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1 | 1118-1121 | The
quantities produced by each factory are represented as matrices given below:
Suppose Fatima wants to know the total production of sport shoes in each price
category 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)
This can be represented in the matrix form as
80
90
60
50
75
70
65
55
90
75
85
75
+
+
+
+
+
+
Rationalised 2023-24
44
MATHEMATICS
This new matrix is the sum of the above two matrices We observe that the sum of
two matrices is a matrix obtained by adding the corresponding elements of the given
matrices |
1 | 1119-1122 | 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)
This can be represented in the matrix form as
80
90
60
50
75
70
65
55
90
75
85
75
+
+
+
+
+
+
Rationalised 2023-24
44
MATHEMATICS
This new matrix is the sum of the above two matrices We observe that the sum of
two matrices is a matrix obtained by adding the corresponding elements of the given
matrices Furthermore, the two matrices have to be of the same order |
1 | 1120-1123 | Rationalised 2023-24
44
MATHEMATICS
This new matrix is the sum of the above two matrices We observe that the sum of
two matrices is a matrix obtained by adding the corresponding elements of the given
matrices Furthermore, the two matrices have to be of the same order Thus, if
11
12
13
21
22
23
A
a
a
a
a
a
a
=
is a 2 × 3 matrix and
11
12
13
21
22
23
B
b
b
b
b
b
b
=
is another
2×3 matrix |
1 | 1121-1124 | We observe that the sum of
two matrices is a matrix obtained by adding the corresponding elements of the given
matrices Furthermore, the two matrices have to be of the same order Thus, if
11
12
13
21
22
23
A
a
a
a
a
a
a
=
is a 2 × 3 matrix and
11
12
13
21
22
23
B
b
b
b
b
b
b
=
is another
2×3 matrix Then, we define
11
11
12
12
13
13
21
21
22
22
23
23
A + B
a
b
a
b
a
b
a
b
a
b
a
b
+
+
+
=
+
+
+
|
1 | 1122-1125 | Furthermore, the two matrices have to be of the same order Thus, if
11
12
13
21
22
23
A
a
a
a
a
a
a
=
is a 2 × 3 matrix and
11
12
13
21
22
23
B
b
b
b
b
b
b
=
is another
2×3 matrix Then, we define
11
11
12
12
13
13
21
21
22
22
23
23
A + B
a
b
a
b
a
b
a
b
a
b
a
b
+
+
+
=
+
+
+
In general, if A = [aij] and B = [bij] are two matrices of the same order, say m × n |
1 | 1123-1126 | Thus, if
11
12
13
21
22
23
A
a
a
a
a
a
a
=
is a 2 × 3 matrix and
11
12
13
21
22
23
B
b
b
b
b
b
b
=
is another
2×3 matrix Then, we define
11
11
12
12
13
13
21
21
22
22
23
23
A + B
a
b
a
b
a
b
a
b
a
b
a
b
+
+
+
=
+
+
+
In general, if A = [aij] and B = [bij] are two matrices of the same order, say m × n Then, the sum of the two matrices A and B is defined as a matrix C = [cij]m × n, where
cij = aij + bij, for all possible values of i and j |
1 | 1124-1127 | Then, we define
11
11
12
12
13
13
21
21
22
22
23
23
A + B
a
b
a
b
a
b
a
b
a
b
a
b
+
+
+
=
+
+
+
In general, if A = [aij] and B = [bij] are two matrices of the same order, say m × n Then, the sum of the two matrices A and B is defined as a matrix C = [cij]m × n, where
cij = aij + bij, for all possible values of i and j Example 6 Given
3
1
1
A
2
3
0
−
=
and
2
5
1
B
1
2
3
2
=
−
, find A + B
Since A, B are of the same order 2 × 3 |
1 | 1125-1128 | In general, if A = [aij] and B = [bij] are two matrices of the same order, say m × n Then, the sum of the two matrices A and B is defined as a matrix C = [cij]m × n, where
cij = aij + bij, for all possible values of i and j Example 6 Given
3
1
1
A
2
3
0
−
=
and
2
5
1
B
1
2
3
2
=
−
, find A + B
Since A, B are of the same order 2 × 3 Therefore, addition of A and B is defined
and is given by
2
3
1
5
1 1
2
3
1
5
0
A+B
1
1
2
2
3
3
0
0
6
2
2
+
+
−
+
+
=
=
−
+
+
ANote
1 |
1 | 1126-1129 | Then, the sum of the two matrices A and B is defined as a matrix C = [cij]m × n, where
cij = aij + bij, for all possible values of i and j Example 6 Given
3
1
1
A
2
3
0
−
=
and
2
5
1
B
1
2
3
2
=
−
, find A + B
Since A, B are of the same order 2 × 3 Therefore, addition of A and B is defined
and is given by
2
3
1
5
1 1
2
3
1
5
0
A+B
1
1
2
2
3
3
0
0
6
2
2
+
+
−
+
+
=
=
−
+
+
ANote
1 We emphasise that if A and B are not of the same order, then A + B is not
defined |
1 | 1127-1130 | Example 6 Given
3
1
1
A
2
3
0
−
=
and
2
5
1
B
1
2
3
2
=
−
, find A + B
Since A, B are of the same order 2 × 3 Therefore, addition of A and B is defined
and is given by
2
3
1
5
1 1
2
3
1
5
0
A+B
1
1
2
2
3
3
0
0
6
2
2
+
+
−
+
+
=
=
−
+
+
ANote
1 We emphasise that if A and B are not of the same order, then A + B is not
defined For example if
2
3
A
1
0
=
,
1
2
3
B
,
1
0
1
=
then A + B is not defined |
1 | 1128-1131 | Therefore, addition of A and B is defined
and is given by
2
3
1
5
1 1
2
3
1
5
0
A+B
1
1
2
2
3
3
0
0
6
2
2
+
+
−
+
+
=
=
−
+
+
ANote
1 We emphasise that if A and B are not of the same order, then A + B is not
defined For example if
2
3
A
1
0
=
,
1
2
3
B
,
1
0
1
=
then A + B is not defined 2 |
1 | 1129-1132 | We emphasise that if A and B are not of the same order, then A + B is not
defined For example if
2
3
A
1
0
=
,
1
2
3
B
,
1
0
1
=
then A + B is not defined 2 We may observe that addition of matrices is an example of binary operation
on the set of matrices of the same order |
1 | 1130-1133 | For example if
2
3
A
1
0
=
,
1
2
3
B
,
1
0
1
=
then A + B is not defined 2 We may observe that addition of matrices is an example of binary operation
on the set of matrices of the same order 3 |
1 | 1131-1134 | 2 We may observe that addition of matrices is an example of binary operation
on the set of matrices of the same order 3 4 |
1 | 1132-1135 | We may observe that addition of matrices is an example of binary operation
on the set of matrices of the same order 3 4 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 | 1133-1136 | 3 4 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 4 |
1 | 1134-1137 | 4 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 4 1) |
1 | 1135-1138 | 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 4 1) Rationalised 2023-24
MATRICES 45
Previously quantities (in standard units) produced by factory A were
Revised quantities produced by factory A are as given below:
Boys
Girls
2 80
2
60
1
2 2
75
2
65
3 2 90
2 85
×
×
×
×
×
×
This can be represented in the matrix form as
160
120
150
130
180
170
|
1 | 1136-1139 | 4 1) Rationalised 2023-24
MATRICES 45
Previously quantities (in standard units) produced by factory A were
Revised quantities produced by factory A are as given below:
Boys
Girls
2 80
2
60
1
2 2
75
2
65
3 2 90
2 85
×
×
×
×
×
×
This can be represented in the matrix form as
160
120
150
130
180
170
We observe that
the new matrix is obtained by multiplying each element of the previous matrix by 2 |
1 | 1137-1140 | 1) Rationalised 2023-24
MATRICES 45
Previously quantities (in standard units) produced by factory A were
Revised quantities produced by factory A are as given below:
Boys
Girls
2 80
2
60
1
2 2
75
2
65
3 2 90
2 85
×
×
×
×
×
×
This can be represented in the matrix form as
160
120
150
130
180
170
We observe that
the new matrix is obtained by multiplying each element of the previous matrix by 2 In general, we may define multiplication of a matrix by a scalar as follows: if
A = [aij] m × n is a matrix and k is a scalar, then kA is another matrix which is obtained
by multiplying each element of A by the scalar k |
1 | 1138-1141 | Rationalised 2023-24
MATRICES 45
Previously quantities (in standard units) produced by factory A were
Revised quantities produced by factory A are as given below:
Boys
Girls
2 80
2
60
1
2 2
75
2
65
3 2 90
2 85
×
×
×
×
×
×
This can be represented in the matrix form as
160
120
150
130
180
170
We observe that
the new matrix is obtained by multiplying each element of the previous matrix by 2 In general, we may define multiplication of a matrix by a scalar as follows: if
A = [aij] m × n is a matrix and k is a scalar, then kA is another matrix which is obtained
by multiplying each element of A by the scalar k 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 |
1 | 1139-1142 | We observe that
the new matrix is obtained by multiplying each element of the previous matrix by 2 In general, we may define multiplication of a matrix by a scalar as follows: if
A = [aij] m × n is a matrix and k is a scalar, then kA is another matrix which is obtained
by multiplying each element of A by the scalar k 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 For example, if
A =
3
1
1 |
1 | 1140-1143 | In general, we may define multiplication of a matrix by a scalar as follows: if
A = [aij] m × n is a matrix and k is a scalar, then kA is another matrix which is obtained
by multiplying each element of A by the scalar k 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 For example, if
A =
3
1
1 5
5
7
3
2
0
5
−
, then
3A =
3
1
1 |
1 | 1141-1144 | 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 For example, if
A =
3
1
1 5
5
7
3
2
0
5
−
, then
3A =
3
1
1 5
9
3
4 |
1 | 1142-1145 | For example, if
A =
3
1
1 5
5
7
3
2
0
5
−
, then
3A =
3
1
1 5
9
3
4 5
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 |
1 | 1143-1146 | 5
5
7
3
2
0
5
−
, then
3A =
3
1
1 5
9
3
4 5
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 We define
–A = (– 1) A |
1 | 1144-1147 | 5
9
3
4 5
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 We define
–A = (– 1) A Rationalised 2023-24
46
MATHEMATICS
For example, let
A =
3
1
5
x
−
, then – A is given by
– A = (– 1)
3
1
3
1
A
( 1)
5
5
x
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 |
1 | 1145-1148 | 5
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 We define
–A = (– 1) A Rationalised 2023-24
46
MATHEMATICS
For example, let
A =
3
1
5
x
−
, then – A is given by
– A = (– 1)
3
1
3
1
A
( 1)
5
5
x
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 In other words, D = A – B = A + (–1) B, that is sum of the matrix
A and the matrix – B |
1 | 1146-1149 | We define
–A = (– 1) A Rationalised 2023-24
46
MATHEMATICS
For example, let
A =
3
1
5
x
−
, then – A is given by
– A = (– 1)
3
1
3
1
A
( 1)
5
5
x
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 In other words, D = A – B = A + (–1) B, that is sum of the matrix
A and the matrix – B Example 7 If
1
2
3
3
1
3
A
and B
2
3
1
1
0
2
−
=
=
−
, then find 2A – B |
1 | 1147-1150 | Rationalised 2023-24
46
MATHEMATICS
For example, let
A =
3
1
5
x
−
, then – A is given by
– A = (– 1)
3
1
3
1
A
( 1)
5
5
x
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 In other words, D = A – B = A + (–1) B, that is sum of the matrix
A and the matrix – B Example 7 If
1
2
3
3
1
3
A
and B
2
3
1
1
0
2
−
=
=
−
, then find 2A – B Solution We have
2A – B = 2 1
2
3
2
3 1
3
1
3
1
0
2
−
−
−
=
2
4
6
3 1
3
4
6
2
1
0
2
−
−
+
−
=
2
3
4
1
6
3
1
5
3
4
1
6
0
2
2
5
6
0
−
+
−
−
=
+
+
−
3 |
1 | 1148-1151 | In other words, D = A – B = A + (–1) B, that is sum of the matrix
A and the matrix – B Example 7 If
1
2
3
3
1
3
A
and B
2
3
1
1
0
2
−
=
=
−
, then find 2A – B Solution We have
2A – B = 2 1
2
3
2
3 1
3
1
3
1
0
2
−
−
−
=
2
4
6
3 1
3
4
6
2
1
0
2
−
−
+
−
=
2
3
4
1
6
3
1
5
3
4
1
6
0
2
2
5
6
0
−
+
−
−
=
+
+
−
3 4 |
1 | 1149-1152 | Example 7 If
1
2
3
3
1
3
A
and B
2
3
1
1
0
2
−
=
=
−
, then find 2A – B Solution We have
2A – B = 2 1
2
3
2
3 1
3
1
3
1
0
2
−
−
−
=
2
4
6
3 1
3
4
6
2
1
0
2
−
−
+
−
=
2
3
4
1
6
3
1
5
3
4
1
6
0
2
2
5
6
0
−
+
−
−
=
+
+
−
3 4 3 Properties of matrix addition
The addition of matrices satisfy the following properties:
(i)
Commutative Law If A = [aij], B = [bij] are matrices of the same order, say
m × n, then A + B = B + A |
1 | 1150-1153 | Solution We have
2A – B = 2 1
2
3
2
3 1
3
1
3
1
0
2
−
−
−
=
2
4
6
3 1
3
4
6
2
1
0
2
−
−
+
−
=
2
3
4
1
6
3
1
5
3
4
1
6
0
2
2
5
6
0
−
+
−
−
=
+
+
−
3 4 3 Properties of matrix addition
The addition of matrices satisfy the following properties:
(i)
Commutative Law If A = [aij], B = [bij] are matrices of the same order, say
m × n, then A + B = B + A 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) |
1 | 1151-1154 | 4 3 Properties of matrix addition
The addition of matrices satisfy the following properties:
(i)
Commutative Law If A = [aij], B = [bij] are matrices of the same order, say
m × n, then A + B = B + A 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) Now
(A + B) + C = ([aij] + [bij]) + [cij]
= [aij + bij] + [cij] = [(aij + bij) + cij]
= [aij + (bij + cij)]
(Why |
1 | 1152-1155 | 3 Properties of matrix addition
The addition of matrices satisfy the following properties:
(i)
Commutative Law If A = [aij], B = [bij] are matrices of the same order, say
m × n, then A + B = B + A 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) Now
(A + B) + C = ([aij] + [bij]) + [cij]
= [aij + bij] + [cij] = [(aij + bij) + cij]
= [aij + (bij + cij)]
(Why )
= [aij] + [(bij + cij)] = [aij] + ([bij] + [cij]) = A + (B + C)
Rationalised 2023-24
MATRICES 47
(iii)
Existence of additive identity Let A = [aij] be an m × n matrix and
O be an m × n zero matrix, then A + O = O + A = A |
1 | 1153-1156 | 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) Now
(A + B) + C = ([aij] + [bij]) + [cij]
= [aij + bij] + [cij] = [(aij + bij) + cij]
= [aij + (bij + cij)]
(Why )
= [aij] + [(bij + cij)] = [aij] + ([bij] + [cij]) = A + (B + C)
Rationalised 2023-24
MATRICES 47
(iii)
Existence of additive identity Let A = [aij] be an m × n matrix and
O be an m × n zero matrix, then A + O = O + A = A In other words, O is the
additive identity for matrix addition |
1 | 1154-1157 | Now
(A + B) + C = ([aij] + [bij]) + [cij]
= [aij + bij] + [cij] = [(aij + bij) + cij]
= [aij + (bij + cij)]
(Why )
= [aij] + [(bij + cij)] = [aij] + ([bij] + [cij]) = A + (B + C)
Rationalised 2023-24
MATRICES 47
(iii)
Existence of additive identity Let A = [aij] be an m × n matrix and
O be an m × n zero matrix, then A + O = O + A = A In other words, O is the
additive identity for matrix addition (iv)
The existence of additive inverse Let A = [aij]m × n be any matrix, then we
have another matrix as – A = [– aij]m × n such that A + (– A) = (– A) + A= O |
1 | 1155-1158 | )
= [aij] + [(bij + cij)] = [aij] + ([bij] + [cij]) = A + (B + C)
Rationalised 2023-24
MATRICES 47
(iii)
Existence of additive identity Let A = [aij] be an m × n matrix and
O be an m × n zero matrix, then A + O = O + A = A In other words, O is the
additive identity for matrix addition (iv)
The existence of additive inverse Let A = [aij]m × n be any matrix, then we
have another matrix as – A = [– aij]m × n such that A + (– A) = (– A) + A= O So
– A is the additive inverse of A or negative of A |
1 | 1156-1159 | In other words, O is the
additive identity for matrix addition (iv)
The existence of additive inverse Let A = [aij]m × n be any matrix, then we
have another matrix as – A = [– aij]m × n such that A + (– A) = (– A) + A= O So
– A is the additive inverse of A or negative of A 3 |
1 | 1157-1160 | (iv)
The existence of additive inverse Let A = [aij]m × n be any matrix, then we
have another matrix as – A = [– aij]m × n such that A + (– A) = (– A) + A= O So
– A is the additive inverse of A or negative of A 3 4 |
1 | 1158-1161 | So
– A is the additive inverse of A or negative of A 3 4 4 Properties of scalar multiplication of a matrix
If A = [aij] and B = [bij] be two matrices of the same order, say m × n, and k and l are
scalars, then
(i)
k(A +B) = k A + kB, (ii) (k + l)A = k A + l A
(ii)
k (A + B) = k ([aij] + [bij])
= k [aij + bij] = [k (aij + bij)] = [(k aij) + (k bij)]
= [k aij] + [k bij] = k [aij] + k [bij] = kA + kB
(iii)
( k + l) A = (k + l) [aij]
= [(k + l) aij] + [k aij] + [l aij] = k [aij] + l [aij] = k A + l A
Example 8 If
8
0
2
2
A
4
2 and B
4
2
3
6
5 1
−
=
−
=
−
, then find the matrix X, such that
2A + 3X = 5B |
1 | 1159-1162 | 3 4 4 Properties of scalar multiplication of a matrix
If A = [aij] and B = [bij] be two matrices of the same order, say m × n, and k and l are
scalars, then
(i)
k(A +B) = k A + kB, (ii) (k + l)A = k A + l A
(ii)
k (A + B) = k ([aij] + [bij])
= k [aij + bij] = [k (aij + bij)] = [(k aij) + (k bij)]
= [k aij] + [k bij] = k [aij] + k [bij] = kA + kB
(iii)
( k + l) A = (k + l) [aij]
= [(k + l) aij] + [k aij] + [l aij] = k [aij] + l [aij] = k A + l A
Example 8 If
8
0
2
2
A
4
2 and B
4
2
3
6
5 1
−
=
−
=
−
, then find the matrix X, such that
2A + 3X = 5B Solution We have 2A + 3X = 5B
or
2A + 3X – 2A = 5B – 2A
or
2A – 2A + 3X = 5B – 2A
(Matrix addition is commutative)
or
O + 3X = 5B – 2A
(– 2A is the additive inverse of 2A)
or
3X = 5B – 2A
(O is the additive identity)
or
X = 1
3 (5B – 2A)
or
2
2
8 0
1
X
5
4
2
2 4
2
3
5 1
3 6
−
=
−
−
−
=
10
10
16
0
1
20
10
8
4
3
25
5
6
12
−
−
+
−
−
−
−
Rationalised 2023-24
48
MATHEMATICS
=
10
16
10
0
1
20
8
10
4
3
25
6
5 12
−
−
+
−
+
−
−
−
=
6
10
1
12
14
3
31
7
−
−
−
−
=
10
2
143
4
3
31
7
3
3
−
−
−
−
Example 9 Find X and Y, if
5
2
X
Y
0
9
+
=
and
3
6
X
Y
0
1
−
=
−
|
1 | 1160-1163 | 4 4 Properties of scalar multiplication of a matrix
If A = [aij] and B = [bij] be two matrices of the same order, say m × n, and k and l are
scalars, then
(i)
k(A +B) = k A + kB, (ii) (k + l)A = k A + l A
(ii)
k (A + B) = k ([aij] + [bij])
= k [aij + bij] = [k (aij + bij)] = [(k aij) + (k bij)]
= [k aij] + [k bij] = k [aij] + k [bij] = kA + kB
(iii)
( k + l) A = (k + l) [aij]
= [(k + l) aij] + [k aij] + [l aij] = k [aij] + l [aij] = k A + l A
Example 8 If
8
0
2
2
A
4
2 and B
4
2
3
6
5 1
−
=
−
=
−
, then find the matrix X, such that
2A + 3X = 5B Solution We have 2A + 3X = 5B
or
2A + 3X – 2A = 5B – 2A
or
2A – 2A + 3X = 5B – 2A
(Matrix addition is commutative)
or
O + 3X = 5B – 2A
(– 2A is the additive inverse of 2A)
or
3X = 5B – 2A
(O is the additive identity)
or
X = 1
3 (5B – 2A)
or
2
2
8 0
1
X
5
4
2
2 4
2
3
5 1
3 6
−
=
−
−
−
=
10
10
16
0
1
20
10
8
4
3
25
5
6
12
−
−
+
−
−
−
−
Rationalised 2023-24
48
MATHEMATICS
=
10
16
10
0
1
20
8
10
4
3
25
6
5 12
−
−
+
−
+
−
−
−
=
6
10
1
12
14
3
31
7
−
−
−
−
=
10
2
143
4
3
31
7
3
3
−
−
−
−
Example 9 Find X and Y, if
5
2
X
Y
0
9
+
=
and
3
6
X
Y
0
1
−
=
−
Solution We have (
)
(
)
5
2
3
6
X
Y
X
Y
0
9
0
1
+
+
−
=
+
−
|
1 | 1161-1164 | 4 Properties of scalar multiplication of a matrix
If A = [aij] and B = [bij] be two matrices of the same order, say m × n, and k and l are
scalars, then
(i)
k(A +B) = k A + kB, (ii) (k + l)A = k A + l A
(ii)
k (A + B) = k ([aij] + [bij])
= k [aij + bij] = [k (aij + bij)] = [(k aij) + (k bij)]
= [k aij] + [k bij] = k [aij] + k [bij] = kA + kB
(iii)
( k + l) A = (k + l) [aij]
= [(k + l) aij] + [k aij] + [l aij] = k [aij] + l [aij] = k A + l A
Example 8 If
8
0
2
2
A
4
2 and B
4
2
3
6
5 1
−
=
−
=
−
, then find the matrix X, such that
2A + 3X = 5B Solution We have 2A + 3X = 5B
or
2A + 3X – 2A = 5B – 2A
or
2A – 2A + 3X = 5B – 2A
(Matrix addition is commutative)
or
O + 3X = 5B – 2A
(– 2A is the additive inverse of 2A)
or
3X = 5B – 2A
(O is the additive identity)
or
X = 1
3 (5B – 2A)
or
2
2
8 0
1
X
5
4
2
2 4
2
3
5 1
3 6
−
=
−
−
−
=
10
10
16
0
1
20
10
8
4
3
25
5
6
12
−
−
+
−
−
−
−
Rationalised 2023-24
48
MATHEMATICS
=
10
16
10
0
1
20
8
10
4
3
25
6
5 12
−
−
+
−
+
−
−
−
=
6
10
1
12
14
3
31
7
−
−
−
−
=
10
2
143
4
3
31
7
3
3
−
−
−
−
Example 9 Find X and Y, if
5
2
X
Y
0
9
+
=
and
3
6
X
Y
0
1
−
=
−
Solution We have (
)
(
)
5
2
3
6
X
Y
X
Y
0
9
0
1
+
+
−
=
+
−
or
(X + X) + (Y – Y) =
8
8
0
8
⇒
8
8
2X
0
8
=
or
X =
8
8
4
4
1
0
8
0
4
2
=
Also
(X + Y) – (X – Y) =
5
2
3
6
0
9
0
1
−
−
or
(X – X) + (Y + Y) =
5
3
2
6
0
9
1
−
−
+
⇒
2
4
2Y
0
10
−
=
or
Y =
2
4
1
2
1
0
10
0
5
2
−
−
=
Example 10 Find the values of x and y from the following equation:
5
3
4
2 7
3
1
2
x
y
−
+
−
=
7
6
15
14
Solution We have
5
3
4
2 7
3
1
2
x
y
−
+
−
=
7
6
15
14
⇒ 2
10
3
4
7
6
14
2
6
1
2
15
14
x
y
−
+
=
−
Rationalised 2023-24
MATRICES 49
or
2
3
10
4
14
1
2
6
2
x
y
+
−
+
−
+
=
7
6
15 14
⇒ 2
3
6
7
6
15
2
4
15
14
x
y
+
=
−
or
2x + 3 = 7
and
2y – 4 = 14
(Why |
1 | 1162-1165 | Solution We have 2A + 3X = 5B
or
2A + 3X – 2A = 5B – 2A
or
2A – 2A + 3X = 5B – 2A
(Matrix addition is commutative)
or
O + 3X = 5B – 2A
(– 2A is the additive inverse of 2A)
or
3X = 5B – 2A
(O is the additive identity)
or
X = 1
3 (5B – 2A)
or
2
2
8 0
1
X
5
4
2
2 4
2
3
5 1
3 6
−
=
−
−
−
=
10
10
16
0
1
20
10
8
4
3
25
5
6
12
−
−
+
−
−
−
−
Rationalised 2023-24
48
MATHEMATICS
=
10
16
10
0
1
20
8
10
4
3
25
6
5 12
−
−
+
−
+
−
−
−
=
6
10
1
12
14
3
31
7
−
−
−
−
=
10
2
143
4
3
31
7
3
3
−
−
−
−
Example 9 Find X and Y, if
5
2
X
Y
0
9
+
=
and
3
6
X
Y
0
1
−
=
−
Solution We have (
)
(
)
5
2
3
6
X
Y
X
Y
0
9
0
1
+
+
−
=
+
−
or
(X + X) + (Y – Y) =
8
8
0
8
⇒
8
8
2X
0
8
=
or
X =
8
8
4
4
1
0
8
0
4
2
=
Also
(X + Y) – (X – Y) =
5
2
3
6
0
9
0
1
−
−
or
(X – X) + (Y + Y) =
5
3
2
6
0
9
1
−
−
+
⇒
2
4
2Y
0
10
−
=
or
Y =
2
4
1
2
1
0
10
0
5
2
−
−
=
Example 10 Find the values of x and y from the following equation:
5
3
4
2 7
3
1
2
x
y
−
+
−
=
7
6
15
14
Solution We have
5
3
4
2 7
3
1
2
x
y
−
+
−
=
7
6
15
14
⇒ 2
10
3
4
7
6
14
2
6
1
2
15
14
x
y
−
+
=
−
Rationalised 2023-24
MATRICES 49
or
2
3
10
4
14
1
2
6
2
x
y
+
−
+
−
+
=
7
6
15 14
⇒ 2
3
6
7
6
15
2
4
15
14
x
y
+
=
−
or
2x + 3 = 7
and
2y – 4 = 14
(Why )
or
2x = 7 – 3
and
2y = 18
or
x = 4
2
and
y = 18
2
i |
1 | 1163-1166 | Solution We have (
)
(
)
5
2
3
6
X
Y
X
Y
0
9
0
1
+
+
−
=
+
−
or
(X + X) + (Y – Y) =
8
8
0
8
⇒
8
8
2X
0
8
=
or
X =
8
8
4
4
1
0
8
0
4
2
=
Also
(X + Y) – (X – Y) =
5
2
3
6
0
9
0
1
−
−
or
(X – X) + (Y + Y) =
5
3
2
6
0
9
1
−
−
+
⇒
2
4
2Y
0
10
−
=
or
Y =
2
4
1
2
1
0
10
0
5
2
−
−
=
Example 10 Find the values of x and y from the following equation:
5
3
4
2 7
3
1
2
x
y
−
+
−
=
7
6
15
14
Solution We have
5
3
4
2 7
3
1
2
x
y
−
+
−
=
7
6
15
14
⇒ 2
10
3
4
7
6
14
2
6
1
2
15
14
x
y
−
+
=
−
Rationalised 2023-24
MATRICES 49
or
2
3
10
4
14
1
2
6
2
x
y
+
−
+
−
+
=
7
6
15 14
⇒ 2
3
6
7
6
15
2
4
15
14
x
y
+
=
−
or
2x + 3 = 7
and
2y – 4 = 14
(Why )
or
2x = 7 – 3
and
2y = 18
or
x = 4
2
and
y = 18
2
i e |
1 | 1164-1167 | or
(X + X) + (Y – Y) =
8
8
0
8
⇒
8
8
2X
0
8
=
or
X =
8
8
4
4
1
0
8
0
4
2
=
Also
(X + Y) – (X – Y) =
5
2
3
6
0
9
0
1
−
−
or
(X – X) + (Y + Y) =
5
3
2
6
0
9
1
−
−
+
⇒
2
4
2Y
0
10
−
=
or
Y =
2
4
1
2
1
0
10
0
5
2
−
−
=
Example 10 Find the values of x and y from the following equation:
5
3
4
2 7
3
1
2
x
y
−
+
−
=
7
6
15
14
Solution We have
5
3
4
2 7
3
1
2
x
y
−
+
−
=
7
6
15
14
⇒ 2
10
3
4
7
6
14
2
6
1
2
15
14
x
y
−
+
=
−
Rationalised 2023-24
MATRICES 49
or
2
3
10
4
14
1
2
6
2
x
y
+
−
+
−
+
=
7
6
15 14
⇒ 2
3
6
7
6
15
2
4
15
14
x
y
+
=
−
or
2x + 3 = 7
and
2y – 4 = 14
(Why )
or
2x = 7 – 3
and
2y = 18
or
x = 4
2
and
y = 18
2
i e x = 2
and
y = 9 |
1 | 1165-1168 | )
or
2x = 7 – 3
and
2y = 18
or
x = 4
2
and
y = 18
2
i e x = 2
and
y = 9 Example 11 Two farmers Ramkishan and Gurcharan Singh cultivates only three
varieties of rice namely Basmati, Permal and Naura |
1 | 1166-1169 | e x = 2
and
y = 9 Example 11 Two farmers Ramkishan and Gurcharan Singh cultivates only three
varieties of rice namely Basmati, Permal and Naura 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 |
1 | 1167-1170 | x = 2
and
y = 9 Example 11 Two farmers Ramkishan and Gurcharan Singh cultivates only three
varieties of rice namely Basmati, Permal and Naura 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 (i)
Find the combined sales in September and October for each farmer in each
variety |
1 | 1168-1171 | Example 11 Two farmers Ramkishan and Gurcharan Singh cultivates only three
varieties of rice namely Basmati, Permal and Naura 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 (i)
Find the combined sales in September and October for each farmer in each
variety (ii)
Find the decrease in sales from September to October |
1 | 1169-1172 | 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 (i)
Find the combined sales in September and October for each farmer in each
variety (ii)
Find the decrease in sales from September to October (iii)
If both farmers receive 2% profit on gross sales, compute the profit for each
farmer and for each variety sold in October |
1 | 1170-1173 | (i)
Find the combined sales in September and October for each farmer in each
variety (ii)
Find the decrease in sales from September to October (iii)
If both farmers receive 2% profit on gross sales, compute the profit for each
farmer and for each variety sold in October Solution
(i)
Combined sales in September and October for each farmer in each variety is
given by
Rationalised 2023-24
50
MATHEMATICS
(ii)
Change in sales from September to October is given by
(iii)
2% of B = 2
100 ×B
= 0 |
1 | 1171-1174 | (ii)
Find the decrease in sales from September to October (iii)
If both farmers receive 2% profit on gross sales, compute the profit for each
farmer and for each variety sold in October Solution
(i)
Combined sales in September and October for each farmer in each variety is
given by
Rationalised 2023-24
50
MATHEMATICS
(ii)
Change in sales from September to October is given by
(iii)
2% of B = 2
100 ×B
= 0 02 × B
= 0 |
1 | 1172-1175 | (iii)
If both farmers receive 2% profit on gross sales, compute the profit for each
farmer and for each variety sold in October Solution
(i)
Combined sales in September and October for each farmer in each variety is
given by
Rationalised 2023-24
50
MATHEMATICS
(ii)
Change in sales from September to October is given by
(iii)
2% of B = 2
100 ×B
= 0 02 × B
= 0 02
=
Thus, in October Ramkishan receives ` 100, ` 200 and ` 120 as profit in the
sale of each variety of rice, respectively, and Grucharan Singh receives profit of ` 400,
` 200 and ` 200 in the sale of each variety of rice, respectively |
1 | 1173-1176 | Solution
(i)
Combined sales in September and October for each farmer in each variety is
given by
Rationalised 2023-24
50
MATHEMATICS
(ii)
Change in sales from September to October is given by
(iii)
2% of B = 2
100 ×B
= 0 02 × B
= 0 02
=
Thus, in October Ramkishan receives ` 100, ` 200 and ` 120 as profit in the
sale of each variety of rice, respectively, and Grucharan Singh receives profit of ` 400,
` 200 and ` 200 in the sale of each variety of rice, respectively 3 |
1 | 1174-1177 | 02 × B
= 0 02
=
Thus, in October Ramkishan receives ` 100, ` 200 and ` 120 as profit in the
sale of each variety of rice, respectively, and Grucharan Singh receives profit of ` 400,
` 200 and ` 200 in the sale of each variety of rice, respectively 3 4 |
1 | 1175-1178 | 02
=
Thus, in October Ramkishan receives ` 100, ` 200 and ` 120 as profit in the
sale of each variety of rice, respectively, and Grucharan Singh receives profit of ` 400,
` 200 and ` 200 in the sale of each variety of rice, respectively 3 4 5 Multiplication of matrices
Suppose Meera and Nadeem are two friends |
1 | 1176-1179 | 3 4 5 Multiplication of matrices
Suppose Meera and Nadeem are two friends Meera wants to buy 2 pens and 5 story
books, while Nadeem needs 8 pens and 10 story books |
1 | 1177-1180 | 4 5 Multiplication of matrices
Suppose Meera and Nadeem are two friends Meera wants to buy 2 pens and 5 story
books, while Nadeem needs 8 pens and 10 story books They both go to a shop to
enquire about the rates which are quoted as follows:
Pen – ` 5 each, story book – ` 50 each |
1 | 1178-1181 | 5 Multiplication of matrices
Suppose Meera and Nadeem are two friends Meera wants to buy 2 pens and 5 story
books, while Nadeem needs 8 pens and 10 story books They both go to a shop to
enquire about the rates which are quoted as follows:
Pen – ` 5 each, story book – ` 50 each How much money does each need to spend |
1 | 1179-1182 | Meera wants to buy 2 pens and 5 story
books, while Nadeem needs 8 pens and 10 story books They both go to a shop to
enquire about the rates which are quoted as follows:
Pen – ` 5 each, story book – ` 50 each How much money does each need to spend Clearly, Meera needs ` (5 × 2 + 50 × 5)
that is ` 260, while Nadeem needs (8 × 5 + 50 × 10) `, that is ` 540 |
1 | 1180-1183 | They both go to a shop to
enquire about the rates which are quoted as follows:
Pen – ` 5 each, story book – ` 50 each How much money does each need to spend Clearly, Meera needs ` (5 × 2 + 50 × 5)
that is ` 260, while Nadeem needs (8 × 5 + 50 × 10) `, that is ` 540 In terms of matrix
representation, we can write the above information as follows:
Requirements
Prices per piece (in Rupees) Money needed (in Rupees)
2
5
8
10
5
50
5
2
5 50
260
8 5
10
50
540
×
+
×
=
×
+
×
Suppose that they enquire about the rates from another shop, quoted as follows:
pen – ` 4 each, story book – ` 40 each |
1 | 1181-1184 | How much money does each need to spend Clearly, Meera needs ` (5 × 2 + 50 × 5)
that is ` 260, while Nadeem needs (8 × 5 + 50 × 10) `, that is ` 540 In terms of matrix
representation, we can write the above information as follows:
Requirements
Prices per piece (in Rupees) Money needed (in Rupees)
2
5
8
10
5
50
5
2
5 50
260
8 5
10
50
540
×
+
×
=
×
+
×
Suppose that they enquire about the rates from another shop, quoted as follows:
pen – ` 4 each, story book – ` 40 each Now, the money required by Meera and Nadeem to make purchases will be
respectively ` (4 × 2 + 40 × 5) = ` 208 and ` (8 × 4 + 10 × 40) = ` 432
Rationalised 2023-24
MATRICES 51
Again, the above information can be represented as follows:
Requirements
Prices per piece (in Rupees) Money needed (in Rupees)
2
5
8
10
404
4
2
40 5
208
8 4
10 4 0
432
×
+
×
=
×
+
×
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)
2
5
8
10
5
4
50
40
5
2
5 50
4
2
40 5
8 5
10 5 0 8 4
10 4 0
×
+ ×
×
+
×
×
+
×
×
+
×
=
260
208
540
432
The above is an example of multiplication of matrices |
1 | 1182-1185 | Clearly, Meera needs ` (5 × 2 + 50 × 5)
that is ` 260, while Nadeem needs (8 × 5 + 50 × 10) `, that is ` 540 In terms of matrix
representation, we can write the above information as follows:
Requirements
Prices per piece (in Rupees) Money needed (in Rupees)
2
5
8
10
5
50
5
2
5 50
260
8 5
10
50
540
×
+
×
=
×
+
×
Suppose that they enquire about the rates from another shop, quoted as follows:
pen – ` 4 each, story book – ` 40 each Now, the money required by Meera and Nadeem to make purchases will be
respectively ` (4 × 2 + 40 × 5) = ` 208 and ` (8 × 4 + 10 × 40) = ` 432
Rationalised 2023-24
MATRICES 51
Again, the above information can be represented as follows:
Requirements
Prices per piece (in Rupees) Money needed (in Rupees)
2
5
8
10
404
4
2
40 5
208
8 4
10 4 0
432
×
+
×
=
×
+
×
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)
2
5
8
10
5
4
50
40
5
2
5 50
4
2
40 5
8 5
10 5 0 8 4
10 4 0
×
+ ×
×
+
×
×
+
×
×
+
×
=
260
208
540
432
The above is an example of multiplication of matrices We observe that, for
multiplication of two matrices A and B, the number of columns in A should be equal to
the number of rows in B |
1 | 1183-1186 | In terms of matrix
representation, we can write the above information as follows:
Requirements
Prices per piece (in Rupees) Money needed (in Rupees)
2
5
8
10
5
50
5
2
5 50
260
8 5
10
50
540
×
+
×
=
×
+
×
Suppose that they enquire about the rates from another shop, quoted as follows:
pen – ` 4 each, story book – ` 40 each Now, the money required by Meera and Nadeem to make purchases will be
respectively ` (4 × 2 + 40 × 5) = ` 208 and ` (8 × 4 + 10 × 40) = ` 432
Rationalised 2023-24
MATRICES 51
Again, the above information can be represented as follows:
Requirements
Prices per piece (in Rupees) Money needed (in Rupees)
2
5
8
10
404
4
2
40 5
208
8 4
10 4 0
432
×
+
×
=
×
+
×
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)
2
5
8
10
5
4
50
40
5
2
5 50
4
2
40 5
8 5
10 5 0 8 4
10 4 0
×
+ ×
×
+
×
×
+
×
×
+
×
=
260
208
540
432
The above is an example of multiplication of matrices We observe that, for
multiplication of two matrices A and B, the number of columns in A should be equal to
the number of rows in B 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 |
1 | 1184-1187 | Now, the money required by Meera and Nadeem to make purchases will be
respectively ` (4 × 2 + 40 × 5) = ` 208 and ` (8 × 4 + 10 × 40) = ` 432
Rationalised 2023-24
MATRICES 51
Again, the above information can be represented as follows:
Requirements
Prices per piece (in Rupees) Money needed (in Rupees)
2
5
8
10
404
4
2
40 5
208
8 4
10 4 0
432
×
+
×
=
×
+
×
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)
2
5
8
10
5
4
50
40
5
2
5 50
4
2
40 5
8 5
10 5 0 8 4
10 4 0
×
+ ×
×
+
×
×
+
×
×
+
×
=
260
208
540
432
The above is an example of multiplication of matrices We observe that, for
multiplication of two matrices A and B, the number of columns in A should be equal to
the number of rows in B 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 Formally, we define multiplication of matrices as follows:
The product of two matrices A and B is defined if the number of columns of A is
equal to the number of rows of B |
1 | 1185-1188 | We observe that, for
multiplication of two matrices A and B, the number of columns in A should be equal to
the number of rows in B 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 Formally, we define multiplication of matrices as follows:
The product of two matrices A and B is defined if the number of columns of A is
equal to the number of rows of B Let A = [aij] be an m × n matrix and B = [bjk] be an
n × p matrix |
1 | 1186-1189 | 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 Formally, we define multiplication of matrices as follows:
The product of two matrices A and B is defined if the number of columns of A is
equal to the number of rows of B Let A = [aij] be an m × n matrix and B = [bjk] be an
n × p matrix Then the product of the matrices A and B is the matrix C of order m × p |
1 | 1187-1190 | Formally, we define multiplication of matrices as follows:
The product of two matrices A and B is defined if the number of columns of A is
equal to the number of rows of B Let A = [aij] be an m × n matrix and B = [bjk] be an
n × p matrix Then the product of the matrices A and B is the matrix C of order m × p 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 |
1 | 1188-1191 | Let A = [aij] be an m × n matrix and B = [bjk] be an
n × p matrix Then the product of the matrices A and B is the matrix C of order m × p 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 In other words,
if A = [aij]m × n, B = [bjk]n × p, then the ith row of A is [ai1 ai2 |
1 | 1189-1192 | Then the product of the matrices A and B is the matrix C of order m × p 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 In other words,
if A = [aij]m × n, B = [bjk]n × p, then the ith row of A is [ai1 ai2 ain] and the kth column of
B is
1 |
1 | 1190-1193 | 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 In other words,
if A = [aij]m × n, B = [bjk]n × p, then the ith row of A is [ai1 ai2 ain] and the kth column of
B is
1 2 |
1 | 1191-1194 | In other words,
if A = [aij]m × n, B = [bjk]n × p, then the ith row of A is [ai1 ai2 ain] and the kth column of
B is
1 2 k
k
nk
b
b
b
, then cik = ai1 b1k + ai2 b2k + ai3 b3k + |
1 | 1192-1195 | ain] and the kth column of
B is
1 2 k
k
nk
b
b
b
, then cik = ai1 b1k + ai2 b2k + ai3 b3k + + ain bnk =
1
n
ij
jk
j
a b
=∑ |
1 | 1193-1196 | 2 k
k
nk
b
b
b
, then cik = ai1 b1k + ai2 b2k + ai3 b3k + + ain bnk =
1
n
ij
jk
j
a b
=∑ The matrix C = [cik]m × p is the product of A and B |
1 | 1194-1197 | k
k
nk
b
b
b
, then cik = ai1 b1k + ai2 b2k + ai3 b3k + + ain bnk =
1
n
ij
jk
j
a b
=∑ The matrix C = [cik]m × p is the product of A and B For example, if
1
1
2
C
0
−3 4
=
and
2
17
D
1
5
4
= −
−
, then the product CD is defined
Rationalised 2023-24
52
MATHEMATICS
and is given by
2
7
1
1
2
CD
1
1
0
3
4
5
4
−
=
−
−
|
1 | 1195-1198 | + ain bnk =
1
n
ij
jk
j
a b
=∑ The matrix C = [cik]m × p is the product of A and B For example, if
1
1
2
C
0
−3 4
=
and
2
17
D
1
5
4
= −
−
, then the product CD is defined
Rationalised 2023-24
52
MATHEMATICS
and is given by
2
7
1
1
2
CD
1
1
0
3
4
5
4
−
=
−
−
This is a 2 × 2 matrix in which each
entry is the sum of the products across some row of C with the corresponding entries
down some column of D |
1 | 1196-1199 | The matrix C = [cik]m × p is the product of A and B For example, if
1
1
2
C
0
−3 4
=
and
2
17
D
1
5
4
= −
−
, then the product CD is defined
Rationalised 2023-24
52
MATHEMATICS
and is given by
2
7
1
1
2
CD
1
1
0
3
4
5
4
−
=
−
−
This is a 2 × 2 matrix in which each
entry is the sum of the products across some row of C with the corresponding entries
down some column of D These four computations are
Thus
13
2
CD
17
13
−
=
−
Example 12 Find AB, if
6
9
2
6
0
A
and B
2
3
7
9
8
=
=
|
1 | 1197-1200 | For example, if
1
1
2
C
0
−3 4
=
and
2
17
D
1
5
4
= −
−
, then the product CD is defined
Rationalised 2023-24
52
MATHEMATICS
and is given by
2
7
1
1
2
CD
1
1
0
3
4
5
4
−
=
−
−
This is a 2 × 2 matrix in which each
entry is the sum of the products across some row of C with the corresponding entries
down some column of D These four computations are
Thus
13
2
CD
17
13
−
=
−
Example 12 Find AB, if
6
9
2
6
0
A
and B
2
3
7
9
8
=
=
Solution The matrix A has 2 columns which is equal to the number of rows of B |
1 | 1198-1201 | This is a 2 × 2 matrix in which each
entry is the sum of the products across some row of C with the corresponding entries
down some column of D These four computations are
Thus
13
2
CD
17
13
−
=
−
Example 12 Find AB, if
6
9
2
6
0
A
and B
2
3
7
9
8
=
=
Solution The matrix A has 2 columns which is equal to the number of rows of B Hence AB is defined |
1 | 1199-1202 | These four computations are
Thus
13
2
CD
17
13
−
=
−
Example 12 Find AB, if
6
9
2
6
0
A
and B
2
3
7
9
8
=
=
Solution The matrix A has 2 columns which is equal to the number of rows of B Hence AB is defined 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
4
21
12
27
0
24
+
+
+
+
+
+
=
75
117
72
25
39
24
Rationalised 2023-24
MATRICES 53
Remark If AB is defined, then BA need not be defined |
1 | 1200-1203 | Solution The matrix A has 2 columns which is equal to the number of rows of B Hence AB is defined 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
4
21
12
27
0
24
+
+
+
+
+
+
=
75
117
72
25
39
24
Rationalised 2023-24
MATRICES 53
Remark If AB is defined, then BA need not be defined In the above example, AB is
defined but BA is not defined because B has 3 column while A has only 2 (and not 3)
rows |
1 | 1201-1204 | Hence AB is defined 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
4
21
12
27
0
24
+
+
+
+
+
+
=
75
117
72
25
39
24
Rationalised 2023-24
MATRICES 53
Remark If AB is defined, then BA need not be defined In the above example, AB is
defined but BA is not defined because B has 3 column while A has only 2 (and not 3)
rows 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 |
1 | 1202-1205 | 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
4
21
12
27
0
24
+
+
+
+
+
+
=
75
117
72
25
39
24
Rationalised 2023-24
MATRICES 53
Remark If AB is defined, then BA need not be defined In the above example, AB is
defined but BA is not defined because B has 3 column while A has only 2 (and not 3)
rows 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 In particular, if both A and B are square matrices of the
same order, then both AB and BA are defined |
1 | 1203-1206 | In the above example, AB is
defined but BA is not defined because B has 3 column while A has only 2 (and not 3)
rows 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 In particular, if both A and B are square matrices of the
same order, then both AB and BA are defined 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 |
1 | 1204-1207 | 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 In particular, if both A and B are square matrices of the
same order, then both AB and BA are defined 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 Example 13 If
2 3
1
2
3
A
and B
4 5
4
2
5
2 1
−
=
=
−
, then find AB, BA |
1 | 1205-1208 | In particular, if both A and B are square matrices of the
same order, then both AB and BA are defined 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 Example 13 If
2 3
1
2
3
A
and B
4 5
4
2
5
2 1
−
=
=
−
, then find AB, BA Show that
AB ≠ BA |
1 | 1206-1209 | 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 Example 13 If
2 3
1
2
3
A
and B
4 5
4
2
5
2 1
−
=
=
−
, then find AB, BA Show that
AB ≠ BA Solution Since A is a 2 × 3 matrix and B is 3 × 2 matrix |
1 | 1207-1210 | Example 13 If
2 3
1
2
3
A
and B
4 5
4
2
5
2 1
−
=
=
−
, then find AB, BA Show that
AB ≠ BA Solution Since A is a 2 × 3 matrix and B is 3 × 2 matrix Hence AB and BA are both
defined and are matrices of order 2 × 2 and 3 × 3, respectively |
1 | 1208-1211 | Show that
AB ≠ BA Solution Since A is a 2 × 3 matrix and B is 3 × 2 matrix Hence AB and BA are both
defined and are matrices of order 2 × 2 and 3 × 3, respectively Note that
2 3
1
2
3
AB
4 5
4
2
5
2 1
−
=
−
=
2
8
6
3 10
3
0
4
8
8
10
12
10
5
10
3
−
+
−
+
−
=
− +
+
−
+
+
and
2 3
2 12
4
6
6
15
1
2
3
BA
4 5
4
20
8 10
12
25
4
2
5
2 1
2
4
4
2
6
5
−
− +
+
−
=
=
−
− +
+
−
−
− +
+
10
2
21
16
2
37
2
2
11
−
= −
−
−
Clearly AB ≠ BA
In the above example both AB and BA are of different order and so AB ≠ BA |
1 | 1209-1212 | Solution Since A is a 2 × 3 matrix and B is 3 × 2 matrix Hence AB and BA are both
defined and are matrices of order 2 × 2 and 3 × 3, respectively Note that
2 3
1
2
3
AB
4 5
4
2
5
2 1
−
=
−
=
2
8
6
3 10
3
0
4
8
8
10
12
10
5
10
3
−
+
−
+
−
=
− +
+
−
+
+
and
2 3
2 12
4
6
6
15
1
2
3
BA
4 5
4
20
8 10
12
25
4
2
5
2 1
2
4
4
2
6
5
−
− +
+
−
=
=
−
− +
+
−
−
− +
+
10
2
21
16
2
37
2
2
11
−
= −
−
−
Clearly AB ≠ BA
In the above example both AB and BA are of different order and so AB ≠ BA But
one may think that perhaps AB and BA could be the same if they were of the same
order |
1 | 1210-1213 | Hence AB and BA are both
defined and are matrices of order 2 × 2 and 3 × 3, respectively Note that
2 3
1
2
3
AB
4 5
4
2
5
2 1
−
=
−
=
2
8
6
3 10
3
0
4
8
8
10
12
10
5
10
3
−
+
−
+
−
=
− +
+
−
+
+
and
2 3
2 12
4
6
6
15
1
2
3
BA
4 5
4
20
8 10
12
25
4
2
5
2 1
2
4
4
2
6
5
−
− +
+
−
=
=
−
− +
+
−
−
− +
+
10
2
21
16
2
37
2
2
11
−
= −
−
−
Clearly AB ≠ BA
In the above example both AB and BA are of different order and so AB ≠ BA But
one may think that perhaps AB and BA could be the same if they were of the same
order But it is not so, here we give an example to show that even if AB and BA are of
same order they may not be same |
1 | 1211-1214 | Note that
2 3
1
2
3
AB
4 5
4
2
5
2 1
−
=
−
=
2
8
6
3 10
3
0
4
8
8
10
12
10
5
10
3
−
+
−
+
−
=
− +
+
−
+
+
and
2 3
2 12
4
6
6
15
1
2
3
BA
4 5
4
20
8 10
12
25
4
2
5
2 1
2
4
4
2
6
5
−
− +
+
−
=
=
−
− +
+
−
−
− +
+
10
2
21
16
2
37
2
2
11
−
= −
−
−
Clearly AB ≠ BA
In the above example both AB and BA are of different order and so AB ≠ BA But
one may think that perhaps AB and BA could be the same if they were of the same
order But it is not so, here we give an example to show that even if AB and BA are of
same order they may not be same Example 14 If
1
0
A
0
1
=
−
and
0
1
B
1
0
=
, then
0
1
AB
1
0
=
−
|
1 | 1212-1215 | But
one may think that perhaps AB and BA could be the same if they were of the same
order But it is not so, here we give an example to show that even if AB and BA are of
same order they may not be same Example 14 If
1
0
A
0
1
=
−
and
0
1
B
1
0
=
, then
0
1
AB
1
0
=
−
and
0
1
BA
1
−0
=
|
1 | 1213-1216 | But it is not so, here we give an example to show that even if AB and BA are of
same order they may not be same Example 14 If
1
0
A
0
1
=
−
and
0
1
B
1
0
=
, then
0
1
AB
1
0
=
−
and
0
1
BA
1
−0
=
Clearly AB ≠ BA |
1 | 1214-1217 | Example 14 If
1
0
A
0
1
=
−
and
0
1
B
1
0
=
, then
0
1
AB
1
0
=
−
and
0
1
BA
1
−0
=
Clearly AB ≠ BA Thus matrix multiplication is not commutative |
1 | 1215-1218 | and
0
1
BA
1
−0
=
Clearly AB ≠ BA Thus matrix multiplication is not commutative Rationalised 2023-24
54
MATHEMATICS
ANote This does not mean that AB ≠ BA for every pair of matrices A, B for
which AB and BA, are defined |
1 | 1216-1219 | Clearly AB ≠ BA Thus matrix multiplication is not commutative Rationalised 2023-24
54
MATHEMATICS
ANote This does not mean that AB ≠ BA for every pair of matrices A, B for
which AB and BA, are defined For instance,
If
1
0
3
0
A
, B
0
2
0
4
=
=
, then AB = BA =
3
0
0
8
Observe that multiplication of diagonal matrices of same order will be commutative |
1 | 1217-1220 | Thus matrix multiplication is not commutative Rationalised 2023-24
54
MATHEMATICS
ANote This does not mean that AB ≠ BA for every pair of matrices A, B for
which AB and BA, are defined For instance,
If
1
0
3
0
A
, B
0
2
0
4
=
=
, then AB = BA =
3
0
0
8
Observe that multiplication of diagonal matrices of same order will be commutative Zero matrix as the product of two non zero matrices
We know that, for real numbers a, b if ab = 0, then either a = 0 or b = 0 |
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