Vedanta Opt. Math 9 Final (2078)

vedanta Excel in Opt. Mathematics - Book 9 Polynomials

5. (a) p(x) = x2 + x, q(x) = x2 – 4x, then verify that p(x).q(x) = q(x).p(x)
(b) If f(x) = x + 4, g(x) = x + 7, verify that f(x). g(x) = g(x). f(x)

Long Questions
6. (a) If f(x) = 11x2 – 5x + 7, g(x) = 13x2 + 5x – 9 and h(x) = 3x2 – 6x + 1, then find the

value of {f(x) + g(x)} –h(x).
(b) If f(x) = x2 + 4x + 3, g(x) = x2 + 7x + 5 and h(x) = x – 7, find the value of

{f(x) + g(x)} – h(x)
7. (a) What should be subtracted from x3 – 3x2 + 6 to get x3 – 4x2 + x + 2 ?

(b) What should be subtracted from sum of y3 + 2y + 1 and y2 + 6y + 2 to get 6y + 8 ?
8. Find the product of p(x) and q(x) when

(a) p(x) = x2 – 2x + 1, q(x) = x3 – 6x2 + x + 1.
(b) p(x) = x2 + x + 1, q(x) = x2 – x + 1
9. Let p(x) = x2 + x + 2, q(x) = 2x2 + 3x + 5 and r(x) = x + 2, then verify that
{p(x) + q(x)} + r(x) = p(x) + {q(x) + r(x)}.
10. (a) Let f(x) = x – 1, g(x) = x2 + x + 1, h(x) = x2 + 2x – x3, then verify that

{f(x) . g(x)} h(x) = f(x) {g(x). h(w)}
(b) Let p(x) = x – 4, q(x) = x + 4 and r(x) = x2 + 16, then verify that

{p(x). q(x)} r(x) = p(x) {q(x). r(x)}.
11. Find the additive inverse of each of the following given polynomial:

(a) p(x) = x4 + 4x3 + 6x2 + 7x + 8
(b) q(x) = 4x4 – 6x3 + 7x2 + 22x + 2

1. (a) 3x2 + 10 (b) 7x3 – 4x2 + 4x (c) 0

2. (a) 3x4 – 2x2 + 9x – 10 (b) –x2 – 7x – 5

3. (a) x2 – y2 (b) x3 – 1 (c) x3 + y3

6. (a) 21x2 + 6x – 3 (b) 2x2 + 10x + 15 7.(a) x2 – x + 4 (b) y3 + y2 + 2y – 5

8. (a) x5 – 8x4 + 14x3 – 7x2 – x + 1 (b) x4 + x2 + 1

11. (a) – x4 – 4x3 – 6x2 – 7x – 8 (b) – 4x4 + 6x3 – 7x2 – 22x – 2

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Matrix 5

5.0 Review

Observe the adjoint calender of 2020 June and answer the following questions
(a) Name the day of 20th June.
(b) How many Saturdays are there in June month ?
(c) How many days are there in June month ?

5.1 Introduction of Matrix

Observe the marks of three students in the first terminal exam for two subjects; Maths and
Science.

Subject Jeny Jack Jaya
Math 88 75 90
Science 60 63 70

(a) What marks did Jeny obtained in Maths ?

(b) What does 63 represent ?

Above table can be written as,

Jeny Jack Jaya

Maths 88 75 90
Science 60 63 70

This representation of marks is called a matrix. In a matrix, the numbers are arranged in rows
and columns. The numbers in the horizontal lines are called rows and that the numbers in
vertical lines are called columns. The numbers 88, 75, and 90 are in the first row (R1) and
that of 60, 63 and 70 are in the second row. The numbers 88 and 60 are in the first column
(C1) and so on. C1 C2 C3

88 75 90 R1
60 63 70 R2

Definition : A matrix is defined as a rectangular arrangement of numbers (elements) into
rows and columns enclosed by a pair of round or square brackets. Each number of a matrix
is called an element or component or entry.

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Notation : Generally matrices are denoted by capital letters like A, B, C, D,........ X, Y, Z. The
elements are denoted by the corresponding small letters along two suffices. The first suffix
indicates the number of row and the latter one indicates the number of column in which the
element appears. For examples,

(i) aij represents the element of matrix A in the ith row and jth column.

(ii) a13 represents the element of matrix A in the first row and the third column.

Order of a Matrix

Let us consider a matrix A = 1 2 3 .
4 5 6
The matrix A has 2 rows and 3 columns. This matrix is of order 2 × 3 (read as 2 by 3). A

matrix of order 2 × 3 is also denoted by A2 × 3.

Definition : The number of "rows × columns" of a matrix is known as the order of the matrix.

Example : A = 123 is an order of 2 × 3.
321
Let A be a matrix of order 3 × 3, then we write in double suffix notation as below.

a11 a12 a13
A = a21 a22 a23

a31 a32 a33

Worked out Examples

Example 1. Write down the orders of the following matrices :
Solution :
(a) A = [1 2 3] 1
Example 2. (b) B = 7
Solution :
3

(c) P= 123 1 40
478 (d) Q = 2 3 4

6 07

(a) The matrix A has one row and three columns. Hence, the order of
matrix A is 1 × 3

(b) The matrix B has three rows and one column. Hence, it is of order 3 × 1.

(c) The matrix P has two rows and three columns. Hence, matrix P is of
order 2 × 3.

(d) The matrix Q has three rows and three columns. Hence, the order of
matrix Q is 3 × 3.

2 45
If A = 6 7 2 , do the following:

0 41
(a) Write the number of elements of matrix A.

(b) Write down the order of the matrix A.

(a) Here the matrix A has 9 elements.

(b) The order of matrix A is 3 × 3

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Example 3. Let A = 2 4 5 . Write the matrix using double suffix notation like a11, a12,
Solution : 6 7 8

...... a23. Then, write the values of A11m a12,........

Here, A = 245
678
In double suffix notation, we can write matrix A as,

A= a11 a12 a13
a21 a22 a23
Equating the corresponding elements, we get,

a11 = 2, a12 = 4, a13 = 5, a21 = 6, a22 = 7, a23 = 8

Exercise 5.1

Very short Questions :

1. (a) Define a matrix with an example.

(b) What is order of matrix ? Give an example of it.

(c) What is the meaning of a23, an element of matrix A ?

(d) If R = 2 3 4 , what is the number of elements in the matrix R ?
2 4 5
2. Write down the order of the following matrices:

123 246
321
(a) A = 6 4 2 (b) B=

452 m
(e) E = n
(c) C= c11 c12 c13 (d) D = [1 2 3]
c21 c22 c23 p

124

3. if A = 6 7 8 , write down the order of the matrix A. How is it written to show the

234
order of matrix A ? Write it in notation.

Short Questions :

–4 6 2
4. (a) If A = 3 2 1 , write the matrix A using double suffix notation, like

4 21
a11, a12, ........ a33. Write the values of a11, a12, ........., a33.

123 b11 b12 b13

(b) If 4 6 7 = b21 b22 b23 , write the values of b11, b12, ......., b33.
542 b31 b32 b33

Which letter is appropriate to denote the above matrix?

1. (d) 6 2. (a) 3 × 3 (b) 2 × 3 (d) 1 × 3

(e) 3 × 1 3. 3 × 3, A3 × 3

4. (a) a11 = –4, a12 = 6, a13 = 2, a21 = 3, a22 = 2, a23 = 1, a31 = 4, a32 = 2, a33 = 1, A

(b) b11 = 1, b12 = 2, b13 = 3, b21 = 4, b22 = 6, b23 = 7, b31 = 5, b32 = 4, b33 = 2

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5.2 Types of Matrices

On the basis of structures of matrices, they are of different types. Some special types of
matrices are defined below :

(a) Row Matrix :

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

For examples : (i) A = [a b c] is a row matrix of order 1 × 3.

(ii) P = [1 4] is a row matrix of order 1 × 2.

(b) Column Matrix :

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

a

For examples : (i) A = b is a column matrix of order 3 × 1.

c

(ii) P = p is a column matrix of order 2 × 1
q
(c) Zero Matrix :

A matrix of any order whose all elements are zeros is called a zero matrix or null
matrix.

For examples : (i) O = [0 0 0] is a zero matrix of order 1 × 3.

000

(ii) O = 0 0 0 is a zero matrix of order 3 × 3.

(d) Rectangular Matrix : 000

A matrix having unequal number of rows and columns is called a rectangular matrix.

For examples : Let a b d
m p q
Here, the number of rows = 2

The number of columns = 3

? 2≠3

Hence, A is a rectangular matrix of order 2 × 3.

(e) Square Matrix :

A matrix whose number of rows and columns are equal is called a square matrix.

123
For example : P = 4 5 6

789
Here, the number of rows = 3

The number of columns = 3

Hence, the given matrix P is a square matrix of order 3 × 3.

Similarly, R = 1 4 is a square matrix of order 2 × 2.
3 9

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Leading Diagonal :

In a square matrix the diagonal that runs from the top left to right bottom is called
leading diagonal or principal diagonal,

123
Let P = 2 6 7

324

Then the diagonal that runs from top left to right bottom with elements 1, 7, and 4 is a

leading, principal/ major diagonal.

(f) Diagonal Matrix :

A square matrix in which the elements except the leading (or principal) diagonal are
zero is called a diagonal matrix.

For examples : Let A = 2 0 ,B= 1 0 0
0 3 0 2 0
0 0 3

Here A and B are diagonal matrices of order 2 × 2 and 3 × 3 respectively.

(g) Scalar Matrix :

A diagonal matrix in which all the elements in the leading (or principal) diagonal are
equal is called a scalar matrix.

For examples : Let A = 2 0 ,B= a 0 k00
0 2 0 a and C = 0 k 0

Here, A, B and C are scalar matrices. 00k

(h) Unit/ Identity Matrix :

A scalar matrix in which all the leading diagonal elements are unity and others zeros is
called a unit matrix. It is denoted by I.

For examples : I = 1 0 100
0 1 I1 = 0 1 0

001

Here, I and I1 are unit or identity matrices of orders 2 × 2 and 3 × 3 respectively.

(i) Triangular Matrix :

A square matrix whose elements either below or above the leading (or principal)
diagonal are each of zero is called a triangular matrix. Triangular matrices are of two
types.

Upper Triangular Matrix

A square matrix whose elements below the leading diagonal each of zero is called the
upper triangular matrix.

Let P = 1 2 ,Q= a b c 1 4 6
0 3 0 d e ,T= 0 3 2
0 o 0 0 8
g

Here, P, Q and T are upper triangular matrices.

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Lower Triangular Matrix

A square matrix whose elements above the leading diagonal each of zero is called the
lower triangular matrix.

For examples : A = 2 0 ,B= 1 0 0 are lower triangular matrices.
Equal Matrices : 3 4 2 3 0
5 6 7
(j)

Two matrices of the same orders are said to be equal if their corresponding elements are
equal. If P and Q are equal matrices, then we write P = Q.

For examples : (i) P= 1 2 and Q = 1 2 are equal matrices.
3 4 3 4

(ii) If p q = 2 7 , then their corresponding elements are equal.
r s 8 9

? p =2, q = 7, r = 8, s = 9 .

(k) Symmetric Matrix :

A square matrix is said to be symmetric matrix if the interchange of rows or columns
makes no change in the given matrix.

In other words a square matrix A = (aij) is called a symmetric if aij = aji for all i and j.

For Examples : (i) A= 2 4 , a12 = a21 = 4
4 3

123
(ii) A = 2 4 5

358
Here, a12 = a21 = 2, a13 = a31 = 3, a23 =a32 = 5,

? A is a symmetric matrix of order 3 × 3. For a symmetric matrix A, AT = A.

(l) Skew – symmetric Matrix :

A square matrix A = (aij) is called skew–symmetric matrix if aij = – aji for all i, j all
diagonal elements are zero.

0 37

For examples, let A = –3 0 5 is a skew – symmetric matrix of order 3 × 3.

–7 –5 0

Here, a12 = – a21 = 3,

a13 = a31, a23, = – a32 = 5.

For a skew - symmetric matrix, AT = – A

(m) Sub – matrix :

A new matrix obtained by omitting some rows or columns from a given matrix is called
a sub–matrix of the given matrix.

Let A = 1 2 3 1 2 ,C= 1 2 3 are sub matrices of matrix A.
3 4 5 ,B= 3 4 3 4 5
7 8
9

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Worked out Examples

Examples 1. State the types of the following matrices with their orders :

1

(a) A = [a b c ] (b) B = 2

3

(c) C= 2 0 100
2 3 (d) D = 0 1 0

100 001

(d) D = 0 1 0 (e) E= 1 2
2 4
001

123 124
(f) A = 2 5 7 (g) M = 0 5 6

379 007

Solution : (a) A = [a b c]. It is a row matrix of order 1 × 3

Example 2. 1
Solution :
(b) B = 2 . It is a column matrix of order 3 × 1.

3

(c) C= 2 0 . It is a lower triangular matrix of order 2 × 2.
2 4

100

(d) D = 0 1 0 . It is an identity matrix of order 3 × 3.

001

(e) E= 1 2 , It is a symmetric matrix of order. 2 × 2.
2 4

123
(f) A = 2 5 7 . Here, a12 = a21 = 2, a13 = a31, a23 = a32 = 7

379
? A is a symmetric matrix of order 3 × 3.

129
(g) M = 0 5 6 , M is an upper triangular matrix of order 3 × 3.

007
If matrix A = [aij] is of order 3 × 3 and aij = 2i – j, construct the matrix A.

Here, A = [aij] , aij = 2i – j, A is matrix of order 3 × 3.

a11 a12 a13
Let A = a21 a22 a23

a31 a32 a33
Now, a11 = 2.1 – 1 = 1, a12 = 2.1 – 2 = 0, a13 = 2.1 – 3 = –1.

a21 = 2.2 – 1 = 3, a22 = 2.2 – 2 = 2, a23 = 2.2 – 3 = 1
a31 = 2.3 – 1 = 5, a32 = 2.3 – 2 = 4, a33 = 2.3 – 3 = 3

1 0 –1
? $ 3 2 1

54 3

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Example 3. If x–4 2y – 3 = 1 1 , find the values of x, y, p and q.
Solution : p–2 8 – 3q 2 –1

Example 4. Here, x–4 2y – 3 = 1 1
Solution : p–2 8 – 3q 2 –1

Example 5. Equating the corresponding elements of equal matrices, we get
Solution :
x–4=1 or, x = 5

2y – 3 = 1 or, 2y = 4 or y = 2

p–2=2 or, p = 2 + 2 = 4

8 – 3q = – 1 or, –3q = –9 or, q = 3

? p = 4, p = – 3, x = 5 and y = 2

If the matrix a–1 c+1 is an identity matrix, find the values of a, b, c
and d. b–1 d+1

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

So, we can write a–1 c+1 = 1 0
b–1 d+1 0 1

Equating the corresponding elements of equal matrices,

we get, a – 1 = 1 or, a = 2

c+1=0 or, c = – 1

b–1=0 or, b = 1

and d + 1 = 1 or, d = 0

? a = 2, b = 1, c = –1, d = 0

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

Here, x – 2y x – 2 = 31
a + 2b 3a – b 51

Equating the corresponding elements of equal matrices,

we get, x–2=1

or x = 3

? x=3

x – 2y =3

or 3 – 2y = 3

or, y = 0

a + 2b = 5 .............. (i)

3a – b = 1 ............... (ii)

From equation (ii), b = 3a – 1

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

put the value of b in equation (i)
a + 2 (3a – 1) = 5
or, a + 6a – 2 = 5

or, 7a = 7
? a = 1.
Put the value of a in equation (ii), we get,

b = 3.1 – 1 = 2
? a = 1, b = 2, x = 3, y = 0

Exercise 5.2

Very Short Question

1. Give on example of each of the following matrices:

(a) Scalar Matrix (b) Diagonal Matrix

(c) Scalar Matrix (d) Square matrix

(e) Row matrix (f) Column Matrix

2. State the types of the following matrices with their orders.

(a) 14 (b) 00 (c) 20
56 00 02

123 100 123
(e) 3 4 5 (e) 2 1 0 (f) 0 4 5

789 432 007

78 9 (h) 08
(g) 8 4 3 –8 0

9 3 10

Short Questions

3. Write down an example of the each of the matrices of the following order :

(a) Column Matrix of order 3 × 1 (b) Row matrix of order 1 × 3

(c) Unit Matrix of order 3 × 3 (d) Diagonal Matrix of order 3 × 3

(e) Scalar Matrix of order 3 × 3. (f) Null matrix of order 3 × 3

(g) Square matrix of order 3 × 3

4. Construct a 2 × 2 matrix whose elements aij are given by :

(a) aij = i + j (b) aij = 2i + j (c) aij = (i) j

(d) aij = 4i – 3j (e) aij = i.j

5. Find the values of x and y when,

(a) (x 2y) = (5 6) (b) 3x –5 = 6 –5
1 y 1 1

(c) x+y = 5 (d) 2x + y = 5
x–y 1 3x – 2y 4

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6. If a matrix has 6 elements in total, what are the possible orders it can have? Given
reasons.

7. (a) If x–1 p is an identity matrix, find the values of x, y, p, and q.
q y–1

(b) If the matrix 4x – 7 2y – 6 is an identity matrix, find the values of x and y.
y–3 2x – 3

(c) For what values of x and y, the matrix 2 3x – 8 is a scalar matrix.
y–4 2

(d) For what values of x and y, the matrix 8 x–y is a scalar matrix.
x+y 8

Long Questions :

8. Construct a matrix A whose order is 3 × 3 and elements are given by:

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

(c) aij = 2i + 3j (d) aij = 2i – j

(e) aij = (–1)i + j

9. (a) Find the values of x, y and z if x+y z–x = 3 2
y + 2z x 8 1

(b) If p+1 5 = 6 y–3 , find the values of x, y, p and q.
–3 q x –2

Project Work

10. List the trigonometric ratios of standard angles from 0° to 90° in a matrix form.

2. (a) Square, 2 × 2 (b) Null, 2 × 2 (c) Scalar, 2 × 2 (d) Square, 3 × 3

(e) Lower triangular, 3 × 3 (f) Upper triangular, 3 × 3 (g) Symmetric, 3 × 3
(h) Skew-symmetric, 2 × 2

4. (a) 23 (b) 34 (c) 11 (d) 1 –2
34 56 24 52

(e) 12 5.(a) x = 5, y = 3 (b) x = 2, y = 1 (c) x = 3, y = 2
24
(d) x = 2, y = 1 6. 1 × 6, 6 × 1, 2 × 3, 3 × 2 7.(a) x = 2, p = 0, q = 0, y = 2
(b) x = 2, y = 3 (c) x = 38, y = 4
(d) x = 0, y = 0
234
11 1 5 8 11

8. (a) 3 4 5 (b) 2 4 8 (c) 7 10 13

456 3 9 27 9 12 15

1 0 –1 1 –1 1

(d) 3 2 1 (e) –1 1 –1

54 3 1 –1 1 (b) x = –3, y = 8, p = 5, q = –2
9. (a) x = 1, y = 2, z = 3

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5.3 Operations on Matrices

Addition of Matrices.

Let us observe the sales of three types of news papers in two days in two shops M and N:

Shop M Shop N

News papers News papers

Day A B C Day A B C

Sunday 200 150 400 Sunday 150 300 250

Monday 240 160 250 Monday 120 250 260

Now answer the following questions
(a) Find the total sales of each type of newspapers on the two days ?

(b) Which type of newspaper is sold maximum in number ?

To find the sum of newspapers on the two days. Let the sales in M and N be denoted by
matrices P and Q.

P = Sunday 200 150 400 Q = Sunday 150 300 250
Monday 240 160 250 Monday 120 250 260

Matrices P ans Q can added as :

P+Q = 200 150 400 + 150 300 250
Definition 240 160 250 120 250 260

= 200 + 150 150 + 300 400 + 250
240 + 120 160 + 250 250 + 260

= 350 450 650
360 410 510

Let A and B be two matrices of same orders. Then, A and B are said to be conformable for
sum and the sum is denoted by A + B. It is obtained by adding the corresponding elements
of A and B.

For Example : Let A = 1 2 3 and B = 2 3 2
4 6 7 1 2 1
Here, each of A and B are of orders 2 × 3.

Then, A+B = 1 2 3 + 2 32
4 6 7 1 21

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

= 355
588

Properties of Matrix Addition

Addition of matrices satisfies the following properties.

Let A, B, and C be of same order matrices.

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(a) Closure Property :

If A and B are two matrices of same orders, their sum A + B is also of same order as that
of A and B.

Let A = 1 2 and B = 2 6 be two matrices of order 2 × 2.
3 4 3 5

Then A + B = 1 2 + 2 6 = 1+2 2+6
3 4 3 5 3+3 4+5

= 3 8 , which is of order 2 × 2.
6 9

Hence, matrix addition satisfies closure property.

(b) Commutative Property :

If A and B are two matrices of same orders, A+B = B+A.

Let A= 1 2 and B = 1 4
6 7 3 6

Then, A+B= 1 2 + 1 4 = 2 6
6 7 3 6 9 13

and B+A = 14 + 12
36 67

= 2 6 26
9 13 9 13

? A+B=B+A

Hence, the addition of two matrices satisfies commutative properly.

(c) Associative Properly :

If A, B, and C are three matrices of same orders,

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

Let A= 1 2 ,B= 2 1 and C = 2 4
Now, 3 4 5 6 6 2

Again, A+B= 1 2 + 21 = 3 3
3 4 56 8 10
?
(A + B) + C = 3 3 + 2 4 = 5 7
8 10 6 2 14 12

B+C= 2 1 + 24 = 4 5
5 6 62 11 8

A + (B + C) = 1 2 + 4 5 = 5 7
3 4 11 8 14 12

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

Hence, matrix addition satisfies associative properly.

(d) Existence of Identity Element :

If A is any matrix and O be a null matrix of same order as that of A, then

A+O=O+A=A

Let, A= 1 6 and O = 0 0
3 7 0 0

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Now, A+O= 1 6 + 0 0 = 1 6
and 3 7 0 0 3 7
?
O+A = 0 0 + 1 6 = 1 6
0 0 3 7 3 7

A+O=O+A=A

Here, O is called additive identity of matrix A. Hence, null matrix is additive identity .

(e) Existence of Additive Inverse : If A is a matrix of any order, then there exists another
matrix (–A) of same order as that of A. Then A + (–A) = O

Where O is a null matrix of same order as that of A .

Let A= 2 5 and (–A) = –2 –5 = 2–2 5–5 = 0 0 =O
7 8 –7 –8 7–7 8–8 0 0
Here, (–A) is called additive inverse of matrix A.

Subtraction of Matrices

Let A and B be two matrices of same order. Then subtraction of matrix B from A is denoted
by A – B and it is obtained by subtracting the elements of B from the corresponding elements
of A. The order of A – B is same order of A or B.

Let A = a11 a12 – b11 b12 = a11 – b11 a12 – b12
a21 a22 b21 b22 a21 – b21 a22 – b22

For example : Let A = 6 7 and B = 4 6
5 2 7 8

Then, A – B = 6 7 – 4 6 = 6–4 7–6 = 2 1
5 2 7 8 5–7 2–8 –2 –6

Worked Out Examples

Example 1. Let A = a11 a12 and P = p11 p12 , find A + P .
Solution : a21 a22 p21 p22

Example 2. Here, A = a11 a12 and P = p11 p12
Solution : a21 a22 p21 p22

Example 3. Now, A + P = a11 a12 + p11 p12 = a11 + p11 a12 + p12
Solution : a21 a22 p21 p22 a21 + p21 a22 + p22

If P = 4 56 and Q = 2 6 7 , then find P + Q.
5 21 4 2 5

Here, P + Q= 4 5 6 + 2 6 7
5 2 1 4 2 5

= 4+2 5+6 6+7 = 6 11 13
5+4 2+2 1+5 94 6

Let P = 2 6 7 and Q = 1 23
2 3 4 2 56

Here, P = 2 67 and Q = 1 2 3 ,
2 34 2 5 6

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

Now, P – Q = 2 6 7 – 1 2 3
2 3 4 2 5 6

= 2–1 6–2 7–3 = 14 4
2–2 3–5 4–6 0 –2 –2

Example 4. Find the additive inverse of P = 6 5 .
Solution : 2 3
6 5 a b
Example 5. Here, P = 2 3 and let Q = c d be additive inverse of P.
Solution :
By definition of additive inverse, we get

P + Q = O, where O is a null matrix of order 2 × 2

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

or, 6+a 5+b = 0 0
2+c 3+d 0 0
Equating the corresponding elements, we gets

6+a=0 or, a = –6

5+b=0 or, b= –5

2+c=0 or, c = –2 ?Q= –6 –5
3+d=0 or, d = –3 –2 –3

214
Let P = 6 2 3 , find the matrix Q such that P + Q = I, where 1 is a unit

542
matrix of order 3 × 3.

214 100

Here, P = 6 2 3 , I = 0 1 0

542 001
Now, P + Q = I

i.e., Q = I – P

1 0 0 2 1 4 –1 –1 –4
0 1 0 – 6 2 3 = –6 –1 –3

0 0 1 5 4 2 –5 –4 –1

Exercise 5.3

Very Short Questions
1. (a) State the required condition for addition of two matrices with an example.

(b) State the required condition for subtraction of two matrices with an example.
(c) Define additive inverse of a matrix with an example
(d) Define additive identity of a matrix with an example.
(e) State the closure property of matrix addition with an example.

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vedanta Excel in Opt. Mathematics - Book 9 Matrix

2. Which of the following matrices can be added and subtracted? Give your reasons.

A= 2 4 ,B= 7 8 , C = [2 4 6]
6 7 2 3

D = [2 1 7], E = 1 and F = [2 7 2]
2

3. (a) If P = 2 3 4 and Q = 1 2 6 , then find (i) P + Q (ii) P – Q.
2 4 7 7 3 2

(b) If M = 1 2 and N = 3 2 , find (i) M + N (ii) M – N
2 5 5 7

Short Questions

4. Let P = 12 and Q = 2 4 , and R = 21 verify the following. Also state the
34 6 7 46
property of each of them.

(a) P + Q = Q + P (b) Q + R = R + Q

(c) P + (–P) = O, where O is a null matrix of order 2 × 2.

(d) (P + Q) + R = P + (Q + R)

5. (a) If A = p q , find additive inverse of A
r s

(b) If P = 6 7 , then find the additive inverse of P.
5 8

6. (a) If P = p q and Q = 0 0 , then verify that P + Q = Q + P = P. What property
r s 0 0

does it state? What is Q called ?

(b) If M = 2 7 , find the additive identity of matrix M.
8 9

7. (a) If 6 5 + 4 y = z 6 , then find the values of p, x, y and z.
2 x 2 1 p 5

(b) If a 2 + 4 x = 6 3 , find the values of a, b, x and y.
b 5 8 y 10 5

(c) If 4p – 1 8 = p+5 s–4 , find he value of p, q, r and s.
q+3 4 5 r+4

8. (a) Let P = 2 3 6 ,Q= –4 6 7 and R = –4 –8 –4 , verify that
2 4 1 82 1 3 2 1

(i) ( P + Q) + R = P + (Q + R)

(b) Find the matrix X of P = [2 4 6], Q = [7 –4 8] and R = [–10 8 7]
and P + Q = R + X.

9. (a) If P + Q = 5 2 and P – Q = 3 6 , find the matrices P and Q.
0 9 –3 2

(b) If A – B = 6 7 and A + B = 10 4 , find the matrices A and B.
8 2 6 7

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Project Work

10. List the number of boy and girl students in your school in each of the class 8, 9, and 10.
Write them in matrix form. Add them to find the total number of boy and girl students.

2. A and B matrices can be added, subtracted, equal orders.

C, D and F matrices can be added, C – D, C – F, D – C, F – D, F – C, D – F

can be performed, equal orders.

3. (a) (i) 3 5 10 (ii) 1 1 –2 (b) (i) 4 4 (ii) –2 0
9 7 9 –5 1 5 7 12 –3 –2
4. (a) Commutative (b) Commutative

5. (a) –p –q (b) –6 –7 (c) Additive inverse (d) Associative
–r –s –5 –8
00
6. (a) Existence of additive identity, additive identity (b) 00

7. (a) x = 4, y = 1, z = 10, p = 4 (b) a = 2, b = 2, x = 1, y = 0

(c) p = 2, q = 2, r = 0, s = 12

8. (b) P = (19 –8 7) (c) (19 –8 7)

44 1 –2 8 11 2 – 3
–1 2
9. (a) P = 3 11 , Q = 3 7 (b) A= 2 ,B=
– 2 22 2 9 5
7
2 2

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5.4 Transpose of Matrix

a bc
Let A = d e f be a square matrix. Interchange the rows and columns of the matrix and

ghi

adg
denote it by A' = b e h . The matrix A' is called the transpose of matrix A.

cf i

Definition : Let A be a given matrix. Then the matrix obtained by interchanging the rows

and columns of the given matrix A is called the transpose of matrix A. It is denoted by A' or
AT or At.

For example : Let A = 146 then its transpose is given by A' = 1 2
231 4 3 . Here, the order of
matrix A is 2 × 3 and order of A' is 3 × 2. 6
1

Properties of Transpose of Matrix

(a) The transpose of a matrix is the matrix itself. i.e. (AT)T = A.

Let A = a b , then AT = a c and (AT)T = a b . ? (AT)T =A
c d b d c d

(b) The transpose of the sum of two matrices is the sum of their transposes.

i.e. (A + B)T = AT + BT

Let A = 1 2 and B = 6 4
3 4 5 2

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

= 1+6 2+4 = 7 6
3+5 4+2 8 6

(A + B)T = 7 8
6 6

Again, AT = 1 3 and BT = 6 5 .
2 4 4 2

Then AT +BT = 1 3 + 6 5 = 7 8
2 4 4 2 6 6

? (A + B)T = AT + BT

(c) If A is any matrix and k is any scalar (real number), then (kA)T = kAT

Let A= 123
456

Take k = 3, kA = 3 123 = 3 6 9
456 12 15 18

3 12
(kA)T = 6 15

9 18

1 4 3 12
Again kAT = 3 2 5 = 6 15

3 6 9 18

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Hence (kA)T = kAT
i.e. (3A)T = 3AT

Worked out Examples

Example 1. If P = 1 2 3 , find PT and verify that (PT)T = P
Solution : 7 8 10

Example 2. Here, P = 1 2 3
Solution : 7 8 10
Then interchanging the rows and columns of the matrix P, we get,

17
PT = 2 8

3 10

Also, (PT)T = 1 2 3 ? (AT)T = A Proved.
7 8 10

If P = 1 2 and Q = 2 4 , verify that (P + Q)T = PT + QT
4 6 6 9

Here, P = 1 2 and Q = 2 4
4 6 6 9

Then PT = 1 4 and QT = 2 6
2 6 4 9

P+Q= 1 2 + 2 4 = 3 6
4 6 6 9 10 15

(P + Q)T = 3 10
6 15

and PT + QT = 1 4 + 2 6 = 3 10
2 6 4 9 6 15

? (P + Q)T = PT + QT Proved.

Exercise 5.4

Very Short Questions

1. (a) Define transpose of a matrix with an example.

(b) State the properties of transpose of a matrix.

2. Find the transpose of the following matrices:

(a) P = [1 2 3], 7
(b) Q = 8

9

(c) R= 2 4 6 123
8 2 3 (d) M = 4 5 6

789

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Short Questions :

3. (a) If A = 2 6 8 , then final AT and verify that (AT)T = A.
3 2 5

(b) If Q = 1 7 8 , verify that (QT)T
2 3 7

4. If P = 6 7 8 and Q = 3 2 4 , then verity that
2 3 4 6 2 1

(i) (P + Q)T = PT + QT

(ii) (PT)T + (QT)T = P + Q

Long Questions

5. (a) If M = 3 4 , show that M + MT is a symmetric matrix.
1 5

(b) If P = 1 3 , show that P + PT is a symmetric matrix.
2 4

0 2 –45
6. Let A = –2 0 –4 , then

45 4 0

(a) Show that A is a skew - symmetric matrix.

(b) Find transpose of A.

(c) Find A + AT. What type of matrix is formed?

7. Let P = 2 3 for k = 4, verify that (kP)T = kPT.
4 5

0 1 –4
8. If R = –1 0 –5 , then show that RT = –R

450

1 (b) [7 8 9] 28 147
2. (a) 2 (c) 4 2 (d) 2 5 8

3 63 369

0 –2 45 103
6. (b) 2 0 4

45 –4 0

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vedanta Excel In Opt. Mathematics - Book 9 Matrix

5.5 Matrix Multiplication

(I) Multiplication of a Matrix by a scalar

Let A = 2 3 , then we can write,
5 4

A+A = 2 3 + 2 3
5 4 5 4

= 2+2 3+3 = 4 6
3+5 4+2 10 8

Also, we write 2A = 2 2 3 = 2×2 2×3 = 4 6
5 4 2–5 2×4 10 8

Similarly, we can find the values of 3A, 4A and 5A.

Definition : Let A be any matrix and k be a scalar, then the matrix obtained by multiplying
each element of A by k is denoted by kA and it is called scalar multiplication of matrix A by
k. The order of matrix kA is same order as that of A.

Example 1. Let A = 2 4 6 , then find the value of 4A.
Solution : 3 2 1

Example 2. Here, A = 2 4 6 , 4A =4 2 4 6 = 4×2 4×4 4×6
Solution : 3 2 1 3 2 1 4×3 4×2 4×1

Example 3. = 8 16 24
Solution : 12 8 4

If P = 3 2 ,Q= 4 5 , then find the value of 3P + 2Q.
4 5 2 1

Here, P = 32 and Q = 4 5
45 2 1

Now, 3P + 2Q = 3 3 2 +2 4 5
4 5 2 1

= 9 6 + 8 10 = 17 16
12 15 4 2 16 17

Find the matrix X if A = 2 3 , B= 2 3 , and X + 5A = 4B
4 6 3 –1

Here, A = 2 3 ,B= 2 3
4 6 3 –1

Now, X + 5A = 4B

or, X = 4B – 5A = 4 2 3 –5 2 3
3 –1 4 6

= 8 12 – 10 15 = –2 –3
12 –4 20 30 –8 –34

(II) Multiplication of Matrices :

Mr Gyan bought 20 packets of tooth paste, 30 bottles of mineral water, and 25 packets of
tasty biscuits at the rate of Rs. 75, Rs. 20 and Rs. 45 respectively. How much money did he
spend? Can we use matrix method to solve those problem ?

Yes, we can solve it by use of matrix. Let, matrix of toothpaste Mineral water, Biscuit be M.

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i.e., M = [20 30 25] be matrix of goods.

75 Tooth paste
N = 20 water

45 Biscuits

be matrix of rates.

75
Now, MN = [20 30 25] 20

45

= [20 × 75 + 30 × 20 + 25 × 45]

= [1500 + 600 + 1125] = [3225]

Hence, Mr. Gyan spent total amount of Rs. 3225 on purchasing the goods.

Definition : Two matrices are said to be multiplicable if the number of columns of the first
matrix is equal to the number of rows of the second matrix. An element of the product
matrix is obtained by adding the products of the corresponding elements of the first matrix
matrix and column of the second matrix.

Let there be Am ×n and Bn×P . Then, the product AB is defined as the number of the column
of the first matrix A is equal to the number of the rows of the second matrix. The product
matrix AB is of order m × p.

Examples : A = 1 2 and B = 4 2 , then AB is defined.
2 4 3 2

Solution : Now, AB = 1 2 42
2 4 32

= 1×4+2×3 1×2+2×2 = 10 6
2×4+4×3 2×2+4×2 20 12

Process of Matrix Multiplication :

Multiplying a Row by a Column

Let us find the product

x
[a b c] y

z
we have to multiply a 1 × 3 matrix by a 3 × 1 matrix.

The number of the column in the first matrix is the same as the number of rows in the
second matrix; so, they are compatible.

The product is given by,

x
[a b c] y = [ax + by + cz]

z
Multiplying Larger Matrices

We know that if two matrices are multiplicable, we multiply a row by a column, multiplying
each element in the ith row of the first matrix by the corresponding element in the jth column
of the second matrix and adding the result.

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Let us take

A= 1 2 3 3 2
2 3 4 and B = 3 1
5
4
Here A and B matrices are of orders 2 × 3 and 3 × 2 respectively.

The product matrix is of order 2 × 2.

Let AB = 1 2 3 3 2 = e11 e12
2 3 4 3 1 e21 e22
4 5
To get e11, multiply Row 1 of the first matrix by Column 1 of the second matrix.

3
e11 = [1 2 3] 3 = 1 × 3 + 2 × 3 + 3 × 4 = 21

4
To get e12, multi Row 1 of the first matrix by column 2 of the second matrix.

2
e12 = [1 2 3] 1 = 1 × 2 + 2 + 1 + 3 × 5 = 19

5
To get e21, multiply Row 2 of the first matrix by Column 1 of the second matrix.

3
e21 = [2 3 4] 3 = 2 × 3 + 3 × 3 + 4 × 4 = 31

4
To get e22, multiply Row 2 of the first matrix by column 2 of the second matrix.

2
e22 = [2 3 4] 1 = 2 × 2 + 3 × 1 + 4 × 5 = 27

5
writing the product matrix, we get

e11 e12 = 21 19
e21 e22 31 27
Therefore, we have,

At a gl21an23ce,34we334wri215te,= 21 19
31 27

123 32 = 1×3+2×3+3×4 1×2+2×1+3×5
234 31 2×3+3×3+4×4 2×2+3×1+4×4×5
45
21 19
= 31 27

Properties of Matrix Multiplication :

(a) In general matrix multiplication is not commuatative

Let A and B be two matrices such that AB and BA are defined, then AB z BA.

Let A = 2 3 and B = 4 5
3 2 2 3

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Now, AB = 2 3 45
3 2 23

= 2×4+3×2 2×5+3×3 = 14 19
3×4+2×2 3×5+2×3 16 21

Again, BA = 4 5 23
2 3 32

= 4×2+5×3 4×3+5×2 = 23 22
2×2+3×3 2×3+3×2 13 12

? AB z BA

Hence, the commutative property does not hold in matrix multiplication

(b) Associative property of Matrix Multiplication
Let A, B and C be three matrices such that AB and BC are defined Then (AB)C= A(BC)

Let A = 1 0 ,B= 2 1 ,C= –1 2
1 2 1 1 3 2

Now, AB = 1 0 2 1 = 2+0 1+0 = 2 1
1 2 1 1 2+2 1+2 4 3

(AB)C = 2 1 –1 2
4 3 32

= –2 + 3 4+2 = 1 6
–4 + 9 8+6 5 14

Again, BC = 2 1 –1 2
1 1 32

= –2 + 3 4+2 = 1 6
–1 + 3 2+2 2 4

A(BC) = 1 0 16
1 2 24

= 1+0 6+0 = 16
1+4 6+8 5 14

? (AB)C = A(BC)

Hence, associative property holds in matrix multiplication.

(c) Distributive Property :
Let A,B, and C be three matrices such that AB and AC are defined. Then,

A(B + C) = AB + AC.

Let A = 1 2 ,B= 2 1 and C = 2 1
0 3 3 2 4 3

Now, B+C= 2 1 + 2 1 = 4 2
3 2 4 3 7 5

A(B + C) = 1 2 42 = 4 + 14 2 + 10 = 18 12
0 3 75 0 + 21 0 + 15 21 15

Again AB + AC = 1 2 21 + 1 2 21
0 3 32 0 3 43

= 2+6 1+4 + 2+8 1+6
0+9 0+6 0 + 12 0+9

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vedanta Excel In Opt. Mathematics - Book 9 Matrix

= 8 5 + 10 7 = 18 12
9 6 12 9 21 15

? A(B + C) = AB + BC

Hence, distributive property holds for matrix multiplication.

(d) Identity Property (or Identity Element for Multiplication)
Let A be a square matrix and I be an identity matrix of the same order as that of A, then

A×I=A=I×A

Let A = 5 6 ,I= 1 0
7 8 0 1

Now, AI = 5 6 1 0 = 5+0 0+6 = 5 6
7 8 0 1 7+0 0+8 7 8

Again, IA = 1 0 56 = 5+0 6+0 = 5 6
0 1 78 0+7 0+8 7 8

? A.I = A = IA

(e) Transpose of Product Property :

Let A = 2 1 and B = 1 0 , then
3 4 2 3

(AB)T = BTAT

Now, AB = 2 1 1 0 = 2+2 0+3 = 4 3
3 4 2 3 3+8 0 + 12 11 12

(AB)T = 4 11
3 12

AT = 2 3 and BT = 1 2
1 4 0 3

Again, BTAT = 1 2 2 3 = 2+2 3+8 = 4 11
0 3 1 4 0+3 0 + 12 3 12

? (AB)T = BTAT

Hence, the transpose of product of two matrices is equal to the product of transpose of
the matrices in reverse order.

12

Example 4. Let A = –3 4 and B = 1 2 , then find AB and BA if possible.
Solution : 21
–2 4
1 2
Here, A = –3 4 and B = 1 2 . As A3×2 and B2×2, AB is defined
1 –2 4
2
and B2×2 and A3×2 and A3×2, BA is not defined.

Now, AB 1 2 12
= –3 4 –2 4
1
2

1–4 2+8 –3 10
= –3 – 8
–6 + 16 = –11 10
2–2
4+4 08

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vedanta Excel In Opt. Mathematics - Book 9 Matrix

Example 5. If 5 X= 20 15 , find the matrix X.
Solution : 2 8 6

Since 5 is of order 2 × 1 and 20 15 is of order 2 × 2, hence X must be of
2 8 6

order 1 × 2. Let X = [x y]

Now, 5 [x y] = 20 15
or, 2 8 6

5x 5y = 20 15
2x 2y 8 6

Equating the corresponding elements, we get,

5x = 20 or, x = 4

and 5y = 15 or, y = 3

Example 6. If P = 1 2 then, prove that P2 – 2P – 5I = O, where I and 0 are unit matrix
Solution: 3 1

Example 7. and null matrix of order 2 × 2 respectively.
Solution:
Here, P = 1 2
3 1

P2 = P.P = 1 2 12
3 1 31

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

2P = 2 1 2 = 2 4
3 1 6 2

5I = 5 1 0 = 5 0
0 1 0 5

Now, P2 – 2P – 5I = 7 4 – 2 4 – 5 0
6 7 6 2 0 5

= 7 4 – 2+5 4+9
6 7 6+0 2+5

= 7 4 – 7 4
6 7 6 7

= 7–7 4–4 = 0 0 =O
6–6 7–7 0 0

? P2 – 2P – 5I = O Proved.

If P + Q = 5 2 and P – Q = 3 6 , find the matrices P and Q.
0 9 0 –1

Here, P + Q = 5 2 ................ (i)
0 9

and P–Q= 3 6 ............ (ii)
0 –1

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

2P = 8 8
0 8

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vedanta Excel In Opt. Mathematics - Book 9 Matrix

or, P = 1 8 8 (multiplying both sides by 1 )
2 0 8 2

? P= 4 4
0 4

put the value of P in equation (i), we get,

Q = 5 2 –P= 5 2 – 4 4
0 9 0 9 0 4

= 1 –2
0 5

Example 8. If P = 1 2 and Q = 2 1 , then show that (PQ)T = QTPT.
Solution: 3 4 4 2

Example 9. Here, P = 1 2 and Q = 2 1 , then
Solution: 3 4 4 1

PT = 1 3 and QT = 2 4
2 4 1 2

Now, PQ = 1 2 . 2 1 = 2+8 1+4 = 10 5
3 4 4 1 6 + 16 3+8 22 11

LHS = (PQ)T = 10 22
5 11

RHS = QTPT = 2 4 . 1 3 = 2+8 6 + 16 = 10 22
1 2 2 4 1+4 3+8 5 11

? LHS = RHS Proved.

What matrix pre–multiplies –1 3 to get (–9 5)?
7 1

Here, –1 3 is a square matrix of order 2 × 2 and the product matrix is
71

[–9 5], the required matrix must be of order 1 × 2.

Let [x y] be the required matrix.

Now, [x y] –1 3 = [–9 5]
7 1

or, [–x + 7y 3x + y] = [–9 5]

Equating the corresponding elements, we get,

– x + 7y = –9 .............. (i)

3x + y = 5 ................. (ii)

From equation (i) x = 7y + 9

Put the value of x in equation (ii), we get,

3(7y + 9) + y = 5

or, 21y + 27 + y = 5

or, 22y = – 22

? y = –1

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vedanta Excel In Opt. Mathematics - Book 9 Matrix

Put the value of y in equation (i), we get
x = 7 × (–1) + 9 = 2

? x = 2, y = –1

Exercise 5.5

Very Short Questions

1. (a) Under what condition two matrices are multiplicable ?

(b) If Am×p and Bp×n and AB = C, what is the order of product matrix C ?

(c) Are A2×3 and B3×2, multiplicable ?

(d) If A2×3 and B3×3 are AB and BA defined ? Give reasons.

(e) For what types of matrices, A2, A3, A4, are defined ?

(f) If A = 2 4 , find (i) 5A, 1 A (ii) 3 A.
7 5 2 2

2. Find the product of the matrices given below.
2 1
(a) [1 2] 5 (b) 2 [4 5]

2 (d) 23 24
(c) [4 3 2] 1 45 62

3

Short Questions :

3. Find the product of given matrices if possible:

(a) A= 1 2 ,B= 3 2 find AB.
2 4 2 5

(b) P= 1 2 3 ,Q= 1 2
4 5 6 4 3 , find PQ.
2
1

(c) M= 2 1 , N = [2 3], find MN.
3 0

(d) P= 1 2 3 ,Q= 3
4 5 6 1 , find PQ.

2

4. (a) If A = 1 –2 1 and B = 2 1 find AB and BA.
2 1 3 3 2
1 1

3

(b) If A = [1 2 3] and B = 2 , find AB and BA.

1

5. (a) If P = 2 1 , then prove that : P2 = P.
–2 –1

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(b) If P = 1 0 , prove that : P2 = P
0 1

6. If P = 1 2 and Q = 2 1 , find the values of :
3 2 4 5

(i) 2P + 3Q (ii) 3P – 2Q.

Long Questions

7. (a) If A = 1 1 show that A2 – 2A = O, where O is a null matrix of order of 2 × 2.
1 1

(b) If P = 3 –5 , proved that P2 – 5P = 14I where I is a unit matrix of order 2 × 2.
–4 2

(c) If A = 3 1 , prove that A2 – 5A + 7I = O, where I and O are unit and null
–1 2

matrices of order 2 × 2.

(d) If P = 4 2 , then prove that (P – 2I) (P – 3I) = O, where O is a null matrix of
–1 1

order 2 × 2 and I unit matrix.

8. (a) If A + 2B = 3 20 and 2A + B = 3 1 3 , find the matrices A and B.
3 35 0 3 7

(b) If 2P + 3Q = 2 7 12 and P + 2Q = 1 0 –1 . Find the matrices P and Q.
13 12 23 –4 –1 –6

9. (a) If A = 2 4 and B = 2 3 , show that : (AB)T = BTAT
3 1 0 4

(b) If P = 2 4 and Q = –1 5 , show that : (PQ)T = QTPT
1 3 2 1

10. (a) If A = 10 and B = 10 show that : AB = BA
02 04

(b) If A = 53 and B = –1 3 show that AB = BA = I where I is a unit matrix of
21 2 –5
order 2 × 2.

11. (a) If 1 2 1 0 = y 2 , find the values of x and y.
x –3 0 1 4 –3

(b) If 4 1 1 –1 = x 1 , then find the values of x and y.
7 –3 13 4 y

(c) If A = 3 0 ,B= a b and AB = A + B, find the values of a, b and c.
0 4 0 c

(d) If –1 0 x = –2 , find the matrix x .
0 –2 y 4 y

12. (a) If P = 2 1 and Q = 3 4 and PR = Q, find the matrix R.
5 3 2 4

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

13. Which matrix pre–multiplies to matrix 1 1 to get 4 5 ?
3 4 6 2

14. (a) If [a b] 2 = [1 4] a , then prove that : a = b.
3 b

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(b) If [m n] m = [1 3] and m + 1 = n, find the values of m and n.
n
15. (a) If P is a matrix of order (2x + 1) × 2 and Q is another matrix of order (3y – 1) × 3.

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

(b) Matrix M has x-rows and (11 – x) columns. Matrix N has y-rows and (y + 5)
columns. If MN and NM both are defined, find the values of x and y.

16. If A = 1 2 ,B= 2 4 ,C= 5 2 ,I= 1 0 then verity the following :
3 0 3 5 3 4 0 1

(i) AB z BA (ii) (AB)C = A(BC) (iii) A(B + C) = AB + AC

(iv) IA = AI = A (v) (BC)T = CTAT.

Project work
17. Prepare a report on "operations on matrices" and present in your classroom.

1. (a) No. of column of 1st matrix = No. of rows of 2nd matrix. (b) m × n

(c) Yes (d) AB defined but BA not defined (e) square matrix

10 20 12 36
25 35 (iii) 21 15
(f) (i) (ii) 7 5
22
22
(c) [17]
2. (a) [12] (b) 45 (d) 22 14
8 10 38 26

3. (a) 7 12 (b) 15 11 (c) MN is not defined (d) 11
14 24 36 29 29

4. (a) –3 –2 , 4 –3 53 6 9 6. (i) 8 7 (ii) –1 4
10 7 7 –4 9 (b) [10], 2 4 6 18 19 1 –4
3 –1 2 3
41

8. (a) A = 1 0 2 ,B= 1 1 –1 (b) P= 1 14 27 ,Q= 0 –7 –14
–1 1 3 2 1 1 38 27 64 –21 –14 –35

11. (a) x = 4, y = 1 (b) x = 5, y = – 16 (c) a = 3 , b = 0, c = 4
2 3

(d) x = 2, y = – 2 12. (a) 7 8 (b) 2 13. 11
–11 –12 0 18 –4

14. (b) m = 2, –3, n = 3, –2 15. (a) x = 1, y = 1 (b) x = 8, y = 3

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Coordinate Geometry 6

6.0 Review

Group discuss the following questions :
(a) Name the four quardrants in the given graph.

(b) Write down the signs of coordinates in each quadrant.

(c) What are the coordinates of any points in the X-axis and Y-axis?

(d) Write the coordinates of the four points P, Q, R and S in the given graph.

(e) Write the distance formula to find the distance between points P(x1, y1) and Q(x2, y2).
(f) From the given graph, find the distance between the points P and R.

Y

7
6
5
Q 4 P

3
2
1
X' -6 -5 -4 -3 -2 -1-O1 1 2 3 4 5 6 X
-2
-3
-4
R -5
S

Y'

6.1 Locus Bus stop Road

Let us study the following activities : House
(a) Sima draws a map to show the path from her house to AP B

nearest bus stop. d
M Fixed Line N
Sima has to cover fixed distance from her house to the
bus stop to go her school.

(b) Let a point P move such that its distance from a fixed
line is always equal to d. The point P traces out a
straight line AB parallel to the fixed line MN. The
point P moves in a straight line.

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(c) Let a point P move in a plane such that its distance P
from a fixed point O is equal to r. The point P traces r
out a circle with centre at O and radius r.
O
(d) Let us consider a set of points in the X-axis. In the
X-axis, all the y-coordinates one zero. Hence y = 0 Y
represents X-axis.

All the above activities are related to path traced by O y=0 X
a point. The path traced by a point under certain X'

condition is a locus. Y'

Definition : The locus of a point is defined as the path traced out by the moving point under
given geometrical conditions. It may also be defined as a locus is the set of points which
satisfy the given geometrical conditions.

The following points should be taken in mind while dealing with a locus:
(i) Every point in the locus must satisfies the given geometrical conditions.

(ii) A point which does not satisfy the given geometrical conditions cannot be on the locus.

(iii) Every point which lies in the locus satisfies the equation of the locus.

(iv) The points which do not lie in the locus do not satisfy the equation of the locus.

(v) To find the locus of a moving point, plot some points satisfying the given geometrical
conditions, and then join them.

Procedure of finding the equation of a locus

To find the equation of a locus with given geometrical conditions, the following are some
general rules:

(i) Draw a figure according to given geometrical condition and take a point P(x, y) on the locus.

(ii) Write down the geometrical conditions according to the points move.

(iii) Express the geometrical conditions in terms of the coordinates (x, y) by using distance
formula and simplify it. The simplest form of the given condition is the required
equation of the locus.

Worked out Examples

Example 1. Does the point (2, 3) lie in the locus x2 + y2 = 13 ?
Solution : Here, x2 + y2 = 13
Put the point (2, 3) i.e. x = 2, y = 3 in the given equation, we get,

22 + 32 = 13
or, 4 + 9 = 13
? 13 = 13 (True).
Hence, the point (2, 3) lies in equation of locus x2 + y2 = 13.

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Example 2. Find the value of k if the point (4, 3) lies in the locus x2 + y2 + kx + 3y – 16 = 0.
Solution :
The given equation of locus is x2 + y2 + kx + 3y – 16 = 0
Example 3.
Solution : The point (4, 3) lies in the locus.

Example 4. So, put x = 4 and y = 3 in the above equation.
Solution :
42 + 32 + k.4 + 3.3 – 16 = 0

or, 16 + 9 + 4k + 9 – 16 = 0

or, 4k = – 18

? k = – 9
2

If the point (2, 3) lies in the locus whose equation is 5x + by = 16. Show that
(0, 8) also lies in the locus.

Here, 5x + by = 16 .... (i)

The point (2, 3) lies in the locus

5.2 + b.3 = 16

or, 3b = 16 – 10

or, 3b = 6

? b=2

Put the value of b in (i), it becomes,

5x + 2y = 16

Put the point (0, 8) in the locus.

5.0 + 2.8 = 16

? 16 = 16 (True)

Hence the point (0, 8) also lies in the locus.

Find the equation of locus of points which move at equidistant from the

points A(2, 3) and B(4, 5). Y

Let P(x, y) be any point in the locus which is at 6 B(4, 5)
equidistant from the points 5
4
A(3, 2) and B(4, 5) 3 P(x, y)

2 A(3, 2)
1
Then, AP = BP
X' -4 -3 -2 -1-O1 1 2 3 4 5 X

or, AP2 = B2 (Squaring on both sides.) -2
-3

or, (x – 3)2 + (y – 2)2 = (x – 4)2 + (y – 5)2 Y'

or, x2 – 6x + 9 + y2 – 4y + 4 = x2 – 8x + 16 + y2 – 10y + 25

or, 2x + 6y = 28

or, x + 3y = 14

? x + 3y = 14 is the required equation of locus.

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Example 5. Find the equation of locus of a point which moves so that its distance from
Solution:
X–axis is always 6 units. Y
Example 6.
Solution: Let P(x, y) be any point in the locus P(x, y)
and A(x, 0) be any point in the X–axis.
Example 7.
Solution: Then, PA = 6 6

Example 8. or, PA2 = 36 (Squaring on both sides) X' A (x, 0) X
Solution: or, (x – x)2 + (y – 0)2 = 36

or, y2 = 36
? y = ± 6 is the required equation of the locus. Y'

Find the equation of locus of point which moves such that its distance from
X–axis is half of its distance from the origin.

Let P(x, y) be any point in the locus and A(x, 0) be a point in X–axis.

Then by question, Y P(x, y)

PA = 1 OP
2

or, 4PA2 = OP2 (Squaring on both sides.)

or, 4PA2 = OP2 X' O A(x, 0) X
Y'
or, 4{(x – x)2 + (y – 0)2} = (x – 0)2 + (y – 0)2

or, 4y2 = x2 + y2

or, 3y2 = x2

or, x2 = 3y2

? x2 = 3y2 is the required equation of the locus.

Let A(1, 2) and B(5, 3) are two fixed points. Find the equation of locus of

point P so that PA : PB = 3:2. Y

Let P(x, y) be any point on the locus and A(1, 2) P(x, y)

and B(5, 3) are two fixed points. 3 A(5, 3)
2 B(1, 2)
Then, PA = 3 1
PB 2 X' O 1 2 3 4 5 6 X

2PA = 3PB

Squaring on both sides, Y'
4{(x – 1)2 + (y – 2)2} = 9{(x – 5)2 + (y – 3)2}

or, 4{(x2 – 2x + 1) + (y2 – 4y + 4)} = 9{(x2 – 10x + 25) + (y2 –6y + 9)}

or, 4x2 + 4y2 – 8x – 16y + 20 = 9x2 + 9y2 – 90x – 54y + 306

? 5x2 + 5y2 – 82x – 38y + 286 = 0 is the required equation.

Let A(3, 0) and B(–3, 0) be two fixed points. Find the locus of point P which
moves so that the sum of square of distance is 25.

Let P(x, y) be any point in the locus

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and A(3, 0) and B(–3, 0) are two fixed points. Y

Then, by questions, AP2 + BP2 = 25 P(x, y)

or, (x – 3)2 + (y – 0)2 + (x + 3)2 + (y – 0)2 = 25

or, x2 – 6x + 9 + y2 + x2 + 6x + 9 + y2 = 25

? 2x2 + 2y2 = 7 is the required equation. X' B 0 A X
(–3, 0) (3, 0)

Y'

Exercise 6.1

Very Short Questions

1. (a) Write down the formula to find the distance between two points P(x1, y1) and Q(x2, y2).
(b) Define locus of a point.

2. Find the equation of locus of a point which moves so that

(a) Its distance from X-axis is always 4 units.

(b) Its distance from X-axis is always 5 units

(c) Its ordinate is always – 6

(d) Its abscissa is always 8.

Short Questions

3. Find the distance between the following pair of points:

(a) (4, 5) and (6, 8) (b) (–2, –2) and (–4, –4)

(c) (a + b, a – b) and (a – b, a + b) (d) (p – q, p + q) and (p + q, p – q)

4. (a) Show that the points (4, 5), (5, –2) and (1, 1) are the vertices an isosceles triangle.

(b) Show that the points (1, 1), (–1, –1) and (– 3, 3 ) are the vertices of an equilateral
triangle.

5. (a) Does the point (2, 4) lies in the locus whose equation is

(i) x + 2y = 10 (ii) x2 + y2 = 20

(b) Do the points (3, 2), (4, 3), (5, 0), (0, –5), and (–3, –4) lie on the locus whose
equation is x2 + y2 = 25?

6. (a) For what value of k will the point (2, 1) lie in the locus whose equation is

x2 + y2 – 4x + 3y + k = 13 ?

(b) For what value of O will the point (2, 3) lie on the locus whose equation is

x2 + y2+ Ox + 2y – 30 = 0 ?

7. (a) If (4, 4) is a point on the locus whose equation is y2 = ax, prove that (16, 8) is
another point on the locus.

(b) If (2, 3) is a point on locus whose equation is ax + 2y = 16 and also show that
(0, 8) is another point on the locus.

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8. Find the locus of a point which moves so that

(a) Its distance is 4 units from the point (4, –2)

(b) Its distance from (1, 2) and (2, –3) are equal.

(c) It is at equidistant from the points (4, 3) and (5, 4)

9. Find the locus of a point which moves so that

(a) Its distance from X-axis is twice its distance from Y-axis.

(b) Its distance from the point (2, 1) is double its distance from (1, 2)

(c) Its distance from Y-axis is half of its distance from the origin.

(d) Its distance from the point (0, 2) is one third of its distance from (3, 0).

Long Questions

10. Find the locus of a point which moves so that

(a) The ratio of its distance from the point (–5, 0) and (5, 0) is 2:3.

(b) The ratio of its distance from the points (1, 2) and (5, 3) is 4:5.

11. (a) Let M(3, 2) and N(7, –4) are two fixed points and P(x, y) be a variable point in the
locus, then find the equation of locus under the following conditions

(i) MP = NP (ii) MP = 2PN

(b) Let A(a, 0) and B(–a, 0) be two fixed points. Find the locus of a point which moves
such that PA2 + PB2 = AB2.

(c) Let A(– 7, 0) and B(7, 0) be two fixed points on a circle. Find the locus of a moving
point P at which AB subtend a right angle.

12. (a) If a point (x, y) is equidistant from the point (2, 3) and (6, 1). Show that the
equation of locus is given by 2x – y = 6.

(b) Let A(a, b) and B(3a, 3b) are two fixed points and P(x, y) is a point such that
PA = PB, then prove that the equation of locus is given by ax + by = 2(a2 + b2).

13. Find the equation of the locus of a point such that

(a) The sum of square of its distance from (0, 2) and (0, –2) is 6.

(b) The sum of the squares of its distances from A(3, 0) and B(–3, 0) is four times the
distance from A and B.

(c) The difference of square of distance from the points A(0, 1) and B(–4, 3) is 16.

(i.e. PA2 – PB2 = 16)

Project Work

14. In which field of our daily life do we use locus? Group discuss in the class and prepare
a report on it.

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1. (a) PQ = (x2 – x1)2 + (y2 – y1)2 2. (a) y = ±4 (b) y = ±5
3. (a) 13 (b) 2 2
(c) y = –6 (d) x = 8

(c) 2 2 b (d) 2 2 q 5.(a) (i) Yes (ii) Yes
(b) x – 5y = 4 (c) x + y = 8
6. (a) k = 13 (b) O = 11
2

8. (a) x2 + y2 – 8x + 4y + 4 = 0

9. (a) y = 2x (b) 3x2 + 3y2 – 4x – 14y + 15 = 0

(c) 3x2 – y2 = 0 (d) 8x2 + 8y2 + 6x – 36y + 27 = 0

10. (a) x2 + y2 + 26x + 23 = 0 (b) 9x2 + 9y2 + 110x – 4y = 419

11. (a) (i) 2x – 3y – 13 = 0 (ii) 3x2 + 3y2 – 50x + 36y + 247 = 0 (b) x2 + y2 = a2

(c) x2 + y2 = 49 13.(a) x2 + y2 + 1 = 0 (b) x2 + y2 – 3 = 0 (c) 2x – y + 10 = 0

6.2 Section Formula

Review

(a) Discuss the meaning of ratio with examples. B
(b) Find the ratio of AP : PB in the given figure. 3cm

4cm P 3cm B
Fig. (i)
Internal Division : Let AB be a given line segment and
P be a point on it. If P lies within AB, P is said to divide A
AB internally in the ratio of AP : PB. If means that P is an
P
internal point of AB.
6cm
To find the coordinates of a point dividing the line
segment joining two given points internally in the given A Fig. (ii)

ratio.

Let A(x1, y1) and B(x2, y2) be two given points. Let P(x,y) be any point on the line joining A
and B. Let P divide AB in the ratio of m1 : m2 internally. The we write AP : PB = m1 : m2.

Perpendiculars AL, PN, and BM are drawn from the point A,P, and B respectively on the
X-axis. Y

B(x2, y2)

Again perpendiculars AQ, PR are drawn from the

points A and B to PN and BM respectively. P(x, y) R

Then, AQ = LN = ON – OL = x – x1 A(x1, y1) Q
PR = NM = OM – ON = x2 – x1

QP = NP – NQ = NP – LA = y – y1 X' O LN MX

BR = MB – MR = MB – NP =y2 – y Y'

Since QP and BR are parallel, triangles PAQ and BPR are similar.

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The corresponding sides of similar triangles are proportional.

? AP = APRQ = QRBP
BP
or, mm12 xx2––xx1 yy2––yy1

Taking the first two ratio, we get,

mm12 xx2––xx1
or, m1x2 – m1x1 = m2x – m2x1

or, m1x2 + m2x1 = m2x – m2x1

or, m1x2 + m2x1 = x(m1 + m2)

? x = m1x2 + m2x1
m1 + m2

Again, taking the first and third ratios, we get,

mm12 yy2––yy1
or, m1y2 – m1y = m2y – m2y1

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

? y = m1y2 + m2y1
m1 + m2
m1x2 + m2x1 m1y2 + m2y1
Hence, the coordinates of P are m1 + m2 , m1 + m2

Special Cases

(i) If P(x, y) is the mid point of line segment joining A(x1, y1) and B(x2, y2) then AP = BP
and AP : BP = 1:1.

Then, (x, y) = m1x2 + m2x1 , m1y2 + m2y1 = x1 + x2 , y1 + y2
m1 + m2 m1 + m2 2 2

(ii) If P(x, y) is an external point of the line segment joining P(x, y)
B(x2, y2)
the point A(x1, y1) and B(x2, y2), then the coordinates of
P are given by

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

A(x1, y1)

(iii) If the point P(x1, y1) divides the line segment joining the points A(x1, y1) and B(x2, y2) in
the ratio of k : 1 internally, then the coordinates of P are given by,

(x, y) = kx2 + x1 , ky2 + y1
k+1 k+1

For the problems in which the ratio is to be calculated by which a point divides the line
segment, the formula is more convenient.

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Centroid Formula

Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of triangle A(x1,y1)
∆ABC. Let D,E, and F be the mid points of the sides BC, CA and

AB respectively. Then the medians AD, BE and CF are drawn, The 2

medians intersect at G which is called the centroid of the triangle F 1G E
B(x2,y2) D (x3,y3)
ABC. Since D is the mid-point of BC, the coordinates of D are
x2 + x3 y2 + y3
2 , 2

From plane geometry, we know that the centroid of a triangle

divides the median in the ratio 2:1. i.e. G divides AD in 2:1 ratio.

2 × x2 + x3 + 1 × x1 2 × y2 + y3 + 1 × y1
2 + 1 2 + 1
G (x, y) =G 2 , 2

=G x1 + x2 + x3 , y1 + y2 + y3
3 3

Definition : The point of intersection of the medians of a triangle is called centroid of the

triangle.

Worked out Examples

Example 1. Find the coordinates of the points which divide the line segment joining
Solution:
(3, 1) and (6, 7) in the ratio of 2:3 B(6, 7)
Example 2.
Solution : Here, (x1, y1) = (3, 1), (x2, y2) = (6. 7) 3
m1 : m2 = 2 : 3 P(x, y)
2
Let the required point be P(x, y) A(3, 1)

Then, by using formula,

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

= 2 × 6+3 × 3, 2 × 7+3 × 1
2+3 2+3

= 12 + 9, 14 + 3
5 5

? (x, y) = 21 , 17
5 5

Find the coordinates of the point which divides the line segment joining

(2,1) and (8, 2) in ratio of 3:2 externally. P(x, y)

Let given two points be A(2, 1) and B(8, 2) 2
Then, A(x1, y1) = A(2, 1), B(x2 y2) = B(8, 2)
B(8, 2)
AP : BP = m1 : m2 = 3:2 3
Let the required external point be P(x, y)

Then, by using formula, A(2, 1)

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P(x, y) = m1x2 – m2x1 , m1y2 – m2y1 = 3 × 8 – 2 × 2 , 3×2–2×1
m1 – m2 m1 – m2 3 – 2 3–2

= (24 – 4, 6 – 2) = (20, 4)

Example 3. Find the coordinates of the mid-point of the line segment joining the point
Solution: (4, 6) and (–2, –4).

Example 4. Here, (x1, y1) = (4, 6) and (x2, y2) = (–2, –4).
Solution: Let the required mid-point be (x, y).

Then by using formula,

(x, y) = x1 + x2 , y1 + y2 = 4 – 2 , 6–4
2 2 2 2

= 2 , 2 = (1, 1)
2 2

In what ratio does the point (3, 2) divide the line segment joining the points
(1, 4) and (–2, 8).

Here, (x1, y1) = (1, 4)

(x2, y2) = (–2, 8)

Let (3, 2) divides the line segment joining the points (1, 4) and (–2, 8) in

ratio of m1:m2.

Now, x = m1x2 + m2x1
m1 + m2

or, 3 = m1(–2) + m2 . 1 or, 3m1 + 3m2 = –2m1 + m2
m1 + m2
m1 2
or, 5m1 = – 2m1 or, m2 = – 5

? m1 : m2 = – 2:5

Hence, the point (1, 4) divides the join of the given points in 2:5 ratio
externally.

Example 5. The point (2, 4) divides the line segment joining the points (4, –6) and (p, q)
Solution: in the ratio of 1:2 internally. Find the values of p and q.

Here, (x1, y1) = (4, –6) and (x2, y2) = (p, q) m1:m2 = 1:2

(x, y) = (2, 4)

Then, x = m1x2 + m2x1
m1 + m2

or, 2 = 1 . p + 2 . 4 or, 2 = p + 8
1 + 2 3

or, 6 = p + 8 ? p = –2

and y = m1y2 + m2y1
m1 + m2

or, 4 = 1 . q+2. (–6)
1+2

or, 12 = q – 12 ? q = 24

Hence, (p, q) = (–2, 24)

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Example 6. In what ratio is the line joining the points (2, –3) and (5, 8) divided by X-axis
Solution : and Y-axis?

Let the X-axis divide the line segment joining the points (2, –3) and (5, 8) in

ratio of m1:m2 Y (5, 8)
Here, (x1, y1) = (2, –3) 8
7
(x2, y1) = (5, 8) 6
At X–axis, y-coordinate = 0 5
4
3
2
y = m1y2 + m2y1 1 X
m1 + m2 X' -6 -5 -4 -3 -2 -1-1O 1 2 3 4 5 6
-2
or, 0 = m1 . 8 + m2 . (–3) -3 (2, -3)
m1 + m2
or, 8m1 – 3m2 = 0 Y'

or, 8m1 = 3m2

? m1 = 3
m2 8

Hence, X-axis divides the joining of given points in 3:8 internally.

Again, let Y-axis divide the join of given points in m:n ratio, At Y-axis
X-coordinate = 0

x = mx2 + nx1
m + n

or, 0 = m .5 + n . 2 or, 0 = m × 5 + n × 2
m + n

or, m : n = – 2 : 5

Hence, Y–axis divides the joining of given points in 2:5 externally in ratio of 2:5.

Example 7. Find the coordinates of the points which trisect the join of (2, 4) and (8, 9)
Solution: equal parts.

Let given points be A(2, 4) and B(8, 9), P(x', y') and Q(x", y") be the required

points which divide AB in three equal parts. P divides AB is 1:2 ratio,

m1:m2 = 1:2 (x1, y1) = (2, 4), (x2, y2) = (8, 9)

Now, P(x', y') = m1x2 + m2x1 , m1y2 + m2y1 B(8, 9)
m1 + m2 m1 + m2 Q(x", y")

= 1 × 8 + 2 × 2 , 1×9+2×4
1 + 2 1+2

= 12 , 17 = 4, 17 P(x', y')
3 3 3

Q(x", y") is he mid-point of PB. (x1, y1) = 4, 17 ,
3
A(2, 4)
x1 + x2 y1 + y2
(x", y") = 2 , 2 (x2, y2) = (8, 9)

= 4 + 8 , 17 + 9 = 6, 22
2 3 3

2

? P 4, 17 and Q 6, 22 are the required points.
3 3

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Example 8. Show that the points A(1, 2), B(3, 0), C(7, 4), and D(5, 6) are the vertices of a
Solution :
parallelogram. A(1, 2) D(5, 6)
Example 9.
Solution: Here, A(1, 2), B(3, 0), C(7, 4), and D(5, 6) are the
vertices of quadrilateral ABCD.

A quadrilateral is a parallelogram if the diagonals

bisect each other at a point. B(3, 0) C(7, 4)

Mid-point of diagonal BD = 3 + 5 , 0+6 = (4, 3)
2 2

Mid-point of diagonal AC = 1 + 7 , 2+4 = (4, 3)
2 2

Hence, the mid-point of diagonal BD = the mid-point of diagonal AC

? ABCD is a parallelogram.

If P(8, 5), Q(–7, –5), and R(–5, 5) are the three vertices of a parallelogram
PQRS in order. Find the fourth vertex.

Here, P(8,5), Q(–7, –5) R(–5, 5), and S(x', y') are the vertices of a parallelogram.

Now, mid-point of diagonal PR. P(8, 5) S(x', y')

= 8 – 5 , 5+5 = 3 , 5
2 2 2

Q(–7,–5) R(–5, 5)

Again, mid-point of QS = – 7 + x' , – 5+ y'
2 2

since PQRS is a parallelogram,

mid-point of diagonal PR = mid-point of QS

or, 3 , 5 = – 7+ x' , – 5 + y'
2 2 2

? 3 = –7 + x' and 5 = – 5+ y'
2 2 2

or, 3 = – 7 + x' or, 10 = – 5 + y'

? x' = 10 ? y' = 15

Hence, the fourth vertex of the parallelogram is (10, 15).

Example 10. The vertices of a triangle are (3, –5), (–7, 4), and (10, –2). Find the centroid of

the triangle. A(3, –5)

Solution: Let the vertices of triangle be A(3, –5), B(–7, 4) and C(10, –2)

Now, A(x1, y1) = (3, –5) E
C(10, –2)
B(x2, y2) = (–7, 4) G

C(x3, y3) = (10, –2) B(–7, 4)

Let G(x, y) be the centroid of the triangle then by using formula,

G(x, y) = x1 + x2 + x3 , y1 + y2 + y2
3 3

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vedanta Excel In Opt. Mathematics - Book 9 Coordinate Geometry

= 3 + (–7) + 10 , – 5+4+ (–2)
3 3

? G(x, y) = (2, – 1)

Hence, the centroid of triangle is (2, –1).

Exercise 6.2

Very Short Questions

1. Let A(x1, y1) and B(x2, y2) be two given points and P divides AB in the ratio of m1 : m2.
Then, write the following formula.

(a) Section formula (i) Internal division (ii) External division
(b) Mid-point formula
A(x1,y1)

2. In the adjoining figure of triangle F E
(a) What are AD, BE and CF called ? B(x2,y2) G
(b) What is the point G called ? D (x3,y3)
(c) Write down the coordinates of G.

3. (a) Write down the coordinates of the mid-point of the line segment joining A(4, 5) and
B(8, 10).

(b) Find the coordinates of the point which divides the join of (5, 4) and (8, 9) in the
ratio of 1:2 internally.

(c) Find the coordinates of the point that divide the join of (1, 3) and (2, 7) externally
in the ratio of 3:4.

(d) Find the coordinates of centroid of the triangle PQR whose vertices are P(1, 2),
Q(4, 5) and C(4, 2).

Short Questions

4. (a) Find the coordinates of a point dividing internally the line joining the points.

(i) (4, 5) and (7, 8) in the ratio of 2:3

(ii) (–3, 6) and (1, –2) in the ratio of 4:5.

(b) Find the coordinates of a point dividing externally the line joining the points.

(i) (1, 4) and (–2, 16) in the ratio of 1:3.

(ii) (5, –2) and (9, 6) in the ratio of 3:1.

5. (a) A line segment having an end (5, –2) is divided by the point (8, 4) in the ratio of
3:1, find the another end.

(b) If A(2, k) is the mid-point of the line segment PQ joining the points P(4, 3) and
Q(0, 2), find the value of k.

(c) If the coordinates of the mid-point of the line joining the point is (3, 5) and (x, y) is
(1, –1). Find the values of x and y.

(d) The point (–2, –9) divides the line segment joining (a. b) and (2, 7) in the ratio of
3:4, find the values of a and b.

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vedanta Excel In Opt. Mathematics - Book 9 Coordinate Geometry

6. (a) Find the coordinates of the centre of circle from the given circles.

(i) (ii)

A(6, –7) O B(6, 7) P(2, 4) O Q(4, 8)

(b) Find the coordinates of unknown vertices of the given parallelograms.

(i) A(8,5) D(p,q) (ii) P(–2, –1) S(1, 2)

(1,3) (1, 1)

B(9,2) C(x,y) Q(a, b) R(x, y)

7. (a) If the centroid of the triangle is (1, 3), its two vertices are (2, 3) and (–3, 4), find the
third vertex.

(b) The medians of a triangle meet at(1, 1) whose two vertices are (5, 9) and (–4, 1).
Find the third vertex.

Long Question

8. (a) In what ratio does the point –1, 24 divide the line joining the points (–3, 4) and
(2, 6)? 5

(b) In what ratio does the point (3, –2) divide the join of the points (1, 4) and (–3, 16) ?

(c) In what ratio does the point (5, 4) divide the line segment joining the points (2, 1)
and (7, 6) ?

9. (a) Find the coordinates of the points of trisection of the line segment joining the
points (2, –2) and (–1, 4).

(b) Find the coordinates of the points of trisection of the line segment joining the
points P(1, 2) and Q(4, 2).

(c) Find the coordinates of the points of trisection of the line segment joining the
points (1, –2) and (–3, 4).

10. (a) If the point (x, 14) divides the line segment joining the points (7, 11) and (–18, 16).
In which ratio the point divides the line and hence find the value of x.

(b) The point (5, y) divides the line segment joining the points (3, 7) and (8, 9) in a
ratio. Find the ratio and value of y.

11. (a) In what ratio does X-axis divide the line, segment joining the points (2, –4) and
(–3, 6)?

(b) In what ratio does X-axis divide the line segment joining the points (2, –3) and
(5, 6)?

(c) In what ratio does Y-axis divide the line segment joining the points (2, –3) and
(–6, 10) ?

(d) In what ratio does Y-axis divide the line segment joining the point (5, 4) and (–2, –2)?

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vedanta Excel In Opt. Mathematics - Book 9 Coordinate Geometry

(e) Find the ratios in which the line joining (3, –2) and (–3, 6) is cut by the axes of
coordinates .

12. (a) Prove that the following points represent the vertices of a parallelogram.

(i) (–2, –1), (1, 0), (4, 3) and (1, 2)

(ii) (2, –2), (8, 4), (5, 7) and (–1, 1)

(iii) (10, –6), (2, –6), (–4, –2) and (4, –2)

(b) Three consecutive vertices of a parallelogram are respectively P(1, 2), Q(1,0) and
R(4, 0). Find the fourth vertex S.

(c) If (3, 7), (5, –7) and (–2, 5) are the vertices of parallelogram, find the coordinates of
the fourth vertex opposite to (5, –7).

13. (a) If the points (–1, 3), (1, –1) and (5, 1) are the vertices of a triangle, then find the
length of the median drawn through (5, 1).

(b) If P(2, –1), Q(–2, 2) and R(–1, 4) are the mid points of the sides of ∆ABC, find the
vertices of the triangle.

14. (a) P(x, y), Q(0, –2), R(5, 3) and S(3, 7) are the vertices of a parallelogram PQRS. Find
the coordinates of point P.

(b) The two consecutive vertices of a parallelogram are (1, 1) and (3, 6). The diagonals
of the parallelogram cut each other at (7, 7/2). Find the remaining vertices of the
parallelogram.

(c) A right angled triangle has the vertices (1, 1), (1, 7) and (7, 1), prove that the mid
point of the hypotenuse lies at equal distance from each of its vertices.

Project Work

15. List the application of section formula and mid-point formula in our daily life and
discuss it on your class room.

3. (a) 6, 15 (b) 6, 17 (c) (–2, –9) (d) (3, 3)
2 3
(ii) (11, 10)
4. (a) (i) 26 , 31 (ii) – 11 , 22 (b) (i) 5 , –2 (d) a = –5, b = –21
5 5 9 9 2 (ii) D(p, q) = (–7, 4)
5
5. (a) (9, 6) (b) k = 2 (c) x = –1, y = –7

6. (a) (i) (6, 0) (ii) (3, 6) (b) (i) C(x, y) = (–6, 1)

7. (a) (4, 2) (b) (2, –7) 8.(a) 2 : 3 (b) 1 : 3 externally (c) 3 : 2

9. (a) (1, 0), (0, 2) (b) (2, 2), (3, 2) (c) – 1 , 0 and –5 , 2
3 3
39
10. (a) 3 : 2, x = –8 (b) 2 : 3, y = 5 11.(a) 2 : 3 (b) 1 : 2
(c) 1 : 3
(d) 5 : 2 (e) 1 : 3, 1 : 1
12. (b) (4, 2)
14. (a) (–2, 2) (c) (–4, 19) 13.(a) 5 (b) (3, 1), (1, –3), (–5, 7)

(b) (13, 6), (11, 1)

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

Statistics : 14
Partition Values

14.0 Review

The following are the marks obtained by 11 students in optional mathematics:
65, 45, 15, 25, 75, 85, 95, 20, 30, 90, 30. Then,

(i) write above marks in ascending and descending orders.
(ii) find the middle value of the marks.
(iii) find the difference between the maximum and minimum marks.
(iv) is there any difference between mean and median of the data ?
In the above data when the marks are arranged in ascending or descending order, the middle
mark is called median. It divides the given data into two equal parts. The difference of the
maximum and minimum value is called range.

Discuss the answer of the following questions in groups.
(a) What is statistics ?
(b) What is frequency ?
(c) What do cummulative frequencies represent ?
(d) Why is arithmetic mean or average used ?
(e) Define individual, discrete and continuous series.

Partition Values

Partition values are those variate values which divide the whole distribution of data into a
number of equal parts.

For example; Median is a partition value which divides a given distribution of data into two
equal parts.

The following are the types of partition values :

(a) Median (b) Quartiles (c) Deciles (d) Percentiles.

In this section, we discuss about quartiles, deciles, and percentiles only.

We learn to calculate partition values of individual and discrete series.

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

14.1 Quartiles

Quartiles are those partition values which divide the whole distribution of data into four
equal parts. There are three quartiles Q1, Q2 and Q3. The first quartile Q1 is called the lower
quartile and the third quartile Q3 is called the upper quartile. The second quartile Q2 is called
the median.

50% 50%

25% 75% Q2 75% 25%
Q1 Md Q3

The variate value Q1 divides the lower half into two equal parts. It means that Q1 divides the
whole data into lower 25% and upper 75%. Q2 divides the given data 50% and 50%. Q2 is
also called median it divides the whole data into two equal halves. Q3 divides the data into
lower 75% and upper 25%.

Calculation of Quartiles for individual and discrete series:

Individual Series

Let n be the total number of observations.

[ ]Q1 = value ofi(n + 1) th
4 item, i = 1, 2, 3.

( )Then, the first Quartile (Q1) = value of th
n+1 item.
4
( )The second Quartile (Q2) = Median (Md) = value of
( )The third Quartile (Q3) = Value of n + 1 th
2 item.

3(n+1) th item.

4

Worked out Examples

Example 1. Find the three quartiles for each of the following set of observations:
Solution:
(a) 15, 16, 10, 13, 17, 22, 21, 25, 18, 23, 19

(b) 70, 60, 80, 50, 90, 100, 40, 85

(a) Arranging the given data in ascending order, we have 10, 13, 15, 16, 17,

18, 19, 21, 22, 23, 25 number of observations (n) = 11

Lower Quartile (Q1) = value of ( )n + 1 th4 item

( )11+ 1 th item = 3rd item = 15
=4
( )n + 1 th
The second Quartile (Q2) = Median (Md) = value of 2 item

= value of ( )11+ 1 th2 item = 6th item = 18

Upper Quartile (Q3) = value of ( )3(n+1) th4 item

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

= value of ( )3 × 12 th4 item

= Value of 9th item = 22

(b) Arranging the given data in ascending order, 40, 50, 60, 70, 80, 85, 90, 100

Number of observations (n) = 8

( )Now, the first Quartile (Q1) = value of th item
n+1
( )= 4
th item = 2.25th item
8+1
4
= 2nd item + (3rd item – 2nd item) × 0.25

= 50 + (60 – 50) × 0.25

= 50 + 10 × 0.25 = 50 + 2.5 = 52.5

Median (Md) = Second Quartile (Q2)
Upper Quartile (Q3) ( )= Value of
( )= Value ofn+1 th item
2
th
8+1
2 item = 4.5th item

4th item + 5th item = 70 + 80 = 75
2 2
( ) ( )=
3(n+1) th item = 3.9 th
=4 4
item

= (6. 75)th item

= 6th item + (7th item – 6th item) × 0.75

= 85 + (90 – 85) × 0.75

= 85 + 5 × 0.75 = 88.75

Example 2. Calculate lower quartile (Q1) and upper quartile (Q3) from the given data:
Solution:
x 10 15 20 25 30 35 40 45 50
f 4 7 9 15 12 10 7 2 1
To calculate the lower quartile (Q1) and upper quartile (Q3).

x f c.f (cumulative frequency)

10 4 4

15 6 10

20 8 18

25 15 33

30 12 45

35 10 55

40 6 61

45 2 63

50 1 64

∑f= N = 64

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

( )N+1 th
Now, lower quartile (Q1) = Value of 4 item

= Value of 4( )64 +1 th item

= Value of (16.25)th term

From the table, c.f. just greater then 16.25 is 18

whose corresponding value is 20.

? Q1 = 20

Upper Quartile (Q3) = Value of 4( )3(N+1) th item

( )3(64 + 1) th

= Value of 4 item

= Value of 48.75th item

c.f. just greater than 48.75 is 55, whose corresponding value is 35.

? Q3 = 35

Exercise 14.1

Very Short Questions
1. (a) Define partition values.

(b) Define the lower quartile and upper quartile.
(c) Which quartile is equal to median ?
(d) Write formula to calculate Q1 and Q3 for discrete series.
(e) What are the positions of the lower quartile and the upper quartile for a discrete

series with 119 terms?
Short Questions
2. Find the lower quartile, median and the upper quartile of the following data:

(a) 4, 6, 8, 12, 14, 16, 20
(b) 10, 15, 20, 25, 30, 35, 40
(c) 2, 5, 7, 8, 9, 10, 12, 20, 22, 24, 26
(d) 5, 6, 8, 10, 12, 14, 20, 24, 26, 29, 34, 35, 38, 40, 45
3. Find Q1, Q2 and Q3 of the following data:
(a) 11, 15, 8, 9, 12, 13, 14, 18
(b) 20, 25, 15, 10, 30, 45, 22, 35
(c) 45, 50, 15, 20, 25, 35, 40, 42, 41, 18, 17, 44
(d) 11, 21, 14, 16, 18, 20, 24, 27, 30, 35, 40, 25

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vedanta Excel In Opt. Mathematics - Book 9 Statistics: Partition Values

Long Questions

4. Find Q1 and Q3 of the following data :

(a) x 10 20 25 30 35 40 50
f 4 6 10 15 13 10 2

(b) x 16 18 20 10 25 30
f 3 4 10 2 12 8

(c) x 40 20 30 10 60 80
f 12 2 8 1 4 3

5. Find Q1 and Q3 of the following data :

(a) Size 11 13 15 17 19 21 22
5
No. of boxes 2 9 20 25 24 15

(b) Class (marks) 5 15 25 35 45
No. of students 5 8 15 10 6

(c) Age (in years) 10 11 12 13 14 15 16 17
No. of students 4 6 8 10 8 6 4 2

(d) Income (Rs.) 100 150 200 250 300 400

No. of labourers 5 8 10 13 10 4

6. Construct discrete frequency distribution table and find Q1 and Q3:
(a) 50, 20, 60, 30, 48, 60, 70, 20, 30, 50, 40, 46, 30, 40, 60, 70, 70, 37, 40, 50

(b) 15, 12, 23, 35, 46, 57, 18, 12, 39, 41, 32, 43, 25, 49, 18, 38, 45, 40, 32, 33.

7. (a) If 2x + 1, 3x – 1, 3x + 5, 5x – 7, 51, 63 and 70 are in ascending order and Q1 = 20,
find the value of x and Q3.

(b) If 10, x + 4, 20, 25, 30, 35, 40 are in ascending order and Q1 = 15, find the value
of x and Q3.

Project Work

8. List the first 100 natural numbers. Find Q1, Q2 and Q3 from the data.

1. (e) 30, 90 2. (a) 6, 12, 16 (b) 15, 25, 35 (c) 7, 10, 22 (d) 10, 24, 35

3. (a) 9.5, 12.5, 14.75 (b) 16.25, 23.5, 33.75 (c) 18.5, 37.5, 43.5

(d) 16.5, 22.5, 29.25 4.(a) 25, 35 (b) 20, 25 (c) 30, 60

5. (a) 15, 19 (b) 15, 35 (c) 12, 15 (d) 150, 300

6. (a) 37, 60 (b) 23, 43

7. (a) x = 7, Q3 = 63 (b) x = 11, Q3 = 35

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