Infinity Optional mathematics book 9 Final for CTP 2077

–∞ A H B∞
D E FGC

In the above number line, AB = 12cm. The points C, D, E, F, G, H ...... are the
mid points of AB, AC, DC, EC, FC, GC, ...... respectively. So, the length of the line
segments AB, AC, DC, EC, FC, GC ...... form a sequence.

12, 6, 3, 32, 43, 38, ................
In decimal form: 12, 6, 3, 1.5, 0.75, 0.375, ......., .......

In this way, the distance between final mid point and the mid point C almost equals

to zero. Hence, the limiting value of the sequence 12, 6, 3, 32, 43, 3 ...... approaches
8

to zero. Here, the limiting value '0' means nearly equals to '0' but not exactly equals

to '0'. From the above example, it is concluded that

The final term of any sequence of numbers will approach to any certain number
but not exactly equals to that number. That certain number is called the limiting
value of the respective sequence. In a number line, the real numbers in the right
side approach to (+ ∞) but not exactly equals to (+ ∞) and the real numbers in the
left side approach to ( – ∞) but not exactly equals to (–∞).

WORKED OUT EXAMPLES

1. From the sequence given below. 51
1.9, 1.99, 1.999, 1.9999, ......, ......
(a) Find out the 8th term of the sequence.
(b) What is the limiting value of the sequence?
Solution:
Here, the given sequence is 1.9, 1.99, 1.999, 1.9999, ......, ......
(a) First term (t1) = 1.9
Second term (t2) = 1.99
Third term (t3) = 1.999
......... ......... ..........
8th term (t8) = 1.99999999

Infinity Optional Mathematics Book - 9

∴ 8th term of the given sequence is 1.99999999
(b) Since the number of 9 increases after decimal, the value nearly equals to 2 but

not exactly equals to 2. So, the limiting value of the sequence is 2.

2. From the given nth­ term, tn = 1 1 , find the first 6 terms. Also find
2n +
the limiting value of that sequence.

Solution: Here,

tn = 1 1
2n +

t1 = 2 × 1 + 1 = 1
1 3

t2 = 2 × 1 + 1 = 1
2 5

t3 = 2 × 1 + 1 = 1
3 7

t4 = 2 × 1 + 1 = 1
4 9

t5 = 2 × 1 + 1 = 1
5 11

t6 = 2 × 1 + 1 = 1
6 13

∴ The sequence formed from given tn is 31, 15, 71, 91, 111, 1 .......
13
Since the last term of the sequence approaches to 0, the limiting value of the

sequence is 0.

Exercise 2.1

Section 'A'
1. Define the limiting value of numerical sequence.
2. (a) What are the nearest whole number of each term of the sequence
0.2, 0.02, 0.002, 0.0002, ......, ...... ?
(b) What are the nearest whole number of each term of the sequence
3.9, 3.99, 3.999, 3.9999, ......, ...... ?

52 Infinity Optional Mathematics Book - 9

3. From the sequences given below.

(a) 0.01, 0.001, 0.0001, 0.00001, ......, ......

(b) 4.01, 4.001, 4.0001, 4.00001, ......, ......

(c) 2.9, 2.99, 2.999, 2.9999, ......, ......

(d) 9.9, 9.99, 9.999, 9.9999, ......, ......

(e) 150, 1050, 10500, 105000, ......, ......
(f) 12, 41, 18, 116, ......, ......
(i) Find the 8th term from the above sequences.

(ii) What is the limiting value of each of the above sequence?
Section 'B'

4. (a) Bisect the line segment of 10cm eight times successively. Show position
of mid points on a number line. Write the sequence formed by the eight
mid points and find its limiting value.

(b) Bisect the line segment AB of 8cm at 6 times towards B. Write the
sequence formed by the six mid points and find its limiting value.

Section 'C'
5. Find the first six terms of the sequences obtained by given nth term.

Also, find the limiting value of the sequence.

(a) tn = 1 (b) tn = 1
3n 2n2

(c) tn = n 1 1 (d) tn = 1 + 1 1
+ 3n +

(e) tn = 3 + (–1)n (f) tn = 1 + (–1)n
n2 n n

Infinity Optional Mathematics Book - 9 53

Limit of a Sequence of Figures

The figure given below show the regular polygons inscribed in a circle.

Fig. (i) Fig. (ii) Fig. (iii) Fig. (iv) Fig. (v)

Observe the above figures and discuss the following questions.

(a) Write the number of sides of the inscribed polygon of figures (i), (ii), (iii), (iv)
and (v).

(b) Find the area of shaded region from each of the above figures by measuring
the radius of the circle and side of each polygon.

(c) Compare the area of shaded regions.
(d) What happens to the area of shaded region as the number of sides of inscribed

polygon increases?
In the above figure, the number of sides of inscribed polygon in fig. (ii) is more than
in figure (i). So, the area of shaded region is less in fig. (ii) than in fig. (i). Similarly,
the number of sides of inscribed polygon in fig. (iii) is more than in fig. (ii). So, the
area of shaded region is less in fig. (iii) than in fig. (ii) and so on. That is when the
number of sides of inscribed polygon is increasing, the area of shaded region is
decreasing. If the number of sides of inscribed regular polygon is countable infinitely
large, then the area of shaded region is nearly equal to zero. So, the limiting value
of the area of shaded region is 0. In other way, the limiting value of the area of
inscribed regular polygon is the area of circle.

Exercise 2.2

Section 'B'

1. In the figure alongside, the inscribed triangle are
formed by joining the mid points of adjacent sides of
the external triangles. Then
(a) How many triangles can be made in this process?
(b) What is the limiting value of the area of triangle?
(c) What is the limiting value of the perimeter of triangle?

54 Infinity Optional Mathematics Book - 9

2. In the adjoining figure, the inscribed squares are formed by joining
the mid points of adjacent sides of the external squares.
Then

(a) How many squares are formed in this figure?
(b) What is the limiting value of the area of square?

(c) What is the limiting value of the perimeter of square?

3. In the given figure, three types of three parallel
lines are intersected each other. Then

(a) How many triangles are formed in this figure?
(b) When the number of parallel lines are increasing,
what is the limit of the number of triangles?

4. A glass has full of water. A man drinks half of water
at once in one drink, then
(a) How many times that the man can drink next half of
water?
(b) What amount of water remains in the glass at last?

Y

5. (a) In the given graph, when the value of x is increasing,

find

(a) difference of the values of x and y? O X
(b) limiting value of y. X

(b) In the given graph, when the value of x is increasing, find Y
(a) difference of the values of x and y?
(b) limiting value of y.

O

Infinity Optional Mathematics Book - 9 55

Limit of a sum of infinite series

In the figures given below AB = 2 units on real number line.

G

–∞ A C D EF B ∞


The points C, D, E, F, G ...... be the mid points of AB, CB, DB, EB, FB, ......

respectively. Upto how many times can the mid point be found out? Does the final

mid point approach to B or exactly equals to B? Discuss in group.

In the above figure,
AB = 2 units

AC = BC = 1 × 2 = 1 unit
2

CD = BD = 1 × 1 = 1 unit and AD = AC + CD = 1 + 1 units
2 2 2

DE = BE = 1 × 1 = 1 unit and AE = AC + CD + DE = 1 + 1 + 1 units
2 2 4 2 4

EF = BF = 1 × 1 = 1 units and AF = AC + CD + DE + EF = 1 + 1 + 1 + 1 units
2 4 8 2 4 8

FG = BG = 1 × 1 = 1 units and AG = AC + CD + DE + EF + FG
2 8 16

= 1 + 1 + 1 + 1 + 1 units
2 4 8 16

The infinite series formed by this process is 1 + 1 + 1 + 1 + 1 + .......
2 4 8 16

Here, S1 = AC = 1 unit

S2 = AD = 1 + 1 units = 1.5 units
2

S3 = AE = 1 + 1 + 1 units = 1.75 units
2 4

S4 = AF = 1 + 1 + 1 + 1 units = 1.875 units
2 4 8

S5 = AG = 1 + 1 + 1 + 1 + 1 units = 1.9375 units
2 4 8 16

Similarly, we can compute S6, S7, ........ Sn (bisects at n times) and Sn is nearly
equals to 2. So, 2 is the limiting value of the sum of the given infinite series.

If the absolute value of the ratio of a term and its preceding term of an infinite series
is less than 1, we can compute the limiting value of that series as a real number.

56 Infinity Optional Mathematics Book - 9

For example, 4 – 2 + 1 – 1 + 1 – 1 + ......... 1
Here the common ratio is – 2 2 4 1 8 2
4 =– 2 and its absolute value is which is less than

1. So, the limiting value of that series is a real number.

If the absolute value of the ratio of a term and its preceding term of an infinite
series is more than 1, the limiting value of the series is not a real number.

For example, 1 + 2 + 4 + 8 + 16 + ........ . Here the common ratio is 2 = 2 which is
1
greater than 1. So, the limiting value of that series is not a real number.

Consider an infinite series of sequence 0.32 as 0.32 + 0.0032 + 0.000032 + .....

or, 32 + 32 + 32 + ....... or, 32 + 32 + 32 + .......
100 10000 1000000 102 104 106

or, 32 × 10-2 + 32 × 10-4 + 32 × 10-4 + 32 × 10-6 + .......

The last term of the series is nearly equal to 0. So,the limiting value of the last term

of that infinite series is 0.

WORKED OUT EXAMPLES

1. In the figure given below, there is a square having a side 2cm.

2cm 2cm 2cm 2cm

2cm 2cm 2cm 2cm
Fig. (i) Fig. (ii) Fig. (iii) Fig. (iv)

(a) Write down the sequence formed by the area of shaded portion.

(b) Find the limiting value of the shaded portion of the last square.

Solution :

Here,side of a square (a) = 2cm

Area of the square (A) = a2 = 22 = 4cm2

Area of shaded portion in fig. (i) = 1 × 4 = 2cm2
2
1
Area of shaded portion in fig. (ii) = 2 + 2 × 2 = 2 + 1 = 3cm2

Area of shaded portion in fig. (iii) = 2 + 1 + 1 × 1 = 2 + 1 + 1 = 7 cm2
2 2 2
1 1 1 1 1 15
Area of shaded portion in fig. (iv) = 2 + 1 + 2 + 2 × 2 = 2 + 1 + 2 + 4 = 4 cm2

Infinity Optional Mathematics Book - 9 57

(a) The sequence of the area of the shaded portion is 2, 3, 27, 145, .......
(b) By continuing this process, the area of shaded portion is nearly equal to the

area of the square but does not exceed the area of the square. So, the limiting
value of the shaded portion is 4cm2.


Exercise 2.3

Section 'A'

1. Which of the following series has the limiting values as a real number?

(a) 1 + 3 + 6 + 12 + ....... (b) 31 + 16 + 112 + 1 + .......
24

(c) 9 + 3 + 1 + ....... (d) 32 – 16 + 8 – 4 + .......

(e) 6 + 30 + 150 + ....... (f) 10 – 1 + 1 – 1 + .......
10 100

(g) 0.2 + 0.02 + 0.002 + ....... (h) 0.45

Section 'B'

2. Find the limiting value of each of the following infinite series

(a) 0.3 + 0.03 + 0.003 + .... (b) 0.24

(c) 6.13 (d) 1 + 4 + 16 + 64 + ...

(e) 8 + 4 + 2 + 1 + 1 + ... (f) 10 + 1+ 1 + 1 + ...
2 10 100

Section 'C'

3. From the figures given below.
(i) Write down the sequence formed by the area of the shaded portion.
(ii) Find the limiting value of the shaded portion of the last figure.

(a)

4cm 4cm 4cm 4cm

4cm 4cm 4cm 4cm
Fig. (i) Fig. (ii) Fig. (iii) Fig. (iv)

(b) 6cm

6cm 6cm 6cm

2cm 2cm 2cm 2cm
Fig. (i) Fig. (ii) Fig. (iii) Fig. (iv)

58 Infinity Optional Mathematics Book - 9

Limit of a Function

Review

For what value of x, the following functions are undefined?

(a) f(x) = 2x (b) f(x) = 1x + 1 (c) f(x) = x+2
x–1 x2 x2 – 4

(d) f(x) = x2 – 9 (e) f(x) = 1 (f) f(x) = 3 2
x + 3 5x 5x +

Note

If the value of the function is in the form of ∞ or ∞ or ∞ – ∞ or ∞ + ∞ or 0
∞ 0

or 00 etc. The function is said to be undefined.

Consider a function f(x) = x + 2. Complete the following table using

f(x) = x + 2.
x –2 –1 0 1 2 3

f(x) = y 0 ? ? ? ? ?

The function f(x) = x + 2 can be shown in mapping diagram and graph as follows.

Y
f

8

–2 0 7
–1 6
0
1 15
2
3 2 4
3

32

4 1
5 X' -2 -1 0 1 2 3 4 5 6 7 8 X

Y'

In the function, the variable x is called 'input' and f(x) or y is called 'output'
which is the value of the function.

Infinity Optional Mathematics Book - 9 59

x2
x
Consider a function f(x) = – 9 . In this function can Y
– 3
0X
you find the value of f(3)? Let's try, put x = 3 then, f(3) = Y'
32
3 –9 = 0 which is called indeterminate form. Y
–3 0

Discuss if x ≠ 3, can you find the value of f(x)?
Consider a function y = f(x) = x2.

Then,
x –3 –2 –1 0 1 2 3
y9410149

The graph of y = x2 is shown alongside. X'

Study the following figures and discuss the following
questions.

YY

DC

DC

DC

AB

AB

X' AB X X' 0 XX 0 X
0

Y' Y' Y'

Fig. (i) Fig. (ii) Fig. (iii)

In figure (i), taking x1 = 1 and x2 = 3. So, AB = 2 unit and BC = 8 units.

Calculate the area of rectangle ABCD.

In figure, (ii), taking x1 = 1.5 and x2 = 2.5. So, AB = 1 unit and BC = 4 units.
Calculate the area of rectangle ABCD.

In figure, (iii), taking x1 = 1.9 and x2 = 2.1. So, AB = 0.2 unit and BC = 0.8 units.
Calculate the area of rectangle ABCD.

60 Infinity Optional Mathematics Book - 9

Similarly, when the value of x1 and x2 are taken nearer to 2, what is the area of
rectangle ABCD? In this case, the area of rectangle ABCD approaches to 4. So the
limiting value of the function f(x) = x2 is 4 when x approaches to 2.

The limit of a function is the functional value of the fixed point in which x
component is nearer to that point.

WORKED OUT EXAMPLES

In

1. If f(x) = 2x – 1 then find the value of f(2) and f(1.999).

Solution: Here, f(x) = 2x – 1
put x = 2, f(2) = 2 × 2 – 1 = 4 – 1 = 3

put x = 1.999, f(1.999) = 2 × 1.999 – 1 = 3.998 – 1 = 2.998

2. Find the limit of the function f(x) = x2 –4 at x = 2.
x–2

Solution: x2 – 4
x–2
Here, f(x) =

When x = 2, the functional value f(2) does not exist.

So, put the values of x nearer to 2.
x 1.9 1.99 1.999 1.9999 2.1 2.01 2.001 2.0001
f(x) 3.9 3.99 3.999 3.9999 4.1 4.01 4.001 4.0001

The table shows that x approaches 2 from the left and the right. Then the

functional value f(x) also approaches 4 i.e. when x is nearer to 2, f(x) is also
x2 – 4
nearer to 4. Hence, the limit of the function f(x) = x–2 is 4 at x = 2.

Exercise 2.4

Section 'A'

1. (a) What is the value of a function?
(b) Define limit of a function.

Section 'B'

2. (a) If f(x) = 3x – 1 then find the value of f(3) and f(2.99).
(b) If f(x) = 2x +1 then find the value of f(2) and f(2.001).
3. (a) If f(x) = 4x then find the difference of f(2) and f(1.99).

Infinity Optional Mathematics Book - 9 61

(b) If f(x) = 2x2 + 1 then find the difference of f(3) and f(2.99).
4. (a) If f(x) = 3x + 1 then find f(3.999) and f(4.001) and express them in whole

number by rounding off.
(b) If f(x) = 5x – 2 then find f(4.99) and f(5.03). Express the values in whole

number and find their difference.

Section 'C'

5. (a) Given that f(x) = x2 – 1 . Then
x– 1

(i) Does f(1) exists? Give reason.

(ii) What are the values of f(x) if x = 0.9, 0.99, 0.999?

(iii) What are the values of f(x) if x = 1.1, 1.01, 1.001?

(iv) Can the values of f(x) that are calculated in (ii) and (iii) express in
whole number by rounding off?

(b) Given that g(x) = x2 – 4 . Then
x – 2

(i) Does f(2) exists? Give reason.

(ii) What are the values of g(x) if x = 2.1, 2.01, 2.001?

(iii) What are the values of g(x) if x = 1.9, 1.99, 1.999?

(iv) Can the values of g(x) that are calculated in (ii) and (iii) express in

whole number by rounding off?

6. Find the limit of the following functions.

(a) f(x) = 2x + 3 at x = 0 (b) f(x) = 3x – 1 at x = 2

(c) f(x) = x2 – 9 at x = 3 (d) f(x) = x2 – 16 at x = – 4
x–3 x + 4

Notational representation of limit

Consider a function f(x) = x + 4 and fill the table given below.

x 1.9 1.99 1.999 1.9999 2 2.1 2.01 2.001 2.0001

f(x) .......... .......... .......... .......... .......... .......... .......... .......... ..........

From the above table, discuss the questions given below.

(a) Express that value of f(x) in a whole number at x = 1.999.
(b) Express that value of f(x) in a whole number at x = 2.001.

62 Infinity Optional Mathematics Book - 9

(c) What is the value of f(x) at x = 2?

From the above table, when the values of x are nearer to 2 from the left or right, the
value of f(x) approaches to 6.

If x approaches to or x tends to 'a' then f(x) approaches to f(a) or f(x) tends to
f(a). Symbolically, lim x → a, f(x) = f(a) which is read as limit x tends to a, f(x)
equals to f(a).

WORKED OUT EXAMPLES

1. Write the notation of the mathematical sentence x tends to –1.

Solution:
The notation of the mathematical sentence x tends to – 1 is x → – 1.

lim
2. Write the mathematical sentence for the notation x → 2 f(x) = 4.
Solution:

The mathematical sentence for the notation lim f(x) = 4 is limit x tends to
2, f(x) equals to 4.
x→2

lim
3. Evaluate : x → 2 (2x + 5).
Solution: Here,

lim
x → 2 (2x + 5) in which f(x) = 2x + 5 as x → 2.


The values of x nearer to 2 from the left are shown in the given below.
x 1.9 1.99 1.999 1.9999
f(x) 8.8 8.98 8.998 8.9998

The values of x nearer to 2 from the right are shown in the given below.
x 2.1 2.01 2.001 2.0001
f(x) 9.2 9.02 9.002 9.0002

From the above both tables as x → 2 then f(x) → 9.
lim

Hence, x → 2 (2x + 5) = 9.

Infinity Optional Mathematics Book - 9 63

Exercise 2.5

Section 'A'

1. Write the notation of the following mathematical sentences.

(a) x tends to a (b) x approaches to 3.

(c) x is nearer to –2. (d) limit x tends to 1, f(x) equals to 3

(e) limit x tends to 'a', f(x) is equal to f(a)
x2 – 4
(f) limit of a function x–2 is 4 as x tends to 2.

2. Write the mathematical sentences for the following notation.

(a) x → 9 (b) x → a

(c) y → –3 (d) lim x → a, f(x) = f(a)

lim (f) lim x → 2, f(x)
(e) x → 0 (x2 + 3)

lim x2 – 9 = 6 lim
(g) x → 3 x–3 (h) x → 2 (3x) = 6.

Section 'B'

3. Complete the following table.

(a) x 1.9 1.99 1.999 2.1 2.01 2.001 x → .......
....... ....... ....... f(x) → .......
f(x) = 3x – 1 ....... ....... .......
1.1 1.01 1.001 x → .......

(b) x 0.9 0.99 0.999

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



(c) x 2.9 2.99 2.999 3.1 3.01 3.001 x → .......

f(x) = x2 – 9 ....... ....... .... ....... ....... f(x) → .......
x–3

Section 'C'

4. Evaluate

lim lim x+3
(a) x → 2 (2x – 1) (b) x → 1 2

(c) lim x2 – 1 lim x2 – 4
x → -1 x + 1 (d) x → 2 x–2

lim x2 ; x < 2

{ (e) x → 2 f(x) where f(x) = 6 – x; x > 2



64 Infinity Optional Mathematics Book - 9

UNIT

3 MATRIX

3.1 Matrix

Introduction to Matrix

Table given below shows the marks obtained by three students Ram, Sita and Gita
in Nepali, English, Mathematics and Science.

Name of Nepali Subjects Science
Students 70 English Mathematics 80
78 90
Ram 60 95 85 70
Sita 85 96
Gita 73 65

Lets express the above information in the rectangular form by omitting the name of
students and subjects and enclosed by round or squared brackets as shown below.



70 95 85 80 70 95 85 80

78 85 96 90 or 78 85 96 90
60 73 65 70 60 73 65 70

In the above rectangular form, the numbers are arranged in horizontal lines (called

rows) and in vertical lines (called columns). Such rectangular arrangement of

numbers is called a matrix. For example,



70 95 85 80 1st row (R1)
78 85 96 90 2nd row (R2)
60 73 65 70 3rd row (R3)

1st 2nd 3rd 4th

Column Column Column Column

(C1) (C2) (C3) (C4)

A rectangular array of numbers arranged in rows and columns and enclosed by
round (or squared) bracket is called as a matrix.

Infinity Optional Mathematics Book - 9 65

Notation of Matrix

Matrices are denoted by capital letters of English alphabet as A, B, C, ....., X, Y, Z

and their elements by the small letters of English alphabet as a, b, c, ....., x, y, z.

For example, P = a b c
d e f

The elements of the matrix can also be denoted by double suffix notation so that its

position can also be easily located. For example,

A= a11 a12 a13 element of a matrix
a21 a22 a23 aij

ith row jth column

where,

a11 is the element in the 1st row and 1st column.
a12 is the element in the 1st row and 2nd column.
a13 is the element in the 1st row and 3rd column.
a21 is the element in the 2nd row and 1st column.
a22 is the element in the 2nd row and 2nd column.
a23 is the element in the 2nd row and 3rd column.

Order of matrix

Consider a matrix A = 123 .
456
789

In the matrix A, there are 3 rows and 3 columns.

So, the order of matrix A is 3 × 3. It is written as A3 × 3 = a11 a12 a13
a21 a22 a23
a31 a32 a33
3 × 3.

The order of matrix is obtained as no. of rows of the matrix × no. of columns of
the matrix.

For examples, P = 1 –3

5 0 and Q = ( 2 0 –3).
24

In the matrix P, there are 3 rows and 2 columns. So, the order of P is 3 × 2 i.e. P3×2.
In the matrix Q, there are 1 row and 3 columns. So, the order of Q is 1 × 3 i.e. Q1×3.


66 Infinity Optional Mathematics Book - 9

WORKED OUT EXAMPLES

1. If B = 123 then
4 –3 0

(a) What is the order of B?

(b) Find the number of elements of B.

(c) Find the values of b11, b13, b22 and b23.
Solution:

(a) The order of B is 2 × 3 i.e. B2×3
(b) The number of elements of B = 6.

(c) b11 = an element of 1st row and 1st column = 1
b13 = an element of 1st row and 3rd column = 3
b22 = an element of 2nd row and 2nd column = –3
b23 = an element of 2nd row and 3rd column = 0

2. Construct 2 × 3 matrix whose elements aij are given by aij = i + 2j
Solution:

Here, aij = i + 2j

Put i = 1 and j = 1. Then a11 = 1 + 2 × 1 = 3

Put i = 1 and j = 2. Then a12 = 1 + 2 × 2 = 5

Put i = 1 and j = 3. Then a13 = 1 + 2 × 3 = 7

Put i = 2 and j = 1. Then a21 = 2 + 2 × 1 = 4

Put i = 2 and j = 2. Then a22 = 2 + 2 × 2 = 6

Put i = 2 and j = 3. Then a23 = 2 + 2 × 3 = 8
3 5 7
∴ The required 2 × 3 matrix is 4 6 8

Exercise 3.1

Section 'A'

1. (a) Define matrix with an example.

(b) What is order of matrix?

(c) Find the number of elements of A = 1 2 3 .
4 5 6

2. Write down the order of the following matrices.

(a) (1 2 3) (b) –3 (c) 23 52
8

Infinity Optional Mathematics Book - 9 67

12 (e) 5 0 1 (f) a b c
3. 3 3 2 d e f
(d) 3 3 g h i
-2 1 -3 5 , find a11, a12, a22 and a21.
3 -6
(a) Let A =

(b) Let B = 3 4 8 , find b12, b21, b22 and b23.
4 2 1

4. Construct the matrices of the following order.

(a) 2 × 1 (b) 1 × 3 (c) 2 × 2

(d) 3 × 2 (e) 2 × 3 (f) 3 × 3

Section 'B'

5. Construct 2 × 3 matrix A whose elements aij are given by

(a) aij = i + j (b) aij = i.j

(c) aij = 2i + j (d) aij = 3i – 2j

(e) aij = (–i) j (f) aij = (–i)2 + j

Type of Matrices

Row matrix
A matrix having only one row is called a row matrix.

For example, P = (a b)1×2, Q = (1 3 5)1×3 are row matrices.

Column matrix
A matrix having only one column is called a column matrix.

p -3
q
For example, A = 2 × 1, B = 2 3 × 1 are column matrices.
5

Rectangular matrix

A matrix in which the number of rows and number of columns are not equal, is
called a rectangular matrix.

For example,

1 1 2 3 30
4 5 6
A = (a b)1×2, B = 2 3×1, C = 2×3, D = 4 -1 3×2 are the rectangular
3 5 3

matrices.

68 Infinity Optional Mathematics Book - 9

Square matrix
A matrix in which the number of rows and the number of columns are equal, is
called square matrix.

For example,

3 4 abc
6 2
P = (2)1×1, Q = 2×2, R = d e f 3×3 are the square matrices.
g h i

Zero or Null matrix
A matrix with its all elements zero is called a zero or null matrix. For example,

O = (0)1×1, O = 0 2×1, O = 0 0 2×2, O = 0 0 0 2×3 are the zero matrices.
0 0 0 0 0 0

Diagonal matrix

A square matrix in which the elements in the main diagonal are non zero and the
rest are zeros is called a diagonal matrix. For example,

1 0 400
0 2
A= 2×2, B = 0 5 0 3×3 are the diagonal matrices.
0 0 -3

Scaler matrix

A diagonal matrix in which all the diagonal elements are equal is called a scaler
matrix. For example,

3 0 200
0 3
P= 2×2, Q = 0 2 0 3×3 are the scaler matrices.
0 0 2

Unit or Identity matrix

A square matrix in which the elements in the main diagonal are 1 and the rest are
zero, is called an unit or identity matrix. For example,

1 0 100
0 1
I= 2×2, I = 0 1 0 3×3 are the unit or identity matrices.
0 0 1

Infinity Optional Mathematics Book - 9 69

Triangular matrix

A square matrix in which all the elements above or below the main diagonal are
zero, is called triangular matrix.

Upper triangular matrix

A square matrix in which all the elements below the main diagonal are zero, is
called upper triangular matrix. For example,

1 3 234
0 2
M= 2×2, N = 0 -3 1 2×3 are the upper triangular matrices.
0 0 5

Lower triangular matrix

A square matrix in which all the elements above the main diagonal are zero, is
called lower triangular matrix. For example,

5 0 300
3 8
X= 2×2, Y = 4 8 0 3×3 are the lower triangular matrices.
3 9 2

Symmetric matrix

A matrix which remains same when its rows and columns are interchanged is called

symmetric matrix. In other word, a matrix A is said to be symmetric if aij = aji for all
values of i and j. For examples,

1 3 258
3 5
A= 2×2, B = 5 6 -4 3×3 are the symmetric matrices.
8 -4 1

Equal matrices

Two matrices are said to be equal if they are of same order and their corresponding
elements are equal. For examples,

If P = 2 4 2×2 and Q = 2 4 2×2 then P = Q.
Thus, 6 -3 6 -3

a c = 3 -2 if and only if a = 3, b = 0, c = -2 and d = 6.
b d 06

70 Infinity Optional Mathematics Book - 9

WORKED OUT EXAMPLES

1. If a + 2 2b – 1 is an identity matrix then find the values of x and y.
4c – 6 d + 2

Solution:

Here, a + 2 2b – 1 is an identity matrix.
4c – 6 d + 2

So, a + 2 2b – 1 = 1 0
4c – 6 d + 2 0 1

Equating the corresponding elements, we get,

a+2=1 2b – 1 = 0 4c – 6 = 0 d+2=1

∴ a = –1 ∴ b = 1 ∴ c = 6 = 3 ∴ d = –1
2 4 2

Here, the required values are a = -1, b = 21, c = 3 and d = -1.
2
p+q 8 68
2. If -6 p–q = -6 2 then find the values of p and q.

Solution: Here,

p+q 8 = 68
-6 p–q -6 2

Equating the corresponding elements, we get,
p + q = 6 ..... (i)
p – q = 2 ..... (ii)
Adding equations (i) and (ii), we get,
p + q + p – q = 6 + 2
or, 2p = 8
∴ p = 4.
Putting the value of p in equation (i), we get,
4 + q = 6
∴ q = 2
Hence, p = 4 and q = 2

Infinity Optional Mathematics Book - 9 71

Exercise 3.2

Section 'A'

1. Define the following types of matrix with an example.

(a) Rectangular matrix (b) Scaler matrix

(c) Unit matrix (d) diagonal matrix

(e) lower triangular matrix (f) symmetric matrix

2. State the type of the following matrices.

(a) (1 -3 4) (b) -5 (c) 00 00 00
3

(d) 3 0 (e) 19 -53 (f) 03 92
0 5

100 500 2 3 -7

(g) 0 1 0 (h) 6 -2 0 (i) 3 -1 4

001 081 -7 4 8

3. State which of the following matrices are equal.

A= 1 -4 ,B= 5 ,C= 2 ,D= 3 7 9
3 2 2 5 1 2 4

E= 2 0 -1 ,F= 1 -4 ,G= 3 9 7 ,H= 2 0 -1
1 2 5 3 2 1 4 2 1 25

4. Construct the following matrices.

(a) 2 × 2 unit matrix (b) 2 × 2 symmetric matrix
(c) 3 × 2 zero matrix (d) 3 × 3 scaler matrix
(e) 3 × 3 unit matrix (f) 3 × 3 upper triangular matrix

Section 'B'

5. (a) Construct 2 × 2 square matrix if aij = 2i – j.
(b) Construct a rectangular matrix of order 2 × 3 if aij = (i + j)2.

6. Find the values of x and y in each of the following cases.

(a) 42xy = -2 (b) (4x – 3 3) = (9 3y – 4)
8

72 Infinity Optional Mathematics Book - 9

(c) 2xx+–y6 = 4 (d) 2x9+ 5 y –6 8 = 1 6
-2 9 -2

7. (a) If 2 x+5 is a symmetric matrix, find the value of x.
2x + 1 6

5 2x – 1 6

(b) If –3 1 4 is a symmetric matrix, find the values of x and y.
6 2x + y –4

8. (a) If 3a – 2 0 is an identity matrix, find the values of a and b.
0 a + 2b

(b) If a–3 0 is a zero matrix, find the values of a and b.
2b + a 0

(c) If 5 2x – 1 is a scaler matrix, find the values of x and y.
3 – 2y 5

Section 'C'

9. Find the values of a, b, c and d if

(a) 2ac––53 4b – 8 = 1 4 (b) a – 1 3 = –5 a+c
d+2 -7 2 5 b d–b 2

(c) 23aa + b = 8 (d) 2a–+2 b 3a6+ b = 16
– b 7 -2 0

(e) 22a–+2ab 4 2b + c = 4 4 b+3
a–c 0 0 –4 d+2

Operations of Matrix

Here, we discuss mainly the four operations of matrices. They are as follow:
(i) Addition of matrices
(ii) Subtraction of matrices
(iii) Multiplication of matrix by a scaler
(iv) Multiplication of matrices.

Addition of matrices
A table given below shows the marks obtained by Sita and Gita in mathematics and
Science of first and second term.

Infinity Optional Mathematics Book - 9 73

First term Second term Total
Sita
Sita Gita Sita Gita 184 Gita
175 185
Mathematics 95 92 89 93 171
Science
83 86 92 85

The above information can be arranged in matrix form as

Math Sita Gita Sita Gita Sita Gita
Science 95 92 89 93 184 185 Math
83 86 + 92 85 = 175 171 Science

Consider two matrices A = 1 5 3 2×3 and B = 1 2 -4
2 -3 6 1 3 2 2×3

Then A + B = 1 5 3 + 1 2 -4 = 1+1 5+2 3–4 = 2 7 -1
2 -3 6 1 3 2 2+1 -3 + 3 6+2 3 0 8 2×3

Let A and B be two matrices of same order. Then the sum of A and B is a new
matrix whose each element is the sum of corresponding elements of A and B. It
is denoted by A + B.

Subtraction of matrices

The table given below shows the income and expenditure of Ram and Shyam for
the months.

Income Expenditure Saving

Ram Shyam Ram Shyam Ram Shyam

Baisakh 15000 14500 5000 8000 10000 6500

Jestha 18000 16000 9000 5500 9000 10500

The above information can be arranged in matrix form as

Baisakh Ram Shyam Ram Shyam Ram Shyam Baisakh
Jestha 15000 14500 5000 8000 10000 6500 Jestha
18000 16000 – 9000 5500 = 9000 10500

583 461

Consider two matrices M = 9 2 6 3×3 and N = 8 5 3 3×3
6 8 2 3 8 9

Then,

583 461 5–4 8–6 3–1 122

M– N= 926 – 853 = 9–8 2–5 6–3 = 1 -3 3 3×3
682 389 6–3 8–8 2–9 3 0 -7


74 Infinity Optional Mathematics Book - 9

Let A and B be two matrices of same order. Then the difference of A and B is a
new matrix whose each term is obtained by subtracting the element of B from
the corresponding elements of A. It is denoted by A – B.

Properties of Matrix Addition

Closure property

Consider two matrices A = 1 2 2×2 and B = 2 4
5 -3 -3 1 2×2

Then A + B = 1 2 + 2 4 = 3 6 2×2 which is a 2 × 2 matrix.
5 -3 -3 1 2 -2

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

Commutative property

Consider two matrices A = 2 5 and B = 8 9
–1 3 3 2

Then, A + B = 2 5 + 8 9 = 2+8 5+9 = 10 14
–1 3 3 2 –1 + 3 3+2 2 5

B+A = 8 9 + 2 5 = 8+2 9+5 = 10 14
3 2 –1 3 3–1 2+3 2 5

∴ A + B = B + A
If A and B are two matrices of the same order then A + B = B + A.

Associative property

Consider three matrices A = 5 -3 ,B= 1 2 and C = 3 –2
2 1 3 4 76

Then, A + B = 5 –3 + 1 2 = 6 –1
2 1 3 4 55

(A + B) + C = 6 –1 + 3 –2 = 9 –3
5 5 7 6 12 11

Again, B + C = 1 2 + 3 –2 = 40
3 4 7 6 10 10

A + (B + C) = 5 -3 + 4 0 = 9 –3
2 1 10 10 12 11

∴ (A + B) + C = A + (B + C) Infinity Optional Mathematics Book - 9 75

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

Identity property

Consider two matrices A = 1 2 and O = 0 0 . Then
3 4 0 0

A+O= 1 2 + 0 0 = 1 2 =A
3 4 0 0 3 4

O+A= 0 0 + 1 2 = 1 2 =A
0 0 3 4 3 4

∴ A + O = O + A = A.

If A is a matrix of any order and O is the null matrix of same order as A then
A+O=O+A=A

Inverse property

Consider two matrices A = 2 5 . Then –A = –2 –5 . Then
6 9 –6 –9

A + (–A) = 2 5 + –2 –5 = 0 0 =O
6 9 –6 –9 0 0

(–A) + A = –2 –5 + 2 5 = 0 0 =O
–6 –9 6 9 0 0

∴ A + (–A) = (–A) + A = O.

If A is a matrix of any order then their exists – A(negative of A) such that
A + (–A) = (–A) + A = 0. So, –A is the additive inverse of A.

Multiplication of Matrix by a Scaler

Consider a matrix A = a b . Then,
c d

A+A+A= a b + a b + a b
c d c d c d

= a+a+a b+b+b = 3a 3b =3 a b = 3A
c+c+c d+d+d 3c 3d c d

Hence, A + A + A = 3A = 3a 3b .
3c 3d

If a matrix of any order is multiplied by a scaler then every element of the
matrix is multiplied by that scaler.

Consider a matrix A = 1 2 . Then 3A= 3 1 2 = 3×1 3×2 = 36
3 4 3 4 3×3 3×4 9 12

76 Infinity Optional Mathematics Book - 9

WORKED OUT EXAMPLES

1. If A= 6 –2 and B= 3 2 then find the matrix C such that B + C = A.
–5 3 1 –6

Solution: Here, A = 6 -2 and B = 3 2
-5 3 1 -6

Let C = a b
c d

We have, B + C = A

3 2 + a b = 6 -2
1 -6 c d -5 3

Equating the corresponding elements, we get,

3 + a = 6 2 + b = – 2 1 + c = –5 –6 + d = 3

∴a=3 ∴ b = – 4 ∴ c = –6 ∴d=9

∴C= a b = 3 –4
c d -6 9

2. If –3 5 is the additive inverse of x–1 –5 then find the values of
2 –1 y+3 1

x and y.

Solution: Since –3 5 is the additive inverse of x–1 –5 ,
2 –1 y+3 1

–3 5 + x–1 –5 = 0 0
2 –1 y+3 1 0 0

or, yx – 4 0 = 0 0
+ 5 0 0 0

Equating the corresponding elements, we get,

x–4=0 y+5=0

∴x=4 ∴y=–5

Hence, x = 4 and y = –5.

3. If A = 1 –3 and B = 4 5 then find 3A – 2B.
–2 4 6 8
Solution:

Here, A = 1 –3 and B = 4 5
–2 4 6 8

Now, 3A – 2B = 3 1 –3 –2 4 5
–2 4 6 8

Infinity Optional Mathematics Book - 9 77

= 3 –9 – 8 10
–6 12 12 16

= 3–8 –9 – 10 = –5 –19
–6 – 12 12 – 16 –18 –4

Exercise 3.3

Section 'A'

1. (a) Write the condition that two matrices can be added or subtracted.

(b) What are the properties of the matrix addition?

(c) Define commutative property of the matrix addition.

2. Which of the following matrices can be added or subtracted?

A = (1 5 –9), B = 3 1 2 ,D= 2 4 3 ,E= 5 6 ,
5 -3 1 6 -2 3 2
6 ,C=
2
-1
F= 5 , G= 3 2), I = 5 , J = 1 23
6 -3 , H = (4 –3 4 56
2

3. (a) If P = -5 6 then find out the additive inverse of P.
-3 2

(b) If A = 5 then write the additive identity of A.
-3

4. (a) If P = 2 –1 then find 3P.
3 –4

(b) If X = 54 then find 7X.
32

Section 'B'

5. Add the following matrices.

(a) -32 + 6 (b) (1 5 -2) + (3 3 1)
3

(c) 51 2 -6 + -1 3 5 (d) --31 -42 + 5 3
-7 3 2 -4 3 1 8

6 10 42 3 2 -5 103

(e) -5 4 + 3 -8 (f) 7 -3 8 + 5 -2 8

11 12 -6 -5 -9 7 3 -7 6 3

78 Infinity Optional Mathematics Book - 9

6. Subtract the following matrices.

(a) 1-50 – 3 (b) (5 3 8) – (1 2 3)
8

(c) 32 -5 – -5 -2 (d) 43 25 16 – 1 2 3
0 1 3 4 5 6

-2 0 41 3 -2 6 8 –3 5

(e) 3 4 – 5 2 (f) 10 12 2 – 1 0 2

02 -3 -2 6 -9 7 –2 6 0

7. (a) If A = 3 4 and B = 8 -6 then show that matrix (A + B) has an order
7 9 30

of 2 × 2.

(b) If P = 2 -2 and Q = 2 1 then show that P + Q = Q + P.
5 4 -3 6

8. (a) If A = 1 -3 and B = -5 2 then find 4A – 3B.
-2 0 11

(b) If P = 5 -2 6 and Q = 1 2 4 then find 3P + 2Q.
10 2 1 3 4

Section 'C'

9. (a) If 9 3x – 3 2 = 6 1 then find the values of x and y.
10 -7 7 6y 3 2

(b) If 3x 5y + -2 4 = x+2 y–4 then find the values of x,y and z.
2 4 0 2z 2 z–2

10. (a) If the additive inverse of 5x – 4 8 is –6 –8 , then find the values
–6 10 6 2y

of x and y.

(b) If a+3 6 –5 is the additive inverse of 2 c–6 5 then find
–2 2b – 1 3 2 –3 d

the values of a, b, c and d.

4 -3 7

11. (a) If A = 3 , B = 7 and C = -4 then find the matrix P under following
-2 2 9

conditions.

(i) P = A + B – C (ii) A + P = C (iii) A – P = B + C

Infinity Optional Mathematics Book - 9 79

(b) If P = 1 2 ,Q= –3 –5 and R = 9 8 , then find the matrix X
3 –4 6 –1 –6 –4

under the following conditions.

(i) X = P – Q + R (ii) X – Q = R (iii) P + X = Q – R

12. (a) If A = (–3 4 5), B = (–6 –3 1) and C = (9 –8 6) then show that

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

(b) If A = 1 3 2 , B= 1 2 3 and C = -8 1 -9 then show
6 -9 -8 1 2 3 -3 5 -2

that A + (B + C) = (A + B) + C.
2 –3

(c) If A = –9 6 then prove that
2 –5

(i) A + O = O + A = A where O is zero matrix of order 3 × 2.

(ii) A + (–A) = (–A) + A = O where O is zero matrix of order 3 × 2.

13. (a) If 2 4x + 3 2 3 = z –7 , find the values of x, y and z.
01 y 2 10 8

(b) If 5 2 x –2 31 = 4 -7 then find the values of x, y and z.
-3 2y z5 50

14. (a) If A = 1 3 ,B= -3 4 and C = 2 0 then find the matrix
-2 6 7 1 -3 10

(2A + 3B – C).

(b) If P = 2 1 ,Q= 0 1 and R = 5 5 then find the value of X
3 4 7 -2 9 12

such that 3P + Q + X = R.

15. (a) If X + Y = 36 and X – Y = 1 2 , find the matrices X and Y.
69 0 1

(b) If 2A + 3B = 7 15 and 3A – B = 5 6 , find the matrices A and B.
14 22 10 11

80 Infinity Optional Mathematics Book - 9

Transpose of Matrix

Consider a matrix A = 1 -3 9 of order 2 × 3.
60 2

If the rows and columns of the matrix A are interchanged then the new matrix is

16
formed as -3 0 which is of order 3 × 2 called transpose of A. It is denoted by AT

92
16

or A'. So, AT = -3 0 .
92

A new matrix obtained by interchanging the rows and columns of a given matrix
A is called as the transpose of A. It is denoted by AT or A'.

Note

If A = AT then A is symmetric matrix.

Properties of transpose of matrix

(i) Consider a matrix A = 1 2 . So, AT = 1 3
3 4 2 4

Again, (AT)T = 1 2 = A.
3 4

∴ (AT)T = A.

(ii) Consider two matrices A = 1 15 and B = 4 9 .
3 12 16 25

Now, AT = 1 3 and BT = 4 16
15 12 9 25

A+B= 1 15 + 4 9 = 5 24
3 12 16 25 19 37

(A + B)T = 5 19
24 37

AT + BT = 1 3 + 4 16
15 12 9 25

= 1+4 3 + 16 = 5 19
15 + 9 12 + 25 24 37

Infinity Optional Mathematics Book - 9 81

∴ (A + B)T = AT + BT.

(iii) If A = 10 12 and k is a constant then,
35 40

k.A = k 10 12 = 10k 12k
35 40 35k 40k

(k.A)T = 10k 35k
12k 40k

kAT = k 10 35 = 10k 35k
12 40 12k 40k

∴ (kA)T = kAT

WORKED OUT EXAMPLES

1. If A = 3 -5 and B = 1 4 then prove that
4 2 2 3

(a) (AT)T = A (b) (A + B)T = AT + BT

Solution:

Here, A = 3 -5 and B = 1 4
4 2 2 3

(a) AT = 3 4
-5 2

(AT)T = 3 -5 =A
4 2

Hence (AT­)T = A.

(b) AT = 3 4 and BT = 1 2
-5 2 4 3

Now, A + B = 3 -5 + 1 4 = 3+1 -5 + 4 = 4 7
4 2 2 3 4+2 2+3 6 5

∴ (A + B)T = 4 6
-1 5

Again, AT + BT = 3 4 + 1 2 = 3+1 4+2 = 4 6
-5 2 4 3 –5 + 4 2+3 –1 5

∴ AT + BT = 4 6
–1 5

Hence, (A + B)T = AT + BT

82 Infinity Optional Mathematics Book - 9

Exercise 3.4

Section 'A'
1. (a) Define transpose of a matrix with an example.
(b) Mention the properties of transpose of matrix.

2. Find the transpose of the following matrices.

(a) A= 4 (b) B = (6 -3 0)
-8
52
(b) C = 5 -3 (d) D = 1 3
-4 8
69
(e) E = 1 3 4
5 7 9 1 11
(f) E = 5 3 0

-2 -3 -4

Section 'B'

3. If P = 2 -5 and Q = 3 2 then verify the following.
6 3 0 9

(a) (PT)T = P (b) (P + Q)T = PT + QT (c) (kP)T = kPT

4. If A = 5 3 -1 and B = 1 03 then prove the following relations.
-6 4 2 9 -11 8

(a) (AT)T = A (b) (A + B)T = AT + BT (c) (kA)T = k.AT

5. (a) If A = 1 3 then prove that (A + AT) is a symmetric matrix.
2 4

2 –1 0

(b) If X = 3 2 1 then show that (X + XT) is a symmetric matrix.
4 -3 5

6. (a) If M = 5x – 1 6 ,N= –6 –3 and MT = N, then find the values of
–3 y+4 6 –2

x and y.

(b) If X = a–1 2b + 5 ,Y= 4 5 and XT + YT = I then find the values
c+1 4d – 1 3 –2

of a, b, c and d.

Infinity Optional Mathematics Book - 9 83

Multiplication of Matrices

Consider an example which helps to understand the multiplication of matrices.

Let A and B be two shops. The shop A sold 20kg of rice and 10kg of dal and the shop
B sold 15kg of rice and 15kg of dal. The prices of rice and dal in Baisakh are Rs.
40 and Rs. 50, in Jestha are Rs. 38 and Rs. 52 per kg respectively. Which can be
illustrated by the following tables.

Shop Amount in kg Price per kg in Rs.
Rice Dal
A Baisakh Jestha
B 20 10
15 15 Rice 40 38

Dal 50 52

Then, the income of shop A in Baisakh = 20 × 40 + 10 × 50 = 1300
the income of the shop B in Baisakh = 15 × 40 + 15 × 50 = 1350

the income of the shop A in Jestha = 20 × 38 + 10 × 52 = 1280

the income of shop B in Jestha = 15 × 38 + 15 × 52 = 1350

The above information can be arranged in matrix form as

20 10 . 40 38 = 20 × 40 + 10 × 50 20 × 38 + 10 × 52 = 1300 1280 .
15 15 50 52 15 × 40 + 15 × 50 15 × 38 + 15 × 52 1350 1350

which is the product of two matrices.

Thus, two matrices A and B can be multiplied if the number of columns of matrix
A is equal to the number of rows of matrix B. It is denoted by AB.

Let A and B be two matrices of order a × b and b × c then the product matrix AB will
be of order a × c, which can be illustrated by following example.

A= 2 5 2×2 × B= 56 8 2×3
3 8 3 –2 6

No. of No. of
columns of = row of

A B

Order of AB = 2 × 3

84 Infinity Optional Mathematics Book - 9

Steps of multiplication of two matrices.

Let A = a1 b1 2×2 and B = a2 b2 2×2. Then
c1 d1 c2 d2

Step 1: First row of A × first column of B.

A×B= a1 b1 × a2 b2 = a1a2 + b1c2 .......
c1 d1 c2 d2 .............. .......

Step 2: First row of A × second column of B.

A×B= a1 b1 × a2 b2 = a1a2 + b1c2 a1 b2 + b1d2
c1 d1 c2 d2 ................. .................

Step 3: Second row of A × First column of B.

A×B= a1 b1 × a2 b2 = a1a2 + b1c2 a1 b2 + b1d2
c1 d1 c2 d2 c1a2 + d1c2 .................

Step 5: Second row of A × Second column of B.

A×B= a1 b1 × a2 b2 = a1a2 + b1c2 a1 b2 + b1d2
Thus, c1 d1 c2 d2 c1a2 + d1c2 c1b2 + d1d2

A×B= a1 b1 × a2 b2 = a1a2 + b1c2 a1 b2 + b1d2
c1 d1 c2 d2 c1a2 + d1c2 c1b2 + d1d2

Consider two matrices A = 3 2 and B = 5 -3
1 4 2 0

Then, A × B = 3 2 × 5 -3
1 4 20

= 3×5+2×2 3 × (–3) + 2 × 0 = 19 -9
1×5+4×2 1 × (–3) + 4 × 0 13 -3

B×A = 5 -3 × 3 2
2 0 1 4

= 5 × 3 + (–3) × 1 5 × 2 + (–3) × 4 = 12 –2 .
2×3+0×1 2×2+0×4 6 4

WORKED OUT EXAMPLES

2 0 –1 5
3 1
1. If P = and Q = 3 5 then find P × Q and Q × P.
2 –4

2 0 -1 5
3 1
Solution: Here P = 2×2 and Q = 3 5 3×2
2 -4

Infinity Optional Mathematics Book - 9 85

Since, no. of columns of P is not equal to no. of rows of Q, P × Q can not be
obtained. But, no. of columns of Q is equal to no. of rows of P, Q × P can be
obtained.

So, Q × P = -1 5 × 2 0
3 5 3 1
2 -4

(–1) × 2 + 5 × 3 (–1) × 0 + 5 × 1

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

-2 + 15 0+5 13 5

= 6 + 15 0 + 5 = 21 5
4 – 12
0–4 -8 -4

2. If –1 2 .X = -2 , find the matrix X.
2 –2 4

Solution: x
y
Let the matrix X = . So,

–1 2 2×2 . x 2×1 = –2
2 –2 y 4 2×1



or, (2-1×) ×x +2 × y = -2
x+ (-2) × y 4

or, 2-xx + 2y = -2
– 2y 4

Equating the corresponding elements, we get,

–x + 2y = – 2 .......(i)

2x – 2y = 4 ......... (ii)

Solving equations (i) and (ii), we get,

x = 2 and y = 0.

Hence, A = x = 2 .
y 0

3. If A = 12 , show that A2 – 2A – 5I = O where I and O are 2 × 2 identity
31

and null matrices respectively.

Solution: 1 2 10 00
3 1 01 00
Here, A = ,I= and O =

86 Infinity Optional Mathematics Book - 9

Now, A2 =A×A= 1 2 . 12
3 1 31

= 1×1+2×3 1×2+2×1 = 74
3×1+1×3 3×2+1×1 67

2A =2 12 = 24
31 62

5I =5 10 = 50
01 05

A2 –2A – 5I = 74 – 2 4 – 5 0
67 6 2 0 5

= 7–2–5 4–4–0 = 00 =O
6–6–0 7–2–5 00

Hence, A2 – 2A – 5I = O

Exercise 3.5

Section 'A'

1. (a) Under what condition, two matrices can be multiplied?

(b) If order of A is 2 × 3 and order of B is 3 × 1 then what is the order of
(A × B)?

(c) If M is a matrix of order 5 × 2 and N is a matrix of order 2 × 4, (i) Is the
matrix product MN defined? (ii) Is the matrix product NM defined?

(d) If A = 2 then is A2 defined?
3
2. Which of the following matrices can be multiplied?

A= 1 2 ,B= 2 ,C= 2 4 5 , D = (5 0 –1),
3 4 3 1 3 –3
0

E= 1 11 –1 2 ,H= 35
–1 2 0 -1
5 3 0 , F = (8, –3), G = 2 -3
-2 -3 -4
Section 'B'

3. Find the product of the following matrices.

(a) (3 4) -1 2
2
(b) 3 . (1 5 6)
0

Infinity Optional Mathematics Book - 9 87

(c) -11 2 . 1 3 (d) -12 03 52 . 3 5
-1 1 2 0 -1
2 -3

4. If P = 3 1 and Q = -1 -2 , find the following matrices.
1 3 3 4

(a) P. Q (b) Q.P (c) P2 (d) Q2

5. (a) A= 4 –1 and B = 2 1 then show that A.B is a unit matrix.
–7 2 7 4

(b) If X = 1 –1 and Y = –1 2 , show that XY is a null matrix.
–1 1 –1 2

6. (a) If 7x + 1 = 2 3 . 0 then find the values of x and y.
6y – 2 -1 8 2

(b) If 2 –1 –3 x = –8 6 then find the values of x and y.
5 –2 y –4 –19 13

(c) If 4 1 2 –1 = x –1 then find the values of x and y.
7 –3 1 3 11 y

(d) If (2x –1). –3 = (10), find the value of x.
–x

7. (a) If A = 2 0 ,B= 4 0 and A2 = B, find the value of a.
–3 a –9 1

(b) If P = 2 a ,Q= 4 36 and P2 = Q, find the value of a.
0 1 0 1

(c) If A = 2 –3 and A2 = I then find the values of a and b.
a b

8. (a) If A = 5 –2 then find the value of AAT.
3 –3

(b) If P = 2 4 and Q = 3 6 then find (PQ)T.
5 3 2 8

9. (a) If A 1 0 = (9 15) then find the matrix A.
1 3

(b) Which matrix pre- multiplies to the matrix 1 1 to get a matrix
3 4

4 5 ?
6 2

88 Infinity Optional Mathematics Book - 9

10. (a) If –1 2 ×P= –2 , find the matrix P.
2 –2 4

(b) If A = 2 1 ,B= 2 4 and AC = B, find the matrix C.
5 3 3 3

(c) Which matrix post multiplies to –1 0 to get a matrix –2 ?
0 –2 4

11. (a) Matrix P has x rows and x + 5 columns. Matrix Q has y rows and 11 – y

columns. If PQ and QP both are defined, find the values of x and y.

(b) Let, A(2x – y) × y and B2 × 4 be any two matrices. If AB and BA are defined,
find the values of x and y.
Section 'C'

12. (a) If A = 4 2 , prove that A2 – 5A + 6I = O where I and O are 2 × 2
-1 1

identity and null matrix respectively.

(b) If B = 3 –5 , show that A2 – 5A = 14I where, I is an identity matrix
–1 2

of order 2 × 2.

(c) If X = 4 2 ,I= 1 0 and O = 0 0 then prove that
-1 2 0 1 0 0

(X – 2I).(X – 3I) = O.

13. (a) If P = 4 0 , Q= x y and PQ = P + Q, find the values of x, y and z.
0 5 0 z

(b) If X = 2 -1 , Y= a b and XY = X + Y, find the values of a, b and c.
0 3 0 c

3x + 2 5 -6 . 5 then find the values of x, y and z.
(c) If y – 4 = 12 20 2

5z – 4 -4 3

Infinity Optional Mathematics Book - 9 89

Properties of Matrix Multiplication

Associative property

Consider three matrices A = 1 –2 ,B= 2 0 and C = –3 –1 , then
3 5 1 3 0 2

AB = 1 –2 . 2 0 = 1 × 2 + (–2) × 1 1 × 0 + (–2) × 3 = 0 –6
3 5 1 3 3×2+5×1 3×0+5×3 11 15

(AB).C = 0 –6 . –3 –1 = 0 × (–3) + (–6) × 0 0 × (–1) + (–6) × 2
11 15 0 2 11 × (–3) + 15 × 0 11 × (–1) + 15 × 2

= 0 –12
–33 19

Again,

BC = 2 0 . –3 –1 = 2 × (–3) + 0 × 0 2 × (–1) + 0 × 2 = –6 –2
1 3 0 2 1 × (–3) + 3 × 0 1 × (–1) + 3 × 2 –3 5

A(BC) = 1 –2 . –6 –2
3 5 –3 5

= 1 × (–6) + (–2) × (–3) 1 × (–2) + (–2) × 5 = 0 –12
3 × (–6) + 5 × (–3) 3 × (–2) + 5 × 5 –33 19

∴ (AB).C = A.(BC)

Hence, multiplication of matrices is associative.

Distributive property

Consider three matrices of A = 2 0 ,B= 3 5 and C = 2 2
1 1 0 2 1 1

Then, B + C = 3 5 + 2 2 = 3+2 5+2 = 5 7
0 2 1 1 0+1 2+1 1 3

A(B + C) = 2 0 . 5 7 = 2×5+0×1 2×7+0×3 = 10 14
1 1 1 3 1×5+1×1 1×7+1×3 6 10

Again, AB = 2 0 . 3 5 = 2×3+0×0 2×5+0×2 = 6 10
1 1 0 2 1×3+1×0 1×5+1×2 3 7

AC = 2 0 . 2 2 = 2×2+0×1 2×2+0×1 = 4 4
1 1 1 1 1×2+1×1 1×2+1×1 3 3

90 Infinity Optional Mathematics Book - 9

AB + AC = 6 10 + 4 4 = 10 14
3 7 3 3 6 10

∴ A(B + C) = AB + AC.

Hence, matrix multiplication is distributive under addition.
Identity property

Consider a matrix A = 1 2 . Then
3 4

Now, A.I = 1 2 . 1 0 = 1×1+2×0 1×0+2×1 = 1 2 =A
3 4 0 1 3×1+4×0 3×0+4×1 3 4

Again, I.A = 1 0 1 2 = 1×1+0×3 1×2+0×4 = 1 2 =A
0 1 3 4 0×1+1×3 0×2+1×4 3 4

∴ A.I = I.A = A
Hence, there is existence of identity matrix in matrix multiplication.

Transpose of product property

Consider two matrices A = –1 2 and B = 0 –2 . Then
4 3 1 3

AB = –1 2 . 0 –2 = (–1) × 0 + 2 × 1 (–1) × (–2) + 2 × 3 = 2 8
4 3 1 3 4×0+3×1 4 × (–2) + 3 × 3 3 1

(AB)T = 2 3 .
8 1

Again, AT = –1 4 and BT = 0 1
2 3 –2 3

BT.AT = 0 1 . –1 4 = 0 × (–1) + 1 × 2 0×4+1×3 = 2 3
–2 3 2 3 (–2) × (–1) + 3 × 2 (–2) × 4 + 3 × 3 8 1

∴ (AB)T = BT.AT
Hence, the transpose of the product of two matrices is equal to the product of the
transpose of the opposite matrices.

Infinity Optional Mathematics Book - 9 91

WORKED OUT EXAMPLES

1. If A = 1 2 and B = 0 -2 then show that the matrix multiplication
3 4 1 -1

is not commutative.

Solution:

Here, A = 1 2 and B = 0 –2 .
3 4 1 –1

Then, A.B = 1 2 . 0 –2 = 1×0+2×1 1 × (–2) + 2 × (–1) = 2 –4
3 4 1 –1 3×0+4×1 3 × (–2) + 4 × (–1) 4 –10

Again, B.A = 0 –2 . 1 2 = 0 × 1 + (–2) × 3 0 × 2 + (–2) × 4 = –6 –8
1 –1 3 4 1 × 1 + (–1) × 3 1 × 2 + (–1) × 4 –2 –2

Since, AB ≠ B.A, the matrix multiplication is not commutative.

2. If A = 0 –2 and B = 1 1 then show that (AB)T = BT.AT
1 3 2 2
Solution:

Here, A = 0 –2 and B = 1 1 .
1 3 2 2

Then, A.B = 0 –2 . 1 1 = 0 × 1 + (–2) × 1 0 × 1 + (–2) × 2 = –4 –4
1 3 2 2 1×1+2×2 1×1+3×2 7 7

\ (AB)T = –4 7
–4 7

Again, AT = 0 1 and BT = 1 2
–2 3 1 2

\ BT.AT = 0 1 . 1 2 1 × 0 + 2 ×(–2) 1×1+2×3
–2 3 1 2 = 1 × 0 + 2 × (–2) 1×1+2×3

= –4 7
–4 7

Hence, (AB)T = BT.AT

92 Infinity Optional Mathematics Book - 9

Exercise 3.6

Section 'A'

1. (a) What are the properties of matrix multiplication?
(b) State associative property for matrix multiplication?
(c) What is the 2 × 2 identity matrix for multiplication?

Section 'B'

2. (a) If P = -1 3 and Q = 5 2 then show that the multiplication of
2 1 4 3

matrices is not commutative.

(b) If A = 1 2 and B = -3 4 then prove that AB ≠ BA.
3 0 1 3

3. (a) If A = -2 5 then show that A.I = I.A. = A where I is the unit matrix of
1 4

2 × 2 order.

(b) If P = -3 0 then show that P.I = I.P = P where I is the unit matrix of
-5 2

2 × 2 order.

Section 'C'

4. (a) If A = 1 2 ,B= –4 y and AB = BA then find the values of x and y.
x 4 3 –1

(b) If P = 1 –2 ,Q= 1 4 and PQ=QP then find the values of a and b.
0 a b 2

5. (a) If A = 1 0 ,B = 2 –3 and C = 4 2 . Then show the following
4 –2 –1 2 1 1

relations.

(i) A(BC) = (AB).C (ii) A(B + C) = AB + AC (iii) (AB)T = BT.AT

(b) If P = 6 8 ,Q= 3 4 and R = 0 1 , then show the following
9 7 5 6 1 3

relations.

(i) (PQ). R = P.(QR) (ii) P(Q + R) = PQ + PR (iii) (PQ)T = QT.PT



Infinity Optional Mathematics Book - 9 93

UNIT

4 COORDINATE GEOMETRY

4.1 Locus

Review: Y

Let us observe the given figure and discuss the

direction of x-axis and y-axis. Also, discuss on the (–, +) (+, +)

following questions. X' OX
(i) How many quadrants are there? (–, –) (+, –)

(ii) In which quadrants both coordinates are

positive and negative? Y'

(iii) Which quadrant contains x-coordinate negative

and y-coordinate positive?

(iv) Which quadrants are the points (2, 3), (-4, 3), (-5, -2) and (7, -4) belongs

to?

Again, Y P
From the given graph. Q
(v) Write the coordinate of the points P, Q, R

and S.

(vi) Find the distance between P and Q. X' O X
R S
(vii) Find the distance between two points
Y'
A(x1, y1) and B(x2, y2)
(viii) Write the coordinates of origin.

(ix) Why the distance between any two points

is always positive?

Locus
Let, us do the following activities.
(i) Draw a circle, by a pencil with the help of a compass as shown in

figure.

94 Infinity Optional Mathematics Book - 9

What is the locus? Is the circle a locus?
(ii) Have you seen river?
How it is flowing? Can you draw its picture from its

starting point to its final point?

(iii) Have you seen Javelin (a long stick with a pointed

end)? How it is thrown? Can you draw its picture when
it is thrown?


Definition
The path traced by a moving point under certain condition/conditions is known as
the locus of a point.

What is/are the condition/conditions?

What is the difference between satisfying points and not satisfying points? Discuss
on above questions.

Equation of a locus:

Let, O be the fixed point and P be the moving point. P moves such that OP is always

at a constant distance. P (x, y)

∴ OP = r(radius of circle) r

Let, O(h, k) and P(x, y) then by distance formula O(h,k)

d = (x2 – x1)2 + (y2 – y1)2
or, OP = r = (x – h)2 + (y – k)2

or, r2 = (x – h)2 + (y – k)2 is the equation of circle. It is also called the equation of

locus.

Some facts about equation of locus :

A locus is a continuous curve or a straight line.

A locus can be represented by an equation.

Equation of a locus is the equation which is satisfied by the coordinates of any point
on the locus.

If coordinates of a point satisfies the equation it lies on the locus.

Infinity Optional Mathematics Book - 9 95

Method of finding the equation of a locus
To find the equation of the locus of a moving point we generally proceed as follows:
(i) Draw the figure according to the given condition and consider a moving point

P(x, y) on the locus.
(ii) Write down the given condition or conditions.
(iii) Explain the condition algebraically in terms of x and y.
(iv) Simplify the algebraic expression and the result is the required equation of

locus.

WORKED OUT EXAMPLES

1. Which of the points (1, 4) and (-4, 3) lies on the locus represented by
x2 + y2 = 25?

Solution: Here,
The given points are (1, 4) and (–4, 3)
The given equation of locus is x2 + y2 = 25 ...........(i)
Taking (1, 4). Where, x = 1, and y = 4 then from equation (i)
x2 + y2 = 25
or, (1)2 + (4)2 = 25
or, 1 + 16 = 25
or, 17 = 25 which is not true.
∴ (1, 4) does not lie on the given locus.
Similarly, taking (–4, 3), where x = -4 and y = 3 then from equation (i)
x2 + y2 = 25
or, (-4)2 + (3)2 = 25
or, 16 + 9 = 25
or, 25 = 25 which is true
∴ (-4, 3) lie on the curve.

2. Find the equation of the locus of a point if its distance from the y-axis
is twice its distance from the x-axis.

Solution: Here,
Let, P(x, y) be the moving point of the locus: From P draw PM⊥OC and PN⊥OY.

96 Infinity Optional Mathematics Book - 9

∴ PM = y, PN = OM = x Y

To find: The equation of locus.

By the question N P(x, y)
O X
PN = 2PM
Y' M
or, x = 2y X'

or, x – 2y = 0 is the required equation of

locus.

3. If (4, 4) is a point on the locus having equation y2 = ax, then prove that
(16, 8) is another point which lies on the locus.

Solution: Here,

The point P(4, 4) lies on the locus having equation; y2 = ax ............(i)

Where, x = 4 and y = 4 then from equation (i).

y2 = ax

or, (4)2 = a × 4

or, 16 = 4a

or, 146 = a
or, a = 4

Substituting a = 4 in equation (i), we get

y2 = 4x ....... (ii) is the required equation of locus.

Again, if the point (16, 8) lies on the locus it must satisfy equation (ii)

x = 16 and y = 8 then

or, y2 = 4x

or, (8)2 = 4 × 16

or, 64 = 64 which is true.

Hence, it is proved that (16, 8) lies on the locus.

4. Find the equation of the locus of a moving point such that its distance

from the point (-1, 5) is double its distance from the point (2, -3). What

curve does the locus represent? P(x, y)
Solution: Here, P(x, y) be a moving point of the locus.

Let, A (–1, 5) and B (2, -3) be two given points.

By using distance formula A(–1, 5) B(2, –3)

AP2 = (x2 – x1)2 + (y2 – y1)2

= (x + 1)2 + (y – 5)2

Infinity Optional Mathematics Book - 9 97

= x2 + 2x + 1 + y2 – 10y + 25

AP2 = x2 + y2 + 2x – 10y + 26 ............ (i)

Similarly,

BP2 = (x – 2)2 + (y + 3)2

or, BP2 = x2 – 4x + 4 + y2 + 6y + 9

or, BP2 = x2 + y2 – 4x + 6y + 13 .......... (ii)

By the question,

AP = 2BP

or, AP2 = 4BP2 (Squaring on both sides)

or, x2 + y2 + 2x – 10y + 26 = 4(x2 + y2 – 4x + 6y + 13)

or, x2 + y2 + 2x – 10y + 26 = 4x2 + 4y2 – 16x + 24y + 52

or, 4x2 – x2 + 4y2 – y2 – 16x – 2x + 24y + 10y + 52 – 26 = 0

or, 3x2 + 3y2 – 18x + 34y + 26 = 0 is the required equation of locus.

Since, the equation is 2nd degree equation in x and y, hence, the locus of P is
a circle.

5. If (4, 0) and (-4, 0) are two fixed points and P(x, y) is a moving point
which moves such that AP2 + BP2 = AB2, obtain the equation of the
locus of the point P.

Solution: Here,

P(x, y) be a moving point of the locus. Let, A(4, 0) and B(–4, 0) are two fixed

points. Y
By using distance formula

AP2 = (x2 – x1)2 + (y2 – y1)2 P(x, y)

or, AP2 = (x – 4)2 + (y – 0)2

or, AP2 = x2 – 8x + 16 + y2 ......... (i) X' X
Similarly, BP2 = (x + 4)2 + (y – 0)2 B(–4, 0) O A(4, 0)
BP2 = x2 + 8x + 16 + y2 .........(ii)
and AB2 = (4 + 4)2 + (0 – 0)2 Y'

= (8)2 + 0

AB2 = 64 ......... (iii)

By the question

AP2 + BP2 = AB2

or, x2 – 8x + 16 + y2 + x2 + 8x + 16 + y2 = 64

98 Infinity Optional Mathematics Book - 9

or, 2x2 + 2y2 + 32 = 64

or, 2x2 + 2y2 – 16 = 0

or, 2(x2 + y2 – 8) = 0

or, x2 + y2 = 8 is the required equation of the locus.

6. A and B are two given point whose coordinates are (-5, 3) and (2, 4)
respectively. A point P moves in such a way that PA:PB = 3:2. Find the
equation of the locus of the point P. What curve does it represent?

Solution: Here,

Let, P(x, y) be moving point of the locus. A(–5, 3) and B(2, 4) are two given
fixed points.

By using distance formula

PA2 = (x + 5)2­ + (y – 3)2 [∴ d2 = (x2 – x1)2 + (y2 – y1)2]

= x2 + 10x + 25 + y2 – 6y + 9 P(x, y)

or, PA2 = x2 + y2 + 10x – 6y + 34 ......... (i)

Similarly, PB2 = (x – 2)2 + (y – 4)2 A(–5, 3) B(2, A)
= x2 – 4x + 4 + y2 – 8y + 16
PB2 = x2 + y2 – 4x – 8y + 20 .......... (ii)

By the question

PA = 3
PB 2

or, 3PB = 2PA

or, 9PB2 = 4PA2 [ Squaring on both sides we get]

or, 9(x2 + y2 – 4x – 8y + 20) = 4(x2 + y2 + 10x – 6y + 34)

or, 9x2 + 9y2 – 36x – 72y + 180 = 4x2 + 4y2 + 40x – 24y + 136

or, 9x2 – 4x2 + 9y2 – 4y2 – 36x – 40x – 72y + 24y + 180 – 136 = 0

or, 5x2 + 5y2 – 76x – 48y + 44 = 0 .......... (iii) is the required equation of locus.

Since, we see that equation (iii) is a second degree equation in x, y and its
coefficient of x2 and y2 are equal and coefficient of xy is zero. Therefore,

equation (iii) represents a circle. Therefore, the locus of P represents the

equation of locus.


Infinity Optional Mathematics Book - 9 99

Exercise 4.4

Section 'A'

1. (a) Write the formula to find the distance between two points A(x1, y1) and
B(x2, y2).

(b) Define locus of a moving point with example.

(c) Write the formula to find the distance between origin and the point
P(x, y).

2. M is a point (3, 4) and N is a point (–3, –4)
(i) Find the length of MN if M is (x1, y1) and N is (x2, y2).
(ii) Find the length of MN if M is (x2, y2) and N is (x1, y1).
(iii) Why there is no change in the length of MN from results of (a) and (b)?

3. (a) How far is the point (5, 0) from the origin?

(b) Find the distance between the following pair of points.

(i) (–1, 3) and (2, 4) (ii) (6, – 2) and (7, 3).

4. (a) Examine which of the points (–4, 3), (3, 4), (3, –2), (5, 0) and (0, –5) lie on
the locus with equation x2 + y2 – 25 = 0.

(b) Does the points (–1, 2) and (1, –1) belong to the locus 3x – 5 = 2y? Check
it.

5. (a) Find the value of k so that the point (2, –3) lies on the locus x + ky = –1.

(b) For what value of 'a' the point (1, 2) lies on the locus ax + ay = 3? Find it.
6. Find the equation of the locus of a point which moves so that

(i) its abscissa is always 4 units.
(ii) its ordinate (y-coordinate) is –3 units.
(iii) its distance from the x-axis is 5 units.
(iv) its distance from the y-axis is 3 units.

Section 'B'

7. (a) Find the value of 'k' so that the point (2, –1) lies on the locus with
equation –2y2 + kx2 – 2x + 3y – 3 = 0.

(b) If the point (2, 5) lies on the locus with equation ax + 8y = 50. Find the
value of 'a'. Also, show that the point (10, 0) lies on the same locus.

8. (a) Find the equation of the locus of a point which moves such that its
distance from (0, 0) is 5 units.

100 Infinity Optional Mathematics Book - 9