algebra of matrices.pdf

7/27/2019 algebra of matrices.pdf

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7

ALGEBRA OF MATRICES

7.1 INTRODUCTION

Rajesh has two factories, one at Delhi and the other at

Bombay. Each factory produces two items of garments for ladies

and gents. The quantities produced by each factory is given in

the matrices below:

Factory at Delhi Factory at Bombay

ITEM I ITEM II ITEM I ITEM II

Ladies 600 550 Ladies 450 600

Gents 300 450 Gents 250 350

We are interested in finding out the total production of

items. So what do we do? Or, we may be interested in finding

the total cost of producing these items if cost per item is

given for each type.

In this lesson, we will be finding ways of answering such

questions by going into addition, multiplication and

algebra of matrices in general.

7.2 OBJECTIVES

After going through this lesson, you should be able to:

state the condition for equality of two matrices

multiply a matrix by a scalar

find the sum of two matrices of the same order


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find the difference of two matrices of the same order

state the condition for multiplication of two matrices

multiply two matrices, if possible

7.3 PREVIOUS KNOWLEDGE

Concept of a matrix

Order of a matrix

Four fundamental operations and their properties

Solution of problems using these properties

7.4 EQUALITY OF MATRICES

Consider the matrix

2 3

A

3 1

Its transpose will be

A 2 3

3 1

Observe that

1 . Order of matrix A = Order of matrix A', i.e., 2 2

2. Every element of A is same as the corresponding elemen

of A'.

In such a case, we say that matrix A = matrix A'.

Consider another example.

2 1 3

Let A = 1 6 4

3 4 5


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Algebra of Matrices :: 153


2 1 3

A' = 1 6 4

3 4 5

Again observe that

( i) Order of matrix A = Order of matrix A' i.e., 3 3

( ii ) Every element of A is same as the corresponding element

of A'.

we say that matrix A = matrix A'

Now consider the matrix

A = 2 3 6

1 4 7


2 1

A' = 3 4

6 7

Are the two matrices equal ?

We observe that

( i ) Order of A is 23 whereas order of A' is 32 hence orderof A order of A'

( i i ) Every element of A is not equal to the correspondingelement of A'.

Therefore we can say that

matrix A matrix A'

Thus, we can define the equality of two matrices as:


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Two matrices A and B are said to be equal if

( i) they are of the same order

and

(i i ) each element of A is equal to the

corresponding element of B.

For example:

Consider two matrices of the same order

x 2

5 and 5

When will the two matrices be equal?

Matrices x 2

=

5 5

if x = 2, since the two matrices are of the same order

Let us take some more examples

Example A:

Find the values of a and b if [a 3] = [4 b]

Solution :

If [a 3] = [4 b] then the corresponding elements of the matrices

will be equal.

a = 4 and b = 3

Example B:

Find the values of x and y if

x 2 1 2

=

3 y 3 5


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Solution:

If x 2 1 2

=

3 y 3 5

then their corresponding elements will be equal.

x = 1 , y = 5

Example C:

For what value of a, b, c and d will the two matrices

a 2 2b 1 2 4

6 3 d and 6 5c 2

be equal

Solution :

Matrix a 2 2b 1 2 4

=

6 3 d 6 5c 2

if the corresponding element of the two matrices will be equal.

i.e. i f a = 1

2b = 4

5c = 3

and d = 2

i f a = 1

b = 2

c =3

5

and d =2

Thus, for a = 1, b = 2, c = and d = 2, the matrix

{

{


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a 2 2b 1 2 4

=

6 3 d 6 5c 2

Example D:

For what value of a, b, c and d will the matrices

a b2d 5 1

3 2b = 3 6

a+c 7 4 7

Solution:

The matrix a b2d 5 1

3 2b = 3 6

a+c 7 4 7

If their corresponding elements are equal

i .e. i f a = 5

a+c = 4

b2d = 1

and 2b = 6

i f a = 5

b = 3

c = 1

and d = 1

Thus, for a = 5, b = 3, c = 1 and d = 1 the two given

matrices will be equal.

Checkpoint

Tick the right choice for Q.1 and Q.2

1. Two matrices can be compared for equality if

{{


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( i ) they are of the same order

(ii ) they are of a different order

2. Two matrices are said to be equal if

( i ) some of the corresponding elements match

(ii) all of the corresponding elements match

3. For what value of a, b, c and d, the matrix

2a b 3 2

=

4 6 d 3c

{ Ans:

1. (i)

2. ( i i )

3. a = b = 2 c =2 d = 4

7.5 SCALAR MULTIPLICATION

Let us consider the following situation :

The marks obtained by three students in English, Hindi

and Maths are as follows:

English Hindi Maths

Elizabeth 20 10 15

Usha 22 25 27

Shabnam 17 25 21

It is also given that these marks are out of 30 in each

case. In matrix form, the above information can be wirtten as

20 10 15

22 25 27

17 25 21

(It is understood that rows corres

pond to the names and columns

correspond to the subjects).


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If the maximum marks are doubled in each case, then

the marks, obtained by these girls will also be doubled. In

matrix form, the new marks can be given as:

2 20 2 10 2 15

2 22 2 25 2 27

2 17 2 25 2 21

40 20 30

= 44 50 54

34 50 42

So, we write that

20 10 15 2 20 2 10 215

2 22 25 27 = 2 22 2 25 227

17 25 21 2 17 2 25 221

40 20 30

= 44 50 54

34 50 42

Now consider the matrix

3 2

A = 2 0

1 6

Let us see what happens, when we multiply the matrix A

by 5

3 2 5 3 5 2

i .e ., 5 A = 5 2 0 = 5 C2 5 0

1 6 5 1 5 6


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15 10

= 10 0

5 30

Thus, we can say that

When a matrix is multiplied by a scalar, then

each of its elements is multiplied by the scalar.

Example E : If A = 2 3 4

1 0 1 , find 2A.

Solution: 2A = 2 (2) 2 (3) 2 (4)

2 (1) 2 (0) 2 (1)

= 4 6 8

2 0 2

Example F: If 1 4

A = 2 1 , find (1) A

3 2

Solution: (1) A = (1) (1) (1) (4)

(1) (2) (1) (1)

(1) (3) (1) (2)

1 4

= 2 1

3 2


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Note: Multiplying a matrix by (1) is same as

writing negative of the same matrix.

Thus, the matrix (1)A is same as A.

Example G: If A = 3 12 15

6 3 18 , find A

Solution: A = (3) (12) (15)

(-6) (3) (-18)

= 2 8 10

4 2 12

Checkpoint:

Tick the right choice

1. When a matrix is multiplied by a scalar then:

(i) elements of row one are multiplied by the scalar

(i i) elements of one column are multiplied by the scala

(ii i) elements of the diagonal are multiplied by the scala

(iv) each element of the matrix is multiplied by the scala

2. What will happen to A if A is multiplied by 3.

where A = 2 5

1 4

Ans : 1 (iv), 2 6 15

3 12{ }


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INTEXT QUESTIONS 7.1

1. ( a) For what values of x and y, the matrices

3 y=

x 2

(b) Find the values of a and b if

[ a 2 1 ] = [ 3 2 b ]

2. Find the values of x and y if

3 2x 3 4

=

y 1 5 1

3. For what values of a, b, c and d, the matrices

2a 5 6 10b

=

4 3d c 15

4. ( a) Find the values of w, x, y and z if

w 2x 2 6

=

x+y w3z 2 1

(b) For what values of a, b, c and d, the matrices

xy 3 1 z

=

0 y+x w+z 3

5. (a) If A = 0 1 2

3 1 4

find 5 A ?


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(b) If A = 1 1

3 0 , find 2 A ?

1 5

6. (a) If A = 1 3 1

0 2 0 , find (1) A ?

2 4 6

(b) If A = 1 0 2

4 2 3 , find (3) A

0 0 1

7. (a) If A = 2 4

6 2 , find A ?

(b) If A = 3 0 1

4 2 0 , find

3

4

A ?

1 0 5

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

2a b+2c 4 5

=

c ad 1 3

9. 2 5 4

If A = 8 3 2 , find2

5

A ?

0 1 6


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7.6 ADDITION OF MATRICES

Two students Mahesh and William compare their performances

in two tests in Mathematices, Physics and English. The

maximum marks in each test in each subject are 50. Themarks scored by them are as follows:

Test 1

Maths Physics English

Mahesh 50 38 33

William 47 40 36

Matrix I

Test 2


Mahesh 45 32 30

William 42 30 39

Matrix II

How can we find their total marks in each subject in the

two tests taken together?

The total marks obtained by :

Mahesh : In Maths = 50 + 45 = 95 out of 100

In Physics = 38 + 32 = 70 out of 100

In English = 33 + 30 = 63 out of 100

Wi ll iam : In Maths= 47 + 42 = 89 out of 100

In Physics = 40 + 30 = 70 out of 100

In English = 36 + 39 = 75 out of 100

In the matrix form, this information can be represented as:


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Mahesh 95 70 63

Williams 89 70 75

Matrix III

Note: ( i ) Matrices I and II are of the same order.

(ii) The elements of this new matrix III are the sum

of the corresponding elements of the Matrix I and

Matrix II.

(iii) The Matrix III is also of order 2 3

Now consider another situation

Let us consider the following matrices :

1 6 4 0 3

A = 3 8 , and B = 5 2 9

2 10 8 4 6

Now the question is can we obtain a matrix whos

elements will be the sum of the corresponding elements o

matrices A and B ?.

Obviously not. The reason is that the order of matrix A

is (32) and the order of matrix B is (33). Therefore, we ar

unable to find a column in the matrix A which corresponds to

the column 3 of matrix B.

Thus, we conclude that

1. The sum of two matrices is defined if they are of

some order.

2. The sum of two matrices is a matrix obtained by

adding the corresponding elements of the given

matrices.


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Example H:

If A = 1 3 and B = 5 2

4 2 1 0

then find A + B

Solution :

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

4+1 2+0 5 2

Example I:

If A = 0 1 1 and B = 3 0 4

2 3 0 1 2 1

find A + B ?

Solution:

A + B = 0+3 1+0 1+4 = 3 1 3

2+(1) 3+2 0+1 1 5 1

Checkpoint

Tick the correct answer

1 Two matrices can be added if

i ) their orders are same

i i) their orders are different

2 When we add two matrices, then

i ) We add the cor responding e lements of the fi rs t

row.

i i ) We add the corresponding elements of the f irst

column.


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{ }

i i i ) We add the corresponding elements o f the tw

given matrices.

3. If A = 2 1 3 and B = 0 1 4 then

5 0 6 2 1 5

find A + B

Ans: 1.(i), 2.(iii), 3. 2 0 1

7 1 11

7.6.1 Properties of Addition

Recall that in case of numbers, we have

( i ) x + y = y + x, i.e. addition is commutative.

( i i ) x + (y + z) = (x + y) + z, i.e., addition is associative

( i i i ) x + 0 = x i.e., addition identity exists

(iv) x + (x) = 0, i.e., addition inverse exists

Let us check these properties in case of matrices.

Let A = 1 2 and B = 0 2

1 3 1 3

thenA + B = 1 + 0 2 + (2) = 1 0

1 + 1 3 + 3 0 6

and B + A = 0 + 1 2 + 2 = 1 0

1 + (1) 3 + 3 0 6

Thus, matrix B + A is same as the matrix A + B.

So we write A + B = B + A

This is called commutative property of addition.


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In general,

For any two matrices A and B of the

same order A + B = B + A

i.e., matrix addition is commutative.

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

2 1 0 2 2 3

then A + (B + C) = 0 3 + 1+1 4+0

2 1 0+2 2+3

= 0 3 2 4

+

2 1 2 5

= 0+2 3+(4)

2+2 1+5

= 2 1

0 6

and (A + B) + C= 0+1 3+(4) 1 0

+

2+0 1+2 2 3

= 1 1 1 0

+

2 3 2 3

= 1+1 1+0

2+2 3+3

= 2 1

0 6

Thus, A + (B + C) = (A + B) + C


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This is called the associative property of addition.

In general,

For any three matrices A,B and C of the

same order, A + (B + C) = (A + B) + C

i.e., matrix addition is associative.

Checkpoint :

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

0 1 2 1 3 2

i ) Find A + B and (A + B) + C.

i i ) Is A + (B + C) = (A + B) + C?

iii ) Is (A + B) + C = (C + A) + B?

iv) Is A + (B + C) = (A + B) + C = (C + A) + B?

{Ans: Ans: ( ii) Yes, (iii) Yes, (iv) Yes.}

Note:

The sum of two matrices, added to the third matrix will give

the same result.

Recall that we have talked about zero matrix. A zero matrix

is that matrix all of whose elements are zero. It can be of any

order.

Let A = 2 2 and O = 0 0

4 5 0 0

then A + O = 2 2 + 0 0

4 5 0 0


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= 2+0 2+0

4+0 5+0

= 2 2

4 5

= A

Thus, we find that A + O = A, where O is a zero matrix.

The matrix O, which is a zero matrix, is called the

additive identity.

In general,

Additive identity is a zero matrix,

which when added to a given matrix,

gives the same given matrix, i.e. A +

O = A.

Check point:

Additive identity of a matrix A is

( i) Any matrix of the same order

(i i ) Zero matrix

(ii i ) Zero matrix of the same order.

{ Ans: (iii) }

7.7 SUBTRACTION OF MATRICES

Consider the matrices

A = 2 1 and B = 1 5

3 4 2 3

then A + (1) B = 2 1 +(1) 1 5

3 4 2 3


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= 2 1 + 1 5

3 4 2 3

= 2 + (1) 1 + (5)

3 + (2) 4 + (3)

= 2 1 1 5

3 2 4 3

= A B

Thus, A + (1)B = A B

Therefore

Given two matrices A and B of the same

order, their difference A B is defined as

the addition of the matrix A with the

negative of the matrix B,i.e., AB = A+(1)B.

Now, Let A = 2 3

1 4

It is possible to find another matrix B of the same ordersuch that

A + B = O

where O is a zero matrix of the order same as that of A or

B?

Obviously if A + B = O, (then)

B = O A

= O + (1) (A)

= 0 0 + 2 3

0 0 1 4

= 0+(2) 0+(3)

0+(1) 0+(4)


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= 2 3

1 4

= (1) 2 3

1 4

= (1) A = A

Thus B = (1) A = A

B is called the additive inverse of the matrix A.

In general,

Given a matrix A, there exists anothermatrix B = (1) A such that A + B =

O, then such a matrix B is called the

additive inverse of the matrix A.

Clearly, A is also the additive inverse of B.

So, B is written as A and A is written as B.

Note:

The elements of A are the negative of the corresponding

element of A.

Checkpoint:

Tick ( ) the right choice.

The additive inverse of a matrix is

( i) the same matrix

( i i ) the zero matrix

(iii) negative of the given matrix

{ Ans: (iii) }

Example J:

If A = 1 0 and B = 3 2

2 1 1 4


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then find A B

Solution:

Since B = 3 2

1 4

therefore, A B = A + (B)

= 1 0 3 2

+

2 1 1 4

= 1 3 0 2

2 1 1 4

= 2 2

1 5

Note : AB can also be obtained by subtracting the elements

of B from corresponding elements of A.

Example K: If A = 2 4 0 3

5 7 and B = 5 2

1 4 3 1

then find

(i) A B (ii) B A

Solution :

(i) A B = 2 4 0 3

5 7 5 2

1 4 3 1

= 2 0 4 3 2 1

5 5 7 2 = 0 9

1 3 4 1 4 3


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{ }

( i i ) Similarly, B A = 2 1

0 9

4 3

Note:

From (i) and (ii), we observe that matrix subtraction is not

commutative.

Checkpoint:

Tick ( ) the right choice

1. What do you understand by subtraction of A from B.

( i ) A + (B) ( i i ) B + (1) A ( i ii ) (1) (A + B)

2. If A = 0 5 1 and B = 1 2 5

2 6 3 3 4 1

then find B A ?

Ans: 1. (ii) 2. 1 7 4

1 2 4


1. [a] If A = 3 1 and B = 0 1 then

5 2 3 2

find A + B

(b) If A = 2 1 and B = 3 2 then

0 5 4 5

find A + 2B

2. (a) If A = 2 5 3 and B = 1 2 3

1 4 0 4 1 5


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then find A + B

(b) If A= 2 3 1 and B = 0 1 2

0 5 2 3 1 2 then

find 3A+B.

3. If A = 1 2 and B = 1 4

4 1 2 6 then

f ind (a) AB (b) A2B

4. If A= 1 2 1 1 0 2

2 3 4 and B = 4 2 3

3 4 5 0 0 1

f ind (a) AB (b) A+B

5. If A = 2 1 3 3 5 2

1 2 1 , B = 3 2 1

0 5 2 1 0 0

and C = 3 0 0

0 4 0

0 0 5 , find

(a) 3A+ 3B+ 3C (b) 3(A+ B + C) (c) 3A + 5A (d) 8A

Note:

From the results of (a) and (b) above, w

observe that 3 (A + B + C) = 3A + 3B + 3C and from (c) and

(d), 3A + 5A = 8A

In general, it is true that

k(A+B+C) = kA+kB+kC

and (k+l)A = kA+lA


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7.8 MATRIX MULTIPLICATION

Saleem and Girdhar are two friends. Saleem wants to buy 3

cows and 2 buffaloes while Girdhar wants to buy 5 cows and

3 buffaloes.

They go to a dairy which quotes the following prices :

1 cow Rs.2000

1 buffalo Rs.15,000

How much money will each spend?

Clearly, the money needed by Saleem and Girdhar will be:

Saleem

3 cows Rs.(3 x 2000) = Rs.6000

2 buffaloes Rs.(2 x 15,000)= Rs.30,000

Total Rs.6000 + Rs.30,000= Rs.36,000.

and Girdhar

5 cows Rs.5 x 2000 = Rs.10,000

3 buffaloes Rs.(3 x 15,000)= Rs.45,000

Total Rs.10,000 + Rs.45,000=Rs.55,000

In matrix form, the above information can be represented

as follows:

Requirements Prices Money Needed

C ows Buffaloes (in Rs.) (in RS.)

3 2 2000 3 2000+215,000 36,000

=

5 3 15,000 5 2000+315,000 55,000

Another dairy in the same locality quotes the following

prices:

1 cow Rs.2500


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1 buffalo Rs.17,000

The money needed by Saleem and Girdhar to buy th

required number of cows and buffaloes from this dairy will be

Saleem

3 cows Rs.3 2500 = Rs.7500

2 buffaloes Rs.2 17000 = Rs.34000

Total Rs.7500 + Rs.34,000 = Rs.41,500

Girdhar

5 cows Rs.5 2500 = Rs.12,500

3 buffaloes Rs.3 17000 = Rs.51,000

Total Rs.12,500 + Rs.51,000 = Rs.63,500

In matrix form, the above information can be written a

follows:

Requirements Prices Money Needed

Cows Buffaloes (in Rs.) (in RS.)

3 2 2,500 3 2500 + 2 17,000 41,500

=

5 3 17,000 5 2500 + 3 17,000 63,000

To have a comparative study, the two informations can

be combined in the following way :

Requirement Prices Money Needed at the two Dairies

Cows Buffaloes (in Rs.) (in Rs.)

3 2 2000 2500 3 2000 + 2 15000 3 2500 + 2 17000

5 3 15000 17000 5 2000 + 3 15000 5 2500 + 3 17000

36,000 41,500

==55,000 63,500


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Consider another example:

Susie and Tina are two friends. Susie wants to buy 3 kg

potatoes and 2 kg onions while Tina wants to buy 4 kg potatoes

and 3 kg onions. They go to shop where the following priceswere quoted.

Potato Rs.10/kg

Onion Rs. 7/kg

They go to another shop, where the following prices were

quoted:

Potato Rs.9/kg

Onion Rs.6/kg

To have a comparative study, let us write these informa-

tions in matrix form and find out how much money will be

needed by them to buy these groceries from these two shop.

Requirement Prices Money Needed at the two shops.

Potato Onion (in Rs.) (in Rs.)

3 2 10 9 (3 10) + (2 7) (3 9) + (2 6) 44 39 =

4 3 7 6 (4 10) + (3 7) (4 9) + (3 6) 61 54

The above example il lustrates mult iplication of matrices.

In general;

If A = a1

b1

and B = B1

a2

b2

B2

then A B = a1

1 + b

1

2+ a

1

1+ b

1b

2

a2

1 + b

2

2a

2

1+ b

2

2

Pictorially, this can be shown as :


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(R1

C1) R

1 C

2)

a1

1 + b

1

2a

1

1+b

1

2

a1

b1

1

a1

b1

1

2

2

(R2

C1) (R

2C

2)

a2

1+ b

2

2a

2

1+b

2

2

1

b1

a2

b2

2

a2 b

2 b

2

We observe that:

1. Matrix A can be multiplied by matrix B because th

order of matrix A is (2 2) and the order of the matrixB is (2 2) i.e., the number of columns in matrix A

is equal to the numbers of rows in matrix B.

2. The number of elements in the resultant matrix will b

the product of the number of rows of matrix A and th

number of columns of matrix B, which in this case i

(2 2) = 4.

3. The elements of the resultant matrix can be obtained bmultiplying the rows of A with the columns of B

elementwise, and then taking their sum.

Example L :

If A = [1 1 2] and B = 2

0

2

find AB and BA. Is AB = BA?

Solution:

Order of matrix A = 1 3

Order of matrix B = 3 1


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number of columns of A = number of rows of B.

AB exists

Now AB = [1 1 2] 2

0

2

= [(1 (2)] + (1 0) + (2 2)]

= [2 + 0 + 4]

= [2]

Thus, AB = [2] is of order (1 1) and the number of

elements in AB = 1 1 = 1

Again, number of columns of B = no. of rows of A

BA exists.

Now BA = 2 [1 1 2]

0

2

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

(0 1) (0 1) (0 2)

(2 1) (2 (1)) (2 2)

= 2 2 4

0 0 0

2 2 4

Thus, BA is of order 3 3 and the number of elementsin BA= 3 3 = 9.

Also, we find that

AB BA

In fact, AB and BA are of different orders.


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Note: Matrix multiplication is not commutative.

Check-point:

Product AB of two matrices A and B is defined if the

number of __________ of matrix A is equal to the number o

___________ of matrix B.

{ Ans: Columns, rows. }

Example M

If A = 1 2 and B = 3

3 1 1 , find AB

Solution

Order of matrix A = 2 2


Q the number of columns of A = number of rows of B

AB exists.

Now AB = 1 2 3 = (1 3) + (2 1)

3 1 1 (3 3) + (1 1)

= 5

8 2 1

Thus AB = 5

8 is of order 2 1 and the number of elements in this matrix AB = 2 1 = 2.

Can we find BA for the above two matrices? The answeis No because the number of columns in B is one which i

not equal to the number of rows in A, which is 2.

Thus, in this example, we find that AB exists but BA doe

not exist.

Remark : If AB exist, then BA need not exist.


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Checkpoint :

Fill in the blanks:

The number of elements in the product of a matrix A of

order (3 2) and matrix B of order (3 3) will be ___________.

{ Ans: 3 x 3 = 9 }

Example N : If A = 2 0 1

0 1 and B = 2 , find

3

AB and BA

Solution : Order of matrix A is (2 2)

and order of matrix B is (3 1)

Now, since the number of columns of matrix the number ofrows of matrix B,

AB does not exist.

Also, since the number of columns of matrix B number ofrows of matrix A,

BA does not exist.

Thus, matrix multiplication can be defined as :

Given two matrices A and B of order m n andn p respectively their product is a matrix of orderm p and the number of elements in this newmatrix will be equal to m p. Also, the elements of

this new matrix can be obtained by multiplying the

rows of A with the columns of B, elementwise, and

then taking their sum.

Example O If A = 1 2 and B = 2 1

1 0 2 2 , then

find AB and BA. Is AB = BA?


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Solution : Order of matrix A = 2 2


Both AB and BA will exist

Now AB = 1 2 2 1 = 2 + 4 1 + 4

1 0 2 2 2 + 0 1 + 0

= 6 5

2 1 2 2

and BA = 2 1 1 2 = 21 4+0

2 2 1 0 22 4+0

= 1 4

0 4 2 2

Thus, we find that AB and BA are of the same order (2 2)but still AB BA.

Thus, we conclude that matrix multiplication is not commu-

tative.

Example P : If A = 2 0 and B = 4 0

0 3 0 1 ,

find AB and BA ? Is AB = BA ?

Solution Order of matrix A = 2 x 2

Order of matrix B = 2 x 2

Both AB and BA exist

Now AB = 2 0 4 0 = 8 + 0 0 + 0

0 3 0 1 0 + 0 0 3

= 8 0

0 3 2 2


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and BA = 4 0 2 0 8 + 0 0 + 0

=

0 1 0 3 0 + 0 0 3

8 0

0 3 2 2

Here, we find that both AB and BA are of the same order

i.e., (2 2) and Also AB = BA.

Thus, we conclude that even though matrix multiplication

is not commutative, but sometimes, it may so happen that

AB = BA.

Hence, if two metrices A and B are multiplied, then the

following five cases are possible;

( i ) Both AB and BA exist but are of different orders.

(ii) Only one of the products AB or BA exists and the other

does not.

( ii i) Neither AB nor BA exists.

(iv) Both AB and BA exist and are of same order, but

AB BA

(v) AB = BA

Checkpoint

Tick () the right choice:

If matrix A is multiplied with matrix B, then

(i ) AB = BA

( i i ) AB BA

(iii) AB may or may not be equal to BA.

{ Ans: : (iii) }

Example Q: If A = 3 0 and I = 1 0

0 3 0 1 , verify that


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184 :: Mathematics

= A22A3I = 0

Solution : Now, A2 = 3 0 3 0 = 9 + 0 0 + 0

0 3 0 3 0 + 0 0 + 9

= 9 0

0 9

2A = 2 3 0 = 6 0

0 3 0 6

3I = 3 1 0 = 3 0

0 1 0 3

A22A3I = 9 0 6 0 3 0

0 9 0 6 0 3

= 9 0 6 0 + 3 0

0 9 0 6 0 3

= 9 0 6+3 0+0

0 9 0+0 6+3

= 9 0 9 0

0 9 0 9

= 99 00

00 99

= 0 0

0 0

= 0


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Hence verified

Checkpoint:

If A = 2 0 and I = 1 0

0 2 0 1 , verify

that A2 A 2I = 0

Example R: Solve the matrix equation

2 3 x = 1

1 1 y 3

Solution 2 3 x = 1

1 1 y 3

2x 3y = 1

x + y 3

2x3y = 1 and x+y = 3

Solving these two equations we get x = 2, y = 1

Check-point:

Fill in the blanks:

1. Given 1 3 x = 5

2 0 y 2

To find the value of x, we need to

( i ) multiply the 1st row with ___________

( i i ) multiply the 2nd

row with __________

(i ii) Equate the two to ____________ respectively.

{ Ans: (i) x & y, (ii) x & y (iii) 5 & 2 }

Example S : If A = 1 1 and B = 1 1

1 1 1 1 , find AB.


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186 :: Mathematics

Solution : AB = 1 1 1 1 = 1 + 1 1 1

1 1 1 1 1 + 1 1 1

= 0 0

0 0

Thus, we find that

Matrix A 0 and matrix B 0, stillthe product AB = 0

Hence, we conclude that the product of two nonzero matrix

can be a zero matrix, whereas in numbers, the product of two

nonzero numbers is always nonzero.

Example T If A = 1 1

1 1 , find A2

Solution A2 = 1 1 1 1

1 1 1 1

= 1 1 1 1

1 + 1 1 + 1

= 0 0

0 0

Thus, we find that

Matrix A 0, still A2 = 0

Hence we conclude that the square of a nonzero matrix

may or may not be equal to zero whereas in numbers, the

square of a nonzero number is always nonzero.

Example U If A = 1 2 , B = 4 0

3 5 1 2

and C = 1 0

0 3


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f ind ( i) (AB)C(ii) A(BC)

Solution (i) AB = 1 2 4 0

3 5 1 2

= 4+2 04

125 0+10

= 6 4

7 10

(AB)C = 6 4 1 0

7 10 0 3

= 6+0 012

7+0 0+30

= 6 12

7 30

(ii) BC = 4 0 1 0

1 2 0 3

= 4+0 0+0

1+0 0+6

= 4 0

1 6

A(BC) = 1 2 4 0

3 5 1 6

= 42 012

12+5 0+30


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188 :: Mathematics

= 6 12

7 30

Thus from (i) and (ii), we find that

(AB) C = A(BC)

i.e. matrix multiplication is associative.

Checkpoint :

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

0 3 2 4 2 0

then

(i) (AB)C = A(BC)

(ii) A(BC) = (CA) B

(iii) (AB)C = (BA) C

Tick the right choice

{ Ans : (i) }

Let A = 2 1 Let us final a matrix

3 4

I = a b

c d such that A I = A

i.e., 2 1 a b=

2 1

3 4 c d 3 4

i.e., 2a+ c 2b + d

=

2 1

3a4c 3b 4d 3 4

i.e., 2a+c = 2 2b+d = 1

and

3a 4c = 3 3b4d = 4


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Solving these equations, we get

a = 1, b = 0, c = 0, d = 1

I = a b 1 0=

c d 0 1

Thus, we have

2 1 1 0 2 1

=

3 4 0 1 3 4

Also, we can verify that

1 0 2 1=

2 1

0 1 3 4 3 4


1 (a) of A = [2 3 0 ] and B = 0

2

1 , find

AB and BA. Is AB = BA ?

(b) if A = 0 2 2 and B = 2 3

1 3 0 1 1

0 2 ,

find AB and BA. Is AB = BA ?

2. (a) If A = a and B = [x y z],

b , find AB ?

Can you find BA?


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190 :: Mathematics

(b) If A = 1 and B = 2 0

0 1 3 , find BA.

Does AB exist ?

3. If A = 2 3 and B = 0

0 1 1

2

(a) Does AB exist? Why?

(b) Does BA exist? Why?

4. (a) If A = 2 1 and B = 1 0

0 3 2 5

find AB and BA. Is AB = BA ?

(b) If A = 1 2 0 and B = 2 3 1

3 5 4 1 0 3

5 3 1 1 2 3

find AB = BA. Is AB = BA?

5. If A = 2 0 and B = 5 0

0 1 0 1

then find AB and BA. Is AB = BA ?

6. If A = 2 5

1 3

verify that A 5A + I = 0

7. (a) Find the values of x and y if

1 1 x = 2

4 5 y 7


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(b) Find the values of a and b if

2 3 a = 4

1 1 b 3

8. If A = 2 0 and B = 0 0

1 0 3 4 ,

verify that AB = 0

9. If A = 3 3

3 3 find A

10.If A = 1 3 B = 2 2 and C = 4 3

2 1 , 1 1 2 3 ,

Find (a) A(BC), (b) (AB)C, (c) Is A(BC)= (AB)C ?

11.If A = 1 3 B = 2 2 and C = 4 3

2 1 , 1 1 2 3 ,

Find (a) (A+B)C, (b) AC+BC, (c) AB, (d) (AB)(A+B)

WHT YOU HAVE LEARNT

Two matrices are said to be equal if they are of the same

order and the corresponding elements are equal.

Scalar multiple of a matrix is obtained by multiplying

each element of the matrix by the scalar.

The sum of two matrices (of same order) is a matrix

obtained by adding the corresponding elements of the

given matrices.

Difference of two matrices A and B is nothing but the

sum of matrix A and the negative of matrix B.


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192 :: Mathematics

Product of two matrices A of order m n and B of order

n p is a matrix of order m p whose elements can be

obtained by multiplying the rows of A with the columns o

B elementwise and then taking their sum.

TERMINAL QUESTIONS

1. Find the value of x and y if

3 2 3 2

xy 5 = 1 x+y

0 1 0 1

2. For what values of x and y, will the matrices

1 2 = x 2

3 x2y2 3 x2+y2

3. What wil l be the values of x, y, z and w if

xy zw = 1 2

x+3z 2xy 5 3

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

0 2 4 1 2 3 ,

f ind (a) A2B , (b) (A+B)3C, (c) C+2A.

5. If S = cos sin and T = cos 2 sin 2

sin cos sin 2 cos 2

find ST and TS when

(a) = 30, =15 (b) = 90, = 45

6 If A= 3 1 and I is the identity matrix of

2 4 order 2, verify that

AI = A = IA


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7 Find the values of x and y if

1 3 x = 3

2 5 y 4

8. Two friends Lisa and Promila go to a shop to buy a few

articles. Lisa purchases 6 pens and 3 notebooks whereas

Promila purchases 2 pens and 5 notebooks. If the cost of 1 pen

is Rs.20/ and the cost of 1 notebook is Rs.7/ find the

matrix AB, where, the matrix A represents the purchases

made by these friends using one row for one friend and the

matrix B represents the prices.

ANSWERS TO INTEXT QUESTIONS

7.1

1. (a) x = 2 , y = 3 (b) a = 3, b = 1

2. x = 2, y = 5 3. a = 3, b = , c = 4, d = 5

4. (a) w = 2, x = 3, y = 5, z = 1

(b) w = 3, x = 2, y = 1, z = 3

5. (a) 0 5 10 (b) 2 2

15 5 20 6 0

2 10

6. (a) 1 3 1 (b) 3 0 6

0 2 0 12 6 9

2 4 6 0 0 3

7. (a) 1 2 (b) 0

3 1 -3 0

0


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194 :: Mathematics

8. a = 2, b = 3, c = 1 d = 5

9. -2

7.2

1 (a) 3 2 (b) 8 5

8 4 8 15

2. (a) 3 7 0 (b) 6 8 1

3 5 5 3 16 4

3. (a) 2 2 (b) 3 6

2 7 0 13

4. (a) 2 2 3 (b) 2 2 3

2 5 1 2 5 1

3 4 4 3 4 4

5. (a) 24 12 15 (b) 24 12 15

12 12 6 12 12 6

3 15 21 3 15 21

(c) 16 8 24 (d) 16 8 24

8 16 8 8 16 8

0 40 16 0 40 16


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7.3

1. (a) AB =(6), BA = 0 0 0

4 6 0

2 3 0

AB BA

(b) AB = 2 2 , BA = 3 13 4

1 6 1 1 2

2 6 0

AB BA

2. (a) AB = ax ay az

bx by bz

BA does not exist because number of columns of Bnumber of rows of A.

(b) BA = 2

1 , AB does not exist.

3. (a) AB does not exist because number of columns of A

number of rows of B.

(b) BA does not exist because number of columns of B

number of rows of A.

4. (a) AB = 0 5 , BA = 2 1 ,

6 15 4 17

AB BA

(b) AB = 4 3 7 BA = 16 8 11

3 17 24 16 11 3

14 13 17 , 10 21 11


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196 :: Mathematics

AB BA

5. AB = 10 0 BA = 10 0

0 1 , 0 1

AB = BA

7. (a) x = 3, y = 1 (b) a = 1, b = 2

9. 0 0

0 0

10. (a) 2 0 (b) 2 0 (c) A(BC)=(AB)C

6 0 6 0

11. (a) 2 9 (b) 2 6 (c) 5 4

4 3 4 3 5 8

(d) 2 5

11 15

ANSWERS TO TERMINAL QUESTIONS

1. x = 3, y = 2

2. x = 1, y = 0

3. w = 1, x =2, y =1, z =1

4. (a) 3 7 (b) 3 11 (c) 2 5

8 0 10 6 2 7


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5. (a) ST = TS = 1 0

0 1

(b) ST = 1 0 TS = 1 0

0 1 0 1

7. x = 27, y = 10

8. AB = 6 3 20 = 141

2 5 7 75

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