Notesale: Turn your study into money

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56

MATHEMATICS

Chapter

3

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MATRICES

3
...
— CANTOR

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The knowledge of matrices is necessary in various branches of mathematics
...
This mathematical tool simplifies
our work to a great extent when compared with other straight forward methods
...
Matrices are not only used as a
representation of the coefficients in system of linear equations, but utility of matrices
far exceeds that use
...
Also, many physical operations such as magnification, rotation and
reflection through a plane can be represented mathematically by matrices
...
This mathematical tool is not only used in certain branches
of sciences, but also in genetics, economics, sociology, modern psychology and industrial
management
...


3
...
We may
express it as [15] with the understanding that the number inside [ ] is the number of
notebooks that Radha has
...
We may express it as [15 6] with the understanding that first number
inside [ ] is the number of notebooks while the other one is the number of pens possessed
by Radha
...


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Radha
has
15
notebooks
and
Fauzia
has
10
notebooks
and
Simran
has
13
notebooks
and
Now this could be arranged in the tabular form as follows:
Notebooks
Pens
Radha
15
6
Fauzia
10
2
Simran
13
5
and this can be expressed as

or

Radha
Notebooks
15
Pens
6
which can be expressed as:

Fauzia
10
2

Simran
13
5

In the first arrangement the entries in the first column represent the number of
note books possessed by Radha, Fauzia and Simran, respectively and the entries in the
second column represent the number of pens possessed by Radha, Fauzia and Simran,

58

MATHEMATICS

he

respectively
...
The
entries in the second row represent the number of pens possessed by Radha, Fauzia
and Simran, respectively
...
Formally, we define matrix as:
Definition 1 A matrix is an ordered rectangular array of numbers or functions
...

We denote matrices by capital letters
...
5 –1 2 ⎥ , C = ⎢
⎥
⎣ cos x sin x + 2 tan x ⎦
⎢
⎥
5 ⎥
6⎦
⎢ 3 5
⎥
7 ⎦
⎣

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⎡– 2
⎢
A=⎢ 0
⎢3
⎣

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In the above examples, the horizontal lines of elements are said to constitute, rows
of the matrix and the vertical lines of elements are said to constitute, columns of the
matrix
...

3
...
1 Order of a matrix
A matrix having m rows and n columns is called a matrix of order m × n or simply m × n
matrix (read as an m by n matrix)
...
We observe that A has
3 × 2 = 6 elements, B and C have 9 and 6 elements, respectively
...
, ain, while the jth column
consists of the elements a1j, a2j, a3j,
...
We can also call
it as the (i, j)th element of A
...


MATRICES

59

Note In this chapter
1
...


he

2
...


We can also represent any point (x, y) in a plane by a matrix (column or row) as

bl

⎡0 ⎤
P = ⎢ ⎥ or [0 1]
...
For example point P(0, 1) as a matrix representation may be given as
⎣ ⎦

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Observe that in this way we can also express the vertices of a closed rectilinear
figure in the form of a matrix
...

Now, quadrilateral ABCD in the matrix form, can be represented as
A B C D
⎡1 3 1 −1⎤
X=⎢
or
⎥
⎣ 0 2 3 2⎦ 2 × 4

A⎡ 1
B⎢ 3
Y= ⎢
C⎢ 1
⎢
−
D⎣ 1

0⎤
2⎥
⎥
3⎥
⎥
2 ⎦ 4× 2

Thus, matrices can be used as representation of vertices of geometrical figures in
a plane
...


Example 1 Consider the following information regarding the number of men and women
workers in three factories I, II and III
Men workers

Women workers

I

30

25

II

25

31

III

27

26

Represent the above information in the form of a 3 × 2 matrix
...

Example 2 If a matrix has 8 elements, what are the possible orders it can have?

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Hence, possible orders are 1 × 8, 8 ×1, 4 × 2, 2 × 4

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Solution We know that if a matrix is of order m × n, it has mn elements
...

Thus, all possible ordered pairs are (1, 8), (8, 1), (4, 2), (2, 4)

Example 3 Construct a 3 × 2 matrix whose elements are given by aij =

⎡ a11 a12 ⎤
Solution In general a 3 × 2 matrix is given by A = ⎢ a21 a22 ⎥
...

Now
2

a11 =

1
|1 − 3 × 1| = 1
2

a12 =

1
5
|1 − 3 × 2 | =
2
2

a21 =

1
1
| 2 − 3 × 1| =
2
2

a22 =

1
| 2 − 3× 2 | = 2
2

a31 =

Therefore

1
| 3 − 3 × 1| = 0
2

a32 =

1
3
| 3 − 3× 2 | =
2
2

⎡1
⎢
⎢1
Hence the required matrix is given by A = ⎢
⎢2
⎢0
⎣

5⎤
2⎥
⎥
2⎥
...

2

MATRICES

61

3
...

(i) Column matrix

is

⎡ 0 ⎤
⎢ ⎥
⎢ 3⎥
For example, A = ⎢ −1 ⎥ is a column matrix of order 4 × 1
...


In general, A = [aij] m × 1 is a column matrix of order m × 1
...


⎤
5 2 3⎥ is a row matrix
...


(iii) Square matrix

A matrix in which the number of rows are equal to the number of columns, is
said to be a square matrix
...

⎡ 3 −1
⎢3
A=⎢
3 2
For example
⎢2
⎢
3
⎣4

0⎤
⎥
1 ⎥ is a square matrix of order 3
...


Note If A = [aij] is a square matrix of order n, then elements (entries) a11, a22,
...
Thus, if A = ⎢ 2 4 −1⎥
...


62

MATHEMATICS

(iv) Diagonal matrix
A square matrix B = [bij] m × m is said to be a diagonal matrix if all its non
diagonal elements are zero, that is a matrix B = [bij] m × m is said to be a diagonal
matrix if bij = 0, when i ≠ j
...
1 0 0 ⎤
⎡ −1 0 ⎤
⎢
2 0 ⎥ , are diagonal matrices
, C=⎢ 0
For example, A = [4], B = ⎢
⎥
0 2⎥
⎣
⎦
⎢ 0
0 3⎥
⎣
⎦

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of order 1, 2, 3, respectively
...

For example

A = [3],

⎡ −1 0 ⎤
B=⎢
⎥,
⎣ 0 −1⎦

⎡ 3
⎢
C=⎢ 0
⎢
⎣0

0

3

0

0⎤
⎥
0⎥
⎥
3⎦

are scalar matrices of order 1, 2 and 3, respectively
...
In other words, the square matrix A = [aij] n × n is an
⎧1 if i = j
identity matrix, if aij = ⎨

...
When order is clear from the
context, we simply write it as I
...

Observe that a scalar matrix is an identity matrix when k = 1
...


MATRICES

63

(vii) Zero matrix
A matrix is said to be zero matrix or null matrix if all its elements are zero
...
We denote
⎣0 0 ⎦ ⎣0 0 0 ⎦
zero matrix by O
...


3
...
1 Equality of matrices

is

Definition 2 Two matrices A = [aij] and B = [bij] are said to be equal if
(i) they are of the same order

bl

(ii) each element of A is equal to the corresponding element of B, that is aij = bij for
all i and j
...
Symbolically, if two matrices A and B are equal, we write A = B
...
5
⎢
If ⎢ z a ⎥ = ⎢ 2
⎢
⎥
⎢b c ⎥ ⎢3
⎣
⎦ ⎣

0 ⎤
⎥
6 ⎥ , then x = – 1
...


Solution As the given matrices are equal, therefore, their corresponding elements
must be equal
...
In the matrix A = ⎢ 35 −2
2
⎢
⎥
⎢
⎥
⎣ 3 1 −5 17 ⎦

is

EXERCISE 3
...

2
...
If a matrix has 18 elements, what are the possible orders it can have? What, if it
has 5 elements?
4
...
Construct a 3 × 4 matrix, whose elements are given by:
(i) aij =

1
| −3i + j |
2

(ii) aij = 2i − j

6
...
Find the value of a, b, c and d from the equation:

⎡ a − b 2 a + c ⎤ ⎡ −1 5 ⎤
⎢ 2a − b 3c + d ⎥ = ⎢ 0 13⎥
⎣
⎦ ⎣
⎦

⎡ x + y + z ⎤ ⎡9 ⎤
⎢ x + z ⎥ = ⎢5⎥
⎢
⎥ ⎢ ⎥
⎢ y + z ⎥ ⎢7 ⎥
⎣
⎦ ⎣ ⎦

MATRICES

65

(A) x =

−1
, y=7
3

(B) Not possible to find

he

8
...
Which of the given values of x and y make the following pair of matrices equal
5 ⎤ ⎡0 y − 2⎤
⎡3 x + 7
⎢ y + 1 2 − 3 x ⎥ , ⎢8
4 ⎥
⎦
⎣
⎦ ⎣

1
2
−2
, y
(D) x
3
3
3
10
...
4 Operations on Matrices

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(C) y = 7, x =

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In this section, we shall introduce certain operations on matrices, namely, addition of
matrices, multiplication of a matrix by a scalar, difference and multiplication of matrices
...
4
...
Each factory produces sport
shoes for boys and girls in three different price categories labelled 1, 2 and 3
...
Then the total production
In category 1 : for boys (80 + 90), for girls (60 + 50)
In category 2 : for boys (75 + 70), for girls (65 + 55)
In category 3 : for boys (90 + 75), for girls (85 + 75)

⎡80 + 90
This can be represented in the matrix form as ⎢ 75 + 70
⎢
⎢90 + 75
⎣

60 + 50 ⎤
65 + 55 ⎥
...
We observe that the sum of
two matrices is a matrix obtained by adding the corresponding elements of the given
matrices
...


he

⎡ a11 a12 a13 ⎤
⎡b11 b12 b13 ⎤
Thus, if A = ⎢
⎥ is a 2 × 3 matrix and B = ⎢
⎥ is another
⎣ a21 a22 a23 ⎦
⎣b21 b22 b23 ⎦

is

⎡ a11 + b11 a12 + b12 a13 + b13 ⎤
2×3 matrix
...

⎣ a21 + b21 a22 + b22 a23 + b23 ⎦

bl

In general, if A = [aij] and B = [bij] are two matrices of the same order, say m × n
...


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⎡2
5 1⎤
⎡ 3 1 − 1⎤
⎢
⎥
Example 6 Given A = ⎢
⎥ and B = ⎢
1 ⎥ , find A + B
2 3 0⎦
−2 3
⎣
⎢
⎣
⎦
2⎥

Since A, B are of the same order 2 × 3
...
We emphasise that if A and B are not of the same order, then A + B is not
⎡ 2 3⎤
⎡1 2 3⎤
defined
...

⎣1 0 ⎦
⎣1 0 1 ⎦

2
...


3
...
2 Multiplication of a matrix by a scalar

Now suppose that Fatima has doubled the production at a factory A in all categories
(refer to 3
...
1)
...
We observe that
⎥
170 ⎥
⎦

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⎡160
This can be represented in the matrix form as ⎢150
⎢
⎢180
⎣

the new matrix is obtained by multiplying each element of the previous matrix by 2
...

In other words, kA = k [aij] m × n = [k (aij)] m × n, that is, (i, j)th element of kA is kaij
for all possible values of i and j
...
5⎤
⎢
⎥
A = ⎢ 5 7 −3 ⎥ , then
⎢2 0 5⎥
⎣
⎦

3 4
...
5⎤ ⎡ 9
⎢
⎥ ⎢
⎥
3A = 3 ⎢ 5 7 −3 ⎥ = ⎢3 5 21 −9 ⎥
⎢2 0 5⎥ ⎢ 6
0 15 ⎥
⎣
⎦ ⎣
⎦

Negative of a matrix The negative of a matrix is denoted by – A
...


68

MATHEMATICS

⎡ 3 1⎤
A= ⎢
⎥ , then – A is given by
⎣ −5 x ⎦

For example, let

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⎡ 3 1 ⎤ ⎡ −3 − 1 ⎤
– A = (– 1) A = (−1) ⎢
⎥=⎢
⎥
⎣ −5 x ⎦ ⎣ 5 − x ⎦
Difference of matrices If A = [aij], B = [bij] are two matrices of the same order,
say m × n, then difference A – B is defined as a matrix D = [dij], where dij = aij – bij,
for all value of i and j
...


Solution We have
1 2 3
2 3 1

3

1 3
1 0 2

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2A – B = 2

bl

⎡1 2 3⎤
⎡ 3 −1 3 ⎤
Example 7 If A = ⎢
⎥ and B = ⎢ −1 0 2 ⎥ , then find 2A – B
...
4
...

Now
A + B = [aij] + [bij] = [aij + bij]
= [bij + aij] (addition of numbers is commutative)
= ([bij] + [aij]) = B + A
(ii) Associative Law For any three matrices A = [aij], B = [bij], C = [cij] of the
same order, say m × n, (A + B) + C = A + (B + C)
...
In other words, O is the
additive identity for matrix addition
...
So
– A is the additive inverse of A or negative of A
...
4
...


Solution We have 2A + 3X = 5B
or
2A + 3X – 2A = 5B – 2A
or
2A – 2A + 3X = 5B – 2A
or
O + 3X = 5B – 2A
or
3X = 5B – 2A

(Matrix addition is commutative)
(– 2A is the additive inverse of 2A)
(O is the additive identity)

1
(5B – 2A)
3

or

X=

or

⎛ ⎡ 10 −10 ⎤
⎛ ⎡ 2 −2 ⎤
⎡8 0 ⎤ ⎞
1 ⎜⎢
1⎜ ⎢
⎥ − 2 ⎢ 4 −2 ⎥ ⎟
20 10 ⎥ +
X = ⎜5 ⎢ 4 2 ⎥
⎥
⎢
⎥⎟ = 3 ⎜⎢
3⎜
⎜ ⎢ −25 5 ⎥
⎢3 6 ⎥ ⎟
⎦
⎣
⎦
⎣
⎦⎠
⎝⎣
⎝ ⎢ −5 1 ⎥

⎡ −16 0 ⎤ ⎞
⎢ −8 4 ⎥ ⎟
⎢
⎥⎟
⎢ −6 −12 ⎥ ⎟
⎣
⎦⎠

70

MATHEMATICS

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⎡5 2⎤ ⎡3 6 ⎤
Solution We have ( X + Y ) + ( X − Y ) = ⎢
⎥+⎢
⎥
...

⎣0 9 ⎦
⎣ 0 −1⎦

he

−10 ⎤
⎡
⎢ −2 3 ⎥
⎡ 10 − 16 −10 + 0 ⎤
⎡ − 6 −10 ⎤ ⎢
⎥
14 ⎥
1⎢
1⎢
⎥
20 − 8 10 + 4 ⎥ = ⎢ 12 14 ⎥ = ⎢ 4
= ⎢
⎥ ⎢
3 ⎥
3
3
⎢ −25 − 6 5 − 12 ⎥
⎢ −31 −7 ⎥ ⎢
⎥
⎣
⎦
⎣
⎦
⎢ −31 −7 ⎥
⎢ 3
3 ⎥
⎣
⎦

⎡8 8⎤
⎡8 8⎤
(X + X) + (Y – Y) = ⎢
⎥ ⇒ 2X = ⎢ 0 8⎥
⎣ 0 8⎦
⎣
⎦

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or

or

X=

1
2

⎡8 8⎤ ⎡ 4 4 ⎤
⎢0 8⎥ = ⎢ 0 4 ⎥
⎣
⎦ ⎣
⎦

Also

⎡5 2⎤ ⎡3 6 ⎤
(X + Y) – (X – Y) = ⎢
⎥−⎢
⎥
⎣ 0 9 ⎦ ⎣0 −1⎦

or

⎡5 − 3 2 − 6 ⎤
⎡ 2 −4 ⎤
(X – X) + (Y + Y) = ⎢
⎥ ⇒ 2Y = ⎢ 0 10 ⎥
9 + 1⎦
⎣ 0
⎣
⎦

or

Y=

1
2

⎡ 2 − 4 ⎤ ⎡ 1 −2 ⎤
⎢ 0 10 ⎥ = ⎢ 0 5 ⎥
⎣
⎦ ⎣
⎦

Example 10 Find the values of x and y from the following equation:
⎡x
2⎢
⎣7

5 ⎤ ⎡3 −4 ⎤
⎡7 6⎤
⎥ + ⎢1 2 ⎥ = ⎢15 14 ⎥
y − 3⎦ ⎣
⎣
⎦
⎦

Solution We have
⎡x
2⎢
⎣7

5 ⎤ ⎡3 −4 ⎤
10 ⎤ ⎡3 − 4 ⎤ ⎡ 7 6 ⎤
⎡7 6⎤
⎡2 x
+
= ⎢
=
⎥ ⇒ ⎢
⎥+⎢
y − 3 ⎥ ⎢1 2 ⎥
⎦ ⎣
⎦
2 ⎥ ⎢15 14 ⎥
⎣15 14 ⎦
⎣14 2 y − 6 ⎦ ⎣1
⎦ ⎣
⎦

MATRICES

or
or
or
i
...


10 − 4 ⎤
⎡2x + 3
6 ⎤ ⎡7 6⎤
⎡2 x + 3
⎡7 6⎤
=
⎢ 14 + 1 2 y − 6 + 2 ⎥ = ⎢
⎥ ⇒ ⎢ 15
2 y − 4 ⎥ ⎢15 14 ⎥
⎣
⎦
⎣
⎦ ⎣
⎦
⎣15 14 ⎦

2x + 3 = 7
2x = 7 – 3

4
2
x =2
x=

and
and
and
and

2y – 4 = 14
2y = 18

(Why?)

18
2
y = 9
...
The sale (in Rupees) of these
varieties of rice by both the farmers in the month of September and October are given
by the following matrices A and B
...

(ii) Find the decrease in sales from September to October
...


Solution
(i) Combined sales in September and October for each farmer in each variety is
given by

72

MATHEMATICS

2
× B = 0
...
02

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=

Thus, in October Ramkishan receives Rs 100, Rs 200 and Rs 120 as profit in the
sale of each variety of rice, respectively, and Grucharan Singh receives profit of Rs
400, Rs 200 and Rs 200 in the sale of each variety of rice, respectively
...
4
...
Meera wants to buy 2 pens and 5 story
books, while Nadeem needs 8 pens and 10 story books
...

How much money does each need to spend? Clearly, Meera needs Rs (5 × 2 + 50 × 5)
that is Rs 260, while Nadeem needs (8 × 5 + 50 × 10) Rs, that is Rs 540
...

Now, the money required by Meera and Nadeem to make purchases will be
respectively Rs (4 × 2 + 40 × 5) = Rs 208 and Rs (8 × 4 + 10 × 40) = Rs 432

MATRICES

73

Again, the above information can be represented as follows:
Requirements Prices per piece (in Rupees) Money needed (in Rupees)
⎡2 5 ⎤
⎢ 8 10 ⎥
⎣
⎦

⎡ 4 × 2 + 40 × 5 ⎤ ⎡ 208 ⎤
⎢ 8 × 4 + 10 × 4 0⎥ = ⎢ 432 ⎥
⎣
⎦ ⎣
⎦

⎡4⎤
⎢ 40 ⎥
⎣ ⎦

⎡5 4⎤
⎢50 40 ⎥
⎣
⎦

⎡ 5 × 2 + 5 × 50 4 × 2 + 40 × 5 ⎤
⎢ 8 × 5 + 10 × 5 0 8 × 4 + 10 × 4 0⎥
⎣
⎦

is

⎡2 5 ⎤
⎢ 8 10 ⎥
⎣
⎦

he

Now, the information in both the cases can be combined and expressed in terms of
matrices as follows:
Requirements Prices per piece (in Rupees) Money needed (in Rupees)

bl

⎡ 260 208 ⎤
= ⎢
⎥
⎣ 540 432 ⎦

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The above is an example of multiplication of matrices
...
Furthermore for getting the elements of the product matrix,
we take rows of A and columns of B, multiply them element-wise and take the sum
...
Let A = [aij] be an m × n matrix and B = [bjk] be an
n × p matrix
...

To get the (i, k)th element cik of the matrix C, we take the ith row of A and kth column
of B, multiply them elementwise and take the sum of all these products
...
ain] and the kth column of

⎡ b1k ⎤
⎢b ⎥
⎢ 2k ⎥
B is ⎢
...
+ ain bnk =
⎢
...

j =1

The matrix C = [cik]m × p is the product of A and B
...
This is a 2 × 2 matrix in which each
and is given by CD = ⎢
⎥
0 3 4⎥ ⎢
⎣
⎦ ⎢ 5 − 4⎥
⎣
⎦

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entry is the sum of the products across some row of C with the corresponding entries
down some column of D
...

⎣ 2 3⎦
⎣
⎦

Solution The matrix A has 2 columns which is equal to the number of rows of B
...
Now
⎡ 6(2) + 9(7) 6(6) + 9(9) 6(0) + 9(8) ⎤
AB = ⎢
⎥
⎣ 2(2) + 3(7) 2(6) + 3(9) 2(0) + 3(8) ⎦

⎡12 + 63 36 + 81 0 + 72 ⎤
⎡ 75 117 72 ⎤
=⎢
⎥ = ⎢
⎥
⎣ 4 + 21 12 + 27 0 + 24⎦
⎣ 25 39 24 ⎦

MATRICES

75

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Remark If AB is defined, then BA need not be defined
...
If A, B are, respectively m × n, k × l matrices, then both AB and BA are defined
if and only if n = k and l = m
...

Non-commutativity of multiplication of matrices
Now, we shall see by an example that even if AB and BA are both defined, it is not
necessary that AB = BA
...
Show that
Example 13 If A = ⎢
⎥ and B = ⎢
⎥
⎣− 4 2 5 ⎦
⎢ 2 1⎥
⎣
⎦
AB ≠ BA
...
Hence AB and BA are both
defined and are matrices of order 2 × 2 and 3 × 3, respectively
...
But
one may think that perhaps AB and BA could be the same if they were of the same
order
...

⎡1 0 ⎤
⎡0 1 ⎤
⎡ 0 1⎤
Example 14 If A = ⎢
⎥ and B = ⎢1 0 ⎥ , then AB = ⎢
⎥
...
Clearly AB ≠ BA
...


and

76

MATHEMATICS

Note This does not mean that AB ≠ BA for every pair of matrices A, B for
which AB and BA, are defined
...


bl

⎡3 5⎤
⎡ 0 −1⎤
Example 15 Find AB, if A = ⎢
⎥ and B = ⎢ 0 0 ⎥
...
This need
not be true for matrices, we will observe this through an example
...

⎣0 2 ⎦ ⎣0 0⎦ ⎣0 0 ⎦
Thus, if the product of two matrices is a zero matrix, it is not necessary that one of
the matrices is a zero matrix
...
4
...

1
...
We have
(AB) C = A (BC), whenever both sides of the equality are defined
...
The distributive law For three matrices A, B and C
...


3
...

Now, we shall verify these properties by examples
...


MATRICES

77

⎡1 1 −1⎤ ⎡ 1 3 ⎤ ⎡ 1 + 0 + 1 3 + 2 − 4 ⎤ ⎡ 2 1 ⎤
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎥
Solution We have AB = ⎢ 2 0
3 ⎥ ⎢ 0 2 ⎥ = ⎢ 2 + 0 − 3 6 + 0 + 12 ⎥ = ⎢−1 18⎥
⎢
⎢ 3 −1 2 ⎥ ⎢−1 4 ⎥ ⎢3 + 0 − 2 9 − 2 + 8 ⎥ ⎢ 1 15⎥
⎣
⎦ ⎣
⎦ ⎣
⎦ ⎣
⎦

is

3⎤
⎡ 1 + 6 2 + 0 3 − 6 −4 + 3 ⎤
⎥ ⎡1 2 3 −4 ⎤ = ⎢ 0 + 4 0 + 0 0 − 4 0 + 2 ⎥
2⎥ ⎢
⎥
2 0 −2 1⎥ ⎢
⎦ ⎢− +
⎥
4⎦ ⎣
⎣ 1 8 −2 + 0 −3 − 8 4 + 4 ⎥
⎦

©
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⎡ 1
BC = ⎢ 0
⎢
⎢
⎣ −1

bl

4 4 −7 ⎤
⎡4
⎢35 −2 −39 22 ⎥
⎥
= ⎢
⎢31
2 −27 11⎥
⎣
⎦

he

4 + 0 6 − 2 − 8 +1 ⎤
⎡2 1⎤
⎡ 2+2
⎢ 1 18⎥ ⎡1 2 3 − 4⎤ = ⎢ −1 + 36 −2 + 0 −3 − 36
(AB) (C) = ⎢
−
4 + 18⎥
⎥ ⎢ 2 0 −2 1⎥ ⎢
⎥
⎦ ⎢ +
⎥
⎢ 1 15⎥ ⎣
⎣
⎦
⎣ 1 30 2 + 0 3 − 30 − 4 + 15⎦

Now

⎡7 2 −3 −1 ⎤
⎢
⎥
= ⎢ 4 0 −4 2 ⎥
⎢7 −2 −11 8 ⎥
⎣
⎦

Therefore

⎡1 1 −1 ⎤ ⎡7 2 −3 −1 ⎤
A(BC) = ⎢ 2 0 3 ⎥ ⎢ 4 0 −4 2 ⎥
⎢
⎥ ⎢
⎥
⎢ 3 −1 2 ⎥ ⎢7 −2 −11 8 ⎥
⎣
⎦ ⎣
⎦

⎡ 7 + 4 − 7 2 + 0 + 2 −3 − 4 + 11 −1 + 2 − 8 ⎤
⎢
⎥
= ⎢14 + 0 + 21 4 + 0 − 6 −6 + 0 − 33 −2 + 0 + 24 ⎥
⎢ 21 − 4 + 14 6 + 0 − 4 −9 + 4 − 22 −3 − 2 + 16 ⎥
⎣
⎦

4 4 −7 ⎤
⎡4
⎢35 −2 −39 22 ⎥
= ⎢
⎥
...
Also, verify that (A + B)C = AC + BC

⎡2
⎢2
−
⎢
⎢3
⎣

is

8⎤
10 ⎥
⎥
0⎥
⎦

⎤ ⎡ 0 − 14 + 24 ⎤ ⎡10 ⎤
⎥ = ⎢ 10 + 0 + 30 ⎥ = ⎢ 20 ⎥
−
⎥ ⎢
⎥ ⎢ ⎥
⎥ ⎢ 16 + 12 + 0 ⎥ ⎢ 28⎥
⎦ ⎣
⎦ ⎣ ⎦

bl

So

⎡ 0 7
⎢
(A + B) C = ⎢−5 0
⎢ 8 −6
⎣

he

⎡ 0 7 8⎤
⎢
⎥
Solution Now, A + B = ⎢−5 0 10 ⎥
⎢
⎣ 8 −6 0⎥
⎦

and

⎡0 1 1 ⎤ ⎡ 2
⎢
⎥ ⎢
−
BC = ⎢1 0 2 ⎥ ⎢ 2
⎢1 2 0 ⎥ ⎢ 3
⎣
⎦ ⎣

⎤ ⎡9⎤
⎥ = ⎢12 ⎥
⎥ ⎢ ⎥
⎥ ⎢30 ⎥
⎦ ⎣ ⎦

©
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Further

⎡ 0 6 7 ⎤ ⎡ 2 ⎤ ⎡ 0 − 12 + 21
⎢
⎥ ⎢ ⎥ ⎢
−
−
AC = ⎢ −6 0 8 ⎥ ⎢ 2 ⎥ = ⎢ 12 + 0 + 24
⎢ 7 − 8 0 ⎥ ⎢ 3 ⎥ ⎢ 14 + 16 + 0
⎣
⎦ ⎣ ⎦ ⎣

So

⎤ ⎡ 0 − 2 + 3⎤ ⎡ 1 ⎤
⎥ = ⎢ 2 + 0 + 6⎥ = ⎢ 8 ⎥
⎥ ⎢
⎥ ⎢ ⎥
⎥ ⎢ 2 − 4 + 0⎥ ⎢ 2 ⎥
−
⎦ ⎣
⎦ ⎣ ⎦

⎡9 ⎤ ⎡ 1 ⎤ ⎡10 ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
AC + BC = ⎢12 ⎥ + ⎢ 8 ⎥ = ⎢ 20 ⎥
⎢30 ⎥ ⎢−2 ⎥ ⎢ 28⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Clearly,

(A + B) C = AC + BC

⎡1 2
⎢
Example 18 If A = ⎢ 3 −2
⎢4 2
⎣

3⎤
1⎥ , then show that A3 – 23A – 40 I = O
⎥
1⎥
⎦

⎡1 2
Solution We have A = A
...
The
cost per contact (in paise) is given in matrix A as
Cost per contact

⎡
⎢
A= ⎢
⎢
⎣

40
100
50

⎤ Telephone
⎥ Housecall
⎥
⎥ Letter
⎦

The number of contacts of each type made in two cities X and Y is given by
Telephone Housecall Letter

⎡1000
B=⎢
⎣3000

5000 ⎤ → X

...


500

80

MATHEMATICS

Solution We have
⎡ 40,000 + 50,000 + 250,000 ⎤ → X
BA = ⎢
⎥
⎣120,000 + 100,000 +500,000 ⎦ → Y

he

⎡340,000 ⎤ → X
= ⎢
⎥
⎣ 720,000 ⎦ → Y
So the total amount spent by the group in the two cities is 340,000 paise and
720,000 paise, i
...
, Rs 3400 and Rs 7200, respectively
...
Let A = ⎢
⎥ , B = ⎢−2 5⎥ , C = ⎢ 3
⎣3 2⎦
⎣
⎦
⎣
Find each of the following:
(i) A + B
(ii) A – B
(iv) AB
(v) BA
2
...
2

⎡a
(i) ⎢
⎣−b

⎡a 2 + b2
(ii) ⎢ 2
2
⎢a + c
⎣

b ⎤ ⎡a b ⎤
+
a ⎥ ⎢b a ⎥
⎦ ⎣
⎦

b 2 + c 2 ⎤ ⎡ 2ab
2bc ⎤
⎥+⎢
⎥
a 2 + b 2 ⎥ ⎣−2ac −2ab ⎦
⎦

−
⎡1
⎢8
(iii) ⎢
⎢2
⎣

4 −6⎤ ⎡12 7 6 ⎤
⎡cos 2 x sin 2 x ⎤ ⎡ sin 2 x cos 2 x ⎤
5 16 ⎥ + ⎢ 8 0 5 ⎥ (iv) ⎢
⎥+⎢ 2
⎥
⎥ ⎢
⎥
2
2
2
⎢ sin x cos x ⎥ ⎢cos x sin x ⎥
⎣
⎦ ⎣
⎦
8 5 ⎥ ⎢ 3 2 4⎥
⎦ ⎣
⎦
3
...

⎡a
(i) ⎢
⎣−b

b ⎤ ⎡ a −b ⎤
a ⎥ ⎢b a ⎥
⎦⎣
⎦

⎡1 ⎤
⎢ ⎥
(ii) ⎢ 2⎥ [2 3 4]
⎢ 3⎥
⎣ ⎦

⎡ 2 3 4 ⎤ ⎡1 −3
(iv) ⎢ 3 4 5 ⎥ ⎢ 0 2
⎢
⎥ ⎢
⎢4 5 6⎥ ⎢3 0
⎣
⎦ ⎣

⎡ 2 −3⎤
⎡ 3 −1 3 ⎤ ⎢
⎥
(vi) ⎢
⎥ 1 0⎥
−1 0 2 ⎦ ⎢
⎣
⎢3 1⎥
⎣
⎦

5⎤
4⎥
⎥
⎥
5⎦

⎡ 1 −2 ⎤ ⎡1 2 3⎤
(iii) ⎢
⎥⎢
⎥
⎣ 2 3 ⎦ ⎣ 2 3 1⎦

⎡2
⎢
(v) ⎢ 3
⎢−1
⎣

1⎤
⎡ 1 0 1⎤
2⎥ ⎢
⎥ −1 2 1⎥
⎦
1⎥ ⎣
⎦

MATRICES

⎡1 2 −3⎤
⎡ 3 −1
⎢5 0 2 ⎥ , B = ⎢ 4 2
4
...

5⎥
⎥
2⎥
5⎥
⎦

is

1

bl

⎡2
⎢3
⎢
1
5
...
Also, verify that A + (B – C) = (A + B) – C
...
Simplify cosθ ⎢
⎥ + sinθ ⎢ cos θ
sin θ⎥
⎣ − sin θ cos θ ⎦
⎣
⎦
7
...
Find X, if Y = ⎢
⎥ and 2X + Y =
⎣1 4 ⎦

⎡ 1 0⎤
⎢ −3 2 ⎥
⎣
⎦

⎡ 1 3 ⎤ ⎡ y 0 ⎤ ⎡5 6 ⎤
9
...
Solve the equation for x, y, z and t, if 2 ⎢
⎣y
⎡ 2⎤
11
...

⎣ 1 ⎦ ⎣5 ⎦

x + y⎤
⎡x y ⎤ ⎡ x 6 ⎤ ⎡ 4
12
...

⎥ = ⎢ −1 2 w ⎥ + ⎢ z + w
3 ⎥
⎣ z w⎦ ⎣
⎦ ⎣
⎦

82

MATHEMATICS

⎡cos x − sin x 0 ⎤
13
...

⎢
⎥
⎢ 0
0
1⎥
⎣
⎦
⎡ 5 −1⎤ ⎡ 2 1 ⎤ ⎡ 2 1 ⎤ ⎡ 5 −1⎤
(i) ⎢
⎥⎢
⎥≠⎢
⎥⎢
⎥
⎣6 7 ⎦ ⎣ 3 4⎦ ⎣ 3 4⎦ ⎣6 7 ⎦

⎡2 0
15
...
Show that

⎡1 0 2⎤
16
...
If A = ⎢
⎥ and I= ⎢ 0 1 ⎥ , find k so that A = kA – 2I
⎣ 4 −2 ⎦
⎣
⎦

⎡
⎢ 0
18
...
A trust fund has Rs 30,000 that must be invested in two different types of bonds
...
Using matrix multiplication, determine how to divide Rs 30,000 among
the two types of bonds
...
5
...
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen
physics books, 10 dozen economics books
...
Find the total amount the bookshop will receive
from selling all the books using matrix algebra
...
Choose the correct answer in Exercises 21 and 22
...
The restriction on n, k and p so that PY + WY will be defined are:
(A) k = 3, p = n
(B) k is arbitrary, p = 2
(C) p is arbitrary, k = 3
(D) k = 2, p = 3
22
...

Definition 3 If A = [aij] be an m × n matrix, then the matrix obtained by interchanging
the rows and columns of A is called the transpose of A
...
In other words, if A = [aij]m × n, then A′ = [aji]n × m
...
5
...
These
may be verified by taking suitable examples
...


84

MATHEMATICS

Solution
(i) We have

he

⎡ 3 4⎤
⎡3
⎢
⎥
3 2⎤
3 2⎤
′ ⎡3
A= ⎢
⎥ ⇒ A′ = ⎢ 3 2 ⎥ ⇒ ( A′ ) = ⎢
⎥=A
4 2 0⎦
4 2 0⎦
⎣
⎣
⎢ 2 0⎥
⎣
⎦
Thus
(A′)′ = A
(ii) We have

⎡ 5
5⎤
⎢
⎥
(A + B)′ = ⎢ 3 − 1 4 ⎥
⎢ 4
4⎥
⎣
⎦

is

3 − 1 4⎤
⎥
4 4⎦

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Therefore

⎡5
⎡ 2 −1 2 ⎤
⎢1
⎥⇒A+B=⎢
2 4⎦
⎣
⎣5

bl

⎡3
3 2⎤
A= ⎢
⎥, B =
0⎦
⎣4 2

Now

So

Thus
(iii) We have

⎡ 3 4⎤
⎡ 2 1⎤
⎢
⎥
⎢
⎥
A′ = ⎢ 3 2 ⎥ , B′ = ⎢−1 2 ⎥ ,
⎢ 2 0⎥
⎢ 2 4⎦
⎥
⎣
⎦
⎣

⎡ 5
5⎤
⎢
⎥
A′ + B′ = ⎢ 3 −1 4 ⎥
⎢ 4
4⎥
⎣
⎦

(A + B)′ = A′ + B′

⎡ 2 −1 2 ⎤ ⎡ 2 k
kB = k ⎢
=
2 4 ⎥ ⎢k
⎣1
⎦ ⎣

−k
2k

Then

⎡ 2k k ⎤
⎡ 2 1⎤
⎢ −k 2k ⎥ = k ⎢−1 2⎥ = kB′
(kB)′ = ⎢
⎥
⎢
⎥
⎢ 2k 4k ⎥
⎢ 2 4⎥
⎣
⎦
⎣
⎦

Thus

(kB)′ = kB′

2k ⎤
4k ⎥
⎦

MATRICES

85

⎡ −2⎤
⎢ ⎥
Example 21 If A = ⎢ 4 ⎥ , B = [1 3 −6] , verify that (AB)′ = B′A′
...
6 Symmetric and Skew Symmetric Matrices

Definition 4 A square matrix A = [aij] is said to be symmetric if A′ = A, that is,
[aij] = [aji] for all possible values of i and j
...
5 −1 ⎥ is a symmetric matrix as A′ = A
⎢ 3 −1
1⎥
⎣
⎦

Definition 5 A square matrix A = [aij] is said to be skew symmetric matrix if
A′ = – A, that is aji = – aij for all possible values of i and j
...
Therefore 2aii = 0 or aii = 0 for all i’s
...


86

MATHEMATICS

⎡ 0
For example, the matrix B = ⎢ −e
⎢
⎢− f
⎣

e
0
−g

f⎤
g ⎥ is a skew symmetric matrix as B′= –B
⎥
0⎥
⎦

he

Now, we are going to prove some results of symmetric and skew-symmetric
matrices
...

Proof Let B = A + A′, then
B′ = (A + A′)′

bl

= A′ + (A′)′ (as (A + B)′ = A′ + B′)
= A′ + A (as (A′)′ = A)

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= A + A′ (as A + B = B + A)
= B

Therefore

B = A + A′ is a symmetric matrix

Now let

C = A – A′

C′ = (A – A′)′ = A′ – (A′)′
= A′ – A

(Why?)

(Why?)

= – (A – A′) = – C

Therefore

C = A – A′ is a skew symmetric matrix
...

Proof Let A be a square matrix, then we can write

1
1
A = (A + A′) + (A − A′)
2
2
From the Theorem 1, we know that (A + A′) is a symmetric matrix and (A – A′) is

a skew symmetric matrix
...
Thus, any square
2
matrix can be expressed as the sum of a symmetric and a skew symmetric matrix
...


he

Solution Here

−3 ⎤
2⎥
⎥
1 ⎥,
⎥
⎥
−3 ⎥
⎥
⎦

bl

−3
2
3

1

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Let

⎡ 4 −3
1
1⎢
P = (B + B′) = − 3 6
2
2⎢
⎢ −3 2
⎣

⎡
⎢ 2
−3⎤ ⎢
⎢ −3
2⎥ = ⎢
⎥
2
−6 ⎥ ⎢ −3
⎦
⎢
⎢2
⎣

is

⎡ 2 −1 1 ⎤
B′ = ⎢− 2 3 −2 ⎥
⎢
⎥
⎢ −4 4 −3⎥
⎣
⎦

−3
2

Now

Thus

Also, let

Then

−3 ⎤
2⎥
⎥
1 ⎥= P
⎥
⎥
−3 ⎥
⎥
⎦

⎡
⎢ 2
⎢
−3
P′ = ⎢
3
⎢2
⎢
⎢ −3 1
⎢2
⎣
1
P = (B + B′) is a symmetric matrix
...

2

Q=

Now

⎡
⎢2
⎢
−3
P+Q=⎢
⎢2
⎢
⎢ −3
⎢2
⎣

−3
2
3
1

−3 ⎤ ⎡
0
2⎥ ⎢
⎥ ⎢
1
1 ⎥+⎢
⎥ ⎢2
⎥ ⎢
5
−3 ⎥ ⎢
⎥ ⎢2
⎦ ⎣

−1
2
0
−3

−5 ⎤
2 ⎥ ⎡2
⎥
⎢
3 ⎥ = ⎢−1
⎥
⎥ ⎢1
⎣
0⎥
⎥
⎦

−2 −4 ⎤
⎥
3 4⎥ = B
−2 −3⎥
⎦

he

Thus

is

88

EXERCISE 3
...


1
...
If A =
⎢
⎥
⎢
⎥
⎢ −2 1 1 ⎥
⎢ 1 3 1⎥
⎣
⎦
⎣
⎦
(i) (A + B)′ = A′ + B′,
(ii) (A – B)′ = A′ – B′
⎡ 3 4⎤
⎡ −1 2 1⎤
3
...
If A′ = ⎢
⎥ and B = ⎢ 1 2 ⎥ , then find (A + 2B)′
⎣ 1 2⎦
⎣
⎦
5
...
If (i) A = ⎢
⎥ , then verify that A′ A = I
⎣ − sin α cos α ⎦

−1 5 ⎤
2 1 ⎥ is a symmetric matrix
...


he

⎡ sin α cos α ⎤
(ii) If A = ⎢
⎥ , then verify that A′ A = I
⎣ − cos α sin α ⎦

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bl

⎡ 0 1 −1 ⎤
(ii) Show that the matrix A = ⎢ −1 0 1 ⎥ is a skew symmetric matrix
...
For the matrix A = ⎢
⎥ , verify that
⎣6 7 ⎦

(i) (A + A′) is a symmetric matrix

(ii) (A – A′) is a skew symmetric matrix

⎡ 0 a
1(
) and 1 ( A − A′ ) , when A = ⎢ −a 0
A + A′
9
...
Express the following matrices as the sum of a symmetric and a skew symmetric
matrix:
⎡ 3 5⎤
(i) ⎢
⎥
⎣1 −1⎦

3 −1⎤
⎡ 3
⎢−2 −2 1⎥
(iii) ⎢
⎥
⎢ −4 −5 2⎥
⎣
⎦

⎡ 6
⎢
(ii) ⎢−2
⎢ 2
⎣

−2 2 ⎤
3 −1 ⎥
⎥
−1 3 ⎥
⎦

⎡ 1 5⎤
(iv) ⎢
⎥
⎣ −1 2 ⎦

90

MATHEMATICS

⎡ cos α − sin α ⎤
12
...

11
...
7 Elementary Operation (Transformation) of a Matrix

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There are six operations (transformations) on a matrix, three of which are due to rows
and three due to columns, which are known as elementary operations or
transformations
...
Symbolically the interchange
of ith and jth rows is denoted by Ri ↔ Rj and interchange of ith and jth column is
denoted by Ci ↔ Cj
...

6 7⎥
⎦

(ii) The multiplication of the elements of any row or column by a non zero
number
...

The corresponding column operation is denoted by Ci → kCi
⎡ 1
1
For example, applying C3 → C3 , to B = ⎢
7
⎣ −1

⎡
⎢ 1
1⎤
⎥ , we get ⎢
⎢ −1
3 1⎦
⎢
⎣

2

2
3

1⎤
7⎥
⎥
1⎥
7⎥
⎦

(iii) The addition to the elements of any row or column, the corresponding
elements of any other row or column multiplied by any non zero number
...


MATRICES

91

The corresponding column operation is denoted by Ci → Ci + kCj
...

⎣ 2 −1⎦
⎣ 0 −5 ⎦

3
...

⎣1 2 ⎦
⎣ −1 2 ⎦
⎡ 2 3 ⎤ ⎡ 2 −3 ⎤
AB = ⎢
⎥⎢
⎥
⎣ 1 2 ⎦ ⎣ −1 2 ⎦

bl

Now

is

For example, let

he

Definition 6 If A is a square matrix of order m, and if there exists another square
matrix B of the same order m, such that AB = BA = I, then B is called the inverse
matrix of A and it is denoted by A– 1
...


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⎡ 4 − 3 −6 + 6 ⎤ ⎡ 1 0 ⎤
= ⎢
⎥=⎢
⎥=I
⎣ 2 − 2 −3 + 4 ⎦ ⎣ 0 1 ⎦

⎡1 0 ⎤
BA = ⎢
⎥ = I
...
e
...
A rectangular matrix does not possess inverse matrix, since for products BA
and AB to be defined and to be equal, it is necessary that matrices A and B
should be square matrices of the same order
...
If B is the inverse of A, then A is also the inverse of B
...

Proof Let A = [aij] be a square matrix of order m
...
We shall show that B = C
...
(1)

AC = CA = I


...


92

MATHEMATICS

Proof From the definition of inverse of a matrix, we have
(AB) (AB)–1 = 1
A–1 (AB) (AB)–1 = A –1 I

(Pre multiplying both sides by A–1)

or

(A–1A) B (AB)–1 = A –1

(Since A–1 I = A–1)

or

IB (AB)–1 = A –1

or

B (AB)–1 = A –1

or

B–1 B (AB)–1 = B –1 A –1

or

I (AB)–1 = B –1 A –1
(AB)–1 = B –1 A –1

Hence

bl

3
...
1 Inverse of a matrix by elementary operations

is

he

or

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Let X, A and B be matrices of, the same order such that X = AB
...

Similarly, in order to apply a sequence of elementary column operations on the
matrix equation X = AB, we will apply, these operations simultaneously on X and on the
second matrix B of the product AB on RHS
...
The matrix B will be the
inverse of A
...

Remark In case, after applying one or more elementary row (column) operations on
A = IA (A = AI), if we obtain all zeros in one or more rows of the matrix A on L
...
S
...

Example 23 By using elementary operations, find the inverse of the matrix
⎡1 2 ⎤
A=⎢
⎥
...


or

⎡ 1 2 ⎤ ⎡1 0 ⎤
⎡1 2 ⎤ ⎡ 1 0 ⎤
⎢ 2 −1⎥ = ⎢ 0 1 ⎥ A, then ⎢ 0 −5⎥ = ⎢ −2 1 ⎥ A (applying R2 → R2 – 2R1)
⎣
⎦ ⎣
⎦
⎣
⎦ ⎣
⎦

MATRICES

0⎤
1
−1 ⎥ A (applying R2 → – R2)
⎥
5
5⎦

or

⎡1
⎢5
⎡1 0 ⎤
= ⎢
⎢0 1 ⎥
⎢2
⎣
⎦
⎢5
⎣

2⎤
5⎥
⎥ A (applying R1 → R1 – 2R2)
−1 ⎥
5⎥
⎦

he

or

⎡1
⎡1 2 ⎤
⎢
⎢0 1 ⎥ = ⎢ 2
⎣
⎦
⎣5

93

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⎡1 2 ⎤
⎡1 0 ⎤
⎢ 2 −1⎥ = A ⎢ 0 1 ⎥
⎣
⎦
⎣
⎦
Applying C2 → C2 – 2C1, we get

bl

is

⎡1 2 ⎤
⎢5 5 ⎥
⎥
Thus
A–1 = ⎢
⎢ 2 −1 ⎥
⎢5 5 ⎥
⎣
⎦
Alternatively, in order to use elementary column operations, we write A = AI, i
...
,

⎡1 0 ⎤
⎡ 1 −2 ⎤
⎢ 2 −5 ⎥ = A ⎢ 0 1 ⎥
⎣
⎦
⎣
⎦

1
Now applying C2 → − C2 , we have
5

⎡
⎢1
⎡1 0 ⎤
⎢2 1⎥ = A ⎢
⎢0
⎣
⎦
⎢
⎣
Finally, applying C1 → C1 – 2C2, we obtain

2⎤
5⎥
⎥
−1 ⎥
5⎥
⎦

⎡1
⎢5
⎡1 0 ⎤
⎢0 1 ⎥ = A ⎢ 2
⎢
⎣
⎦
⎢5
⎣

2⎤
5⎥
⎥
−1 ⎥
5⎥
⎦

Hence

⎡1
⎢5
A–1 = ⎢
⎢2
⎢5
⎣

2⎤
5⎥
⎥
−1 ⎥
5⎥
⎦

94

MATHEMATICS

Example 24 Obtain the inverse of the following matrix using elementary operations

⎡1 0 0 ⎤
⎡0 1 2⎤
⎢
⎥
⎢
⎥
Solution Write A = I A, i
...
, ⎢1 2 3 ⎥ = ⎢0 1 0 ⎥ A
⎢0 0 1 ⎥
⎢3 1 1 ⎥
⎣
⎦
⎣
⎦

he

⎡0 1 2⎤
A = ⎢1 2 3 ⎥
...
e
...

⎣ −5 1 ⎦
⎡10 −2 ⎤ ⎡1 0 ⎤
Solution We have P = I P, i
...
, ⎢
⎥=⎢
⎥ P
...
Therefore, P–1 does not exist
...
4

Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17
...
⎢
⎥
⎣2 7⎦

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⎡ 2 1⎤
2
...
⎢
⎥
⎣ 2 3⎦

4
...
⎢
⎥
⎣7 4 ⎦

⎡ 2 5⎤
6
...


⎡3 1 ⎤
⎢5 2 ⎥
⎣
⎦

⎡4 5⎤
8
...
⎢
⎥
⎣2 7 ⎦

⎡ 3 −1⎤
10
...
⎢
⎥
⎣ 1 −2 ⎦

⎡ 6 −3 ⎤
12
...
⎢
⎥
...
⎢ 2 2 3 ⎥
⎢ 3 −2 2 ⎥
⎣
⎦

13
...
⎢−3
⎢
⎣2

3 −2 ⎤
0 −5 ⎥
⎥
5 0⎥
⎦

⎡ 2 0 −1 ⎤
⎢
⎥
17
...
Matrices A and B will be inverse of each other only if
(A) AB = BA

(B) AB = BA = 0

(C) AB = 0, BA = I

(D) AB = BA = I

98

MATHEMATICS

Miscellaneous Examples
⎡ cos θ sin θ ⎤
⎡ cos nθ sin nθ ⎤
n
Example 26 If A = ⎢
⎥ , then prove that A = ⎢ − sin nθ cos nθ ⎥ , n ∈ N
...


⎡ cos θ sin θ ⎤
⎡ cos nθ sin nθ ⎤
n
P(n) : If A = ⎢
⎥ , then A = ⎢ − sin nθ cos nθ ⎥ , n ∈ N
⎣ − sin θ cos θ ⎦
⎣
⎦

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⎡ cos θ sin θ ⎤
⎡ cos θ sin θ ⎤
1
P(1) : A = ⎢
⎥ , so A = ⎢ − sin θ cos θ ⎥
⎣
⎦
⎣ − sin θ cos θ ⎦
Therefore,
the result is true for n = 1
...
So

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⎡ cos k θ sin k θ ⎤
⎡ cos θ sin θ ⎤
k
P(k) : A = ⎢
⎥ , then A = ⎢ − sin k θ cos k θ ⎥
⎣
⎦
⎣ − sin θ cos θ ⎦

Now, we prove that the result holds for n = k +1

Now

⎡ cos θ sin θ ⎤ ⎡ cos k θ sin k θ ⎤
k
Ak + 1 = A ⋅ A = ⎢
⎥⎢
⎥
⎣ − sin θ cos θ ⎦ ⎣ − sin k θ cos k θ ⎦

cos θ sin k θ + sin θ cos k θ ⎤
⎡ cos θ cos k θ – sin θ sin k θ
= ⎢
⎥
⎣ − sin θ cos k θ + cos θ sin k θ − sin θ sin k θ + cos θ cos k θ ⎦

⎡ cos (θ + k θ) sin (θ + k θ) ⎤ ⎡ cos ( k + 1)θ sin ( k + 1)θ ⎤
= ⎢
⎥=⎢
⎥
⎣ − sin (θ + k θ) cos (θ + k θ) ⎦ ⎣ − sin ( k + 1)θ cos ( k + 1)θ ⎦

Therefore, the result is true for n = k + 1
...

cos n θ ⎥
⎦

Example 27 If A and B are symmetric matrices of the same order, then show that AB
is symmetric if and only if A and B commute, that is AB = BA
...

Let
AB be symmetric, then (AB)′ = AB

MATRICES

99

he

But
(AB)′ = B′A′= BA (Why?)
Therefore
BA = AB
Conversely, if AB = BA, then we shall show that AB is symmetric
...


bl

is

⎡ 2 −1⎤
⎡5 2⎤
⎡ 2 5⎤
Example 28 Let A = ⎢
⎥ , B = ⎢ 7 4 ⎥ , C = ⎢ 3 8⎥
...


Solution Since A, B, C are all square matrices of order 2, and CD – AB is well
defined, D must be a square matrix of order 2
...
Then CD – AB = 0 gives
⎣c d ⎦

or

⎡ 2 5⎤ ⎡ a b ⎤ ⎡ 2 −1⎤ ⎡ 5 2 ⎤
⎢ 3 8⎥ ⎢ c d ⎥ − ⎢ 3 4⎥ ⎢7 4⎥ = O
⎣
⎦⎣
⎦ ⎣
⎦⎣
⎦

or

⎡ 2a + 5c 2b + 5d ⎤ ⎡ 3 0 ⎤ ⎡ 0 0 ⎤
⎢ 3a + 8c 3b + 8d ⎥ − ⎢ 43 22 ⎥ = ⎢
⎥
⎣
⎦ ⎣
⎦ ⎣0 0 ⎦

or

2b + 5d ⎤ ⎡ 0 0 ⎤
⎡ 2a + 5c − 3
⎢3a + 8c − 43 3b + 8d − 22 ⎥ = ⎢ 0 0 ⎥
⎦
⎣
⎦ ⎣

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Let

By equality of matrices, we get
2a + 5c – 3 = 0

...
(2)
2b + 5d = 0

...
(4)
Solving (1) and (2), we get a = –191, c = 77
...

Therefore

⎡ a b ⎤ ⎡ −191 −110 ⎤
D= ⎢
⎥=⎢
44 ⎥
⎣ c d ⎦ ⎣ 77
⎦

100

MATHEMATICS

Miscellaneous Exercise on Chapter 3

3n

1

3n

1

3n

1

3n

1

3n

1

3n

1

n 1

3

n 1

3

, n N
...
If A = ⎢1 1 1⎥ , prove that A n
⎢
⎥
⎢1 1 1⎥
⎣
⎦

he

⎡0 1 ⎤
1
...


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bl

⎡ 3 −4 ⎤
⎡1 + 2n −4n ⎤
n
3
...


4
...

5
...


z⎤
⎡0 2 y
⎢ x y − z ⎥ satisfy the equation
6
...


⎡1 2 0 ⎤ ⎡ 0 ⎤
7
...
If A = ⎢
⎣−1

1⎤
, show that A2 – 5A + 7I = 0
...
Find x, if [ x −5 −1] ⎢ 0 2 1 ⎥ ⎢ 4 ⎥ = O
⎢
⎥⎢ ⎥
⎢2 0 3⎥ ⎢1 ⎥
⎣
⎦⎣ ⎦

MATRICES

101

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...

Annual sales are indicated below:
Market
Products
I
10,000
2,000
18,000
II
6,000
20,000
8,000
(a) If unit sale prices of x, y and z are Rs 2
...
50 and Rs 1
...

(b) If the unit costs of the above three commodities are Rs 2
...
00 and 50
paise respectively
...


bl

⎡1 2 3⎤ ⎡ −7 −8 −9 ⎤
11
...
If A and B are square matrices of the same order such that AB = BA, then prove
by induction that ABn = BnA
...

Choose the correct answer in the following questions:
13
...
If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
2
15
...

A matrix having m rows and n columns is called a matrix of order m × n
...

[aij]1 × n is a row matrix
...


A = [aij]m × m is a diagonal matrix if aij = 0, when i ≠ j
...

A = [aij]n × n is an identity matrix, if aij = 1, when i = j, aij = 0, when i ≠ j
...


he

A = [aij] = [bij] = B if (i) A and B are of same order, (ii) aij = bij for all
possible values of i and j
...


k(A + B) = kA + kB, where A and B are of same order, k is constant
...


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102

n

If A = [aij]m × n and B = [bjk]n × p , then AB = C = [cik]m × p, where cik = ∑ aij b jk
j =1

(i) A(BC) = (AB)C, (ii) A(B + C) = AB + AC, (iii) (A + B)C = AC + BC
If A = [aij]m × n, then A′ or AT = [aji]n × m

(i) (A′)′ = A, (ii) (kA)′ = kA′, (iii) (A + B)′ = A′ + B′, (iv) (AB)′ = B′A′

A is a symmetric matrix if A′ = A
...

Any square matrix can be represented as the sum of a symmetric and a
skew symmetric matrix
...

Inverse of a square matrix, if it exists, is unique

---