LIMIT 2UNIT
Estimated Teaching Periods : 10
Competency
To demonstrate the concept of the limit of the function.
Unit
LIMIT
Learning Outcomes
At the end of this unit, the students will be able to:
give simple concept of limit (from numerical sequence, based on figure sequence, based on the sum of infinite
series) and write it in symbol.
What do you learn?
Simple Concept on Limit
� Numerical Sequence
� Sequence of Figures
� Sum of Infinite Series
� Value of Function
� Concept of Limit of Function
� Meaning and Introduction to x → a
Are you ready? � Ascending and descending number patterns
� Sequence and its types
� Rounding numbers � Relation and function with their components
� Number line � Value of function
� Patterns of figures
� Set of natural numbers
Specification Grid
Chapter Cognitive Knowl- Under- Applica- Higher Total Total
Domain edge standing tion ability Ques- Marks
tions
Topic 1 mark 2 marks 4 marks 5 marks
2 Limit From All Topics 1 - 1 - 25
51
21.1 NUMERICAL SEQUENCE
LIMIT Learning Objectives
At the end of this topic, the students will be able to:
• write the sequence and estimate the last term • write the limit of the sequence
Let us study the following patterns of the numbers:
(a) 2, 7, 12, 17, ............................... (b) 4, 8, 16, 32, ...............................
(c) 24, 13, 2, – 9, ............................... (d) 21, 61, 118, 514, ...............................
(e) 53, 130, 135, 330, ............................... (f) 0.1, 0.01, 0.001, ...............................
Can you add three more terms in each of the above number pattern? What are they? These number patterns
are called sequence. How many terms are there in each sequence? How can you find the 25th term in each
sequence?
Let us consider a general term 2n – 1 of a sequence. Then write the sequence formed by the general term.
5n
The sequence becomes, 15, 235, 1525, 6275, ............................... and so on. It is easily computed
by using scientific
Let us find out the 15th term. Can you find? calculate.
15th term = 2 × 15 – 1 = 29
515 30517578125
= 0.000000000950272 (Decimal form)
= 9.50272 × 10– 10 (Scientific Notation)
Can you find the hundredth term? Write in scientific notation.
It is noticed that if the denominator of the fraction or rational number is greater than its numerator and
both are simultaneously increasing, it is approaching to the lesser number.
In other hand, let us consider a number line of real number which extends up to + ∞ (positive infinite
countable number) at right side and – ∞ (negative infinite countable number) at left side.
–∞ –2 –1 AE B +2 +∞
0F D C +1
What is noticed by you?
Table a line segment AB = 1 unit from O. Find out the midpoint C of AB. Similarly, find the midpoints D,
E, F, ..... of AC, AD, AE, ................ towards A.
52 Illustrated Optional Mathematics-9
In the same manner, which point is nearer to the last midpoint?
The length of the line segments AB, AC, AD, AE, AF, .......... form a sequence as 1, 21, 14, 18, 116, .............. .
In decimal form, 1.0, 0.5, 0.25, 0.125, 0.0025, ............ .
Which is the tenth midpoint?
At first, compute the general term.
Its general term is 1 1. LIMIT
2n –
So, its tenth midpoint = 1 1 = 1 = 1 = 0.001953125
210 – 29 512
Similarly, the 20th midpoint = 1 = 1 = 1.907348633 × 10– 6 = 0.000001907348633
220 – 524288
1
When the new line segment is further more bisected at countable infinitely times, the length of the last
line segment is approximately approached to zero (0), but not exactly zero (0). This zero (0) is called the
limit of the sequence 1, 21, 41, 18, 116, ............... .
Let us take another general term 2n – 1. Then we easily find the following terms:
n
10th term = 2 × 10 – 1 = 1.9
10
100th term = 2 × 100 – 1 = 1.999
100
1000th term = 2 × 1000 – 1 = 1.9999
1000
10000th term = 2 × 10000 – 1 = 1.99999 and so on.
10000
When n is increasing up to positively countable infinite, the value of the term is nearer to 2. This 2 is limit
2n – 1.
of the sequence having general term n
Example 1 : Add three more terms in the sequence 1210, 12010, 120100, .............. and find its 10th term in
decimal form and scientific notation.
Solution: The given sequence is 2101, 12010, 120100, ....................
In the sequence, the numerator is constant and denominator is increased by 10 times.
The general term of the sequence is 1201n. Therefore the 10th term is
120110 = 21
10000000000
= 0.0000000021 = 2.1 × 10– 9
Numerical Sequence 53
Example 2 : Find the tenth term of the sequence 190, 19090, 999 ............. . Convert the sequence into
1000
decimal form and write its limit when its nth term is countable infinitely large.
Solution: Here, the given sequence is 190, 19090, 1909090, ....................
In the sequence, the numerator of each term is added by one 9 in each place and the
denominator is multiplied by 10.
LIMIT So, its tenth term is 190909090909090909090.
The sequence in decimal form as,
0.9, 0.99, 0.999, .................., 0.9999999999, ...............
When the nth term is countable infinitely large, the term will approximately be one (1).
Hence, the limit of the sequence is 1.
EXERCISE – 2.1
1. What are next three more terms in each sequence given below? Write them.
(a) 2, 8, 14, 20, ........................ (b) – 23, – 17, – 11, – 5, ........................
(c) 14, 6, – 2, – 10, ........................ (d) 3, 6, 12, 24, ........................
(e) 14, 2, 27, 429, ........................ (f) 15, 110, 115, 210, ........................
(g) 170, 1070, 10700, 107000, ........................ (h) 110, 1030, 10500, ........................
2. Find the general term of the following sequences and also find their twentieth term.
(a) – 25, – 18, – 11, ........................ (b) 14, 21, 1, 2, ........................
(c) 130, 1030, 10300, 100300, ........................ (d) 9, 190, 1900, 10900, ........................
3. Write the 7th term of the following sequences and find the approximately last term when their
number of terms are increased.
(a) 2.1, 2.01, 2.001, ........................ (b) 0.9, 0.99, 0.999, ........................
(c) 0.1, 0.01, 0.001, ........................ (d) 3.9, 3.99, 3.999, ........................
4. (a) What are the nearest whole number of each term of the sequence 5.1, 5.01, 5.001, ................
by rounding off?
(b) What are the approximate whole number of each term of the sequence 0.1, 0.01, 0.001, .............
by rounding off?
5. What are the limiting values of the given sequences?
(a) 1.9, 1.99, 1.999, ........................ (b) 1.01, 1.001, 1.0001, ........................
(c) 2.1, 2.01, 2.001, ........................ (d) 9.9, 9.99, 9.999, ........................
54 Illustrated Optional Mathematics-9
6. Write the sequences obtained from the following general terms and convert into decimal form.
Write their limit when n is countable infinitely large number.
(a) 31n (b) 2 1 (c) 2n + 1 (d) 4n – 1
5n – n n
(e) n12 (f) (n 2 1)2 (g) n2 3 1 (h) 2 – 7n
+ + n
7. (a) Bisect line segment of the length 2 units towards 0 by 6 times and show them in a number line. LIMIT
When it bisects by 100 times, what is the limit of the 100th midpoint?
(b) Take a line segment PQ of the length 10 cm. Bisect it 8 times towards P and write the all
midpoints in sequence. When it bisects at 100 times, find its value in scientific notation and
write its limit.
(c) Draw a line segment AB of the length 5 cm. Bisect it at 7 times towards B. Write the sequence
of the length of segments towards B formed by the seven midpoints. When it is bisected at 32
times, what is the 32th midpoint and its limit?
Hints: The sequence becomes 5 × 1, 5 × 3, 5 × 7, 5 × 15, ........... .
2 4 8 16
ANSWERS
1. (a) 26, 32, 38 (b) 1, 7, 13 (c) – 18, – 26, – 34 (d) 48, 96, 192
(e) 3243, 24201, 162807 (f) 215, 310, 315 (g) 1007000, 10007000, 100070000 (h) 100700, 1009000, 11
1000000
2. (a) 7n – 32, 108 (b) 2n – 3, 131072 (c) 3 , 3 × 10– 20 (d) 9 1, 9 × 10– 19
10n 10n
–
3. (a) 2.0000001, 2 (b) 0.9999999, 1 (c) 0.0000001, 0 (d) 3.9999999, 4
4. (a) 5, 5, 5, 5, ....... (b) 0, 0, 0, 0, .......
5. (a) 2 (b) 1 (c) 2 (d) 10
6. (a) 31, 91, 217, 811, .....; 0.3333, 0.1111, 0.037037, 0.012346789, ...... ; 0
(b) 2, 25, 225, 1225, .......; 2, 0.4, 0.08, 0.016, ........; 0 (c) 3, 52, 73, 49, .......; 3, 2.5, 2.333, 2.25, ........; 2
(d) 3, 27, 131, 145, .......; 3, 3.5, 3.67, 3.75, .........; 4 (e) 1, 14, 19, 116, .......; 1, 0.25, 0.11, 0.0625, .......; 0
(f) 42, 92, 126, 225, .......; 0.5, 0.22, 0.125, 0.08, .......; 0 (g) 23, 53, 130, 137, .......; 1.5, 0.6, 0.3, 0.18, .........; 0
(h) – 5, – 212, – 319, – 426, .......; – 5, – 6, – 6.33, – 6.5; – 7
7. (a) 1, 0.5, 0.25, 0.125, 0.0625, ..............; 0
(b) 5, 2.5, 1.25, 0.625, 0.3125, 0.15625, 0.078125, .......; 7.888609052 × 10 – 30; 0
(c) 2.5, 3.75, 4.375, 4.6875, 4.84375, 4.921875, 4.9609375; ............, 4.999999999; 5.
Numerical Sequence 55
21.21 FIGURE SEQUENCE
LIMIT Learning Objectives
At the end of this topic, the students will be able to:
• find the limit of the sequence of the given figure patterns
A polygon is a closed plane figure containing three Triangle Quadrilateral Pentagon Hexagon
or more sides. A regular polygon is a polygon that Polygons Circle
is equilateral and equiangular. The triangle has three
sides, the quadrilateral has 4 sides, the pentagon has Chiliagon Myriagon Megagon
5 sides. Similarly, the hectagon has 100 sides, the Polygons
chiliagon has 1,000 sides, the myriagon has 10,000
sides, the megagon has 10,00,000 sides and so on.
When the sides of the regular polygon are increasing,
it approximates a circle or cyclic polygon.
Let us construct the regular polygons in the circles having radius 3 cm each as follows:
3 cm
3 cm
3 cm
3 cm
5.2 cm 4.2 cm 3.5 cm 3 cm 2.6 cm 2.3 cm
Fig. (i) Fig. (ii) Fig. (iii) Fig. (iv) Fig. (v) Fig. (vi)
Fig. Area of shaded region Formula for finding the
3 22 3 length of the side of an
4 a2 7 4 (5.2)2 = 28.29 – 11.71 = 16.58 cm2
(i) A = πr2 – = × 32 – inscribed regular polygon,
22 a = 2r sin π
7 n
(ii) A = πr2 – a2 = × 32 – (4.2)2 = 28.29 – 17.64 = 10.65 cm2
(iii) A = πr2 – 5 × b 4a2 – b2 = 22 × 32 – 5 × 3.5 4 × 32 – 3.52 = 28.29 – 21.32 = 6.97 cm2
4 7 4
(iv) A = πr2 – 6 × 3 = 22 × 32 – 6 × 3
4 a2 7 4 32 = 28.29 – 22.38 = 5.91 cm2
(v) A = πr2 – 7 × b 4a2 – b2 = 22 × 32 – 7 × 2.6 4 × 32 – 2.62 = 28.29 – 24.60 = 3.69 cm2
4 7 4
(vi) A = πr2 – 8 × b 4a2 – b2 = 22 × 32 – 8 × 2.3 4 × 32 – 2.32 = 28.29 – 25.49 = 2.8 cm2
4 7 4
56 Illustrated Optional Mathematics-9
Observe the difference between the area of the circle having radius 3 cm and inscribed regular polygon
in each figure. When the number of sides of the polygon is increasing, the difference of their areas is
decreasing. In which condition, the area of the inscribed regular polygon and circle are equal?
The inscribed regular polygon has n-gon where n is countable infinitely large, it approximates a circle.
In this case, the difference between the area of the n-gon and circle is nearer to zero (0). Therefore, the
area of the polygon when the number of sides is increasing maximum, the limiting value of the area of the
polygon is the area of the inscribed circle.
LIMIT
Example 1 : The outer square has 4 cm side in the given figure. Find the area and
perimeter of the sixth square made by joining the midpoints of any pair of
adjacent sides of the outer squares simultaneously as shown in the figure.
Write the limiting value of the area and perimeter of the square so formed
when the number of midpoints are countable infinitely large. 4 cm
Solution: From the given figure,
The side of the outer square (l1) = 4 cm 4
2
The side of the square formed by joining the midpoints of 1st square (l2) = = 2 cm
The side of the square formed by joining the midpoints of 2nd square (l3) = 2 = 1 cm
2
i.e. it forms the sequence as
4, 2, 1, .................
or, 14, 24, 44, ...............
or, 240, 241, 242, ..............
or, 214– 1, 224– 1, 4
23 – 1, ..............
∴ Its general term is 4 1.
Now, 2n –
∴ the side of the square formed by 18jo2in=in6g14thcme s2i=xt0h.0s1q5u6ar2e5=cm2642– = 4 = 4 = 1 cm
= 1 25 32 8
The area of the sixth square
The perimeter of the sixth square = 4 × 1 = 1 cm = 0.5 cm
8 2
When the number of the midpoints of the new squares is countable infinitely large, the limiting
value of the area and perimeter of the square so formed are approximately both zero (0).
EXERCISE – 2.2
1. (a) How many triangles having unit sides are there in the given equilateral triangle XYZ with XY
X DC
= YZ = XZ = 4 cm. Colour the alternative triangles with
two different colours. 4 cm
4 cm
(b) How many small rhombus are there in the given 6 cm
rhombus ABCD having side 6 cm and what is the
length of each side of the small rhombus? Colour the Y 4 cm Z A 6 cm B
alternative rhombus by two different colours. Q. No. 1 (a) Q. No. 1 (b)
Figure Sequence 57
2. (a) In the given figure, three pairs of parallel lines intersect each n1 n2 m1 m2
other. How many triangles are formed in the figure? When
the number of parallel lines are increasing, what is the limit l1
of the number of triangles? Discuss it. l2
(b) Observe the given weaving 'Nanglo'. How many rectangles
are there in the figure? When the number of bamboo strips Q. No. 2 (a) Q. No. 2 (b)
are increasing, what is the limit of the number of rectangles? A
LIMIT Discuss it. B 8 cm C
Q. No. 3 (b)
3. (a) Write the sequences of the sides, perimeters and areas of the S R
squares formed by joining the midpoints of the successive
sides of the given square PQRS having sides 4 cm. Write
their limits.
(b) Find the sequences of the sides, perimeters and areas of the P 4 cm Q
triangles obtained by joining the midpoints of the successive Q. No. 3 (a)
sides of the given equilateral triangle ABC having the sides
8 cm. Write their limits.
4. (a) A glass has full of juice. A boy drinks half of the juice at once
in each drink. What amount of juice remains in the glass at
last? At what times does he finish the glass of juice? What is
the limiting value of the quantity of juice?
(b) A girl cuts half of a cucumber. She cuts again and again half
in each cut. What amount of cucumber will left at last? At
what times does she finish to cut the cucumber? What is the
limiting value of the quantity of cucumber?
5. (a) In the adjoining graph, when the values of x are increasing, Y
what is the difference of the value of y from x? For what Y
value of x the value of y is 0?
(b) In the given figure, when the values of y are increasing, what
is the difference between the values of y and x? For what XX
value of y the value of x is 0? OO
Q. No. 5 (a) Q. No. 5 (b)
ANSWERS
1. (a) 16, 23 (b) 16, 1.5, 26 2. (a) 4, countable infinite (b) 15, count. infin.
3. (a) 4, 2 2, 2, 2, ........; 16, 8 2, 8, 4 2, .......; 16, 8, 4, 2, .......; 0, 0, 0
(b) 8, 4, 2, 1, .........; 24, 12, 6, 3, ........; 16 3, 4 3, 3, 43, .........; 0, 0, 0
4. (a) Approximately zero, Countable infinite times, 0 (b) Approximately zero, Countable infinite times, 0
5. (a) Approximately zero, countable infinite large number or x →
(b) Approximately zero, countable infinite large number or y →
Project Work
Draw a rectangle of the length 6 cm and breadth 4 cm and bisect it along the breadth. Bisect again and
again the half rectangle 5 times by drawing. Write down the sequences formed by the sides, perimeters
and areas of the rectangles. What is the limit of each sequence?
58 Illustrated Optional Mathematics-9
21.31 SUM OF INFINITE SERIES
Learning Objectives LIMIT
At the end of this topic, the students will be able to:
• find the sum of countable infinite series.
An infinite series is the sum of the terms of an infinite sequence. An example of an infinite sequence is 1,
12, 41, 81, 1 1 1
..........., and then the series obtained from this sequence would be 1 + 2 + 4 + 8 + ..............
with sum going on forever.
Now, we know, 1 + 1 + 1 + 1 + .............. up to 1 1,
20 21 22 23 2n –
where n is countable infinite large. The sum of infinite series is denoted as
∞ 1 = 1 + 1 + 1 + 1 + ..............
Σ 2n – 1 2 4 8
n=1
Where Σ is the sigma and works the sum of all terms as summation.
Consider a line segment AB having length 2 units and bisect it at C. Again, BC is bisected at D, for more
over bisect BD at E and bisect BE at F as shown in the given figure.
In the above figure, F
AB = 2 units A C DEB
AC = BC = 1 unit
CD = BD = 1 BC = 1 × 1 = 1 cm and AD = AC + CD = 1 + 1 units
2 2 2 2
DE = BE = 1 BD = 1 × 1 = 1 units and AE = AD + DE = 1 + 1 + 1 units
2 2 2 4 2 4
EF = BF = 1 BE = 1 × 1 = 1 units and AF = AE + EF = 1 + 1 + 1 + 1 units
2 2 4 8 2 4 8
In the same manner, we can bisect AB continuously and the last midpoint is nearer to the point B.
Now, we write
Sum of the 1st piece (S1) = AC = 1 unit
Sum of the 1st two pieces (S2) = AD = 1 + 1 = 1.5 units
2
Sum of the 1st three pieces (S3) = AE = 1 + 1 + 1 = 7 = 1.75 units
2 4 4
Sum of the 1st four pieces (S4) = AF = 1 + 1 + 1 + 1 = 15 = 1.875 units
2 4 8 8
Sum of Infinite Series 59
Similarly, we can compute S5, S6, S7, .........., Sn (bisects at n times) where Sn is approximately to 2. Here,
2 is the limiting value of the sum of the given infinite series. Therefore,
∞ 1
Σ 2n – 1
Sn = n=1 ≈2
LIMIT Note: If the absolute value of the ratio of the succeeding term and preceding term of a geometric series
is less them 1, we can compute the limiting value of the infinite series as a real number and it converges
otherwise diverses.
Absolute Value of the nth Term:
Consider an infinite series of the sequence 0.63 as
0.63 + 0.0063 + 0.000063 + ..............
or, 16030 + 63 + 63 + ..................
10000 1000000
or, 16032 + 63 + 63 + ..................
104 106
or, 63 × 10– 2 + 63 × 10– 4 + 63 × 10– 6 + ........... up to 63 × 10– 2n.
When n is countable infinitely large number, the last term of the series is a positive number nearer to zero
(0). That is the absolute value of the nth term of the series is 0.
Example 1 : The given square of the length 4 cm is bisected as 4 cm 4 cm
shown in the shaded portion. Write the sequences Fig (i)
formed by the areas of the shaded portions. Fig (ii) Fig (iii)
Find the limiting value of the sum of all shaded
portions.
Solution: From the given square having length 4 cm,
Area of square = l2 = 42 = 16 cm2
Area of the shaded portion in fig (i) = 1 × 16 = 8 cm2
Area of the shaded portion in fig (ii) = 2 +
Area of the shaded portion in fig (iii) = 8 + 1 × 8 = 8 + 4 = 12 cm2 14 cm2 and so on.
+ 1 × 4 = 8 + 4+2=
8 2
4 2
∴ The required sequence of the area of the shaded portion in the given square
is 8, 12, 14, ................
Since the area of the square of the area is 16 cm2, so the area of shaded portion does not
exceed 16 cm2. If the shaded area is continuously increasing by half of previous area, the
value of the shaded area is nearer to 16 cm2. Hence, the limiting value of the sum of the
shaded portions in the last square is 16 cm2.
Example 2 : Write down the first five terms of the sequence having the nth term tn = 1 + (–n1)n. Then find
the absolute value of last term. 2n
Solution: Given, tn = 1 + (– 1)n
2n n
If n = 1, then t1 = 1 + (– 1)1 = 1 – 1
21 1 2
60 Illustrated Optional Mathematics-9
If n = 2, then t2 = 1 + (– 1)2 = 1 + 1
22 2 4 2
If n = 3, then t3 = 1 + (– 1)3 = 1 – 1
23 3 8 3
If n = 4, then t4 = 1 + (– 1)4 = 1 + 1
24 4 16 4
If n = 5, then t5 = 1 + (– 1)5 = 1 – 1
25 5 32 5
4 cm
Therefore, the required first five terms are2 cm 1 – 1 , 1 + 1 , 1 – 1 , 1 + 1 and 1 – 1 .
LIMIT24283164325
Again,
The series formed by the nth term is 1 – 1 + 1 + 1 + 1 – 1 + 1 + 1 + .......... and the
2 4 2 8 3 16 4
limiting value of its final term is nearer to 0. Hence, the absolute value of the last term of
the series is 0.
EXERCISE – 2.3
1. Which of the following geometric series has the limiting value as a fixed real number?
(a) 31 + 1 + 1 + ..................... (b) 2 + 3 + 4.5 + ....................
6 12
1 1
(c) 32 – 16 + 8 – 4 + .................... (d) 5 – 1 + 5 – 25 + ......................
(e) 0.3 + 0.03 + 0.003 + ................... (f) 2.9 + 5.8 + 11.6 + .......................
(g) 0.35 (h) 0.27
2. (a) The given equilateral triangle of the side 2 cm is bisected by its median as shown in the shaded
portion below. Write down the sequence formed by the area of the shaded portions. Also, find its
20th and 50th terms and find the limiting value of the sum of all shaded portions of the last triangle.
2 cm
2 cm Fig (ii) Fig (iii) Fig (iv)
Fig (i)
(b) The given rectangle of the length 8 cm and breadth 4 cm is bisected by its diagonal as shown in
the shaded portion below. Write down the sequence formed by the area of the shaded portions.
Also, find the sum of first 30 terms and 100 terms in scientific notation and find limiting value
of the sum of all shaded portions of the last rectangle.
8 cm
Fig (i) Fig (ii) Fig (iii) Fig (iv)
3. Find the first seven terms of the sequence obtained by the following nth terms. What is the
limiting value of the nth term or last term of the sequence when n is countable infinitely greater?
(a) tn = 51n (b) tn = 1 1 (c) an = 1
2n + 2n
Sum of Infinite Series 61
(d) yn = 1 + n12 (e) tn = 1 + (–n12)n (f) tn = 5 + (– 2)n
n n2 +1
4. (a) If tn = 2 + 3 is the nth term of a sequence, find t1, t2, t3, t4, t5 and t6. Also, compute t2 – t1, t3 – t2,
n2
t4 – t3, t5 – t4 and t6 – t5. When the value of n is increasing positive infinitely, what is the value
of tn? Write it.
LIMIT (b) If the nth term of a sequence is 3 – (–3n12)n, find the values of the first seven terms and the sum
of two successive terms. When the value of n is increasing up to + ∞, what is the value of the
nth term? Write it.
5. (a) What is the sum of the even number of the terms of the infinite series 1 – 1 + 1 – 1 + ...........?
(b) Find the sum of the odd number of the terms of the infinite series 1 – 1 + 1 – 1 + ................
6. (a) Find the sum of the odd number of the terms of the infinite series having nth term
tn = (– 1)n + (–21)n.
(b) What is the sum of the even number of the terms of the infinite series having nth term 3 – (–51)n?
7. Find the sum of the following infinite series:
∞ 4 (b) ∞ 7 (c) Σ–∞ (2n) Σ(d) ∞1
2n 10n n=1 3n – 2
(a) nΣ=1 Σ n=0
n=1
ANSWERS
1. (a), (c), (d), (e), (g) and (h) 2. (a) 23, 3 3, 7 3, 15 3, ........., (2n – 1) 3; (220 – 1) 3, (250 – 1) 3, 3 cm2
16 2n 220 250
4 8
(b) 16, 24, 28, 30, ............; 30 32 , 100 32 ; 32 cm2
Σ 2n Σ 2n
n=1 n=1
3. (a) 15, 215, 1125, 6125, 31125, 151625, 781125; 0 (b) 31, 15, 17, 19, 111, 113, 115; 0 (c) 12, 82, 138, 342, 550, 762, 978; 0
(d) (1 + 1), 1 + 1 , 1 + 1 , 1 + 1 , 1 + 1 , 1 + 1 , 1 + 1 ; 0
2 4 3 9 4 16 5 25 6 36 7 49
(e) (1 – 1), 1 + 1 , 1 – 1 , 1 + 1 , 1 – 1 , 1 + 1 , 1 – 1 ; 1
4 9 16 25 36 49
(f) (5 – 1), 5 + 4 , 5 – 8 , 5 + 16 , 5 – 32 , 5 + 64 , 5 – 128 ; 5
6 12 20 30 62 56
4. (a) (2 + 3), 2 + 3 , 2 + 3 , 2 + 3 , 2 + 3 , 2 + 3 ; – 94, – 152, – 478, – 42070, – 31010; 2
4 9 16 25 36
(b) 3 + 1 , 3 – 1 , 3 + 1 , 3 – 1 , 3 + 1 , 3 – 1 , 3 + 1 ; 245, 614083, 2453929, 2430907, 126720101, 351279329; 3
3 12 27 48 75 108 147
5. (a) 0 (b) 1 6. (a) – 23 (b) 3n
7. (a) 4 (b) 79 (c) 2 (d) 4.5
Project Work
Draw six squares with the length 6 cm. Now, bisect it by its diagonal and colour half of it. Each
non-coloured triangle is bisected by its median 5 times. Now colour the half at each time. Write
down the sequence formed by the area of the coloured portions in each square. Also, find the area of
coloured portions in the 50th square in scientific notation and also find limiting value of the sum of all
shaded portions of the last rectangle.
62 Illustrated Optional Mathematics-9
21.41 LIMIT OF FUNCTION
Learning Objectives
At the end of this topic, the students will be able to:
• calculate the limit of the function by using table.
y=x+2
Let us consider a function f(x) = x + 2. Put the values of LIMITf
x from 1 to 5. What are the values of f(x) or y? Y x x+2
=y
Show the function f(x) in the graph and mapping
diagram. The function f(x) = x + 2 gives a straight line f
in the graph. x+2 3
1
In the function, the variable x is called 'input' and its 24
f(x) or y is called 'out put' or the value of the function. 35
Consider another function f(x) = x2 – 24. X' O X4 6
x – Y' 5 7
x2 – 4 Input Output
x – 2
Can you find the value of the function f(x) = at x = 2? What is the value of f(2)? f(2) is meaning less
because f(2) = 00. Other indeterminate forms are ∞∞, 0 × ∞, 1∞, ∞ – ∞, 0°,
This is called indeterminate form.
∞°, etc. These indeterminate forms are the limiting conditions.
If x ≠ 2, can you find the values of the function f(x) = x2 – 24?
x –
Draw the graphs for the function f(x) = x2 and its different domains.
Y Y Y
DC
y = x2 – 4 DC
x – 2 AB
DC
AB
X' A BX X' O X X' O X
O
Y' Y' Y'
Fig (i) Fig (ii) Fig (iii)
In the fig(i), take x1 = 1 unit and x2 = 3 units. What is the area of the rectangle ABCD having length
AB = 2 units and breadth BC = 7 cm?
In the fig (ii), take x1 = 1.5 units and x2 = 2.5 units. What is the area of the rectangle ABCD having length
AB = 1 cm and breadth BC = 2 cm? In the fig (iii), take x1 = 1.9 units and x2 = 2.1 units. What is the area
of the rectangle ABCD with AB = 0.2 units and BC = 0.8 units. Similarly, when the values of x are taken
nearer to 2, the area of the rectangle ABCD approaches at the point 4. What is the value of f(x) at x = – 1.
Therefore f(2) = 4 is the value of the function in f(x) = x2 – 4 and when the values of x are nearer to 2, the
x – 2
Limit of Function 63
position of rectangle has the y-coordinate 4. This is called the limit of f(x) at x = 2. Therefore, the limit
of the function is the y-component of the fixed point in which x-component is nearer to the finite number
for x.
Example 1 : If f(x) = 3x – 1, find the values of f(3) and f(2.999).
Solution: 2 3x – 1
LIMIT Given, f(x) = 2
Now, let's put x = 3 in the given function, we get
f(3) = 3 × 3 – 1 = 9 – 1 = 8 = 4.
2 2 2
Again, let's put x = 2.999, then
f(2.999) = 3 × 2.999 – 1 = 8.999 – 1
2 2
7.997
= 2 = 3.9985
Example 2 : Estimate the limit numerically for the function f(x) = x2 – 16 at x = 4.
x – 4
x2 – 16
Solution: Given, f(x) = x – 4
The domain of f(x) exits the set of all real number except 4. Put the values of x near to 4
from left and right in the given function and fill the results in the table below:
Y x2 – 16
x – 4
x 3.9 3.99 3.999 3.9999 10 f(x) =
f(x) 7.9 7.99 7.999 7.9999 8
6
x 4.1 4.01 4.001 4.0001 4 X
f(x) 8.1 8.01 8.001 8.0001 2
–4 –2 O 2 4 6
These two tables show that x approaches 4 from left or right the values of function f(x)
appear to be approaching 8. i.e. when x is near to 4, f(x) is near to 8 as also shown in the
x2 – 16
graph. Hence the limit of the function f(x) = x – 4 is 8.
Example 3 : Find the limit of the function f(x) = x2, x < 2 at x = 2
6 – x, x > 2
Solution: Given,
f(x) = x2, x < 2 at x = 2
6 – x, x > 2
Put the approximate values of 2 in the given functions:
For f(x) = x2, x < 2 For f(x) = 6 – x, x > 2 2.001
x 1.9 1.99 1.999 x 2.1 2.01 3.999
f(x) 3.61 3.9601 3.996001 f(x) 3.9 3.99
From the above tables, x closes to 2, f(x) is arbitrarily close to 4. Hence, the limit of the
given function is 4.
64 Illustrated Optional Mathematics-9
EXERCISE – 2.4
1. (a) What is limit of a function? (b) What is the value of f(x) = 2x2 – 1 at x = – 1?
2. (a) If the function f(x) = 3x + 1, what are the values of f(2), f(2.99) and f(– 1.01)?
(b) What are the values of the function f(x) = 4x – 3 at x = 2, 1.09 and 2.01?
3
3. (a) In which whole number rounds off the value of f(2.99) in the function f(x) = 4x2 + 5.23?
(b) In which whole number rounds off the value of f(1.001) in f(x) = 4 – 2x2? LIMIT
4. (a) If f(x) = 3x2 + 1, find the difference between f(3.01) and f(2.99).
(b) If f(x) = 5x2 – 10, find the value of f(2.01) + f(1.99) and convert the value in whole number by
rounding off. x3 – 11.
x –
5. (a) It is given that f(x) = Does f(x) have a fixed real number at x = 1? For this, evaluate the
following:
(i) What are the values of f(x) at x = 1.1, 1.01, 1.001 and 1.0001?
(ii) What are the values of f(x) at x = 0.9, 0.99, 0.999 and 0.9999?
(iii) Can the values of f(x) that are computed in (i) and (ii) express in whole numbers by
rounding off? x
+1
(b) Does the value of g(0) denote a fixed real number in the function g(x) = x – 1? For this,
perform the following questions:
(i) What are the values of g(x) at x = – 0.1, – 0.01, – 0.001 and – 0.0001?
(ii) What are the values of g(x) at x = 0.1, 0.01, 0.001 and 0.0001?
(iii) Can the values of g(x) that are calculated in (i) and (ii) express in whole numbers by
rounding off.
6. Estimate the limit numerically for the following functions:
(a) f(x) = x2 – 9 at x = – 3 (b) g(x) = x2 – 1 at x = 1
x+ 3 x – 1
(c) f(x) = 16 – x2 at x = – 4 x+1–2
4 + x (d) h(x) = x – 3 at x = 3
(e) g(x) = x2 – 3x at x = 0 (f) f(x) = x3 – x2 + x – 1 at x = 1
x x–1
7. Find the limit of the following functions:
(a) f(x) = 2x + 1, x < 2 at x = 2 (b) h(x) = 4 – x, x < 1 at x = 1
x + 3, x > 2 4x – x2, x > 1
x2 – x – 2, x < 2 at x = 2 x2 – 25, x < 5 at x = 5
x–2 x–5
(d) g(x) = 2(x3 + 125), x > 5 at x = 5
(c) k(x) = x3 + 4, x > 2 at x = 2 5(x + 5)
x+2
ANSWERS
2. (a) 7, 9.97, – 2.03 (b) 35, 0.453, 1.68 3. (a) 31 (b) 2
4. (a) 0.36
(b) 20.001, 20 5. (a) No, (i) and (ii) show to your teacher, yes, 3
(b) No, (i) and (ii) show to your teacher, yes, 2 6. (a) – 6 (b) 2
7. (a) 5 (c) 3
(c) 8 (d) 1 (e) – 3 (f) 2 (b) 3 (d) 10
4
Limit of Function 65
21.51 NOTATIONAL REPRESENTATION
OF LIMIT
LIMIT Learning Objectives
At the end of this topic, the students will be able to:
• write down the notation of the limit of the given function.
Consider a function f(x) = x2 – 16 and fill the table for the function f(x).
x – 4
x 3.9 3.99 3.999 3.9999 4 4.1 4.01 4.001 4.0001
........ ........ ........ ........ ∞ ........ ........ ........ ........
f(x) = x2 – 16
x – 4
Can you round off the all values of f(x) into whole numbers? What is the finite value by rounding off? The
x2 – 16
finite value 8 is the limit of the function f(x) = x – 4 as x approaches 4.
Suppose f(a) denotes a finite number. The notation of f(x) approaches f(a) as x approaches a number 'a'
can be defined as follows:
If a function f(x) can be made arbitrarily close to the number f(a) taking x sufficiently close to but different
from the number 'a', from both the left and right sides of 'a', then the limit of f(x) as x approaches 'a' is
f(a). In short if x approaches or tends to 'a', then f(x) approaches f(a) or f(x) tends to f(a). It is symbolically
denoted by as follows:
The discussion of the limit concept is facilitated by using a special notation if we consider the arrow head
'→' represents the word 'approach' or 'tend'. The symbol x → a – indicates that x tends to 'a' from the left.
that is , through numbers that are less than 'a' and x → a+ indicates that x tends to a from the right, that is
through numbers that are greater than 'a'. Finally, the notation x → a signifies that x tends to 'a' from both
sides.
Then the notation of the limit of f(x) as x tends to 'a' is symbolised as lim a f(x) = f(a).
x→
In the above table, we are supposing x → 4– (for example, 3.9 3.99, 3.999 are to the left of 4 cm the
number line) and x → 4+ (for example, 4.1, 4.01, 4.001 are to the right of 4 cm on the number line)
Therefore, f(x) = lim x2 – 16 = 8.
x→4 x–4
Example 1 : Write the mathematical sentence for the notation x → 5.
Solution: The mathematical sentence for the notation x → 5 is that x approaches 5. That is the values
of x take less or greater than 5.
66 Illustrated Optional Mathematics-9
Example 2 : Find the value of lim x3 – 11.
x→1 x –
Solution: Given,
x l→im 1 x3 – 1 in which f(x) = x3 – 1 as x → 1.
x – 1 x – 1
Making tables as x → 1– and x → 1+ for f(x),
x → 1– 0.9 0.99 0.999 0.9999 This problem is solved by LIMIT
f(x) 2.71 2.9701 1.997001 2.99970001 factorization method as follow:
lim x3 – 1
x→1 x–1
= lim (x – 1)(x2 + x +1)
x → 1+ 1.1 1.01 1.001 1.0001 x→1 x–1
f(x) 3.31 3.0301 3.003001 3.00030001
= x lim 1 x2 + x + 1 [ x ≠ 1]
→
From the above both tables as x → 1 then = 12 + 1 + 1 [ Applying limit]
f(x) → 3. Hence, lim x3 – 1 = 3. = 3
x→1 x – 1
EXERCISE - 2.5
1. Write the notation of the following mathematical sentences:
(a) x approaches a.
(b) z tends to – 2.
(c) y is nearer to – 1.
(d) t approximates 0.
(e) Limit x tends to a, f(x) is equal to f(a).
(f) Limit y approaches 0, z(y) is equal to 7.
(g) x tends to a from the right.
(h) x approaches 2 from the left.
2. (a) Write the notation of the limit as x tends to 2 of a function f(x).
(b) Write the notation that the limit of a function x2 – 1 as x approaches – 1 is – 2.
x+ 1
3. Write the following limiting notations into mathematical sentences:
(a) x → a (b) y → – 3 (c) z → 4 –
(d) x → 2 + (e) lim 2x2 + 1 (f) lim x2 – 9
x→0 x→5 x–2
(g) lim f(x) = f(a) (h) lim (2x) = 4 (i) x lim 3 x2 – 9 = – 6
x→a x→2 →– x+ 3
Notational Representation of Limit 67
4. Fill the following tables and write the limiting value of the function in symbolic form:
(a) x 1.9 1.99 1.999 2.1 2.01 2.001 x→2
........ ........ ........ ........
f(x) = 2x – 1 ........ ........ f(x) → ........
(b) x – 0.1 – 0.01 – 0.001 0.1 0.01 0.001 x → ........
........ ........ ........ ........ ........ ........ f(x) → ........
LIMIT f(x) = 3x2 + x
x
(c) x 6.9 6.99 6.999 7.1 7.01 7.001 Limit notation
........ ........ ........ ........ ........ ........
g(x) = 2x + 4 x lim g(x) = ........
x+2 →7
(d) x 0.9 0.99 0.999 1.1 1.01 1.001 Limit notation
........ ........ ........ ........ ........ ........ ..........................
h(x) = x2 – 1
x – 1
5. Find the values of the following functions for the given limiting conditions by using tables and
check by factorization method if possible:
(a) lim 2x – 1 (b) lim 4x + 3 (c) lim x2 – 16
x→1 x→3 3 x→–4 x+4
(d) lim x2 – 11 (e) lim x – 1 (f) lim x2 – 25
x→1 x – x→5 x→–5 x+5
(g) lim x2 +x– 6 (h) lim x2 + 5x + 6
x→2 x–2 x → 4 x2 + 2x – 3
(i) lim f(x) where f(x) = x + 3, x < 0 (j) lim f(x) where f(x) = 2x2 – 4, x > 2
x→0 – x + 3, x > 0 x→2 x2 – 4, x < 2
x–2
ANSWERS
1 - 3 Show to your teacher.
4. (a) 3 (b) 1 (c) 2 (d) 2
(c) – 8 (d) 2
5. (a) 1 (b) 5 (j) 2 (i) 3 (e) 2
(j) 4
(f) – 10 (g) 5
Project Work
Consider two functions f(x) = x2 – 25 and g(x) = x2 – 10x + 25. For which values of x, both functions
x – 5 x–5
have indeterminate form that is meaningless? Find the limit of the both functions at the same value
that gives the functions in indeterminate form and check them by factorization method or by using
limiting condition.
68 Illustrated Optional Mathematics-9
MATRIX 3UNIT
Estimated Teaching Periods : 20
Competency
To demonstrate and application apply the concept of matrix and its operations.
UnitAt the end of this unit, the students will be able to: Learning Outcomes
MATRIX
introduce a matrix and tell its types.
solve the problems by using operations of matrices (addition, subtraction, multiplication and
transpose).
write the properties of matrix addition and multiplication.
Introduction to Matrix What do you learn?
� Basic on Matrix � Types of Matrices
Operations on Matrices � Properties of Matrix Addition
� Addition and Subtraction of Matrices
� Transpose of Matrix
Matrix Multiplication � Properties of Matrix Multiplication
� Multiplication of Matrices
Are you ready? � Monthly calendar
� Row and column
� Arrangement of numbers � Round and square brackets
� Mark sheet and Mark ledger � Null element
� Rectangular and square shapes � Equal components
� Counting the number of rows and columns
� Identify element
� Addition and subtraction of numbers
� Interchanging process in rows and columns
Specification Grid
Chapter Cognitive Knowl- Under- Applica- Higher Total Total
Domain edge standing tion Ability Ques- Marks
tions
Topic 1 mark 2 marks 4 marks 5 marks
Introduction to Matrix
3 Matrix Operations of Matrices 1 2 1 - 49
Matrix Multiplication
69
31.1 MATRIX
3.1 (A) Basic on Matrix
Learning Objectives
At the end of this topic, the students will be able to:
• define matrix. • identify the order of the matrix. • write the notation of the matrix.
MATRIX Matrix and Its components
Consider a mark ledger of the marks obtained by the students Nisha Lama, Suman Nepali and Sophia
K.C. in the subjects English, Mathematics and Science is given below:
S.N. Name Subjects English Mathematics Science
1. Nisha Lama 88 92 94
2. Suman Nepali 87 98 92
3. Sophia K.C. 80 87 79
Let’s express the above information, by omitting the name of students and subjects, in the rectangular
form and enclosed by the round (or small) or square (or big) brackets as shown below:
88 92 94 88 92 94
87 98 92 or 87 98 92
80 87 79 80 87 79
This type of the arrangement of numbers in the rectangular form enclosed by round or square brackets is
called a matrix (in plural form, matrices). The numbers arranged in the horizontal lines, are called rows
and the numbers arranged in the vertical lines, are called columns of the matrix. Also, the numbers or
objects that make the matrix are called the elements or components of the matrix. For example, let's see
the following matrix:
88 92 94 1st Row (R1) Oh! There are three
87 98 92 2nd Row (R2) rows, three columns and
80 87 79 3rd Row (R3) 9 elements of component.
1st 2nd 3rd 88 92 94
Column Column Column
(C1) (C2) (C3)
Order of Matrix 87 98 92
In the above matrix, it has 3 rows and 3 columns. So the order of the matrix is 3 by 3. It is 80 87 79 3 × 3
written as 3×3 and is read as 3 cross 3. It is represented in the matrix as shown above.
The number of rows and columns taken together in the form of cross (using‘×’ between them),
is called an order of the matrix. It is also called a dimension or size of the matrix.
70 Illustrated Optional Mathematics-9
Notation of Matrix
The matrices are generally denoted by the capital letters of the English alphabet as A, B, C, ...., X, Y, Z
and their elements or entries or components are represented by the small letters as a, b, c, ……, x, y, z.
Consider a matrix A = a b c
d e f
a b c
The matrix A has 2 rows and 3 columns. So, the order of the matrix A is A 2 × 3= d e f 2×3
denoted by 2 × 3 and is also represented as shown alongside:
Where a is the element in the 1st row and 1st column. A2×3= a b c
b is the element in the 1st row and 2nd column. d e f 2×3
c is the element in the 1st row and 3rd column.
d is the element in the 2nd row and 1st column. MATRIX
e is the element in the 2nd row and 2nd column.
f is the element in the 2nd row and 3rd column.
Generally, the standard form of the matrix in double suffix nutation is as given:
In this way, an element of a matrix, in general, is written as amn, where m is the number of the rows and n
is the number of columns to which it lies.
ith row a11 b12 c13 .... a1j .... a1n
a21 b22 c23 .... a2j .... a2n
Element ai j M m×n = a31 b32 c33 .... a3j .... a3n
jth column
.... .... .... .... .... .... ....
ai1 bi2 ci3 .... aij .... ain
.... .... .... .... .... .... ....
am1 bm2 cm3 .... amj .... amn m × n
Example 1 : If A = 1 2 3 then; (a) find the number of elements of A.
4 5 6
(b) What is the order of A? (c) find the values of a11, a13, a22 and a23.
Solution: Here, A = 1 2 3
4 5 6
Now, (a) There are 6 elements in the matrix A.
(b) The order of A is 2 × 3.
1 2 3 with a11 a12 a13 , we get
(c) Comparing the matrix A = 4 5 6 a21 a22 a23
a11 = 1, a13 = 3, a22 = 5 and a23 = 6.
Example 2 : Construct a 2 × 2 matrix whose elements aij are given by aij = 2i – j.
Solution: Here, aij = 2i – j
If i = 1 and j = 1, then a11 = 2 × 1 – 1 = 1.
If i = 1 and j = 2, then a12 = 2 × 1 – 2 = 0.
If i = 2 and j = 1, then a21 = 2 × 2 – 1 = 3.
If i = 2 and j = 2, then a22 = 2 × 2 – 2 = 2.
∴ The required 2 × 2 matrix is 1 0 .
3 2
Matrix 71
EXERCISE - 3.1(A)
1. (a) What is order of matrix? (b) Define matrix and its notations.
2. Following table displays the bill of the purchased goods by three persons:
Cost of Goods Pencils Copies Books Pens
Name of Person 25 120 235 95
Ram 27 115 185 75
24 135 315 80
Laxman
Bharat
(a) Express the information in matrix form. (b) How many elements are there in the matrix?
MATRIX (c) Find the order of the matrix.
(d) If the matrix is denoted by M, write it in the form of order.
3. Determine the order of the following matrices:
(a) [1] (b) [1 2] (c) 2 (d) [0 1 3]
1
(e) 2 3 (f) –1 2 0 12 1
4 1 –2 3 4 (g) 3 4 (h) 2
0 –1 –3
4. Find the values of a11, a21, a13, a23 and a32 from the following matrices:
1 –3 4 1 0 4 4 10 2 –2 0 6
2 4 1 –2 1 5
(a) (b) (c) – 2 3 4 (d) 0 4 – 4 1
0 –2 2 –1 5 3 4
5. (a) Construct a 2×2 matrix whose elements aij are given below.
i. aij= i × j ii. aij= 2i – 3j iii. aij = ij
(b) Construct a 3×3 matrix whose elements aij are given below.
i. aij = 2i + j ii. aij = i – j iii. aij = 2i2 – j3
2
ANSWERS
2. (b) 12 (c) 3 × 4 (d) M3 ×
4
3. (a) 1 × 1 (b) 1 × 2 (c) 2 × 1 (d) 1 × 3
(h) 3 × 1
(e) 2 × 2 (f) 2 × 3 (g) 3 × 2 (d) 2, 0, 0, - 4, 5
4. (a) 1, 2, 4, 1 (b) 1, - 2, 4, 5 (c) 4, - 2, 0, 4, - 2
iii. 1 1
5. (a) i. 1 2 ii. – 1 – 4
2 4 2×2 1 – 2 2 × 2 2 4 2×2
3/2 2 5/2 0 –1 –2 1 – 6 – 25
b) i. 5/2 3 7/2 ii. 1 0 – 1 iii. 7 0 – 19
7/2 4 9/2 3 × 3 2 1 0 3×3
17 10 – 9 3 × 3
Project Work
Make a matrix from the grade point or obtained marks of three subjects of you and your three friends.
Write the number of components, rows and column in it. Also, find the order of the matrix that you wrote.
72 Illustrated Optional Mathematics-9
3.1 (B) Types of Matrices
Learning Objectives
At the end of this topic, the students will be able to:
• write the types of the matrices. • identify equal matrices and solve the related problems.
(a) Row Matrix
A matrix having only one row is called row matrix. For example, A = [1 2 3] is a row matrix of the order
1 × 3. The general form of the row matrix is A1 × n = [a11 a12 …............….. a1n]1 × n.
(b) Column Matrix a11
A matrix having only one column is called column matrix. a21
For example: M = –1 is a column matrix of the order 2×1. Mm × 1 = ... MATRIX
0 ...
The general form of the column matrix is as shown alongside: am1 m × 1
(c) Null or Zero Matrix
A matrix having all components zero is called a null or zero matrix. It is denoted by O.
For example: O1 × 1 = [0]1 × 1, O1 × 2 = [0 0]1 × 2, O2 × 2 = 0 0 2 × 2 and O2 × 3 = 0 0 0 2 × 3 are the null or zero
matrices. 0 0 0 0 0
(d) Square Matrix
A matrix having equal number of rows and columns is called square matrix.
For example: A1 × 1 = [1]1 × 1, B2 × 2 = 2 4 and C3×3 = a b c
–1 0
2×2 d e f are the square matrices.
g h i 3×3
(e) Rectangular Matrix
A matrix having unequal number of rows and columns is called rectangular matrix. In other words, a
matrix which is not square is called rectangular matrix.
For example: A1 × 2 = [a b]1 × 2, B2 × 1 = a 2 × 1 and C2 × 3 = 1 –1 0 are the rectangular matrices.
b 2 4 –3
2×3
(f) Diagonal Matrix
A square matrix in which all the elements are zeros except the main diagonal, is called a diagonal matrix.
The main diagonal is the diagonal running from the first element a11 to the last element amn (or from left
top to right bottom) in the square matrix. The main diagonal is also called principal or leading diagonal.
Alternatively, a square matrix in which all the off diagonal elements are zeros is known as a diagonal
matrix. For example: 1 0 and –1 0 0
0 2 0 2 0 are the diagonal matrices.
0 0 3
(g) Scalar Matrix
A diagonal matrix, in which the diagonal elements are equal, is called a scalar matrix.
For example: 2 0 , –1 0 3 0 0
0 2 0 –1 and 0 3 0 are the scalar matrices.
0 3
0
Matrix 73
(h) Identity or Unit Matrix
A diagonal matrix, in which the diagonal elements are equal to 1, is called an identity or unit matrix.
For example: I1 = [1], I2 = 1 0 1 0 0
0 1 1 0 are the identity or unit matrices. The identity matrix
and I3 = 0 0 1
is denoted by In. 0
(i) Equal Matrices
Two matrices of the same order are said to be equal if their corresponding elements are equal.
For example: A = a b and B = w x are equal if a = w, b = x, c = y and d = z.
c d y z
(j) Triangular Matrices
MATRIX A square matrix in which all the elements above or below the main diagonal are zeros, is known as
triangular matrix. There are two types of triangular matrices.
(i) Upper Triangular Matrix:
A square matrix in which all the elements below the main diagonal are zeros, is known as upper triangular
matrix. For example: 1 2 and 1 2 3
0 3 0 4 5 are the upper triangular matrices.
0 0 6
(ii) Lower Triangular Matrix:
A square matrix in which all the elements above or the main diagonal are zeros, is known as lower
triangular matrix. For example: 1 0 and 1 0 0
2 3 4 5 0 are the lower triangular matrices.
6 2 9
(k) Symmetric Matrices
If row and column of a square matrix are interchanged, then it remains the same. Such matrix is called
symmetric matrix. In other words, a square matrix A = [aij] is said to be symmetric matrix if aij = aji for all
1 2 3
values of i and j. For example: a b and 2 4 5 are the symmetric matrices.
b d 3 5 6
In a b , a12 = a21 and in 1 2 3 a12 = a21, a13 = a31, a23 = a32.
b d 2 4 5
3 5 6
Example 1 : Classify the following matrices and give suitable reasons:
(a) 2 0 (b) 10
0 2 01
Solution: Here, (a) 2 0 is a scalar matrix because its main diagonal has the same elements.
0 2
(b) 1 0 is an identity matrix because its main diagonal has unit elements.
0 1
74 Illustrated Optional Mathematics-9
Example 2 : Find the values of x, y and z if x–2 y+1 = 2x + 3 2y – 1 .
z–3 0 2 0
Solution: Here, x–2 y+1 = 2x + 3 2y – 1
z–3 0 2 0
Equating corresponding elements of the given equal matrices, we get
x – 2 = 2x + 3; y + 1 = 2y – 1 z–3=2
z=2+3
or, – 2 – 3 = 2x – x; or, 1 + 1 = 2y – y or, z=5
or, x = – 5; or, y = 2 or,
Example 3 : For what values of a and b, the matrix 2a – b 0 is an identity.
3a + b 1
Solution: Since the matrix 2a – b 0 is an identity so, 2a – b 0 = 1 0 MATRIX
3a + b 1 3a + b 1 0 1
∴ 2a – b = 1 ………… (i) and 3a + b = 0 ………… (ii)
Adding the equations (i) and (ii), we get
2a – b = 1
3a + b = 0 1
5
5a = 1 ∴ a =
Putting the value of a in the eqn (ii), we get
3 × 1 + b = 0 or, b = – 3
5 5
Hence, the given matrix 2a – b 0 is the identity for a = 1 and b = – 53.
3a + b 1 5
EXERCISE - 3.1 (B)
1. (a) What is symmetric matrix?
(b) Define with examples ii. Square matrix iii. Diagonal matrix
i. Row matrix v. Equal matrices vi. Triangular matrices
iv. Identity matrix
(c) Write the types of triangular matrices.
2. Identify the types of the following matrices:
(a) [– 1] (b) [ 2 –1] (c) 0 (d) 22 1
0 0 –1 2
(e) 2 2 (f) 1 0 200 200
1 –1 0 1 (g) 0 1 0 (h) – 1 3 0
003 3 4 –2
3. State which of the following matrices are equal:
A= 1 2 B= 3 1 2 C= 1 D= 2
3 –1 1 –1 4 –2 1
E= 3 1 2 F= 1 2 G= 3 1 2 H= 2 0 –1
1 –1 4 3 –1 2 0 –1 3 2 2
Matrix 75
4. Construct the following matrices:
(a) 2 × 4 zero matrix (b) 2 × 2 unit matrix
(c) 3 × 3 scalar matrix (d) 3 × 3 symmetric matrix
5. (a) Construct a 2 × 2 matrix whose element aij = 2i × 3j. Is it symmetric matrix?
(b) Construct a 3 × 3 matrix whose elements aij = 3i2 – 2ji. Is it symmetric matrix?
(c) Construct a matrix N2 × 3 whose elements aij = (i + j)2. Name the type of matrix.
(d) If the matrix M = [aij] and aij = 2i2 – j, construct the matrix M1 × 4.
6. Find the values of x, y and z in each of the following cases:
(a) 2x = 4 (b) 2x + 1 = 3
3y 6 2–y 1
MATRIX (c) 3x + 3 z+2 = 6 2 (d) 2x – y 4 = 1 4
1 2 – 3y 1 1 – 2y 2x z + 2y x+1 1
(e) 2 – 2x 4 2y + z = 4 4 y+2
2x + y x–z 0 0 –1 0
7. (a) For what value of x, y and z, the matrices 2x – 1 0 1 and 3 0 1 are equal?
x+y–z 2y 2 2 y+1 2
(b) For what values of a and b, the matrix 2a – 3 a+b is equal to the matrix 1 2 ?
0 1 0 1
8. (a) For what values of x and y, the matrix 2x – 1 0 is an identity matrix?
3 – 2y 1
(b) If the matrix p–1 q 1 is an identity matrix, find the values of p, q, r and s.
9. (a) r s– y, is a scalar matrix?
3 2x – 4
For what values of x and the matrix 4 – 2y 3
(b) If the matrix 1 2b – c is a null matrix, then find the values of a, b, c and d.
2a + d 2c – 4
ANSWERS
2. (a) square (b) rectangular or row (c) null or column (d) rectangular
(e) square (f) identity (g) diagonal (h) lower triangular
3. A = F, B = E 4. Show to your teacher
5. (a) = 6 12 . Yes, symmetric 1 – 1 –3
12 24 (b) = 10 4 –6 . No,
(c) 4 9 16 . Rectangular 25 11 – 27
9 16 25 (d) [1 0 – 1 – 2]
6. (a) 2, 2 (b) 1, 1 (c) 1, 1, 0 (d) 1, 1, -1 (e) – 1, 2, 0
(b) 2, 0 9. (a) 2, 2 (b) 0, 1, 2, 0
7. (a) 2, 1, 1
8. (a) 1, 3 (b) 2, 0, 0, 2
2
Project Work
Collect buying bills in your home and make the matrixes by keeping the numerical figures only. Write
their order and types.
76 Illustrated Optional Mathematics-9
31.21 OPERATIONS ON MATRICES
3.2 (A) Addition and Subtraction of Matrices
Learning Objectives
At the end of this topic, the students will be able to:
• add the matrices. • subtract the matrices.
There are mainly four operations on matrices. They are as follows:
i) Addition of matrices ii) Subtraction of matrices MATRIX
iii) Multiplication of matrices by a scalar iv) Multiplication of matrices
Addition of Matrices
If A and B are two matrices of the same order, then A and B are said to be conformable or compatible for
addition. The sum of the matrices A and B is denoted by A + B and is obtained by adding the corresponding
elements of A and B.
For example: if A = 1 2 3 and B= 6 5 4
4 5 6 3 2 1 2×3
2×3
Then A + B = 1 2 3 + 6 5 4 = 1+6 2+5 3+4 = 7 7 7
4 5 6 3 2 1 4+3 5+2 6+1 7 7 7 2×3
Subtraction of Matrices
If A and B are two matrices of the same order, then A and B are said to be conformable for subtraction.
The difference of the matrices A and B is denoted by A - B is obtained by subtracting the elements of B
from the corresponding elements of A.
For example; if A = 3 2 and B = 1 0
–1 4 2 – 1 2×2
2×2
Then, A – B = 3 2 – 1 0 = 3–1 2–0 = 2 2
–1 4 2 –1 –1 – 2 4 – (– 1) –3 5 2×2
Example 1 : If A = 1 2 , B = –1 2 and C = 4 0 then find A + B + C and A – B + C.
3 4 0 3 –2 5
Solution: Here, A = 1 2 ,B= –1 2 and C = 4 0
3 4 0 3 –2 5
Now, A + B + C = 1 2 + –1 2 + 4 0 = 1–1+4 2+2+0 = 4 4
3 4 0 3 –2 5 3+0–2 4+3+5 1 12
Again,
A–B+C= 1 2 – –1 2 + 4 0 = 1 – (–1) + 4 2–2+0 = 6 0
3 4 0 3 –2 5 3–0–2 4–3+5 1 6
Operations on Matrices 77
Example 2 : Solve the equation 2 1 –M= –7 4 .
5 0 2 6
Solution: Here, 2 1 –M= –7 4
5 0 2 6
or, 2 1 – –7 4 =M ∴ M= 9 – 3 .
5 0 2 6 3 – 6
Example 3 : If A = 4 4–y ,B= –2 2y – 2 and A + B = 2 3 , find the values of x and y.
2x – 1 1 3 –1 4 0
Solution: Here, A = 4 4–y and B = –2 2y – 2
2x – 1 1 3 –1
Now, A + B = 2 3
4 0
MATRIX
or, 4 4–y + –2 2y – 2 = 2 3
2x – 1 1 3 –1 4 0
or, 2 y+2 = 2 3
2x + 2 0 4 0
∴ 2x + 2 = 4 and y + 2 = 3
or, 2x = 4 – 2 and y = 3 – 2
or, x = 2 and y = 1
2
∴ x = 1 and y = 1.
Example 4 : If 2A – B = 1 2 and B – A = 3 –4 , then find the matrices A and B.
–2 3 2 7
Solution: Here, 2A – B = 1 2 ............ (i) and
–2 3
Oh! These equalities are
B–A= 3 –4 ............ (ii) simultaneous linear equations,
2 7
4 –2 which can be solved as in
By adding equations (i) and (ii), we get A = 0 10 algebra in Compulsory Math.
Putting the value of A in eqn (ii), we get
B– 4 –2 = 3 –4
0 10 2 7
or, B= 3 –4 + 4 –2 = 7 –6
2 7 0 10 2 17
∴ A= 4 –2 and B = 7 –6 .
0 10 2 17
EXERCISE - 3.2 (A)
1. (a) If A = [a] and B = [b], what is the sum of A and B?
(b) If P = [a b] and Q[c d], what is the matrix of P – Q?
2. If A = –1 2 , B = 2 –3 and C = 0 1 , find the following:
0 1 4 1 2 0
(a) A + B (b) A + C (c) A – B (d) A – C
(h) B + C – A
(e) A + B – C (f) A – B – C (g) A – B + C
78 Illustrated Optional Mathematics-9
3. (a) If X = 2 3 , find the matrix Y such that X + Y= 3 –1 .
–1 0 2 1
(b) If A = 2 –5 and B = 4 –1 , find the matrix C such that A = B + C.
–1 4 0 3
4. Solve the following:
(a) X– 1 –2 = 3 4 + I, where I is a 2 × 2 identity matrix
2 4 0 –2
(b) 9 1 + 1 5 – X = O, where O is a 2 × 2 null matrix.
5 3 7 – 11
5. (a) If x 3 + 1 2 = 4 5 , find the values of x and y.
0 2y – x 0 –2 0 3
(b) If 1 x = 7 0 – 6 4 , find the values of x and y. MATRIX
5 1+x 10 5 5 y
6. (a) If A = 5 b , B = 6 4 and C = 1 – 1 , find the values of a and b such that A = B + C.
a 0 2 6 2a – 6
(b) If P = a b , Q = 1 2 and R = 5 6 , find the values of a, b, c and d such that P = Q + R.
c d 3 4 6 8
7. (a) If 2A + B = 1 3 and A – B = –4 6 , find the matrices A and B.
5 –2 –2 –1
(b) Find the matrices C and B if 2C + B = 1 2 and C – 2B = 8 6 .
3 0 4 5
8. (a) If A = 1 2 3 and B = 2 –1 1 , find the matrix C such that A + B + C is zero matrix.
–2 0 1 2 –2 3
(b) If P = 2 4 and Q = –2 2 , find the matrix R such that P – Q + R is an identify matrix.
3 5 4 1
ANSWERS
2. (a) 1 –1 (b) –1 3 (c) –3 5 (d) –1 1 (e) 1 –2 (f) –3 4
4 2 2 1 –4 0 –2 1 2 2 –6 0
(g) –3 6 (h) 3 –4 3. (a) 1 –4 (b) –2 –4 4. (a) 5 2 (b) 10 6
–2 0 60 3 1 –1 1 2 3 12 –8
5. (a) 3, 4 (b) – 4, 8 6. (a) –2, 3 (b) 6, 8, 9, 12
7. (a) –1 3 3 –3 (b) –3 –2 22 8. (a) –3 –1 – 4 (b) –3 –2
1 –1 3 0 –1 –2 21 0 2 – 4 1 –3
Project Work
Collect the information of stock of 5 types of the same kind of fruits from two fruit shops and write them
in matrix form. Find the total stocks in both shops. Also, record the remaining fruits on next days after
selling some fruits and find the number of fruits sold by fruit-sellers.
Operations on Matrices 79
3.2 (B) Properties of Matrix Addition
Learning Objectives
At the end of this topic, the students will be able to:
• prove the properties of matrix addition.
Addition of matrices satisfies the following properties:
(i) Closure Property: If A and B are two matrices of the same order, then A + B also has the same order
as that of A or B. 1 2 2 1
4 5 1 3 2×2
For example; if A = and B =
2×2
Then A + B = 1 2 + 2 1 = 3 3 , which is a 2 × 2 matrix.
4 5 1 3 5 8
2×2
MATRIX Hence, the closure property is satisfied in addition.
(ii) Commutative Property: If A and B are two matrices of the same order, then A + B = B + A.
For example: if A = 1 –1 and B = 0 2 .
3 4 1 2
2×2 2×2
Then A + B = 1 –1 + 0 2 = 1 1 and B + A = 0 2 + 1 –1 = 1 1
3 4 1 2 4 6 1 2 3 4 4 6 2×2
2×2
∴ A + B = B + A.
Hence, the commutative property is satisfied in addition.
(iii) Associative Property: If A, B and C are three matrices of the same order, then (A + B) + C = A + (B + C).
For example, if A = 2 1 , B= –1 2 and C = 0 –1
3 4 1 2 2 –1
Then A + B = 2 1 + –1 2 = 1 3
3 4 1 2 4 6
∴ (A + B) + C = 1 3 + 0 –1 = 1 2 and B + C = –1 2 + 0 –1 = –1 1
4 6 2 –1 6 5 1 2 2 –1 3 1
∴ A + (B + C) = 2 1 + –1 1 = 1 2 ∴ (A + B) + C = A + (B + C).
3 4 3 1 6 5
Hence, the associate property is satisfied in addition.
(iv) Additive Identity Property: If A is a matrix of any order and O is the null matrix of the same order,
then A + O = O + A = A.
For example; If A = 1 2 and O= 0 0 then,
–3 4 0 0
A+O= 1 2 + 0 0 = 1 2 = A, and O + A = 0 0 + 1 2 = 1 2 =A
–3 4 0 0 –3 4 0 0 –3 4 –3 4
∴ A + O = O + A = A.
Hence, O is an additive identity in addition.
(v) Additive Inverse Property: If A is a matrix of any order, then there exists another matrix B of the
same order such that A + B = B + A = O. Then B = – A. So – A is the additive inverse of A.
For example; If A = 2 –1 and then – A = –2 1
–3 4 3 –4
80 Illustrated Optional Mathematics-9
Now, A + (– A) = 2 –1 + –2 1 = 0 0 =O
–3 4 3 –4 0 0
(– A) + A = –2 1 + 2 –1 = 0 0 =O
3 –4 –3 4 0 0
Hence, the addition of matrices satisfies the additive inverse.
Example 1 : If A = 1 3 4 , B = 0 –1 2 and C = –1 0 4 , prove that:
–1 –2 –3 –3 4 1 0 –1 3
(A + B) - C = A + (B - C).
Solution: Here, A = 1 3 4 , B= 0 –1 2 and C = –1 0 4
–1 –2 –3 –3 4 1 0 –1 3
MATRIX
Now, A + B = 1 3 4 + 0 –1 2 = 1 2 6
–1 –2 –3 –3 4 1 –4 2 –2
∴ LHS = (A + B) – C = 1 2 6 – –1 0 4 = 2 2 2
–4 2 –2 0 –1 3 –4 3 –5
Again, B – C = 0 –1 2 – –1 0 4 = 1 –1 –2
–3 4 1 0 –1 3 –3 5 –2
∴ RHS = A + (B – C) = 1 3 4 + 1 –1 –2 = 2 2 2
–1 –2 –3 –3 5 –2 –4 3 –5
Hence, (A + B) – C = A + (B – C)
EXERCISE - 3.2 (B)
1. If A = –1 2 , B = 2 –3 and C = 0 1 , prove the following relations:
0 1 4 1 2 0
(a) A + B = B + A (b) (A + B) + C = A + (B + C) (c) (A + B) – C = A + (B - C)
(d) A – B = – (B – A) (e) (C – A) +B = C – (A – B) (f) (B – A) – C = B – (A + C)
2. (a) If A = 1 2 and B = –4 0 , verify the commutative property of matrix addition.
3 4 –2 1
(b) If P = 1 2 3 , Q = –1 2 –4 and R = 2 4 0 , prove the associative property of
4 5 6 0 –2 3 1 2 –2
matrix addition.
3. If A = 2 3 ,B= 0 –2 and O = 0 0 , prove the following properties.
4 –1 4 5 0 0
(a) A + O = O + A = A (b) B + (– B) = (– B) + B = O
4. (a) Find the additive inverse of the matrix M = 2 –1 .
0 4
(b) If the additive inverse of the matrix a–2 b is 2 0 , find the values of a and b.
1 3 –1 –3
5. (a) If A = 4 5 and B = –2 4 , find A – B and B – A. Is the commutative property over
–1 3 6 –1
subtraction satisfied?
(b) If L = –2 3 0 ,M= 4 5 –3 and N = –1 2 4 , show that (L – M) – N ≠ L – (M – N).
0 4 2 1 2 –2 2 –3 5
Operations on Matrices 81
6. (a) For what values of x and y, the sum of matrices A = 2x – 1 5 and B = 2 2x – 5 will be
identity matrix ? 2y + 1 2 3 x–1
(b) If A = 2x + 1 3 1 , B = 1 2 , C = 4 3–z , D = 6 2 and A + B = C + D, find the
2 2y – 5 5 3 2x – 1 4 5
values of x, y and z.
7. Find the values of x, y, z, p, q, r, s and t if,
(a) x y + 2 x = 2x – 1 2 + 4 3y – 1
z t 1 4 3z – 1 2t – 1 2 0
(b) 2p – 1 2q + 1 2–q = 3p 4 + 4 3q – 1
1 r–s 3s 2r – 1 2s – r 4 2 0
MATRIX ANSWERS
4. (a) –2 1 (b) 0, 0 5. (a) 6 1 , –6 – 1 , No
0 –4 (b) 4, 4, 0 –7 4 7 – 4
6. (a) 0, – 2 7. (a) –1, –1, 0, 5 (b) – 4, – 1, 3, – 1
22 2
3.2 (C) Transpose of Matrix
Learning Objectives
At the end of this topic, the students will be able to:
• transpose a matrix. • prove the properties of the transpose of the matrix.
A new matrix obtained by interchanging the corresponding rows and columns of a matrix is called the
transpose of the matrix. The transpose of a matrix A is denoted by A', AT or At. If a matrix has the order of
m×n, then its transpose has the order n × m which is obtained by interchanging the order of A.
For example; if A = 1 2 3 , then the transpose of A is AT = 1 4
4 5 6 2 5
2×3 3 6 3×2
Properties of Transpose of Matrix
(i) Double Transpose Property
The transpose of the transpose of a matrix is a matrix itself. i.e (AT)T = A.
For example; if A = 1 2 3 then AT = 1 4
4 5 6 2 5
2×3 3 6 3×2
Again, (AT)T = 1 2 3 ∴ (AT)T = A.
4 5 6
2×3
82 Illustrated Optional Mathematics-9
(ii) Transpose Over Addition
The transpose of the sum of two matrices is the sum of their transposes. i.e (A + B)T = AT + BT.
For example;
if A= 1 0 and B = –2 1 then,
2 –1 3 4
A+B= 1 0 + –2 1 = –1 1
2 –1 3 4 5 3
∴ (A + B)T = –1 5
1 3
Again,
AT = 1 2 and BT = –2 3
0 –1 1 4
MATRIX
∴ AT + BT = 1 2 + –2 3 = –1 5
0 –1 1 4 1 3
∴ (A + B)T = AT + BT.
Example 2 : Given that A = x–y –3 and B = 4 3 . Find the values of x and y if AT = B.
5 2y 5 –4
Solution: Given, A = x–y –3 ∴ AT = x–y 5 and B = 4 3
5 2y –3 2y 5 –4
Now, AT = B
or, x––3y 25y = 4 3
5 –4
∴ 2y = – 4
or, y = – 2 and x – y = 4
or, x + 2 = 4
or, x = 4 – 2 = 2
∴ x = 2 and y = – 2.
EXERCISE - 3.2 (C)
1. (a) What is transpose of matrix?
(b) What is the transpose of the matrix AT?
2. Transpose the following matrices: 1
(c) 2
(a) A = [2 4] (b) B= 2 4 (d) 2 4 –1
147 1 2 3 30 5
3. (a) If P = 2 5 8 , find PT and (PT)T. Determine the relation between P and (PT)T.
369
(b) If Q = –1 –2 –3 , find QT and (QT)T. State the relation between Q and (QT)T.
3 2 1
Operations on Matrices 83
4. (a) If M = 2 –1 and N = –3 0 , find M' + N' and (M + N)'. Are M' + N' and (M + N)' equal ?
4 6 4 1
(b) If A = 3 4 5 and B = –1 –2 –3 , find At - Bt and (A – B)t. Can you write At – Bt =
1 2 3 5 4 3
(A – B)t ?
(c) If K = 3 4 5 and c = – 2, prove that (cK)T = cKT
1 2 3
5. If P = 3 –4 ,Q= 3 1 and R = 4 3 , prove that:
2 6 0 –2 –1 2
(a) (PT)T = P (b) (P + Q)t = Pt + Qt
(c) (P – Q)t = Pt – Qt (d) (P + Q + R)t = Pt + Qt + Rt
MATRIX (e) (P – Q – R)t = Pt – Qt – Rt (f) (P – R)t = – (Rt – Pt)
6. If A = 1 2 , B = 1 2 and C = –1 3 , find :
3 4 0 3 2 5
(a) AT + B (b) A – BT (c) AT – BT
(d) AT – BT + C (e) A – BT – CT (f) (A + B)T – CT
7. (a) If A = 1 3 , prove that A + AT is a symmetric matrix.
4 2
234
(b) If M = – 1 2 – 3 , show that M + M T is a symmetric matrix.
015
8. (a) If A = cos θ sin θ and A + AT = I, find the measure of the angle θ for first quadrant.
– sin θ cos θ
(b) If P = sin2 θ –1 and P + PT = I, find the values of the acute angles θ and α.
cot2 α 1/2
9. (a) Given that A = x+1 2 and B = 3 1 , find the values of x and y such that AT = B.
2y – 3 0 2 0
(b) Given that X = 2x – y z–y and Y = x–y 2 , find the values of x, y and z such that X = Y'.
2y – z 1 1 1
ANSWERS
2 2 1 23 3. (a) P = (PT)T (b) Q = (QT)T
4 4 2
2. (a) (b) (c) [1 2 3] (d) 4 0
–1 5
4. (a) Yes (b) Yes 6. (a) 2 5 (b) 0 2 (c) 0 3 (d) –1 6
2 7 1 1 0 1 2 6
(e) 1 0 (f) 31 8. (a) 60° (b) 45°, 45° 9. (a) 2, 2 (b) 0, 3, 4
–2 –4 12
84 Illustrated Optional Mathematics-9
31.31 MATRIX MULTIPLICATION
3.3 (A) Multiplication of Matrices
Learning Objectives MATRIX
At the end of this topic, the students will be able to:
� multiply a matrix by a scalar.
� multiply two matrices.
Multiplication of a Matrix by a Scalar
Consider a matrix A = 1 –1 .
2 4
1 –1 1 –1
Then, A+A= 1 –1 + 1 –1 Similarly, –A–A=– 2 4 – 2 4
2 4 2 4
–1–1 +1+1
or, 2A = 1+1 –1 – 1 or, – 2A = –2–2 –4–4
2+2 4+4
–2 2
or, 2A = 2 –2 or, – 2A = –4 –8
4 8
–2 × 1 – 2 × (– 1)
or, 2A = 2×1 2×–1 or, – 2A = –2 × 2 –2×4
2×2 2×4
1 –1
∴ 2A = 2 1 –1 . ∴ –2A = –2 2 4 .
2 4
a b
From the above examples, we conclude that a matrix A = c d multiplied by a scalar quantity k, the
scalar k multiplies all elements of that matrix and the order of the matrix after multiplication is the same
as the initial matrix.
i.e. kA = k a b = ka kb .
c d kc kc
Properties of multiplication of a matrix by a scalar
Let A and B be two matrixes and m and n be two scalar quantities, then the properties of multiplication of
a matrix by a scalar quantity satisfy the following properties :
(i) (m + n)A = mA + nA.
(ii) m(A + B) = mA + mB, where A and B must have the same order.
(iii) IA = A = AI, where I is an identity matrix.
(iv) If n = 0, then nA = O, where O is a null matrix of the order of A.
Matrix Multiplication 85
Example 1 : If A = 0 1 and B = 2 –2 then find the matrix 3A + 2B.
2 3 4 1
Solution: Given, A = 0 1 , B = 2 –2
2 3 4 1
Now, 3A + 2B = 3 0 1 +2 2 –2 = 3×0 3×1 + 2×2 2 × (– 2)
2 3 4 1 2×3 3×3 2×4 2×1
= 0 3 + 4 –4 = 0+4 3–4 = 4 –1
6 9 8 2 6+8 9+2 14 11
∴ 3A + 2B = 4 –1 .
14 11
MATRIX Example 2 : If 2 2 –3 + 1 y = z –1 , find the values of x, y and z.
1 x 4 –2 6 3x
Solution: Here, 2 2 –3 + 1 y = z –1
1 x 4 –2 6 3x
or, 24 –2x6 + 1 y = z –1 or, 5 y–6 = z –1
4 –2 6 3x 6 2x – 2 6 3x
Equating the corresponding components, we get
z=5
y – 6 = –1 or, y = 6 – 1 = 5
2x – 2 = 3x, or, 3x – 2x = – 2 or, x = – 2
∴ x = – 2, y = 5 and z = 5.
Example 3 : If 2A + B = 4 –3 and 3A - 2B = 6 13 then find the matrix A and B.
5 7 4 –7
Solution: Here, 2A + B = 4 –3 …..…(i) 3A – 2B = 6 13 ……. (ii)
5 7 4 –7
Multiplying by 2 in eqn(i) and adding with eqn(ii), we get
4A + 2B = 8 –6
10 14
(+) 3A – 2B = 6 13
4 –7
7A = 14 7
14 7
14 7
1 14 7 7 7 2 1
or, A = 2 14 7 = 14 7 = 2 1
77
Substituting the matrix A in eqn(i), we get
2 2 1 +B= 4 –3
2 1 5 7
or, 44 22 + B = 4 –3 or, B= 4 –3 – 4 2 = 4–4 –3–2 = 0 –5
5 7 5 7 4 2 5–4 7–2 1 5
∴ A= 2 1 and B = 0 –5 .
2 1 1 5
86 Illustrated Optional Mathematics-9
Multiplication of Two Square Matrices
Let us consider an example,
Hari bought 5 books at the rate of Rs. 150 and 10 copies at the rate of Rs. 20.
Let’s make a bill.
SN Articles Rate Quantity
1 Books Rs. 150 5 pcs
2 Copies Rs. 20 10 pcs
What is the total cost?
∴ Total Price = Rs. 150 ×5 + Rs. 20 × 10 = Rs. 750 + Rs. 200 = Rs. 950.
Again, rewriting the rates in row matrix; A = [150 20]1 × 2
Similarly, rewriting the quantities in column matrix; B = 5 2×1 MATRIX
10
Now, multiplying A and B as given below;
A × B = [150 20] 5 = [150 × 5 + 20 × 10] = [950].
10
∴ A × B = [950]1 × 1
Furthermore, we consider another example,
A car company has produced two models of cars. The number of seats and windows in each car are
shown in the following table:
Model of cars I II
No. of seats 45
No. of windows 24
The target of the production of these two models of cars in the first two months is show in the given table:
Models of cars 1st month 2nd month
I 3 2
II 2 4
What is the total number of seats and windows required to produce the both models of car in the first two
months?
We can write the above information in the form of matrix;
A= 4 5 and B = 3 2
2 4 2 4 2×2
2×2
Total number of seats required in the 1st month = 4 × 3 + 5 × 2 = 22.
Total number of seats required in the 2nd month = 4 × 2 + 5 × 4 = 28.
Total number of windows required in the 1st month = 2 × 3 + 4 × 2 = 14.
Total number of windows required in the 2nd month = 2 × 2 + 4 × 4 = 20.
Now, representing the above results in the form of matrix
A×B= 4 5 × 3 2 = 4×3+5×2 4×2+5×4 = 22 28
2 4 2 4 2×3+4×2 2×2+4×4 14 20 2 × 2
Matrix Multiplication 87
Steps of Multiplication of Two Matrices:
Let A = a1 b1 and B = c1 d1 . Then
a2 b2 c2 d2
2×2 2×2
Step I: First row × First column: A × B = a1 b1 × c1 d1 = a1c1 + b1c2 ... ...
a2 b2 c2 d2 ... ... ... ...
Step II: First row × Second column: A × B = a1 b1 × c1 d1 = a1c1 + b1c2 a1d1 + b1d2
a2 b2 c2 d2 ... ... ... ...
Step III: Second row × First column: A × B = a1 b1 × c1 d1 = a1c1 + b1c2 a1d1 + b1d2
a2 b2 c2 d2 a2c1 + b2c2 ... ...
Step IV: Second row × Second column: A × B = a1 b1 × c1 d1 = a1c1 + b1c2 a1d1 + b1d2
a2 b2 c2 d2 a2c1 + b2c2 a2d1 + b2d2
After observing the above examples, we notice that the number of column of A and the
MATRIX number of rows of B are equal and also the order the multiplication of matrices A × B has the
number of rows of A × the number of column of B. If A m × p and Bp × n are two matrices then
Am×p × Bp×n = (A×B)m×n = Cm×n, where C = A × B.
If the matrices C = 1 2 and D = [2 4]1 × 2 cannot be multiplied because the matrix C
2 3
2×2
has 2 columns and the matrix D has 1 row, which are not equal.
Two matrices are conformable for multiplication (or multiplication of two matrices is
defined) if the number of columns in the pre-matrix (first matrix) is equal to the number of
rows in the post matrix (second matrix). An element of the product matrix is obtained by
adding the products of the corresponding elements of the row of pre-matrix and the column
of post matrix. The order of the product matrix is determined by the number of rows in the
pre-matrix and by the number of columns in the post-matrix.
Example 1 : If P = 3 0 ,Q= a b and P + Q = PQ, find the values of a, b and c.
0 2 0 c
Solution: Given, P= 3 0 , Q = a b
0 2 0 c
Now, P + Q = PQ
or, 30 02 + a b = 3 0 ab
0 c 0 2 0c
or, 3 +0 a 2 b c = 3a 3b
+ 0 2c
Equating the corresponding elements, we get
3 + a = 3a, or, 3a – a = 3 or, 2a = 3 or, a = 32.
or, 2b = 0 or, b = 0.
b = 3b, or, 3b – b = 0 or, c = 2.
∴ 2 + c = 2c or, 2c – c = 2
a = 32, b = 0 and c = 2.
88 Illustrated Optional Mathematics-9
Example 2 : If M = 2 3 , find the matrix N such that MN = 6 11 .
–1 4 8 0
Solution: Let, N= a b , then MN = 6 11
c d 8 0
or, 23 a b = 6 11 or, –2aa ++ 3c 2b + 3d = 6 11
–1 4 c d 8 0 4c – b + 4d 8 0
Comparing the corresponding elements, we have
2a + 3c = 6 …….. (i) 2b + 3d = 11 …….. (ii)
– a + 4c = 8 …….. (iii) – b + 4d = 0 …….. (iv)
Solving eqn(i) and (iii), then a = 0 and c = 2.
Solving eqn(ii) and (iv), then b = 4 and d = 1.
∴ The required matrix N is 0 4 .
2 1
2 1 MATRIX
Example 3 : Which matrix pre-multiples to the matrix –3 –1 to get [6 2]?
Solution: Here, the order of the post matrix 2 1 is 2 × 2 and that of product of matrix [6 2] is
–3 –1
1 × 2. So, the order of the pre-matrix must be 1 × 2. b]1 × 2 2 1 2 × 2 = [6 2]1 × 2
Now, let the required pre-matrix be [a b], then [a –3 –1
[a b] 2 1 = [6 2] or, [2a – 3b a – b] = [6 2]
–3 –1
Equating the corresponding elements, we get
2a – 3b = 6 ………… (i) and a – b = 2 …………… (ii)
Solving eqn (i) and eqn (ii), then a = 0 and b = – 2.
Hence, the required pre-matrix is [0 –2].
Example 4 : If A = 1 2 , then show that A2 – 2A = 5I, where I is unit matrix of order 2×2.
3 1
Solution: Given, A = 1 2 ,I= 1 0
3 1 0 1
Now, A2 – 2A = 5I
or, 31 21 13 2 –2 1 2 =5 1 0 or, 7 4 – 2 4 = 5 0
1 3 1 0 1 6 7 6 2 0 5
or, 05 50 = 5 0 . Proved
0 5
EXERCISE - 3.3 (A)
1. (a) Can A2 × 1 and B 1 × 4 multiply? (b) Are P3 × 4 and B3 × 3 multiped?
2. (a) If A = 1 2 and B = 5 0 , find 3A – 2B.
3 4 –2 3
(b) If A = 2 –1 4 and B = 0 –2 2 , find 2A + 1 B.
3 2 –4 2 1 3 2
3. (a) If 2 x 2 –3 2 z = 1 2 , find the values of x, y and z.
–1 y –2 4 4 0
(b) If P = 2 a ,Q= –1 0 and 2P + 3Q = d 3 , find the values of a, b, c and d.
–1 b c 3 –2 1
Matrix Multiplication 89
4. (a) If A + 2B = 2 –4 and 2A – B = 4 7 , find the matrices A and B.
–2 6 –4 2
(b) If 2P – 3Q = 2 –6 and 3P – 2Q = 3 1 , find the matrices P and Q.
14 –7 16 –8
5. For which of the following product of two matrices is defined or not? State reason:
(a) [2 1] 3 (b) [2 3] 2 4 (c) 2 2 13 –1 (d) 1 1 –2
4 3 –1 –1 4 20 4 2 05
6. If A = 1 –2 and B = 3 4 0 then,
0 3 1 2 –2
(a) find AB. (b) Is BA compatible ? Give reason. Find BA if compatible.
7. Find the product of the following pairs of matrices and write down their order:
MATRIX (a) [1 2] –2 (b) [3 4] 2 –2 1 (c) 1 –2 1
1 1 0 –3 05 2
–2 (e) 14 0 1 –2 (f) 0 22 1
(d) [1 2 0] 0 2 –3 –2 2 3 –1 4 0 –2
01
1
8. If P = 1 2 ,Q= 0 1 and R = 1 –2 then find the following:
3 –1 2 4 4 3
(a) PQ (b) (2Q)R (c) R(3P) (d) R2
9. (a) Which matrix post multiplies 1 –1 to get a matrix 1 1 ?
2 3 –3 2
(b) Which matrix post multiplies –1 to get a matrix 5 4 ?
0 0 0
(c) Which matrix pre-multiplies 1 0 to get a matrix 3 –1 ?
–4 –2 0 –2
(d) Which matrix pre-multiplies to the matrix 2 –1 to get [3 – 4] ?
5 0
ANSWERS
–7 6 4 –3 9 3. (a) 27, 6, 23 (b) 23, – 4, 0, 1
13 6
2. (a) (b) 7 9 – 13
2 2
2 2 0 –3 1 3 0 0
4. (a) –2 2 , 0 2 (b) 4 –2 , –2 1 5. (a), (b), (c) are defined and (d) is not defined.
6. (a) 13 0 4 (b) No 7. (a) [0]1 × 1 (b) [10 – 6 – 9]1 × 3
6 –6
(c) –103 (d) [ – 2]1 × 1 (e) 9 10 (f) 0 –4
–4 – 15 – 2 – 9 2×2
2×1 2×2
8. (a) 4 9 (b) 8 6 (c) – 15 12 (d) –7 –8
–2 –1 36 16 39 15 16 1
1
9. (a) 0 1 (b) [– 5 – 4] (c) 5 2 (d) [4 – 1]
–1 0
41
Project Work
Jenuam bought 5 kg of potato at the rate of Rs. 32 and 3 kg of tomato at the rate of Rs. 65. How much
amount should he pay to the shopkeeper for it? Calculate the total cost by using matrix multiplication.
90 Illustrated Optional Mathematics-9
3.3 (B) Properties of Matrix Multiplication
Learning Objectives
At the end of this topic, the students will be able to:
verify the properties of matrix multiplication.
Property-1: Associative Property
Suppose, A, B and C are the three matrices such that (AB)C and A(BC) are compatible then (AB)C =
A(BC).
Example: If A = 1 2 ,B= 0 1 and C = 2 4 then,
3 4 –2 –1 0 3
(AB)C = 12 0 1 24 MATRIX
34 –2 –1 03
= 1 × 0 + 2 × (– 2) 1 × 1 + 2 × (– 1) 2 4 = –4 –1 24
3 × 0 + 4 × (– 2) 3 × 1 + 4 × (– 1) 0 3 –8 –1 03
= –8+0 – 16 – 3 = –8 – 19
– 16 + 0 – 32 – 3 – 16 – 35
Again, A(BC) = 1 2 0 1 24
3 4 –2 –1 0 3
= 1 2 0 3 = –8 – 19
3 4 –4 – 11 – 16 – 35
∴ (AB)C = A(BC). Proved.
Property-2: Distributive Property over Addition
Suppose, A, B and C are three matrix such that A(B + C) and AB + AC are compatible, then A(B + C) = AB
+ AC
Example: If A = 0 –1 ,B= 1 2 and C = 2 1 then,
2 1 3 0 0 –3
A(B + C) = 0 –1 1 2 + 2 1
2 1 3 0 0 –3
= 0 –1 3 3 = –3 3
2 1 3 –3 9 3
–1
Again, AB + AC = 0 1 1 2 + 0 –1 2 1
2 3 0 2 1 0 –3
= –3 0 + 0 3 = –3 3
5 4 4 –1 9 3
Hence, A(B + C) = AB + AC. Proved.
Property-3: Existence of Identity Property
If A is a matrix of m × n order and I, another matrix of the same order such that AI = A = IA, then I is called
multiplication identity of A.
Example: If A = 1 2 and I = 1 0 , then AI = 1 2 1 0 = 1 2 = A.
0 3 0 1 0 3 0 1 0 3
Similarly, IA = 1 0 1 2 = 1 2 = A.
0 1 0 3 0 3
Hence, AI = A = IA. Proved.
Matrix Multiplication 91
Example 1 If P = 1 0 and Q = –1 2 then prove that: (P + Q)T = PT + QT.
Solution: 2 4 3 5
Given, P = 1 0 and Q = –1 2
2 4 3 5
Now, (P + Q)T = PT + QT
or, 21 04 + –1 2 T 1 0 T –1 2T
3 5 2 4 3 5
= +
or, 0 2 T 1 2 + –1 3
5 9 0 4 2 5
=
MATRIX or, 0 5 = 0 5 . Proved.
2 9 2 9
Example 2 If A = cos θ – sin θ and B = cos φ – sin φ , prove that AB = BA.
sin θ cos θ sin φ cos φ
Solution: Given, A = cos θ – sin θ and
sin θ cos θ
B = cos φ – sin φ
sin φ cos φ
Now, LHS = AB = cos θ – sin θ cos φ – sin φ
sin θ cos θ sin φ cos φ
= cos θ.cos φ – sin θ.sin φ – cos θ.sin φ – sin θ.cos φ
sin θ.cos φ + cos θ. sin φ – sin θ.sin φ + cos θ.cos φ
LHS = BA = cos φ – sin φ cos θ – sin θ
sin φ cos φ sin θ cos θ
cos φ.cos θ – sin φ.sin θ – cos φ.sin θ – sin φ.cos θ
= sin φ.cos θ + cos φ. sin θ – sin φ.sin θ + cos φ.cos θ
= cos θ.cos φ – sin θ.sin φ – cos θ.sin φ – sin θ.cos φ
sin θ.cos φ + cos θ. sin φ – sin θ.sin φ + cos θ.cos φ
Hence, AB = BA. Proved.
EXERCISE - 3.3(B)
1. (a) If A = 2 5 and B = 2 8 then show that: AB ≠ BA.
3 6 1 4
(b) If P = 1 2 and Q = –4 2 then prove that: PQ = QP.
2. (a) 3 4 3 –1
(b)
(c) If A = 1 1 then verify that A2 = – I, where I is an identity matrix.
–2 –1
If M = 3 4 and N = 11 –4 then show that: MN = NM = I.
8 11 –8 3
If A = 2 –2 and B = 3 –6 then show that AB = O, where O is null matrix.
3 –3 3 –6
92 Illustrated Optional Mathematics-9
3. If X = 1 –2 ,Y= 3 –1 and Z = 0 2 then verify the following:
0 –1 2 0 –1 3
(a) (XY)Z = X(YZ) (b) X(Y + Z) = XY + XZ (c) (Y + Z)X = YX + ZX
4. (a) If A = 1 2 then prove that: AI = IA = A.
–4 3
(b) If P = 4 2 then verify that: (P – 2I)(P – 3I) = O.
–1 1
5. (a) If P = 0 4 , Q = 1 4 and R = –2 5 then prove that: PQ = PR.
0 6 3 –5 3 –5
(b) If A = 3 2 , show that: A2 = A.
–3 –2
6. (a) If P = 1 2 then prove that: P2 = 2P + 5I.
3 1
MATRIX
(b) If A = 3 –5 then show that: A2 – 5A = 14I, where I is an identity matrix of order 2 × 2.
–4 2
011
(c) If P = 1 0 1 then prove that: P3 – 3P = 2I, where I is an identity matrix of order 3 × 3.
110
7. (a) If A = 1 2 and B = 0 1 then show that: 3(AB) = (3A)B = A(3B)
–3 –2 3 0
(b) If P = 0 1 and Q = –1 0 then test whether – 2(PQ) = (– 2P)Q = P(– 2Q).
2 –3 1 2
8. (a) If [2x – 3] –2 = [–3], find the value of x.
–x
(b) If [x y] –1 3 = [1 3], find the values of x and y .
0 2
(c) Find the values of a and b if a+b b 2 = 3 .
2a a–b –1 2
(d) If 2 0 x y = 2 5 – 2 3 , find the values of x, y and z.
0 3 0 z 3 4 3 1
1 0 –1 2 x
(e) If 0 0 2 –1 = y then find the values of x, y and z.
1 0 – 1 –2 z
9. (a) If P = 2 –1 and P2 = I, find the values of a and b.
a b
(b) If P = 2 0 ,Q= x 3 ,R= 0 3 and PQ = 2P + QR, find the values of x and y.
1 4 y 1 –4 –1
10. (a) If a = 2 1 –1 , find the values of a and b.
b 4 3 2
(b) If 1 2 x = 2 , find the values of x and y.
2 1 y 4
(c) If x (3) = 6 , find the values of x and y.
y 9
(d) If [a b] –1 4 = [2 3], find the values of a and b.
2 3
(e) If x y –1 = –1 , find the values of x and y.
y x 2 5
Matrix Multiplication 93
11. (a) If A = 2 –2 ,B= 0 0 and AX = B then find the matrix X.
1 3 2 –1
(b) If P = 3 1 ,Q= 2 and PR = Q, find the matrix R.
–2 –3 1
(c) If M = 5 4 ,N= –1 0 and M = LN, find the matrix L.
9 8 3 2
(d) If A = 2 4 , B = [3 1] and CA = B, find the matrix C.
–1 3
(e) If C = sin θ – cos θ , D = sin θ and CE = D, find the matrix E.
cos θ sin θ cos θ
12. (a) If A = 1 0 and B = –2 4 , find the matrix M such that 2MA = B.
2 –2 14 –8
MATRIX
(b) If P = 2 5 and P2 – 2Q = P, find the matrix Q.
–2 –1
(c) Find the matrix A when B = 3 –1 and A + 2B2 = 3B.
2 0
13. If A = 1 2 and B = 6 –2 then show that:
–3 5 3 4
(a) (A + B)T = AT + BT (b) (3A – B)T = 3AT – BT (c) (AB)T = BTAT
ANSWERS
8. (a) 3 (b) – 1, 3 (c) – 1, 5 (d) 0, 1, 1 (e) 4, – 4, 4
(d) 0, 1
9. (a) 3, – 2 (b) 2, – 16 (d) [1 – 1]
10. (a) 0, 2 (b) 2, 0 (c) 2, 3 (e) 3, 1
11. (a) 1 – 1 (b) 1 (c) 1 2 (e) 1
2 – 4 –1 3 4 0
1
1 4
2
12. (a) 1 –1 (b) –4 0 (c) –5 3
3 2 0 –4 –6 4
Project Work
A mobile company has produced two models of mobiles by charging small solar panels. The number of
solar plates and batteries in each mobile set and production of these two models of mobiles in the first
two months are shown in the following table:
Model of Mobile Sets I II Model of Mobile Sets 1st Month 2nd Month
No. of solar panels 34 I 13 16
No. of batteries 22 II 15 20
What is the total number of solar panels and batteries required to produce the both models of mobile sets
in the first two months?
94 Illustrated Optional Mathematics-9
UNIT
4COORDINATE GEOMETRY
Estimated Teaching Periods : 30
Competency
To solve the problems related to locus, line segment, equations of straight lines based on
coordinate geometry.
To solve the problems related to area of triangle and quadrilateral by using coordinates.
At the end of this unit, the students will be able to: Learning Outcomes
introduce the locus with example.
Unit
COORDINATE
GEOMETRY
find the division point of the line segment based on the given ratio and solve the related problems
and find the distance of a straight line from the given point.
find the equation related to straight line in different cases and transform to each other.
solve the problems related to the area of triangle and quadrilateral by using coordinates.
Distance Formula What do you learn?
� Distance between two points
� Applications of distance formula
Equation of Locus � Methods of finding equation of locus
� Fundamental principle of locus
Section Formula � Mid-point formula � Centroid formula
� Internal and external division
Equation of Straight Line Equation of straight line in standard form
� Equation of straight line parallel to axes � Equation of the line in special cases
� Reduction of equation in standard form �
Area of Triangle and Quadrilateral
Are you ready?
� Coordinate plane � Graph or square grid � Properties of geometric figures
� Straight line � Moving point and its path � Parallel and perpendicular lines
� Pythogorean relation � Midpoint and centroid � Equation
� Angle made by straight line
Specification Grid
Chapter Topic Cognitive Knowl- Under- Applica- Higher Total Total
Domain edge standing tion ability Ques- Marks
tions
1 mark 2 marks 4 marks 5 marks
4 Coordinate From All Topics 2 2 1 1 6 15
Geometry
95
41.1 DISTANCE FORMULA AND ITS
APPLICATIONS
4.1 (A) Distance Formula
Learning Objectives
At the end of this topic, the students will be able to:
• derive the distance formula between two points(x1, y1)& (x2, y2). • solve the problems by using distance formula.
Rectangular System of Coordinate System 2nd Quadrant y-axis Y 1st Quadrant
In the adjoining figure, two number lines XX' and YY' (–, +) 5 (+, +)
perpendicularly intersect at O in a plane. These lines are called 4
axes (plural form of axis). The point of intersection of the axes is 3
called an origin and is denoted by O which represents the zero for 2
COORDINATE these axes. The horizontal axis XOX' is called the x-axis and the –5 – 4 –3 –2 –1 1 x-axis
GEOMETRY
vertical axis YOY' is called the y-axis. These two axes divide the X' O 123 45 X
plane into four equal parts, each part is called a quadrant or –1
quarter. These four quarters are XOY, YOX', X' OY' and XOY'. 3rd Quadrant –2 4th Quadrant
–3
They are respectively called the first, second, third and fourth (–, –) –4 (+, –)
quadrants. The right side of the x-axis is positive part and that of –5
left is negative part for x-axis. Similarly, the upper side of the Y'
y-axis is called positive part and that of lower side is called
negative part for y-axis. Hence the first quadrant has both positive values, i.e., (+, +). Similarly, in second,
third and fourth quadrants, there exists (–, +), (– , –) and (+, –) respectively.
Let P be a point on the plane. Draw the perpendiculars PM and PN Y
to the x-axis and y-axis respectively. Suppose, OM = a and ON = b.
i.e., the point P runs a units along x-axis and b units along y-axis. N P (a, b)
Thus, the point P is associated with the numbers a and b. The order
pair (a, b) represents the coordinates of P. Here, a is called the x– b
coordinate or abscissa, which is generally represented by x and b, Rise O aM
Run
the y–coordinate or ordinate, which is generally represented by y. X' X
Hence, this plane represents only length and breadth. So, it is called
the Cartesian plane or xy–plane. Generally, this plane is called the
two dimensional coordinate plane (simply coordinate plane). Y'
The branch of mathematics which deals with the study of geometry in a coordinate system is called the
coordinate geometry. It is also called analytic geometry.
Note: a) If the point P lies on the x-axis, then its coordinate becomes (a, 0) because in the x-axis the ordinate
b) is always zero. That means the point does not move up or down, but only moves left or right. So, the
x-axis is represented by the line y = 0.
If the point lies on the y-axis, then its coordinate becomes (0, b) because in the y-axis, the abscissa
is always zero. That means the point does not move left or right, but only moves up or down. So, the
y-axis is represented by the line x = 0.
96 Illustrated Optional Mathematics-9
Distance Between Two Points Y
In the adjoining graph, we can easily find the length of the line segments AC A(2, 3) B (6, 6)
and BC which are parallel to the axes. But how can we find the length of the M C (6, 3)
line segment AB? NX
From the graph, OM = 2 units and ON = 6 units. Therefore, AC = MN = ON
– OM = 6 – 2 = 4 units. Similarly, AM = 3 units and BN = 6 units. Therefore, X' O
BC = BN – CN = BN – AM = 6 – 3 = 3 units.
Since ∆ABC is a right-angled triangle,AB is hypotenuse, BC is perpendicular Y' B(x2, y2)
and AC is the base. We have the Pythagoras theorem, Y C
h2 = p2 + b2 A(x 1, y 1) d
or, AB2 = BC2 + AC2
or, AB = BC2 + AC2 X' M NX
= 32 + 42 = 9 + 16 = 25 = 5 O
∴ AB = 5 units. Y'
Now, let us find the distance between the points A(x1, y1) and B(x2, y2). Draw the perpendiculars AM and
BN on the x-axis. Again, draw AC⊥BN as shown in the adjoining figure. Y COORDINATE
GEOMETRY
Here, OM = x1, ON = x2, AM = y1 and BN = y2.
∴ AC = MN = ON – OM = x2 – x1 (x1, y1)
and BC = BN – CN = BN – AM = y2 – y1 d
In the figure, ∆ABC is as right–angled triangle. O X
∴ AB = BC2 + AC2 (x2 – x1)2 + (y2 – y1)2 .
Hence the distance between the given two points (x1, y1) and (x2, y2) is given by,
d = (x2 – x1)2 + (y2 – y1)2 units
Note: The distance from the origin to the given point (x1, y1) is given by
d = (0 – x1)2 + (0 – y1)2 = (x1)2 + (y1)2 units
Example 1 : Plot the given points A(2, 5) , B(0, 2) and C(– 5, 0) on the graph. Also, find the length
of the line segments AB, CA and BC. Y
Solution: Plotting the given points on the graph. A (2, 5)
Now, we have,
d = (x2 – x1)2 + (y2 – y1)2 X' C (–5, 0) B (0, 2) X
∴ AB = (0 – 2)2 + (2 – 5)2 = 4 + 9 = 13 units, O
CA = (2 + 5)2 + (5 – 0)2 = 49 + 25 = 74 units, Y'
BC = (–5 – 0)2 + (0 – 2)2 = 25 + 4 = 29 units.
Example 2 : Find the distance of the point (– acos θ, asin θ) from the origin.
Solution: Here the distance of a point (– a cos θ, a sin θ) from the origin is x12 + y12
= (– a cos θ)2 + (a sin θ)2
= a2cos2 θ + a2 sin2 θ = a2(cos2 θ + sin2 θ) = a2 × 1 = a2 = a units.
Distance Formula and its Applications 97
Example 3 : Find the length of the line segment which makes x-intercept Y
4 units and y-intercept 3 units. (0, 3)
Solution: The coordinates of the two end points of the line segment which 3 units
makes x-intercept as 4 units and y-intercept as 3 units are (4, 0)
and (0, 3) respectively, Then the length of the line segment is X' O 4 units (4, 0) X
Y'
(0 – 4)2 + (3 – 0)2 = 16 + 9 = 25 = 5 units
Example 4 : If A(1, 3) and B(4, 7) are two end points of the line AB and P(–2, 0) and Q(4, 8) are two
end points of the line PQ, then prove that PQ = 2AB.
Solution : Here, A(1, 3) and B(4, 7) are two end points of the line AB and P(– 2, 0) and Q(4, 8) are
two points of the line PQ.
Now, we have
d = (x2 – x1)2 + (y2 – y1)2
∴ AB = (4 – 1)2 + (7 – 3)2 = 9 + 16 = 25 = 5 units,
PQ = (4 + 2)2 + (8 – 0)2 = 36 + 64 = 100 = 10 units.
∴ PQ = 2AB.
COORDINATE Example 5 : If the distance between two points A(x, 2) and B(– 2, 3) is 37 units , find the value of x.
GEOMETRY
Solution : Here, the distance between two points A(x, 2) and B(– 2, 3) is 37 units .
i.e., AB = 37 units. or, x2 + 8x – 4x – 32 = 0 Do you know?
or, (–2 – x)2 + (3 – 2)2 = 37 or, x(x + 8) – 4(x + 8) = 0 (–a – b)2 = {–(a + b)}2
= (–1)2 (a + b)2
or, (2 + x)2 + 12 = 37 or, (x + 8) (x – 4) = 0 = (a + b)2
and (b – a)2 = (a – b)2
or, 4 + 4x + x2 + 1 – 37 = 0 Either, x + 8 = 0 ∴ x = – 8
or, x2 + 4x – 32 = 0 OR, x – 4 = 0 ∴ x = 4
Hence, the possible values of x are – 8 and 4.
Example 6 : Prove that the points (9, 7), (7, 5) and (4, 2) are collinear.
Solution : Let A, B and C be the given three points (9, 7), (7, 5) and (4, 2) respectively. Then
AB = (7 – 9)2 + (5 – 7)2 = 4 + 4 = 8 = 2 2 unit
BC = (4 – 7)2 + (2 – 5)2 = 9 + 9 = 18 = 3 2 units
AC = (4 – 9)2 + (2 – 7)2 = 25 + 25 = 50 = 5 2 units
Here, AB + BC = 2 3 + 3 2, = 5 2 units
∴ AB + BC = AC. Hence, the given three points are collinear.
Example 7 : If the points A(p + q, q – p) and B(p – q, p + q) are equi-distant
from the point C(x, y), prove that py = qx. C (x, y)
Solution : Here, the points A(p + q, q – p) and B(p – q, p + q) are
equidistant from the point C(x, y). A (p + q, q – p) B (p – q, p + q)
So, CA = CB
or, (p + q – x)2 + (q – p – y)2 = (p – q – x)2 + (p + q – y)2 [ Distance formula]
or, p2 + q2 + x2 + 2pq – 2qx – 2px + q2 + p2 + y2 – 2 pq + 2py – 2qy = p2 + q2 + x2 – 2pq + 2qx
– 2px + p2 + q2 + y2 + 2pq – 2qy – 2py [ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
or, – 2qx + 2py = 2qx – 2py or, 4py = 4qx ∴ py = qx. Proved
98 Illustrated Optional Mathematics-9
EXERCISE – 4.1 (A)
1. (a) State the distance formula.
(b) Write the distance from origin to any point P(x1, y1).
(c) If the points P(a1, b1) and Q(a2, b2) then what is the distance between P and Q?
2. Plot the following points on the graph paper and join them. Which shapes did you find? Write their
name.
(a) A(3, – 2), B(0, 5) and C(2, 3). (b) P(3, 1), Q(– 3, – 1), R(– 2, 1) and S(– 2, 0)
Also, find the length of the sides of each shape. Y
B
3. Find the coordinates of the point A, B, C and D from the adjoining
figure. Also, find the distance between the following points. A
X' O C X
(a) A and B (b) C and E (c) B and D (d) A and E
4. Find the distance of the following points from the origin. DE
Y'
(a) (6, 8) (b) (–2, 9) (c) (0, 7)
(d) (asin θ + bcos θ, acos θ – bsin θ) (e) (acos α, asin α)
5. Find the distance between the following points.
(a) (a, b) and (2a, 2b) (b) (p, 2q) and (3p, – q)
(c) (a + b, – b) and (a, a – b) (d) (s, t) and (s + t, t – s) COORDINATE
GEOMETRY
6. (a) Find the length of the line segment which makes the x-intercept 6 units and y-intercept 8 units.
(b) The x–coordinate (or abscissa) of the point A on the x-axis is – 4 and the y–coordinate (or
ordinate) of a point B on the y-axis is 5. Find the distance between A and B.
7. (a) If A(–1, 2) and B(3, 0) are two end points of the line segment AB and P(1, 1) and Q(3, 0) are
two end points of the line segment PQ, then prove that AB = 2PQ.
(b) If M(3, – 1) and N(– 2, 0) are two ends of the line segment MN and R(2, 1) and S(5, 3) are the
two ends of the line segment RS, then prove that MN = 2 RS.
8. (a) Find the value of k if the distance between the points (3, k) and (– 3, 2) is 2 10 units.
(b) If the distance between the points (a, 2a) and (4a, 6a) is 20 units, find the value of a.
(c) Find the value of x and y if the distance of the point (x, y) from (– 2, 0) and (2, 0) are both 14 units.
9. (a) Find the coordinates of the points on the x-axis which are at a distance of 5 units from the point (3, 2).
(b) What point on the x-axis is equidistant from the points A(–3, 3) and B(5, 3) ?
10. (a) Find the coordinates of the point on the y-axis whose distance is 13 units from the point (2, – 1).
(b) Find the coordinates of the point on the y-axis which is equidistant from the point (1, 0) and (1, 4).
11. (a) Prove that the points A(2, 3), B(5, 6) and C(8, 9) are collinear.
(b) Show that the points (6, 2), (– 3, – 7) an (3, –1) are collinear.
12. (a) If the points A(a + b, b – a) and B(a – b, a + b) are equidistant from the point C(x, y), prove that bx = ay.
(b) If the point (x, y) is equidistant from (– 3, 1) and (4, – 2), show that 3y = 7x – 5.
ANSWERS
2. (a) Triangle, 58 units, 2 2 units, 26 units (b) Quadrilateral, 2 10 units, 5 units, 1 unit, 26 units
3. A(3, 1), B(– 1, 3), C(1, 0), D(– 2, – 3), E(2, – 3) (a) 2 5 units, (b) 10 units (c) 37 units (d) 17 units
4. (a) 10 units (b) 85 units (c) 7 units (d) a2 + b2 units (e) a unit
5. (a) a2 + b2 units (b) 4p2 + 9q2 units (c) a2 + b2 units (d) s2 + t2 units 6. (a) 10 units (b) 41 units
8. (a) 0 or 4 (b) 4 or – 4 (c) 0; 8 3 or – 8 3 9. (a) (2, 0) or (4, 0) (b) (1, 0) 10. (a) (0, 2) or (0, – 4) (b) (0, 2)
Distance Formula and its Applications 99
4.1 (B) Applications of Distance Formula
Learning Objectives
At the end of this topic, the students will be able to:
• apply the distance formula to prove the geometric properties of triangle, quadrilaterals and circle.
Theverticalpointsofanygeometricfigureareasthecoordinatesofthegraph.Ifweplotthecoordinatesofthepoints
and join them, continuously, then completes the geometric figure. For example, when we plot the pointsA(2, 2),
B( 5, 7) and C(8, 2) and then join them, it becomes a triangle ABC.
d = (x2 – x1)2 + (y2 – y1)2 units
Now, let us measure the length of the sides AB, BC and CA by using distance formula,
∴ AB = (5 – 2)2 + (7 – 2)2 = (3)2 + (5)2 = 9 + 25 = Y C(5, 7)
34 units,
AB = (8 – 5)2 + (2 – 7)2 = (3)2 + (– 5)2 = 9 + 25 =
34 units,
COORDINATE AB = (2 – 8)2 + (2 – 2)2 = (– 6)2 + 02 = 36 = 6 units. X' A(2, 2) B(8, 2)
GEOMETRY O X
Here, we can see AB = BC. Hence ∆ABC is an isosceles triangle and the Y'
given points A(2, 2), B(5, 7) and C(8, 2) are the vertices of the isosceles
triangle.
Similarly, by using the distance formula, we can verify the properties of different triangles and
quadrilaterals. The method of verifying the properties is given below:
SN Name Figure Showing Properties
All sides are not equal in lengths.
1. Scalene A
Triangle i.e., AB ≠ BC ≠ CA.
BC
2. Isosceles P Any two sides are equal in lengths.
Triangle i.e., PQ = PR.
QR
3. Equilateral L All sides are equal in lengths.
Triangle i.e., LM = MN = NL.
MN
4. Right Triangle A Square of longer side is equal to the sum of the
square of other two shorter sides.
5. Isosceles BC i.e., AC2 = AB2 + BC2.
Right Triangle P
Any two sides are equal and square of longer side is
QR equal to the sum of the square of other two sides.
i.e., PQ = QR and PR2 = PQ2 + QR2.
100 Illustrated Optional Mathematics-9
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Final-for-print-Illustrated-OPT-Math-9-2076-2-1
Discover the best professional documents and content resources in AnyFlip Document Base.
Search