33 etc. are called
1 0 0
the diagonal elements. For example: D = 2 0 , E = 0 – 2 0 Matrices
0 – 1 0 5
0

D and E are diagonal matrices of orders 2 and 3 respectively.

g. Scalar Matrix

A diagonal matrix with all its diagonal elements equal is called a scalar matrix.

2 0 – 4 0 0

X = 0 2 , Y = 0 – 4 0 X and Y are scalar matrices of orders 2 and 3 respectively.
h. Unit or Identity Mat r0ix 0 – 4

A square matrix with all diagonal elements unity and the rest zero is called a unit matrix or an identity

1 0 0 1 0
matrix. T = 0 1 0 , S = 0 1 T and S are unit matrices of orders 3 and 2 respectively. A unit
0 1
0

matrix of order 2 is denoted by I2 and I3 means a unit matrix of order 3.

i. Equal Matrices

Two matrices of the same order are said to be equal if their corresponding elements are equal.

2 – 1 2 –1 1 3 – 2 1 3 – 2
For the matrices P = 4 3 , Q = 4 3 ,A= 4 – 5 0
3 – 5 0 , B = 4 –1 6

–1 6 3

P and Q are equal matrices as they are matrices of same order having corresponding equal elements.

Similarly, A and B are equal matrices.

j. Symmetric Matrix

A square matrix is known as symmetric matrix if interchange of rows and columns makes no change
in the given original matrix. For example,

1 2 1 2 – 3

P = 2 5 , A = 2 4 5
– 3 5 6

i.e. having same elements in rows and columns of a square matrix is called symmetric matrix.

Representation and Types of Matrices ~ 73

k. Triangular Matrix

A square matrix in which all the elements above or below (or both) the main diagonal are zero is called
a triangular matrix. There are two types:

i. Upper Triangular Matrix

A square matrix in which all the elements below the main diagonal are zero is called an Upper Triangular

Matrix. For example, 1 – 5 and 1 2 –3
0 6 0 4 5 are upper triangular matrices.
0 0 6

ii. Lower Triangular Matrix

A square matrix in which all the elements above the main diagonal are zero is called a Lower Triangular

Matrix. For example, 5 0 3 0 0
2 3 and 5 – 4 0 are lower triangular matrices.
6 – 6
6

Example 2:

Construct a 2 × 3 matrix whose elements are given by aij = 4i – j.

 As, aij = 4i – j

when i = 1, j = 1: a11 = 4 × 1 – 1 = 3; i = 2, j = 1: a21 = 4 × 2 – 1 = 7

i = 1, j = 2: a12 = 4 × 1 – 2 = 2; i = 2, j = 2: a22 = 4 × 2 – 2 = 6

i = 1, j = 3: a13 = 4 × 1 – 3 = 1; i = 2, j = 3: a23 = 4 × 2 – 3 = 5

Hence, the required 2×3 matrix is

A= a 11 a12 a13 = 3 2 1
a 21 a22 a23 7 6 5

Example 3:

x – 1 3 – 3 x + z
If 5 y = w – y 2 , find the values of w, x, y and z.

 We have, x – 1 3 – 3 x + z
5 y = w – y 2

By the definition of equality of matrices,

x – 1 = – 3 x+z=3
∴ x = – 2 or, – 2 + z = 3

∴ z = 5

Again, y = 2 and w – y = 5
or, w – 2 = 5

∴ w = 7

Hence, w = 7, x = – 2, y = 2 and z = 5.

74 ~ Perfect Optional Mathematics Class 9

EXERCISE : 3.1

1. a. Write an example of row matrix of order 1×4.

b. Write an example of column matrix of order 3×1.

2. a. If the matrix A = x 0 , then write its type.
0 x

1 0 0

b. If the matrix A = 0 2 0 , then write its type.
0 0 3

3. a. Write the order of matrix 1 2 3 .
4 5 6

b. In matrix P = a11 a12 a13 , write its order and type. Matrices
a21 a22 a23

4. a. If matrix A = 2 1 12 16 and a2j = 17, find the value of j.
2 4 13 17

b. If the matrix A (aij)2×3, where aij = i + j, then what is the value of a21? Write it.

a b c

5. a. If matrix A = d e f , then write the number of elements of A.
g h i

b. If 1 2 = 1 2 , then find the value of x and y.
3 4 x y

6. a. A matrix has 8 elements. What are the possible orders can it have?

b. If A is a matrix of order 2×5, find the number of elements of A.

7. What are the orders of the following matrices:

a b 3 1 2
7
a. 81 –3 2 42 b. – p q c. 40 ––  24 3
m – n 9 6 1

8. a. A school has three branches A, B and C. The number of boys and girls in those branches are as
follows:

Branch A has 270 boys and 209 girls,

Branch B has 184 boys and 201 girls,

Branch C has 203 boys and 143 girls.

Express the above information in matrix form.

b. If there are 210 apples, 167 mangoes and 78 oranges in shop A and 198 apples, 200 mangoes and
103 oranges in shop B, represent this information in the form of a matrix.

1 4 3 – 7
9. a. For the given matrix A = 3 6 – 4 8 ,

– 2 7 – 1 2

i. what is the order of the matrix? ii. find the values of a12, a23 and a14.
iii. find the values of a11 + a31 and a24 – a14.

Representation and Types of Matrices ~ 75

b. 2 – 3 7 and B = 2 4 1 ,
If A = 3 8 4 6 – 5 – 3
1
6 – 7

i. write the orders of A and B. ii. find the values of a22, a31, b12 and b23.

iii. find the values of a11 + b23, a23 – b12.

10. State the types of the following matrices:

0 0 1 c. (2 7) d. 10 0
a. 0 0 b. 2 0 – 3
0 4
0 1 0 0
h. 0 1 0
e. –0 4 0 f. –0 2 – 2 g. 23 07 1
– 4 0 0 0

i. –1 6 27 35 2 0 0 1 2 3
j. 0 4 0 k. 2 – 2 4
– 3
0 0 3 4 1

2 4 1 – 2 0 0 b c d

. 0 1 – 2 m. 3 5 0 n. 0 – a e

0 0 4 5 1 6 0 0 – f

a 0 0 a b c
o. 0 b 0 p. b r – d
0 c c – d s
0

11. State which of the following matrices are equal.

A= 1 3 , 0 2 C= 6 1 4 ,
7 – 5 B = 3 7 , 7 4 0

1 – 4

D= 7 1 , E= 1 3 , 0 2
5 6 7 – 5 F = 3 7

1 – 4

12. Construct a 2 × 3 matrix whose elements aij are given by the formula.

a. aij = i + 2j b. aij = j c. aij = 3i – 2j

13. Construct a 3 × 3 matrix whose elements bij are given by the relation.

a. bij = 2i + 3j b. bij = 3j – i c. bij = 4i – 3j

14. a. If 2 x and y – 3 are equal matrices, find the values of x and y.
5 – 2 5 – 2

b. If 3 a = 3 b + 1 , find the values of a and b.
b 4 4 4

c. If x + y 3 = 2 3 , find the alues of x and y.
x y y + 6 – 2

d. If a a + b = 2 5 , find the values of a, b, c and d.
– 3 c + 6 c 1+d

e. If 2p q+1 = p + 2 r+s , find the values of p, q, r and s.
5 q–3 r + 1 p+1



76 ~ Perfect Optional Mathematics Class 9

3.2 Operation of Matrices

Introduction

Here we deal with the following operations on matrices to express two or more matrices into a single matrix. Matrices
a. Addition of matrices
b. Subtraction of matrices
c. Multiplication of a matrix by a scalar
d. Multiplication of matrices
e. Transpose of a matrix

a. Addition of Two Matrices

The sum of two matrices A and B of same order is a new matrix whose each element is the sum of
corresponding elements of A and B. It is denoted by A + B and the order of A + B is same as that of A or B.

Note: If two matrices are of different orders then their sum is not defined.

The following are the matrices of the number of boys and girls in three classes of two schools A and B.

Group A Group B
Boys Girls Boys Girls

Class 10 25 30 Class 10 18 10

P = Class 11 18 21 Q = Class 11 20 18

Class 12 16 18 Class 12 10 12

Now, we want to find the total number of boys and girls classwise.

Total boys in class 10 = 25 + 18 = 43

Total girls in class 10 = 30 + 10 = 40

Total boys in class 11 = 18 + 20 = 38

Total girls in class 11 = 21 + 18 = 39

Total boys in class 12 = 16 + 10 = 26

Total girls in class 12 = 18 + 12 = 30

The above information can be obtained by adding given two matrices P and Q.

Group A Group B

Boys Girls Boys Girls

Class 10 25 30 18 10

P + Q = Class 11 18 21 + 20 18

Class 12 16 18 10 12

Boys Girls
Class 10 25 + 18 30 + 10

P + Q = Class 11 18 + 20 21 + 18

Class 12 16 + 10 18 + 12

Operation of Matrices ~ 77

Boys Girls

Class 10 43 40

P + Q = Class 11 38 39

Class 12 26 30

Example 1:

If A = 1 3 – 4 and B = –  5 3 0 , find A + B.
7 – 3 0 2 4 – 4

 Here, A = 1 3 – 4 and B = – 5 3 0
7 – 3 0 2 4 – 4

Now, A + B = 1 3 – 4 + – 5 3 0
7 – 3 0 2 4 – 4

or, A+B= 1 – 5 3 + 3 – 4 + 0
7 + 2 – 3 + 4 0 – 4

\ A+B= – 4 6 – 4
9 1 – 4

b. Subtraction of Matrices

The subtraction of two matrices A and B of same order is a new matrix whose each element is obtained by
subtracting the elements of B from corresponding elements of A. It is denoted by A – B and the order of
A – B is same as that of A or B.

Note: If two matrices are of different orders then their difference is not defined.

Example 2:

2 – 3 – 2 7

If P = 1 4 and Q = 0 – 5 , find P – Q and Q – P.

6 7 3 8

 2 – 3 – 2 7
Here, P = 1 4 and Q = 0 – 5

6 7 3 8

2 – 3 – 2 7 2 + 2 – 3 – 7 4 – 10

P – Q = 1 4 – 0 – 5 = 1 – 0 4 + 5 = 1 9

6 7 3 8 6 – 3 7 – 8 3 – 1

Similarly,

– 2 7 2 – 3
Q – P = 0 – 5 – 1 4

3 8 6 7

–  2 – 2 7 + 3 – 4 10
= 0 – 1 – 5 – 4 = – 1 – 9

3 – 6 8 – 7 3 1

78 ~ Perfect Optional Mathematics Class 9

Properties of Matrix Addition

Addition of matrices satisfy the following properties.

⇒ Closure Property

If A and B are two matrices of same order, then the sum A + B is also a matrix of the same order as that of

A or B. For example:

A= 2 3 1 and B = 4 – 7 0  , then
2 – 1 7 – 3 – 2 3
 2×3  2×3

then A + B = 2 3 1 + 4 – 7 0 = 6 – 4 1 Matrices
2 – 1 7 – 3 – 2 3 – 1 – 3 10
 2×3

Hence, the order of A + B is 2×3 that is same as the order of A or B.

⇒ Commutative Property

If A and B are two matrices of same order, then A + B = B + A.
For example:

1 3 – 3 1
A = – 2 4 and B = 5 3
5 – 1 0
0

1 3 – 3 1 – 2 4
A + B = – 2 4 + 5 3 = 3 7
5
0 5 – 1 0 – 1
4
– 3 1 1 3 – 2 7
B + A = 5 3 + – 2 4 = 3 5

– 1 0 0 5 – 1

\ A + B = B + A.

⇒ Associative Property

If A, B and C are three matrices of same order, then (A + B) + C = A + (B + C).

For example:

If A = 1 4 ,B= 0 2 and C = 1 5
–  2 3 –  3 4 6 – 1

then A + B = 1 4 + 0 2 = 1 + 0 4+2 = 1 6
–  2 3 –  3 4 – 2 – 3 3+4 – 5 7

Now, (A + B) + C = 1 6 + 1 5
– 5 7 6 – 1

= 1 + 1 6+5 = 2 11
–  5 + 6 7–1 1 6

Operation of Matrices ~ 79

Again, B + C = 0 2 + 1 5 = 0 + 1 2 + 5 = 1 7
–  3 4 6 – 1 –  3 + 6 4 – 1 3 3

Now, A + (B + C) = 1 4 + 1 7 = 1 + 1 4+7 = 2 11
–  2 3 3 3 – 2 + 3 3+3 1 6

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

⇒ Additive Identity

If A is any matrix and O is a null matrix of same order, then A + O = O + A = A.
For example.

If A = 1 3 7 and O = 0 0 0 then
– 4 0 – 2 0 0 0

A + O = 1 3 7 + 0 0 0
– 4 0 – 2 0 0 0

= 1 + 0 3 + 0 7 + 0 = 1 3 7
– 4 + 0 0 + 0 – 2 + 0 – 4 0 – 2

Similarly, O + A = 1 3 7 =A
– 4 0 – 2
Hence, A + O = O + A = A.

⇒ Additive Inverse

If A is any matrix then there exists a negative matrix of A i.e. – A such that

A + (– A) = (– A) + A = O. For example,

If A = 3 – 4 then – A = –  3 4
– 1 5 1 – 5

A + (– A) = 3 – 4 + –  3 4 = 3 – 3 – 4 + 4 = 0 0 =O
– 1 5 1 – 5 – 1 + 1 5–5 0 0

Similarly, (– A) + A = O

∴ A + (– A) = (– A) + A = O.

c. Multiplication of Matrix by a Scalar

If A be a matrix of any order and k be a real number then the scalar multiplication of A by k is a new
matrix obtained by multiplying each element of A by k and is denoted by kA.

For example: If A = 2 – 1 3 , then by addition of matrices
1 4 – 2

2A =A+A= 2 – 1 3 + 2 – 1 3
1 4 – 2 1 4 – 2

= 2 + 2 – 1 + (– 1) 3 + 3 = 4 – 2 6
1 + 1 4 + 4 – 2 + (–  2) 2 8 – 4

2A = 2 × 2 2 × – 1 2 × 3
2 × 1 2 × 4 2 × (– 2)

80 ~ Perfect Optional Mathematics Class 9

2A = 2  2 – 1 3
1 4 – 2

Similarly, 21A = 21  2 – 1 3
1 4 – 2

1 × 2 21 × (– 1) 1 × 3 1 – 12  3
2 2 21 2 2
= =
1 1 1 – 1
2 × 1 2 × 4 2 × (– 2)

and – 3A = – 3 2 – 1 3
1 4 – 2

= – 3×2 – 3×(– 1) – 3×3 = – 6 3 – 9 . Matrices
– 3×1 – 3×4 – 3×(–  2) – 3 – 12 6

Example 3:

If A = 2 1 and B = –  2 3 , find 2A + 3B and A – 4B.
3 4 –  1 4

 Here, A = 2 1 and B = –  2 3 ,
3 4 –  1 4

Now, 2A + 3B = 2  2 1 + 3  –  2 3 = 4 2 + –  6 9
3 4 –  1 4 6 8 –  3 12

= 4 + (– 6) 2+9 = –  2 11
6 + (– 3) 8 + 12 3 20

A – 4B = 2 1 –4 –  2 3
3 4 –  1 4

= 2 1 – –  8 12
3 4 –  4 16

= 2 – (– 8) 1 – 12
3 – (– 4) 4 – 16

= 10 – 11
7 – 12

Example 4:

If X + Y = –  1 9 and 2X – Y = 7 0 , find the matrices X and Y.
7 4 2 8
 The given equations are

X+Y= –  1 9 ... ... ... (1)
7 4 ... ... ... (2)

2X – Y = 7 0
2 8

Adding equations (1) and (2).

Operation of Matrices ~ 81

(X + Y) + (2X – Y) = –  1 9 + 7 0
7 4 2 8

or, 3X = –  1 9 + 7 0
7 4 2 8

or, 3X = – 1 + 7 9+0
7 + 2 4+8

or, 3X = 6 9
9 12

or, X = 1 6 9 = 2 3
3 9 12 3 4

Now from equation (1), we have

X+Y= –  1 9
7 4

or, 2 3 +Y= –  1 9
3 4 7 4

or, Y= –  1 9 – 2 3
7 4 3 4

or, Y= –  1 – 2 9–3
7 – 3 4–4

∴ Y= –  3 6
4 0

Example 5:

If 2 x 3 +4 1 y = 2 10 , find the values of w, x, y and z.
–  1 4 1 + x – 3 z y+w

 Here, 2 x 3 +4 1 y = 2 10
–  1 4 1 + x – 3 z y+w

or, –2 x2 6 + 4 4y = 2 10
8 4 + 4x – 12 z y+w

or, 22x++4x4 6 + 4y = 2 10
– 4 z y+w

or, 2x + 4 = 2, or, 6 + 4y = 10,

or, 2x = – 2, or, 4y = 4

\ x = – 1 \ y = 1

2 + 4x = z, – 4 = y + w

or, 2 + 4 × – 1 = z or, – 4 = 1 + w

\ z = – 2 \ w = – 5

\ w = – 5, x = – 1, y = 1 and z = – 2.

82 ~ Perfect Optional Mathematics Class 9

EXERCISE : 3.2A

1. Find the values of a, b, c and d.

a. 2a – 3 = b c b. b 1– 3a +2 2 36 = 12 c +5 4 65
4 –  4 4

c. a + 2 b + 2a = 3 – 3 d. a +d b a–b = 4 2
7 c–4 d – 2 d–7 7 a + 1 c+b

2. Add the following matrices:

a. –1  4 3 + –  2 1 b. 14 65 –5 6 + –0 2 – 1 0
2 – 7 2 2 4

1 – 3 1 4 1 2 – 3 1 2 – 3 Matrices
c. 2 0 + 2 – 3 d. 4 – 5 6 + 4 – 5 6

5 – 4 0 7 – 7 8 9 – 7 8 9

3. Subtract the following matrices:

a. 23 – – 7 b. 1 2 7 – 3 1 4
0 – 7 4 0 – 7 6 0

c. – 3 3a b b – 2a – b 8 6 10 0
– 2a 2b a d. 4 7 – – 2 8
5
3 – 2 3

4. If A = –  2 3 and B = 8 8 , find:
4 0 10 – 6

a. A + B b. A – B c. 2A d. – 12 B

5. A= 1 3 ,B= –  2 1 and C = –  3 1 , then find
5 4 0 3 –  1 4

a. A + B + C b. A + 2B c. 12 A – 2B d. A – B + 2C

e. 2A – 21 B + C f. A – 31 B + 21 C

6. a. Solve for the matrix X from equation X + 1 3 = 4 7 .
5 – 3 8 9

b. If A = 1 – 2 , find the matrix X such that A – 2X = 3 4 .
–3 4 5 2

c. If x + 2 = 5 , find the values of x and y.
– 3 y 7

d. If 2 1 + 1 x+y = 3 8 , find the values of x and y.
x 7 2 1 5 8

Operation of Matrices ~ 83

7. Find the additive inverse of the following matrices.

a. –2 4 – 1 – 2 1 c. 12 43 –4 3 1 – 4 0
5 b. 3 4 d. 2 – 5 7
– 6 9
6 – 7 3

8. If A = 1 3 and B = –  3 2 , then find:
0 4 1 – 5

a. A – 2B b . 2A + B c. 2A – 3B d. 4A + 2B

9. a. If P+Q= 3 3 and 3P – Q = –  3 5 , find P and Q.
2 5 6 7

b. If 2A –B= 0 4 and A + 2B = 5 2 , find A and B.
2 1 11 3

10. Find the values of a, b and c from the following equations.

a. 2 a 3 +3 3 1 = 8 9 b. 3 a 4 +2 1 b = 5 11
1 0 2 b 8 9 0 3 – 2 – 3 – 2c – 3

11. If A = 1 2 – 3 ,B= 0 2 – 1 and C = 2 3 1 then find:
– 2 0 4 2 4 6 5 6 3

a. 2A + B – C b. A + 2B + 3C c. 21 A + B – 13 C d. 32 A – B + 2C

12. a. If A = 1 3 and B = –  3 5 , then find a matrix X such that A – 2B – 3X = 10 2 .
– 4 0 6 7 2 1

b. If P = 1 2 ,Q= 1 0 and R = 6 6 , then find a matrix Y such that 3P + Q + 2Y = R.
4 3 – 2 7 8 10

13. a. If 1 a + b 2 = 5 15 , then find the values of a, b, c and d.
– 1 – 2 3 c 2d 3

b. If x – 2 5 + 1 5 = 1 2 , then find the values of x, y and z.
2z 7 – 3 y+2 3 4

c. If x 7 +2 y w = 3 – 5z + 3 , find the values of w, x, y and z.
y z –  x 2w 4 10

14. a. If A + B = –  1 9 and 2A – B = 7 0 , find the matrices A and B.
8 4 4 17

b. If X + 3Y = 1 – 1 and 2X – 4Y = – 8 3 , find the matrices X and Y.
15 – 4 –  10 – 8

c. If I is a unit matrix of order 2×2, find the matrix A such that 2A – 3I = –  1 4 .
6 5



84 ~ Perfect Optional Mathematics Class 9

Multiplication of Matrices Table II

Let us see the following two tables of certain information.
Table I

Families Quantity in kg Families Price in Rs. per kg
Rice Potato Poush Magh
X 14 2 Rice
Y 18 3 Potato 25 26
12 14

The table I shows the amount of rice and potato consumed by two families X and Y and table II shows the
price of rice and potato per kg in two different months Poush and Magh.

How can we find the total expenditure of families X and Y for the months Poush and Magh? It can be shown Matrices

in the following table. Table III

Families Poush Magh
X 14 × 25 + 2 × 12 14 × 26 + 2 × 14
Y 18 × 25 + 3 × 12 18 × 26 + 3 × 14

Above multiplication operation can be presented in the matrix form as follows:

14 2  + 25 26  = 14 × 25 + 2 × 12 14× 26 + 2 × 14  
18 3 12 14 18 × 25 + 3 × 12 18 × 26 + 3 ×14 2×2
2×2 2×2

In above multiplication of matrices, the elements of the first row of the first matrix are treated with the

elements of the first column of the second matrix. So the number of elements in the first row of the first

matrix and the number of elements in the first column of the second matrix must be equal. i.e. number of
columns of first matrix = number of rows of the second matrix. The multiplication of two matrices will be
clear from following example.

If A = 1 2 and B = 1 2 – 3 , the product of two matrices A and B is denoted by AB and is
–  3 4 4 0 – 5

calculated by the following process.

Here, A = 1 2 and B = 1 2 – 3 .
–  3 4 4 0 – 5

  1 2 1 2 – 3 = 1×1+2×4 ... ... ... ... ... ...
– 3 4 4 0 – 5 ... ... ... ... ... ... ... ... ...

  1 2 1 2 – 3 = 1×1+2×4 1×2+2×0 ... ... ...
– 3 4 4 0 – 5 ... ... ... ... ... ... ... ... ...

  1 2 1 2 – 3 = 1×1+2×4 1 × 2 + 2 × 0 1× (– 3) + 2 × (– 5)
– 3 4 4 0 – 5 ... ... ... ... ... ... ... ... ...


Operation of Matrices ~ 85

  1 2 1 2 – 3 = 1×1+2×4 1 × 2 + 2 × 0 1× (– 3) + 2 × (– 5)
– 3 4 4 0 – 5 – 3 × 1 + 4 × 4 ... ... ... ... ... ...



  1 2 1 2 – 3 = 1×1+2×4 1 × 2 + 2 × 0 1× (– 3) + 2 × (– 5)
– 3 4 4 0 – 5 – 3 × 1 + 4 × 4 – 3 × 2 + 4 × 0 ... ... ...



  1 2 1 2 – 3 = 1×1+2×4 1 × 2 + 2 × 0 1× (– 3) + 2 × (– 5)
– 3 4 4 0 – 5 – 3 × 1 + 4 × 4 – 3 × 2 + 4 × 0 – 3 × (– 3) + 4 × (– 5)



\ AB = 1 2   1 2 – 3  
–  3 4 2 × 2 4 0 – 5 2 × 3

= –1 3××11++24××44 1×2+2×0 1× (– 3) + 2 × (– 5)  
– 3 × 2 + 4 × 0 – 3 × (– 3) + 4 × (– 5)
2×3

= 193 2 – 13  
– 6 – 11 2×3

Hence, if A and B are two matrices of order m × n and n × p respectively then the product AB is a matrix of

order m × p.

Properties of Matrix Multiplication

If A, B and C are the matrices which are comfortable for multiplication with each other. The matrices B and
C are of same order then the following properties hold for matrix multiplication.
a. In general, AB ≠ BA (Non commutative)
b. A(BC) = (AB)C (Associative property)
c. A(B + C) = AB + AC (Distributive property)
d. AI = IA = A (Multiplicative identity)
Where I is a unit matrix comfortable for AI and IA. The above properties can easily be verified.

Example 1:

Find the product of following matrices.

1 b. 1 (1 5)
a. (1 3 2) – 2 3

– 3

 a. 1
(1 3 2) – 2 = (1 × 1 + 3 × (– 2) + 2 × (– 3))

– 3

= (1 – 6 – 6) = (– 11)

b. 31  2 × 1 (1 5)1 × 2 = 1 × 1 1 ×5  
3 × 1 3 × 5 2×2

= 1 5 
3 15 2×2

86 ~ Perfect Optional Mathematics Class 9

Example 2:

If A = 1 – 4 and B = –  1 2 , find AB. Does BA define? If BA is defined, find BA and show
7 3 3 0

that AB ≠ BA.

 Here, A = 1 – 4 2 × 2 and B = –  1 2 2 × 2
7 3 3 0

AB = 1 – 4 –  1 2
7 3 3 0

= 1 × – 1 + (– 4) × 3 1× 2 + (– 4) × 0 = –   13 2 and
7 × – 1 + 3 × 3 7×2+3×0 2 14

BA = –  1 2 1 – 4 Matrices
3 0 7 3

= – 1 × 1 + 2 × 7 – 1× – 4 + 2 × 3 = 13 10
7 × – 1 + 3 × 3 3 × – 4 + 0 × 3 3 – 12

Since, Corresponding elements of AB and BA are not equal, AB ≠ BA.

Example 3:

If A = 1 0 , B = 1 2 and C = 2 0 verify that:
2 – 1  –  2 3 3 – 1

a. A(B + C) = AB + AC b. A(BC) = (AB)C

 a. B+C= 1 2 + 2 0 = 3 2
–  2 3 3 – 1 1 2

A(B + C) = 1 0 × 3 2 = 3 + 0 2+0 = 3 2
2 – 1 1 2 6 – 1 4–2 5 2

AB = 1 0 1 2 = 1 – 0 2+0 = 1 2
2 – 1 –  2 3 2 + 2 4–3 4 1

AC = 1 0 2 0 = 2 + 0 0+0 = 2 0
2 – 1 3 – 1 4 – 3 0+1 1 1

∴ AB + AC = 1 2 + 2 0 = 3 2
4 1 1 1 5 2

∴ A(B + C) = AB + AC.

b. BC = 1 2 × 2 0 = 2 + 6 0–0 = 8 – 2
–  2 3 3 – 1 – 4 + 9 0–3 5 – 3

A(BC) = 1 0 8 – 2 = 8 + 0 – 2 + 0 = 8 – 2
2 – 1 5 – 3 16 – 5 – 4 + 3 11 – 1

AB = 1 0 × 1 2 = 1 + 0 2–0 = 1 2
2 – 1 –  2 3 2 + 2 4–3 4 1

(AB)C = 1 2 × 2 0 = 2 + 6 0–2 = 8 – 2
4 1 3 – 1 8 + 3 0–1 11 – 1

∴ A(BC) = (AB)C.

Operation of Matrices ~ 87

Example 4:

If A = 1 2 show that A2 – 2A – 5I = O, where I and O are idendity and null matrices of order 2×2.
3 1

 Here, A = 1 2 ,I= 1 0 and O = 0 0 , then
3 1 0 1 0 0

A2 = 1 2 1 2 = 1 × 1 + 2 × 3 1×2+2×1
3 1 3 1 3 × 1 + 1 × 3 3 × 2 + 1 ×1

1 + 6 2+2 = 7 4
= 3 + 3 6+1 6 7

Now, A2 – 2A – 5I = 7 4 – 2× 1 2 – 5× 1 0
6 7 3 1 0 1

= 7 4 – 2 4 – 5 0
6 7 6 2 0 5

= 7 – 2 – 5 4–4–0 = 0 0 proved.
6 – 6 – 0 7–2–5 0 0

Example 5:

If A = 1 2 , find a matrix X such that AX = 5 6 .
–  3 0 –  3 0

 A is a 2×2 matrix and the matrix on the right hand side is also a 2×2 matrix, so X must be a 2×2 matrix.

Let X = a b and AX = 5 6
c d – 3 0

or, –1 3 2 a b = 5 6
0 c d – 3 0

or, –a 3+a 2c b + 2d = 5 6
+ 0 – 3b + 0 – 3 0

or, – 3a = – 3 ∴a=1

or, a + 2c = 5 ∴c=2

or, – 3b = 0 ∴b=0

or, b + 2d = 6 ∴d=3

∴ X = a b = 1 0
c d 2 3

Example 6:

If 7 x + 1 = 2 3 0 , find the values of x and y.
6y – 2 – 1 8 2

 We have, 7x + 1 = 2 3 0
6y – 2 –  1 8 2

or, 67 yx + 1 = 2 × 0 + 3 × 2
– 2 – 1 × 0 + 8 × 2

or, 67 xy + 1 = 6
– 2 16

88 ~ Perfect Optional Mathematics Class 9

Now, comparing corresponding elements, we have

7x + 1 = 6 and 6y – 2 = 16

or, 7x = 6 – 1 or, 6y = 16 + 2

or, 7x = 5 or, 6y = 18

∴ x = 57 or, y = 18 = 3
6

∴ y = 3
5
Hence, x = 7 and y = 3

Transpose of a Matrix

A matrix obtained by interchanging the rows and columns of a given matrix is known as transpose of the

1 0 Matrices

matrix. If A is given matrix its transpose is denoted by AT or A’. For example if A = 3 – 2 then its

transpose is AT = 1 3 5 . 5 4
0 – 2 4
Example 7:

If A = 2 3 and B = 0 – 1 then prove that (AB)T = BTAT.
–  1 4 2 3

 Here, A = 2 3 then AT = 2 – 1 and B = 0 – 1 then BT = 0 2 .
–  1 4 3 4 2 3 –  1 3

Now, AB = 2 3 0 – 1 = 0 + 6 – 2 + 9 = 6 7 and (AB)T = 6 8
–  1 4 2 3 0 + 8 1 + 12 8 13 7 13

Again BTAT = 0 2 2 – 1 = 0 + 6 0+8 = 6 8
–  1 3 3 4 – 2 + 9 1 + 12 7 13

∴ (AB)T = BTAT. EXERCISE : 3.2B

1. Find the product of the following matrices.

a. (2 3) 4 b. (2 3 4) 4 c. 10 2 4 d. 10 0 2 0
5 5 1 5 1 0 3

6

2. a. If A = 2 0 3 1 0
1 4 0 and B = 2 4 , find AB and BA.

3 – 1

b. If A = 2 3 and B = a b and AB = 4 8 , find the values of a and b.
1 4 0 2 2 9

3. Find the value of x:

a. (x – 3) 2 = (4) b. (3 x) 2x = (20)
4 4

c. 10 0 x+1 = 2 d. 25 ––  21 –y 3 x = – 8 6
– 1 y – 2 3 – 4 –   19 13

e. 2b a – 2 = 6 f. –0 2 –0 5 yx = –– 1 20
4 5 0

Operation of Matrices ~ 89

4. a. If A = 3 4 and B = 3 – 4  , then prove that AB is an identity matrix.
2 3 –  2 3

b. If P = 2 m 7 ,Q= 9 n and PQ is an identity matrix, then find the values of m and n.
5 9 –  5 4

5. If P = 1 3 1 4
1 4
and Q = 0 – 2  , find
a. PT 3 5 c. PQT
b. QT

d. PTQT e. (PQT)T f. (QP)T

6. If A = 1 2 ,B= 5 – 2 and C = 1 3  , then show that:
3 0 0 3 6 – 2

a. AB ≠ BA b. (AB)C = A(BC)
c.
7. a. A(B + C) = AB + AC d. B(C – A) = BC – BA

b. If A = 4 2 1 and B = 2 – 2
3 4 – 5 0 3 , find AB. Does BA define? If BA is defined, find BA.
1 – 4
Are AB and BA equal?

If A = 0 4 ,B= 1 4 and C = –  2 5 , then show that AB = AC.
0 6 3 – 5 3 – 5

8. In each of the following cases, prove that AB = BA = I.

a. A= 7 4 and B = 3 – 4 b. A= 11 – 4 and B = 3 4
5 3 – 5 7 –  8 3 8 11

9. Solve the matrix equations if X is a matrix of order 2×1.

a. 12 3 X= – 4 b. –5 4 07 X = 49
4 – 7

10. a. Which matrix pre-multiplies to the matrix 1 1 give matrix 4 6 ?
3 4 5 2

b. Which matrix pre-multiplies to the matrix 1 – 2 to get the matrix (4 – 1)?
2 3

11. If A = 1 3 and B = – 2 – 3 , show that:
5 6 0 4

a. (AT)T = A b. (A + B)T = AT + BT c. (AB)T = BTAT d. (2A)T = 2AT

12. a. If A = 4 2 , find A2 and show that A2 – 5A + 6I = O, where I and O are the identity and the
–  1 1

zero matrices of order 2×2.

b. If B = 2 2 , find B2 and show that B2 – 5B + 8I = O, where I and O are the identity and the
1 2

zero matrices of order 2×2.

c. If A = 1 4 and B = –  2 4 then show that A2 – 8B – 17I = O where I and O are unit and
–  2 5 – 2 0

zero matrices of order 2×2.



90 ~ Perfect Optional Mathematics Class 9

4 CO-ORDINATE GEOMETRY

Curriculum Contents Included Periods

a. Locus and its introduction 20
b. Division of line segment in the given ratio
c. Equation of straight lines Co-ordinate
- Parallel to axes
- Slope - intercept form
- Intercepts form
- Perpendicular form
- Reduction to standard forms
- Slope point form
- Two points form
d. Distance between a point and a line
e. Area of triangle and quadrilateral

Objectives

On completion of this unit, students will be able to:

solve the equation of locus of a moving point under the given condition.

find the point equidistant from given three points.

find the co-ordinates of point which divides the given line segment in given ratio.

find the mid-point of a line segment joining the given points.

obtain the equation of straight lines in slope intercept form, double intercepts form, perpendicular form,
point slope form and two points form.

reduce the given equation in slope intercept form, intercepts form and perpendicular form.

find the distance between a straight line and a point.

solve the problems related to area of triangle and quadrilateral.

Materials

Y Y Y YY

B B (0,b) M B B (x2,y2)

c b p (x1,y1) (x1,y1)

X' q X X' O (a,0) X X' O a X X' q X X' O A X

AO aA AO

Y' Y' Y' Y' Y'

y = mx + c x + y = 1 x cos a + y sin a = p y – y1 = m(x – x1) y – y1 = xy22  ––  xy11 (x – x1)
a b

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Co-ordinate 1 mark each 5 marks each Questions Marks
2 marks each 4 marks each
2 1 6 15
21

 ~ 91

4.1 Section Formula

Introduction

In this section, we can find the co-ordinates of a point which divides internally or externally the given line
segment joining the given two points in the given ratios as follows.

Internal Division

To find the co-ordinates of a point which divides internally the line joining two points (x1, y1) and (x2, y2) in
the given ratio m1:m2.

Let P(x1, y1) and Q(x2, y2) be the given two points and the point R(x, y) divides PQ in the ratio m1:m2.

i.e. PR = m1 Y Q(x2,y2)
RQ m2 y2 – y
m2
Draw perpendiculars PA, RB and QC on X– axis and PE⊥RB and R(x,y)

RD⊥QC. m1 x2 – x D
(x1,y1) y – y1
Here, OA = x1, PA = y1
OB = x, RB = y P E
x – x1

OC = x2, QC = y2 and X' O A BC X
PE = AB = OB – OA = x – x1

RD = BC = OC – OB = x2 – x Y'

RE = RB – BE = RB – PA = y – y1

QD = QC – DC = QC – RB = y2 – y

Now, ∆PRE and ∆RQD are similar as they are equiangular triangles. The corresponding sides of similar

triangles are proportional.

i.e. PR = PE = RE
RQ RD RQ

i.e. m1 = x – x1 = y – y1
m2 x2 – x y2 – y
m1 x – x1
From the first and the second ratio, m2 = x2 – x

or, m2x – m2x1 = m1x2 – m1x

or, m1x + m2x = m1x2 + m2x1

or, x = m1x2 + m2x1
m1 + m2

Similarly, we can have, y = m1y2 + m2y1
m1 + m2

Hence, the co-ordinates of point R is

(x, y) = m1x2 + m2x1 , m1y2 + m2y1
m1 + m2 m1 + m2

92 ~ Perfect Optional Mathematics Class 9

Note: 1. Among given two points the first given point must be considered as (x1, y1) and next point as (x2, y2).
AP
2. If A(x1, y1) and B(x2, y2) are two points and P(x, y) is a point on line joining A and B, then the ratio PB

can be given by either m1 = x – x1 or m1 = yy2––yy1.
m2 x2 – x m2


Cor 1. If m1 = m2 then R(x, y) is the mid-point of PQ.

Here, for the co-ordinates of the mid-point R(x, y) of the line Q(x2,y2)

joining two points P(x1, y1) and Q(x2, y2) is R(x,y)

R(x, y) = m1x2 + m2x1 , m1y2 + m2y1 P(x1,y1)
m1 + m2 m1 + m2

= m1x2 + m1x1 , m1y2 + m1y1
m1 + m1 m1 + m1

= m1(x2 + x1), m1(y2 + y1) Co-ordinate
2m1 2m1

(x, y) = x1 + x2, y1 + y2
2 2

External Division

To find the co-ordinates of a point which divides externally the line joining two points (x1, y1) and (x2, y2)
in the given ratio m1:m2.

Let P(x1, y1) tahnedfQig(uxr2e,,yi2.)e.bQePRtRhe=gmmiv12en two points. The point R(x, y) divides PQ externally in the ratio m1:m2
as shown in Y

Draw perpendiculars PA, QB and RC on X– axis and again draw R(x,y)

PN⊥RC which cuts QB at M. Q(x2,y2) m1

Here, OA = x1, PA = y1 m2 y2 – y
OB = x2, QB = y2
OC = x, RC = y and (x1,y1) y 2 – y1 N
P
x2 – x1 M

x – x1

PM = AB = OB – OA = x2 – x1 X' O A BC X

PN = AC = OC – OA = x – x1 Y'

QM = QB – MB = QB – PA = y2 – y1

RN = RC – NC = RC – PA = y – y1

Now, ∆PRN and ∆PQM are similar as they are equiangular triangles. The corresponding sides of similar

triangles are proportional.

i.e. PPQR = PN = RN i.e. m1 = x – x1 = y – y1
PM QM m1 – m2 x2 – x y2 – y1

Section Formula ~ 93

From the first and the third ratio,

m1m–1m2 = y – y1
y2 – y1

or, y(m1 – m2) – m1y1 + m2y1 = m1y2 – m1y1

or, y(m1 – m2) = m1y2 – m2y1

or, y = m1y2 – m2y1
m1 – m2

Similarly, x = m1x2 – m2x1
m1 – m2

Hence, the co-ordinates of R is

(x, y) = m1x2 – m2x1 , m1y2 – m2y1
m1 – m2 m1 – m2

Note: The point R divides line segment PQ in the ratio – 2:3 has same meaning as in the ratio 2:3 externally.

Example 1:

Find the co-ordinates of the point which divides the line joining the points (– 3, – 2) and (2, 8) in
the ratio 3:2

a. internally b. externally

 a. Let R(x, y) divides the line joining given two points A(– 3, – 2) and B(2, 8) in the ratio 3:2

internally. Then the co-ordinates of R is given by B(2, 8)

(x, y) = m1x2 + m2x1 , m1y2 + m2y1 R(x, y) 2
m1 + m2 m1 + m2
3
Here, m1 : m2 = 3 : 2, (x1, y1) = (– 3, – 2), (x2, y2) = (2, 8)

(x, y) = 3 × 2 + 2 × (– 3), 3 × 8 + 2 × (– 2) A(– 3, – 2)
3 + 2 3 + 2

= 0 , 20 = (0, 4)
55

Hence, the required co-ordinates are (0, 4).

b. Let M(x, y) divides the line joining given two points A(– 3, – 2) and B(2, 8) in the ratio 3:2 externally.

Then the co-ordinates of M is

(x, y) = m1x2 – m2x1 , m1y2 – m2y1 R(x, y)
m1 – m2 m1 – m2
3

Here, m1 : m2 = 3 : 2, (x1, y1) = (– 3, – 2), (x2, y2) = (2, 8) 2

(x, y) = 3 × 2 – 2 × (– 3) , 3 × 8–2× (– 2) B(2, 8)
3 – 2 3–2
A(– 3, – 2)

= 6 + 6, 24 + 4 = 12, 28 = (12, 28)
1 1 1 1

Hence, the required co-ordinates are (12, 28).

94 ~ Perfect Optional Mathematics Class 9

Example 2:

Find the mid-point between (2, 4) and the origin.

 Let A(2, 4) = (x1, y1) and the origin O(0, 0) = (x2, y2) and B(x, y) be the mid-point of OA then by using

mid-point formula,

B(x, y) = x1 + x2, y1 + y2
2 2

= 2 + 0, 4 + 0 = 22, 4 = (1, 2)
2 2 2

\ Mid-point of OA is B(1, 2).

Example 3:

If Q(2, 4) is the mid-point of the line segment joining the points P(a, b) and R(– 1, – 3) then find the

co-ordinates of point P. Co-ordinate

 Let P(a, b) = (x1, y1) and the origin R(– 1, – 3) = (x2, y2) and Q(2, 4) be the mid-point of PR then by

using mid-point formula,

x = x1 + x2 and y = y1 + y2
2 2
or,
or, 2 = a – 1 or, 4 = b – 3
2 or, 2
\ 8=b–3
or, 4 = a – 1

or, 4 + 1 = a 8+3=b

\ a = 5 b = 11

\ The co-ordindates of point P is (5, 11).

Example 4:

If A(4, 5), B(0, 3) and C(2, – 1) are the vertices of a triangle ABC, find the length of the median

drawn through the first vertex.

 Let D(x, y) be the mid-point of side BC of triangle ABC having vertices at the points A(4, 5), B(0, 3)

and C(2, – 1). A(4, 5)

The co-ordinates of mid-point D of BC is

x1 + x2, y1 + y2 = 0 + 2, 3 – 1
2 2 2 2

i.e. (1, 1) B(0, 3)
For the length of the median AD, A(4, 5) = (x1, y1) and D(1, 1) = (x2, y2), D(x, y)
d = (x2 – x1)2 + (y2 – y1)2
= (1 – 4)2 + (1 – 5)2 C(2, – 1)

= 9 + 16 = 5 units
Hence, the length of median drawn through the first vertex is 5 units.

Section Formula ~ 95

Example 5:

Find the ratio in which the line segment AB having A(– 5, 2) and B(2, 3) is divided by the Y-axis.
 Here, let the line segment AB having A(– 5, 2) and B(2, 3)
Y
is divided by the Y-axis at point C(0, b) in the ratio m1:m2.
Then considering A(– 5, 2) = (x1, y1), B(2, 3) = (x2, y2) and
C(0, b) = (x, y), we can use the section formula for x co-ordinate. B(2, 3)
C(0, b)
Alternately,
m1x2 + m2x1 A(– 5, 2)
x = m1 + m2 From the ratio of x co-ordinate X' X
O
m1 × 2+ m2 × (– 5) mm12 = x1 – x
or, 0 = m1 + m2 x – x2
Y'
or, mm12
or, 0 = 2m1 – 5m2 = – 5 – 0
0–2

or, 2m1 = 5m2 or, mm21 = – 5
– 2
m1 5
\ m2 = 2 \ m1 = 5
m2 2

Example 6:

If the points A(6, 4) , B(a, 2) and C(1, – 1) are three points on the same straight line. Find the ratio

in which the point B divides the line segment AC and also find the value of a.

 Suppose the point B(a, 2) divides the line segment joining A(6, 4) and C(1, – 1) in the ratio k : 1.

So, taking A(6, 4) = (x1, y1), C(1, – 1) = (x2, y2) and B(a, 2) = (x, y), the y co-ordinate of B is

y = m1y2 + m2y1 C(1, –  1) Now, the x co-ordinate of point B is
m1 + m2 B(a, 4)
m1x2 + m2x1
or, 2 = k × (– 1) + 1 × 4 x = m1 + m2

k+1

or, 2k + 2 = – k + 4 A(6, 4) or, a = 2 × 1 + 3 × 6 = 2 + 18
or, 2k + k = 4 – 2 2 + 3 5

or, 3k = 2 or, a = 20
5
2
∴ k = 3 ∴a=4

Hence, the required ratio is m1:m2 = k : 1 = 23 : 1 = 2:3.

Example 7:

Prove that the points (– 3, 1), (– 2, 3), (4, 4) and (3, 2) are the vertices of a parallelogram.
 Let the given four points are A(– 3, 1), B(– 2, 3), C(4, 4) and
C(4, 4)

D(3, 2). As the quadrilateral having diagonals bisecting each other B(– 2, 3)

is a parallelogram, if diagonals AC and BD bisect each other, then

ABCD will be a parallelogram.

Here, the mid-point of diagonal AC is x1 + x2, y1 + y2 D(3, 2)
2 2
– 3 + 4, 1 + 4 12, 5
2 2 i.e. 2 A(– 3, 1)

96 ~ Perfect Optional Mathematics Class 9