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1 | 1218-1221 | 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 This need
not be true for matrices, we will observe this through an example |
1 | 1219-1222 | 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 This need
not be true for matrices, we will observe this through an example Example 15 Find AB, if
0
1
A
0
−2
=
and
3
5
B
0
0
=
|
1 | 1220-1223 | 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 This need
not be true for matrices, we will observe this through an example Example 15 Find AB, if
0
1
A
0
−2
=
and
3
5
B
0
0
=
Solution We have
0
1
3
5
0
0
AB
0
2
0
0
0
0
−
=
=
|
1 | 1221-1224 | This need
not be true for matrices, we will observe this through an example Example 15 Find AB, if
0
1
A
0
−2
=
and
3
5
B
0
0
=
Solution We have
0
1
3
5
0
0
AB
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 |
1 | 1222-1225 | Example 15 Find AB, if
0
1
A
0
−2
=
and
3
5
B
0
0
=
Solution We have
0
1
3
5
0
0
AB
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 3 |
1 | 1223-1226 | Solution We have
0
1
3
5
0
0
AB
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 3 4 |
1 | 1224-1227 | Thus, if the product of two matrices is a zero matrix, it is not necessary that one of
the matrices is a zero matrix 3 4 6 Properties of multiplication of matrices
The multiplication of matrices possesses the following properties, which we state without
proof |
1 | 1225-1228 | 3 4 6 Properties of multiplication of matrices
The multiplication of matrices possesses the following properties, which we state without
proof 1 |
1 | 1226-1229 | 4 6 Properties of multiplication of matrices
The multiplication of matrices possesses the following properties, which we state without
proof 1 The associative law For any three matrices A, B and C |
1 | 1227-1230 | 6 Properties of multiplication of matrices
The multiplication of matrices possesses the following properties, which we state without
proof 1 The associative law For any three matrices A, B and C We have
(AB) C = A (BC), whenever both sides of the equality are defined |
1 | 1228-1231 | 1 The associative law For any three matrices A, B and C We have
(AB) C = A (BC), whenever both sides of the equality are defined 2 |
1 | 1229-1232 | The associative law For any three matrices A, B and C We have
(AB) C = A (BC), whenever both sides of the equality are defined 2 The distributive law For three matrices A, B and C |
1 | 1230-1233 | We have
(AB) C = A (BC), whenever both sides of the equality are defined 2 The distributive law For three matrices A, B and C (i) A (B+C) = AB + AC
(ii) (A+B) C = AC + BC, whenever both sides of equality are defined |
1 | 1231-1234 | 2 The distributive law For three matrices A, B and C (i) A (B+C) = AB + AC
(ii) (A+B) C = AC + BC, whenever both sides of equality are defined 3 |
1 | 1232-1235 | The distributive law For three matrices A, B and C (i) A (B+C) = AB + AC
(ii) (A+B) C = AC + BC, whenever both sides of equality are defined 3 The existence of multiplicative identity For every square matrix A, there
exist an identity matrix of same order such that IA = AI = A |
1 | 1233-1236 | (i) A (B+C) = AB + AC
(ii) (A+B) C = AC + BC, whenever both sides of equality are defined 3 The existence of multiplicative identity For every square matrix A, there
exist an identity matrix of same order such that IA = AI = A Now, we shall verify these properties by examples |
1 | 1234-1237 | 3 The existence of multiplicative identity For every square matrix A, there
exist an identity matrix of same order such that IA = AI = A Now, we shall verify these properties by examples Example 16 If
1
1
1
1 3
1
2
3
4
A
2
0
3 , B
0 2
and C
2
0
2
1
3
1
2
1 4
−
−
=
=
=
−
−
−
, find
A(BC), (AB)C and show that (AB)C = A(BC) |
1 | 1235-1238 | The existence of multiplicative identity For every square matrix A, there
exist an identity matrix of same order such that IA = AI = A Now, we shall verify these properties by examples Example 16 If
1
1
1
1 3
1
2
3
4
A
2
0
3 , B
0 2
and C
2
0
2
1
3
1
2
1 4
−
−
=
=
=
−
−
−
, find
A(BC), (AB)C and show that (AB)C = A(BC) Rationalised 2023-24
MATRICES 55
Solution We have
1
1
1
1 3
1
0 1 3
2
4
2
1
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
−
+
+
+
−
=
=
+
−
+
+
= −
−
−
+
−
−
+
(AB) (C)
2
2
4
0
6
2
8
1
2
1
1
2
3
4
1 18
1
36
2
0
3
36
4
18
2
0
2
1
1 15
1
30
2
0
3
30
4
15
+
+
−
−
+
−
= −
= − +
− +
− −
+
−
+
+
−
−
+
=
4
4
4
7
35
2
39
22
31
2
27
11
−
−
−
−
Now
BC =
1
6
2
0
3
6
4
3
1 3
1
2
3
4
0 2
0
4
0
0
0
4
0
2
2
0
2
1
1 4
1
8
2
0
3
8
4
4
+
+
−
− +
−
=
+
+
−
+
−
−
− +
− +
− −
+
=
7
2
3
1
4
0
4
2
7
2
11
8
−
−
−
−
−
Therefore
A(BC) =
7
2
3
1
1
1
1
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
4
7
35
2
39
22
31
2
27
11
−
−
−
−
|
1 | 1236-1239 | Now, we shall verify these properties by examples Example 16 If
1
1
1
1 3
1
2
3
4
A
2
0
3 , B
0 2
and C
2
0
2
1
3
1
2
1 4
−
−
=
=
=
−
−
−
, find
A(BC), (AB)C and show that (AB)C = A(BC) Rationalised 2023-24
MATRICES 55
Solution We have
1
1
1
1 3
1
0 1 3
2
4
2
1
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
−
+
+
+
−
=
=
+
−
+
+
= −
−
−
+
−
−
+
(AB) (C)
2
2
4
0
6
2
8
1
2
1
1
2
3
4
1 18
1
36
2
0
3
36
4
18
2
0
2
1
1 15
1
30
2
0
3
30
4
15
+
+
−
−
+
−
= −
= − +
− +
− −
+
−
+
+
−
−
+
=
4
4
4
7
35
2
39
22
31
2
27
11
−
−
−
−
Now
BC =
1
6
2
0
3
6
4
3
1 3
1
2
3
4
0 2
0
4
0
0
0
4
0
2
2
0
2
1
1 4
1
8
2
0
3
8
4
4
+
+
−
− +
−
=
+
+
−
+
−
−
− +
− +
− −
+
=
7
2
3
1
4
0
4
2
7
2
11
8
−
−
−
−
−
Therefore
A(BC) =
7
2
3
1
1
1
1
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
4
7
35
2
39
22
31
2
27
11
−
−
−
−
Clearly, (AB) C = A (BC)
Rationalised 2023-24
56
MATHEMATICS
Example 17 If
0
6
7
0
1
1
2
A
6
0
8 , B
1
0
2 , C
2
7
8
0
1
2
0
3
= −
=
= −
−
Calculate AC, BC and (A + B)C |
1 | 1237-1240 | Example 16 If
1
1
1
1 3
1
2
3
4
A
2
0
3 , B
0 2
and C
2
0
2
1
3
1
2
1 4
−
−
=
=
=
−
−
−
, find
A(BC), (AB)C and show that (AB)C = A(BC) Rationalised 2023-24
MATRICES 55
Solution We have
1
1
1
1 3
1
0 1 3
2
4
2
1
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
−
+
+
+
−
=
=
+
−
+
+
= −
−
−
+
−
−
+
(AB) (C)
2
2
4
0
6
2
8
1
2
1
1
2
3
4
1 18
1
36
2
0
3
36
4
18
2
0
2
1
1 15
1
30
2
0
3
30
4
15
+
+
−
−
+
−
= −
= − +
− +
− −
+
−
+
+
−
−
+
=
4
4
4
7
35
2
39
22
31
2
27
11
−
−
−
−
Now
BC =
1
6
2
0
3
6
4
3
1 3
1
2
3
4
0 2
0
4
0
0
0
4
0
2
2
0
2
1
1 4
1
8
2
0
3
8
4
4
+
+
−
− +
−
=
+
+
−
+
−
−
− +
− +
− −
+
=
7
2
3
1
4
0
4
2
7
2
11
8
−
−
−
−
−
Therefore
A(BC) =
7
2
3
1
1
1
1
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
4
7
35
2
39
22
31
2
27
11
−
−
−
−
Clearly, (AB) C = A (BC)
Rationalised 2023-24
56
MATHEMATICS
Example 17 If
0
6
7
0
1
1
2
A
6
0
8 , B
1
0
2 , C
2
7
8
0
1
2
0
3
= −
=
= −
−
Calculate AC, BC and (A + B)C Also, verify that (A + B)C = AC + BC
Solution Now,
0
7
8
A+B
5
0
10
8
6
0
= −
−
So
(A + B) C =
0
7
8
2
0
14
24
10
5
0
10
2
10
0
30
20
8
6
0
3
16
12
0
28
−
+
−
−
= −
+
+
=
−
+
+
Further
AC =
0
6
7
2
0 12
21
9
6
0
8
2
12
0
24
12
7
8
0
3
14
16
0
30
−
+
−
−
= −
+
+
=
−
+
+
and
BC =
0
1
1
2
0
2
3
1
1
0
2
2
2
0
6
8
1
2
0
3
2
4
0
2
−
+
−
=
+
+
=
−
+
−
So
AC + BC =
9
1
10
12
8
20
30
2
28
+
=
−
Clearly,
(A + B) C = AC + BC
Example 18 If
1
2
3
A
3
2
1
4
2
1
=
−
, then show that A3 – 23A – 40 I = O
Solution We have
2
1
2
3
1
2
3
19
4
8
A
A |
1 | 1238-1241 | Rationalised 2023-24
MATRICES 55
Solution We have
1
1
1
1 3
1
0 1 3
2
4
2
1
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
−
+
+
+
−
=
=
+
−
+
+
= −
−
−
+
−
−
+
(AB) (C)
2
2
4
0
6
2
8
1
2
1
1
2
3
4
1 18
1
36
2
0
3
36
4
18
2
0
2
1
1 15
1
30
2
0
3
30
4
15
+
+
−
−
+
−
= −
= − +
− +
− −
+
−
+
+
−
−
+
=
4
4
4
7
35
2
39
22
31
2
27
11
−
−
−
−
Now
BC =
1
6
2
0
3
6
4
3
1 3
1
2
3
4
0 2
0
4
0
0
0
4
0
2
2
0
2
1
1 4
1
8
2
0
3
8
4
4
+
+
−
− +
−
=
+
+
−
+
−
−
− +
− +
− −
+
=
7
2
3
1
4
0
4
2
7
2
11
8
−
−
−
−
−
Therefore
A(BC) =
7
2
3
1
1
1
1
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
4
7
35
2
39
22
31
2
27
11
−
−
−
−
Clearly, (AB) C = A (BC)
Rationalised 2023-24
56
MATHEMATICS
Example 17 If
0
6
7
0
1
1
2
A
6
0
8 , B
1
0
2 , C
2
7
8
0
1
2
0
3
= −
=
= −
−
Calculate AC, BC and (A + B)C Also, verify that (A + B)C = AC + BC
Solution Now,
0
7
8
A+B
5
0
10
8
6
0
= −
−
So
(A + B) C =
0
7
8
2
0
14
24
10
5
0
10
2
10
0
30
20
8
6
0
3
16
12
0
28
−
+
−
−
= −
+
+
=
−
+
+
Further
AC =
0
6
7
2
0 12
21
9
6
0
8
2
12
0
24
12
7
8
0
3
14
16
0
30
−
+
−
−
= −
+
+
=
−
+
+
and
BC =
0
1
1
2
0
2
3
1
1
0
2
2
2
0
6
8
1
2
0
3
2
4
0
2
−
+
−
=
+
+
=
−
+
−
So
AC + BC =
9
1
10
12
8
20
30
2
28
+
=
−
Clearly,
(A + B) C = AC + BC
Example 18 If
1
2
3
A
3
2
1
4
2
1
=
−
, then show that A3 – 23A – 40 I = O
Solution We have
2
1
2
3
1
2
3
19
4
8
A
A A
3
2
1
3
2
1
1
12
8
4
2
1
4
2
1
14
6
15
=
=
−
−
=
Rationalised 2023-24
MATRICES 57
So
A3 = A A2 =
1
2
3
19
4
8
63
46
69
3
2
1
1
12
8
69
6
23
4
2
1
14
6
15
92
46
63
−
=
−
Now
A3 – 23A – 40I =
63
46
69
1
2
3
1
0
0
69
6
23 – 23 3
2
1 – 40 0
1
0
92
46
63
4
2
1
0
0 1
−
−
=
63
46
69
23
46
69
40
0
0
69
6
23
69
46
23
0
40
0
92
46
63
92
46
23
0
0
40
−
−
−
−
−
+ −
−
+
−
−
−
−
−
=
63
23
40
46
46
0
69
69
0
69
69
0
6
46
40
23
23
0
92
92
0
46
46
0
63
23
40
−
−
−
+
−
+
−
+
− +
−
−
+
−
+
−
+
−
−
=
0
0
0
0
0
0
O
0
0
0
=
Example 19 In a legislative assembly election, a political group hired a public relations
firm to promote its candidate in three ways: telephone, house calls, and letters |
1 | 1239-1242 | Clearly, (AB) C = A (BC)
Rationalised 2023-24
56
MATHEMATICS
Example 17 If
0
6
7
0
1
1
2
A
6
0
8 , B
1
0
2 , C
2
7
8
0
1
2
0
3
= −
=
= −
−
Calculate AC, BC and (A + B)C Also, verify that (A + B)C = AC + BC
Solution Now,
0
7
8
A+B
5
0
10
8
6
0
= −
−
So
(A + B) C =
0
7
8
2
0
14
24
10
5
0
10
2
10
0
30
20
8
6
0
3
16
12
0
28
−
+
−
−
= −
+
+
=
−
+
+
Further
AC =
0
6
7
2
0 12
21
9
6
0
8
2
12
0
24
12
7
8
0
3
14
16
0
30
−
+
−
−
= −
+
+
=
−
+
+
and
BC =
0
1
1
2
0
2
3
1
1
0
2
2
2
0
6
8
1
2
0
3
2
4
0
2
−
+
−
=
+
+
=
−
+
−
So
AC + BC =
9
1
10
12
8
20
30
2
28
+
=
−
Clearly,
(A + B) C = AC + BC
Example 18 If
1
2
3
A
3
2
1
4
2
1
=
−
, then show that A3 – 23A – 40 I = O
Solution We have
2
1
2
3
1
2
3
19
4
8
A
A A
3
2
1
3
2
1
1
12
8
4
2
1
4
2
1
14
6
15
=
=
−
−
=
Rationalised 2023-24
MATRICES 57
So
A3 = A A2 =
1
2
3
19
4
8
63
46
69
3
2
1
1
12
8
69
6
23
4
2
1
14
6
15
92
46
63
−
=
−
Now
A3 – 23A – 40I =
63
46
69
1
2
3
1
0
0
69
6
23 – 23 3
2
1 – 40 0
1
0
92
46
63
4
2
1
0
0 1
−
−
=
63
46
69
23
46
69
40
0
0
69
6
23
69
46
23
0
40
0
92
46
63
92
46
23
0
0
40
−
−
−
−
−
+ −
−
+
−
−
−
−
−
=
63
23
40
46
46
0
69
69
0
69
69
0
6
46
40
23
23
0
92
92
0
46
46
0
63
23
40
−
−
−
+
−
+
−
+
− +
−
−
+
−
+
−
+
−
−
=
0
0
0
0
0
0
O
0
0
0
=
Example 19 In a legislative assembly election, a political group hired a public relations
firm to promote its candidate in three ways: telephone, house calls, and letters The
cost per contact (in paise) is given in matrix A as
A =
40
Telephone
100
Housecall
50
Letter
Cost per contact
The number of contacts of each type made in two cities X and Y is given by
Telephone
Housecall
Letter
1000
500
5000
X
B
Y
3000
1000 10,000
→
=
→
|
1 | 1240-1243 | Also, verify that (A + B)C = AC + BC
Solution Now,
0
7
8
A+B
5
0
10
8
6
0
= −
−
So
(A + B) C =
0
7
8
2
0
14
24
10
5
0
10
2
10
0
30
20
8
6
0
3
16
12
0
28
−
+
−
−
= −
+
+
=
−
+
+
Further
AC =
0
6
7
2
0 12
21
9
6
0
8
2
12
0
24
12
7
8
0
3
14
16
0
30
−
+
−
−
= −
+
+
=
−
+
+
and
BC =
0
1
1
2
0
2
3
1
1
0
2
2
2
0
6
8
1
2
0
3
2
4
0
2
−
+
−
=
+
+
=
−
+
−
So
AC + BC =
9
1
10
12
8
20
30
2
28
+
=
−
Clearly,
(A + B) C = AC + BC
Example 18 If
1
2
3
A
3
2
1
4
2
1
=
−
, then show that A3 – 23A – 40 I = O
Solution We have
2
1
2
3
1
2
3
19
4
8
A
A A
3
2
1
3
2
1
1
12
8
4
2
1
4
2
1
14
6
15
=
=
−
−
=
Rationalised 2023-24
MATRICES 57
So
A3 = A A2 =
1
2
3
19
4
8
63
46
69
3
2
1
1
12
8
69
6
23
4
2
1
14
6
15
92
46
63
−
=
−
Now
A3 – 23A – 40I =
63
46
69
1
2
3
1
0
0
69
6
23 – 23 3
2
1 – 40 0
1
0
92
46
63
4
2
1
0
0 1
−
−
=
63
46
69
23
46
69
40
0
0
69
6
23
69
46
23
0
40
0
92
46
63
92
46
23
0
0
40
−
−
−
−
−
+ −
−
+
−
−
−
−
−
=
63
23
40
46
46
0
69
69
0
69
69
0
6
46
40
23
23
0
92
92
0
46
46
0
63
23
40
−
−
−
+
−
+
−
+
− +
−
−
+
−
+
−
+
−
−
=
0
0
0
0
0
0
O
0
0
0
=
Example 19 In a legislative assembly election, a political group hired a public relations
firm to promote its candidate in three ways: telephone, house calls, and letters The
cost per contact (in paise) is given in matrix A as
A =
40
Telephone
100
Housecall
50
Letter
Cost per contact
The number of contacts of each type made in two cities X and Y is given by
Telephone
Housecall
Letter
1000
500
5000
X
B
Y
3000
1000 10,000
→
=
→
Find the total amount spent by the group in the two
cities X and Y |
1 | 1241-1244 | A
3
2
1
3
2
1
1
12
8
4
2
1
4
2
1
14
6
15
=
=
−
−
=
Rationalised 2023-24
MATRICES 57
So
A3 = A A2 =
1
2
3
19
4
8
63
46
69
3
2
1
1
12
8
69
6
23
4
2
1
14
6
15
92
46
63
−
=
−
Now
A3 – 23A – 40I =
63
46
69
1
2
3
1
0
0
69
6
23 – 23 3
2
1 – 40 0
1
0
92
46
63
4
2
1
0
0 1
−
−
=
63
46
69
23
46
69
40
0
0
69
6
23
69
46
23
0
40
0
92
46
63
92
46
23
0
0
40
−
−
−
−
−
+ −
−
+
−
−
−
−
−
=
63
23
40
46
46
0
69
69
0
69
69
0
6
46
40
23
23
0
92
92
0
46
46
0
63
23
40
−
−
−
+
−
+
−
+
− +
−
−
+
−
+
−
+
−
−
=
0
0
0
0
0
0
O
0
0
0
=
Example 19 In a legislative assembly election, a political group hired a public relations
firm to promote its candidate in three ways: telephone, house calls, and letters The
cost per contact (in paise) is given in matrix A as
A =
40
Telephone
100
Housecall
50
Letter
Cost per contact
The number of contacts of each type made in two cities X and Y is given by
Telephone
Housecall
Letter
1000
500
5000
X
B
Y
3000
1000 10,000
→
=
→
Find the total amount spent by the group in the two
cities X and Y Rationalised 2023-24
58
MATHEMATICS
Solution We have
BA =
40,000
50,000
250,000
X
Y
120,000 +100,000 +500,000
+
+
→
→
=
340,000
X
Y
720,000
→
→
So the total amount spent by the group in the two cities is 340,000 paise and
720,000 paise, i |
1 | 1242-1245 | The
cost per contact (in paise) is given in matrix A as
A =
40
Telephone
100
Housecall
50
Letter
Cost per contact
The number of contacts of each type made in two cities X and Y is given by
Telephone
Housecall
Letter
1000
500
5000
X
B
Y
3000
1000 10,000
→
=
→
Find the total amount spent by the group in the two
cities X and Y Rationalised 2023-24
58
MATHEMATICS
Solution We have
BA =
40,000
50,000
250,000
X
Y
120,000 +100,000 +500,000
+
+
→
→
=
340,000
X
Y
720,000
→
→
So the total amount spent by the group in the two cities is 340,000 paise and
720,000 paise, i e |
1 | 1243-1246 | Find the total amount spent by the group in the two
cities X and Y Rationalised 2023-24
58
MATHEMATICS
Solution We have
BA =
40,000
50,000
250,000
X
Y
120,000 +100,000 +500,000
+
+
→
→
=
340,000
X
Y
720,000
→
→
So the total amount spent by the group in the two cities is 340,000 paise and
720,000 paise, i e , ` 3400 and ` 7200, respectively |
1 | 1244-1247 | Rationalised 2023-24
58
MATHEMATICS
Solution We have
BA =
40,000
50,000
250,000
X
Y
120,000 +100,000 +500,000
+
+
→
→
=
340,000
X
Y
720,000
→
→
So the total amount spent by the group in the two cities is 340,000 paise and
720,000 paise, i e , ` 3400 and ` 7200, respectively EXERCISE 3 |
1 | 1245-1248 | e , ` 3400 and ` 7200, respectively EXERCISE 3 2
1 |
1 | 1246-1249 | , ` 3400 and ` 7200, respectively EXERCISE 3 2
1 Let
2
4
1
3
2
5
A
, B
, C
3
2
2
5
3
4
−
=
=
=
−
Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
2 |
1 | 1247-1250 | EXERCISE 3 2
1 Let
2
4
1
3
2
5
A
, B
, C
3
2
2
5
3
4
−
=
=
=
−
Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
2 Compute the following:
(i)
a
b
a
b
b
a
b
a
+
−
(ii)
2
2
2
2
2
2
2
2
2
2
2
2
a
b
b
c
ab
bc
ac
ab
a
c
a
b
+
+
+
−
−
+
+
(iii)
1
4
6
12
7
6
8
5
16
8
0
5
2
8
5
3
2
4
−
−
+
(iv)
2
2
2
2
2
2
2
2
cos
sin
sin
cos
sin
cos
cos
sin
x
x
x
x
x
x
x
x
+
3 |
1 | 1248-1251 | 2
1 Let
2
4
1
3
2
5
A
, B
, C
3
2
2
5
3
4
−
=
=
=
−
Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
2 Compute the following:
(i)
a
b
a
b
b
a
b
a
+
−
(ii)
2
2
2
2
2
2
2
2
2
2
2
2
a
b
b
c
ab
bc
ac
ab
a
c
a
b
+
+
+
−
−
+
+
(iii)
1
4
6
12
7
6
8
5
16
8
0
5
2
8
5
3
2
4
−
−
+
(iv)
2
2
2
2
2
2
2
2
cos
sin
sin
cos
sin
cos
cos
sin
x
x
x
x
x
x
x
x
+
3 Compute the indicated products |
1 | 1249-1252 | Let
2
4
1
3
2
5
A
, B
, C
3
2
2
5
3
4
−
=
=
=
−
Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
2 Compute the following:
(i)
a
b
a
b
b
a
b
a
+
−
(ii)
2
2
2
2
2
2
2
2
2
2
2
2
a
b
b
c
ab
bc
ac
ab
a
c
a
b
+
+
+
−
−
+
+
(iii)
1
4
6
12
7
6
8
5
16
8
0
5
2
8
5
3
2
4
−
−
+
(iv)
2
2
2
2
2
2
2
2
cos
sin
sin
cos
sin
cos
cos
sin
x
x
x
x
x
x
x
x
+
3 Compute the indicated products (i)
a
b
a
b
b
a
b
a
−
−
(ii)
1
2
3
[2 3 4]
(iii)
1
2
1
2
3
2
3
2
3
1
−
(iv)
2
3
4
1
3
5
3
4
5
0
2
4
4
5
6
3
0
5
−
(v)
2 1
1
0
1
3 2
1
2
1
1 1
−
−
(vi)
2
3
3
1
3
1
0
1
0
2
3
1
−
−
−
Rationalised 2023-24
MATRICES 59
4 |
1 | 1250-1253 | Compute the following:
(i)
a
b
a
b
b
a
b
a
+
−
(ii)
2
2
2
2
2
2
2
2
2
2
2
2
a
b
b
c
ab
bc
ac
ab
a
c
a
b
+
+
+
−
−
+
+
(iii)
1
4
6
12
7
6
8
5
16
8
0
5
2
8
5
3
2
4
−
−
+
(iv)
2
2
2
2
2
2
2
2
cos
sin
sin
cos
sin
cos
cos
sin
x
x
x
x
x
x
x
x
+
3 Compute the indicated products (i)
a
b
a
b
b
a
b
a
−
−
(ii)
1
2
3
[2 3 4]
(iii)
1
2
1
2
3
2
3
2
3
1
−
(iv)
2
3
4
1
3
5
3
4
5
0
2
4
4
5
6
3
0
5
−
(v)
2 1
1
0
1
3 2
1
2
1
1 1
−
−
(vi)
2
3
3
1
3
1
0
1
0
2
3
1
−
−
−
Rationalised 2023-24
MATRICES 59
4 If
1
2
3
3
1
2
4
1
2
A
5
0
2 , B
4
2
5
and C
0
3
2
1
1
1
2
0
3
1
2
3
−
−
=
=
=
−
−
, then compute
(A+B) and (B – C) |
1 | 1251-1254 | Compute the indicated products (i)
a
b
a
b
b
a
b
a
−
−
(ii)
1
2
3
[2 3 4]
(iii)
1
2
1
2
3
2
3
2
3
1
−
(iv)
2
3
4
1
3
5
3
4
5
0
2
4
4
5
6
3
0
5
−
(v)
2 1
1
0
1
3 2
1
2
1
1 1
−
−
(vi)
2
3
3
1
3
1
0
1
0
2
3
1
−
−
−
Rationalised 2023-24
MATRICES 59
4 If
1
2
3
3
1
2
4
1
2
A
5
0
2 , B
4
2
5
and C
0
3
2
1
1
1
2
0
3
1
2
3
−
−
=
=
=
−
−
, then compute
(A+B) and (B – C) Also, verify that A + (B – C) = (A + B) – C |
1 | 1252-1255 | (i)
a
b
a
b
b
a
b
a
−
−
(ii)
1
2
3
[2 3 4]
(iii)
1
2
1
2
3
2
3
2
3
1
−
(iv)
2
3
4
1
3
5
3
4
5
0
2
4
4
5
6
3
0
5
−
(v)
2 1
1
0
1
3 2
1
2
1
1 1
−
−
(vi)
2
3
3
1
3
1
0
1
0
2
3
1
−
−
−
Rationalised 2023-24
MATRICES 59
4 If
1
2
3
3
1
2
4
1
2
A
5
0
2 , B
4
2
5
and C
0
3
2
1
1
1
2
0
3
1
2
3
−
−
=
=
=
−
−
, then compute
(A+B) and (B – C) Also, verify that A + (B – C) = (A + B) – C 5 |
1 | 1253-1256 | If
1
2
3
3
1
2
4
1
2
A
5
0
2 , B
4
2
5
and C
0
3
2
1
1
1
2
0
3
1
2
3
−
−
=
=
=
−
−
, then compute
(A+B) and (B – C) Also, verify that A + (B – C) = (A + B) – C 5 If
2
5
2
3
1
1
3
3
5
5
1
2
4
1
2
4
A
and B
3
3
3
5
5
5
7
2
7
6
2
2
3
3
5
5
5
=
=
, then compute 3A – 5B |
1 | 1254-1257 | Also, verify that A + (B – C) = (A + B) – C 5 If
2
5
2
3
1
1
3
3
5
5
1
2
4
1
2
4
A
and B
3
3
3
5
5
5
7
2
7
6
2
2
3
3
5
5
5
=
=
, then compute 3A – 5B 6 |
1 | 1255-1258 | 5 If
2
5
2
3
1
1
3
3
5
5
1
2
4
1
2
4
A
and B
3
3
3
5
5
5
7
2
7
6
2
2
3
3
5
5
5
=
=
, then compute 3A – 5B 6 Simplify
cos
sin
sin
cos
cos
+ sin
sin
cos
cos
sin
θ
θ
θ
−
θ
θ
θ
−
θ
θ
θ
θ
7 |
1 | 1256-1259 | If
2
5
2
3
1
1
3
3
5
5
1
2
4
1
2
4
A
and B
3
3
3
5
5
5
7
2
7
6
2
2
3
3
5
5
5
=
=
, then compute 3A – 5B 6 Simplify
cos
sin
sin
cos
cos
+ sin
sin
cos
cos
sin
θ
θ
θ
−
θ
θ
θ
−
θ
θ
θ
θ
7 Find X and Y, if
(i)
7
0
3
0
X + Y
and X – Y
2
5
0
3
=
=
(ii)
2
3
2
2
2X + 3Y
and 3X
2Y
4
0
1
−5
=
+
=
−
8 |
1 | 1257-1260 | 6 Simplify
cos
sin
sin
cos
cos
+ sin
sin
cos
cos
sin
θ
θ
θ
−
θ
θ
θ
−
θ
θ
θ
θ
7 Find X and Y, if
(i)
7
0
3
0
X + Y
and X – Y
2
5
0
3
=
=
(ii)
2
3
2
2
2X + 3Y
and 3X
2Y
4
0
1
−5
=
+
=
−
8 Find X, if Y =
3
2
1
4
and 2X + Y =
1
0
3
2
−
9 |
1 | 1258-1261 | Simplify
cos
sin
sin
cos
cos
+ sin
sin
cos
cos
sin
θ
θ
θ
−
θ
θ
θ
−
θ
θ
θ
θ
7 Find X and Y, if
(i)
7
0
3
0
X + Y
and X – Y
2
5
0
3
=
=
(ii)
2
3
2
2
2X + 3Y
and 3X
2Y
4
0
1
−5
=
+
=
−
8 Find X, if Y =
3
2
1
4
and 2X + Y =
1
0
3
2
−
9 Find x and y, if
1
3
0
5
6
2 0
1
2
1
8
y
x
+
=
10 |
1 | 1259-1262 | Find X and Y, if
(i)
7
0
3
0
X + Y
and X – Y
2
5
0
3
=
=
(ii)
2
3
2
2
2X + 3Y
and 3X
2Y
4
0
1
−5
=
+
=
−
8 Find X, if Y =
3
2
1
4
and 2X + Y =
1
0
3
2
−
9 Find x and y, if
1
3
0
5
6
2 0
1
2
1
8
y
x
+
=
10 Solve the equation for x, y, z and t, if
1
1
3
5
2
3
3
0
2
4
6
x
z
y
t
−
+
=
11 |
1 | 1260-1263 | Find X, if Y =
3
2
1
4
and 2X + Y =
1
0
3
2
−
9 Find x and y, if
1
3
0
5
6
2 0
1
2
1
8
y
x
+
=
10 Solve the equation for x, y, z and t, if
1
1
3
5
2
3
3
0
2
4
6
x
z
y
t
−
+
=
11 If
2
1
10
3
1
5
x
y −
+
=
, find the values of x and y |
1 | 1261-1264 | Find x and y, if
1
3
0
5
6
2 0
1
2
1
8
y
x
+
=
10 Solve the equation for x, y, z and t, if
1
1
3
5
2
3
3
0
2
4
6
x
z
y
t
−
+
=
11 If
2
1
10
3
1
5
x
y −
+
=
, find the values of x and y 12 |
1 | 1262-1265 | Solve the equation for x, y, z and t, if
1
1
3
5
2
3
3
0
2
4
6
x
z
y
t
−
+
=
11 If
2
1
10
3
1
5
x
y −
+
=
, find the values of x and y 12 Given
6
4
3
1
2
3
x
y
x
x
y
z
w
w
z
w
+
=
+
−
+
, find the values of x, y, z and w |
1 | 1263-1266 | If
2
1
10
3
1
5
x
y −
+
=
, find the values of x and y 12 Given
6
4
3
1
2
3
x
y
x
x
y
z
w
w
z
w
+
=
+
−
+
, find the values of x, y, z and w Rationalised 2023-24
60
MATHEMATICS
13 |
1 | 1264-1267 | 12 Given
6
4
3
1
2
3
x
y
x
x
y
z
w
w
z
w
+
=
+
−
+
, find the values of x, y, z and w Rationalised 2023-24
60
MATHEMATICS
13 If
cos
sin
0
F ( )
sin
cos
0
0
0
1
x
x
x
x
x
−
=
, show that F(x) F(y) = F(x + y) |
1 | 1265-1268 | Given
6
4
3
1
2
3
x
y
x
x
y
z
w
w
z
w
+
=
+
−
+
, find the values of x, y, z and w Rationalised 2023-24
60
MATHEMATICS
13 If
cos
sin
0
F ( )
sin
cos
0
0
0
1
x
x
x
x
x
−
=
, show that F(x) F(y) = F(x + y) 14 |
1 | 1266-1269 | Rationalised 2023-24
60
MATHEMATICS
13 If
cos
sin
0
F ( )
sin
cos
0
0
0
1
x
x
x
x
x
−
=
, show that F(x) F(y) = F(x + y) 14 Show that
(i)
5
1
2
1
2
1
5
1
6
7
3
4
3
4
6
7
−
−
≠
(ii)
1
2
3
1
1
0
1
1
0
1
2
3
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
2
3
4
2
3
4
1
1
0
−
−
−
≠
−
15 |
1 | 1267-1270 | If
cos
sin
0
F ( )
sin
cos
0
0
0
1
x
x
x
x
x
−
=
, show that F(x) F(y) = F(x + y) 14 Show that
(i)
5
1
2
1
2
1
5
1
6
7
3
4
3
4
6
7
−
−
≠
(ii)
1
2
3
1
1
0
1
1
0
1
2
3
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
2
3
4
2
3
4
1
1
0
−
−
−
≠
−
15 Find A2 – 5A + 6I, if
2
0
1
A
2
1
3
1
1
0
=
−
16 |
1 | 1268-1271 | 14 Show that
(i)
5
1
2
1
2
1
5
1
6
7
3
4
3
4
6
7
−
−
≠
(ii)
1
2
3
1
1
0
1
1
0
1
2
3
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
2
3
4
2
3
4
1
1
0
−
−
−
≠
−
15 Find A2 – 5A + 6I, if
2
0
1
A
2
1
3
1
1
0
=
−
16 If
1
0
2
A
0
2
1
2
0
3
=
, prove that A3 – 6A2 + 7A + 2I = 0
17 |
1 | 1269-1272 | Show that
(i)
5
1
2
1
2
1
5
1
6
7
3
4
3
4
6
7
−
−
≠
(ii)
1
2
3
1
1
0
1
1
0
1
2
3
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
2
3
4
2
3
4
1
1
0
−
−
−
≠
−
15 Find A2 – 5A + 6I, if
2
0
1
A
2
1
3
1
1
0
=
−
16 If
1
0
2
A
0
2
1
2
0
3
=
, prove that A3 – 6A2 + 7A + 2I = 0
17 If
3
2
1
0
A
and I=
4
2
0
1
−
=
−
, find k so that A2 = kA – 2I
18 |
1 | 1270-1273 | Find A2 – 5A + 6I, if
2
0
1
A
2
1
3
1
1
0
=
−
16 If
1
0
2
A
0
2
1
2
0
3
=
, prove that A3 – 6A2 + 7A + 2I = 0
17 If
3
2
1
0
A
and I=
4
2
0
1
−
=
−
, find k so that A2 = kA – 2I
18 If
0
tan 2
A
tan
0
2
α
−
=
α
and I is the identity matrix of order 2, show that
I + A = (I – A)
cos
sin
sin
cos
α
−
α
α
α
19 |
1 | 1271-1274 | If
1
0
2
A
0
2
1
2
0
3
=
, prove that A3 – 6A2 + 7A + 2I = 0
17 If
3
2
1
0
A
and I=
4
2
0
1
−
=
−
, find k so that A2 = kA – 2I
18 If
0
tan 2
A
tan
0
2
α
−
=
α
and I is the identity matrix of order 2, show that
I + A = (I – A)
cos
sin
sin
cos
α
−
α
α
α
19 A trust fund has ` 30,000 that must be invested in two different types of bonds |
1 | 1272-1275 | If
3
2
1
0
A
and I=
4
2
0
1
−
=
−
, find k so that A2 = kA – 2I
18 If
0
tan 2
A
tan
0
2
α
−
=
α
and I is the identity matrix of order 2, show that
I + A = (I – A)
cos
sin
sin
cos
α
−
α
α
α
19 A trust fund has ` 30,000 that must be invested in two different types of bonds The first bond pays 5% interest per year, and the second bond pays 7% interest
per year |
1 | 1273-1276 | If
0
tan 2
A
tan
0
2
α
−
=
α
and I is the identity matrix of order 2, show that
I + A = (I – A)
cos
sin
sin
cos
α
−
α
α
α
19 A trust fund has ` 30,000 that must be invested in two different types of bonds The first bond pays 5% interest per year, and the second bond pays 7% interest
per year Using matrix multiplication, determine how to divide ` 30,000 among
the two types of bonds |
1 | 1274-1277 | A trust fund has ` 30,000 that must be invested in two different types of bonds The first bond pays 5% interest per year, and the second bond pays 7% interest
per year Using matrix multiplication, determine how to divide ` 30,000 among
the two types of bonds If the trust fund must obtain an annual total interest of:
(a)
` 1800
(b)
` 2000
Rationalised 2023-24
MATRICES 61
20 |
1 | 1275-1278 | The first bond pays 5% interest per year, and the second bond pays 7% interest
per year Using matrix multiplication, determine how to divide ` 30,000 among
the two types of bonds If the trust fund must obtain an annual total interest of:
(a)
` 1800
(b)
` 2000
Rationalised 2023-24
MATRICES 61
20 The bookshop of a particular school has 10 dozen chemistry books, 8 dozen
physics books, 10 dozen economics books |
1 | 1276-1279 | Using matrix multiplication, determine how to divide ` 30,000 among
the two types of bonds If the trust fund must obtain an annual total interest of:
(a)
` 1800
(b)
` 2000
Rationalised 2023-24
MATRICES 61
20 The bookshop of a particular school has 10 dozen chemistry books, 8 dozen
physics books, 10 dozen economics books Their selling prices are ` 80, ` 60 and
` 40 each respectively |
1 | 1277-1280 | If the trust fund must obtain an annual total interest of:
(a)
` 1800
(b)
` 2000
Rationalised 2023-24
MATRICES 61
20 The bookshop of a particular school has 10 dozen chemistry books, 8 dozen
physics books, 10 dozen economics books Their selling prices are ` 80, ` 60 and
` 40 each respectively Find the total amount the bookshop will receive from
selling all the books using matrix algebra |
1 | 1278-1281 | The bookshop of a particular school has 10 dozen chemistry books, 8 dozen
physics books, 10 dozen economics books Their selling prices are ` 80, ` 60 and
` 40 each respectively Find the total amount the bookshop will receive from
selling all the books using matrix algebra Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively |
1 | 1279-1282 | Their selling prices are ` 80, ` 60 and
` 40 each respectively Find the total amount the bookshop will receive from
selling all the books using matrix algebra Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively Choose the correct answer in Exercises 21 and 22 |
1 | 1280-1283 | Find the total amount the bookshop will receive from
selling all the books using matrix algebra Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively Choose the correct answer in Exercises 21 and 22 21 |
1 | 1281-1284 | Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively Choose the correct answer in Exercises 21 and 22 21 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 |
1 | 1282-1285 | Choose the correct answer in Exercises 21 and 22 21 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 If n = p, then the order of the matrix 7X – 5Z is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n
3 |
1 | 1283-1286 | 21 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 If n = p, then the order of the matrix 7X – 5Z is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n
3 5 |
1 | 1284-1287 | 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 If n = p, then the order of the matrix 7X – 5Z is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n
3 5 Transpose of a Matrix
In this section, we shall learn about transpose of a matrix and special types of matrices
such as symmetric and skew symmetric matrices |
1 | 1285-1288 | If n = p, then the order of the matrix 7X – 5Z is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n
3 5 Transpose of a Matrix
In this section, we shall learn about transpose of a matrix and special types of matrices
such as symmetric and skew symmetric matrices 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 |
1 | 1286-1289 | 5 Transpose of a Matrix
In this section, we shall learn about transpose of a matrix and special types of matrices
such as symmetric and skew symmetric matrices 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 Transpose of the matrix A is
denoted by A′ or (AT) |
1 | 1287-1290 | Transpose of a Matrix
In this section, we shall learn about transpose of a matrix and special types of matrices
such as symmetric and skew symmetric matrices 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 Transpose of the matrix A is
denoted by A′ or (AT) In other words, if A = [aij]m × n, then A′ = [aji]n × m |
1 | 1288-1291 | 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 Transpose of the matrix A is
denoted by A′ or (AT) In other words, if A = [aij]m × n, then A′ = [aji]n × m For example,
if
2
3
3
2
3
5
3
3
0
A
3
1
, then A
1
5
1
0
1
5
5
×
×
=
′ =
−
−
3 |
1 | 1289-1292 | Transpose of the matrix A is
denoted by A′ or (AT) In other words, if A = [aij]m × n, then A′ = [aji]n × m For example,
if
2
3
3
2
3
5
3
3
0
A
3
1
, then A
1
5
1
0
1
5
5
×
×
=
′ =
−
−
3 5 |
1 | 1290-1293 | In other words, if A = [aij]m × n, then A′ = [aji]n × m For example,
if
2
3
3
2
3
5
3
3
0
A
3
1
, then A
1
5
1
0
1
5
5
×
×
=
′ =
−
−
3 5 1 Properties of transpose of the matrices
We now state the following properties of transpose of matrices without proof |
1 | 1291-1294 | For example,
if
2
3
3
2
3
5
3
3
0
A
3
1
, then A
1
5
1
0
1
5
5
×
×
=
′ =
−
−
3 5 1 Properties of transpose of the matrices
We now state the following properties of transpose of matrices without proof These
may be verified by taking suitable examples |
1 | 1292-1295 | 5 1 Properties of transpose of the matrices
We now state the following properties of transpose of matrices without proof These
may be verified by taking suitable examples (i)For any matrices A and B of suitable orders, we have
(A′)′ = A,
(ii)
(kA)′ = kA′ (where k is any constant)
(iii)
(A + B)′ = A′ + B′
(iv)
(A B)′ = B′ A′
Example 20 If
2
1
2
3
3
2
A
and B
1
2
4
4
2
0
−
=
=
, verify that
(i)
(A′)′ = A,
(ii)
(A + B)′ = A′ + B′,
(iii)
(kB)′ = kB′, where k is any constant |
1 | 1293-1296 | 1 Properties of transpose of the matrices
We now state the following properties of transpose of matrices without proof These
may be verified by taking suitable examples (i)For any matrices A and B of suitable orders, we have
(A′)′ = A,
(ii)
(kA)′ = kA′ (where k is any constant)
(iii)
(A + B)′ = A′ + B′
(iv)
(A B)′ = B′ A′
Example 20 If
2
1
2
3
3
2
A
and B
1
2
4
4
2
0
−
=
=
, verify that
(i)
(A′)′ = A,
(ii)
(A + B)′ = A′ + B′,
(iii)
(kB)′ = kB′, where k is any constant Rationalised 2023-24
62
MATHEMATICS
Solution
(i)
We have
A =
(
)
3
4
3
3
2
3
3
2
A
3 2
A
A
4
2
0
4
2
0
2
0
′
′
′
⇒
=
⇒
=
=
Thus
(A′)′ = A
(ii)
We have
A = 3
3
2 ,
4
2
0
B =
2
1
2
5
3
1
4
A
B
1
2 4
5
4
4
−
−
⇒
+
=
Therefore
(A + B)′ =
5
5
3
1
4
4
4
−
Now
A′ =
3
4
2 1
3 2 , B
1 2 ,
2
0
2 4
′ = −
So
A′ + B′ =
5
5
43 1 4
4
−
Thus
(A + B)′ = A′ + B′
(iii)
We have
kB = k 2
1
2
2
2
1
2
4
2
4
k
k
k
k
k
k
−
−
=
Then
(kB)′ =
2
2 1
2
1 2
B
2
4
2 4
k
k
k
k
k
k
k
k
′
−
=
−
=
Thus
(kB)′ = kB′
Rationalised 2023-24
MATRICES 63
Example 21 If
[
]
2
A
4 , B
1
3
6
5
−
=
=
−
, verify that (AB)′ = B′A′ |
1 | 1294-1297 | These
may be verified by taking suitable examples (i)For any matrices A and B of suitable orders, we have
(A′)′ = A,
(ii)
(kA)′ = kA′ (where k is any constant)
(iii)
(A + B)′ = A′ + B′
(iv)
(A B)′ = B′ A′
Example 20 If
2
1
2
3
3
2
A
and B
1
2
4
4
2
0
−
=
=
, verify that
(i)
(A′)′ = A,
(ii)
(A + B)′ = A′ + B′,
(iii)
(kB)′ = kB′, where k is any constant Rationalised 2023-24
62
MATHEMATICS
Solution
(i)
We have
A =
(
)
3
4
3
3
2
3
3
2
A
3 2
A
A
4
2
0
4
2
0
2
0
′
′
′
⇒
=
⇒
=
=
Thus
(A′)′ = A
(ii)
We have
A = 3
3
2 ,
4
2
0
B =
2
1
2
5
3
1
4
A
B
1
2 4
5
4
4
−
−
⇒
+
=
Therefore
(A + B)′ =
5
5
3
1
4
4
4
−
Now
A′ =
3
4
2 1
3 2 , B
1 2 ,
2
0
2 4
′ = −
So
A′ + B′ =
5
5
43 1 4
4
−
Thus
(A + B)′ = A′ + B′
(iii)
We have
kB = k 2
1
2
2
2
1
2
4
2
4
k
k
k
k
k
k
−
−
=
Then
(kB)′ =
2
2 1
2
1 2
B
2
4
2 4
k
k
k
k
k
k
k
k
′
−
=
−
=
Thus
(kB)′ = kB′
Rationalised 2023-24
MATRICES 63
Example 21 If
[
]
2
A
4 , B
1
3
6
5
−
=
=
−
, verify that (AB)′ = B′A′ Solution We have
A =
[
]
2
4 , B
1
3
6
5
−
=
−
then
AB =
[
]
2
4
1
3
6
5
−
−
=
2
6
12
4
12
24
5
15
30
−
−
−
−
Now
A′ = [–2 4 5] ,
1
B
3
6
′ =
−
B′A′ =
[
]
1
2
4
5
3
2
4
5
6
12
15
(AB)
6
12
24
30
−
′
−
= −
=
−
−
−
Clearly
(AB)′ = B′A′
3 |
1 | 1295-1298 | (i)For any matrices A and B of suitable orders, we have
(A′)′ = A,
(ii)
(kA)′ = kA′ (where k is any constant)
(iii)
(A + B)′ = A′ + B′
(iv)
(A B)′ = B′ A′
Example 20 If
2
1
2
3
3
2
A
and B
1
2
4
4
2
0
−
=
=
, verify that
(i)
(A′)′ = A,
(ii)
(A + B)′ = A′ + B′,
(iii)
(kB)′ = kB′, where k is any constant Rationalised 2023-24
62
MATHEMATICS
Solution
(i)
We have
A =
(
)
3
4
3
3
2
3
3
2
A
3 2
A
A
4
2
0
4
2
0
2
0
′
′
′
⇒
=
⇒
=
=
Thus
(A′)′ = A
(ii)
We have
A = 3
3
2 ,
4
2
0
B =
2
1
2
5
3
1
4
A
B
1
2 4
5
4
4
−
−
⇒
+
=
Therefore
(A + B)′ =
5
5
3
1
4
4
4
−
Now
A′ =
3
4
2 1
3 2 , B
1 2 ,
2
0
2 4
′ = −
So
A′ + B′ =
5
5
43 1 4
4
−
Thus
(A + B)′ = A′ + B′
(iii)
We have
kB = k 2
1
2
2
2
1
2
4
2
4
k
k
k
k
k
k
−
−
=
Then
(kB)′ =
2
2 1
2
1 2
B
2
4
2 4
k
k
k
k
k
k
k
k
′
−
=
−
=
Thus
(kB)′ = kB′
Rationalised 2023-24
MATRICES 63
Example 21 If
[
]
2
A
4 , B
1
3
6
5
−
=
=
−
, verify that (AB)′ = B′A′ Solution We have
A =
[
]
2
4 , B
1
3
6
5
−
=
−
then
AB =
[
]
2
4
1
3
6
5
−
−
=
2
6
12
4
12
24
5
15
30
−
−
−
−
Now
A′ = [–2 4 5] ,
1
B
3
6
′ =
−
B′A′ =
[
]
1
2
4
5
3
2
4
5
6
12
15
(AB)
6
12
24
30
−
′
−
= −
=
−
−
−
Clearly
(AB)′ = B′A′
3 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 |
1 | 1296-1299 | Rationalised 2023-24
62
MATHEMATICS
Solution
(i)
We have
A =
(
)
3
4
3
3
2
3
3
2
A
3 2
A
A
4
2
0
4
2
0
2
0
′
′
′
⇒
=
⇒
=
=
Thus
(A′)′ = A
(ii)
We have
A = 3
3
2 ,
4
2
0
B =
2
1
2
5
3
1
4
A
B
1
2 4
5
4
4
−
−
⇒
+
=
Therefore
(A + B)′ =
5
5
3
1
4
4
4
−
Now
A′ =
3
4
2 1
3 2 , B
1 2 ,
2
0
2 4
′ = −
So
A′ + B′ =
5
5
43 1 4
4
−
Thus
(A + B)′ = A′ + B′
(iii)
We have
kB = k 2
1
2
2
2
1
2
4
2
4
k
k
k
k
k
k
−
−
=
Then
(kB)′ =
2
2 1
2
1 2
B
2
4
2 4
k
k
k
k
k
k
k
k
′
−
=
−
=
Thus
(kB)′ = kB′
Rationalised 2023-24
MATRICES 63
Example 21 If
[
]
2
A
4 , B
1
3
6
5
−
=
=
−
, verify that (AB)′ = B′A′ Solution We have
A =
[
]
2
4 , B
1
3
6
5
−
=
−
then
AB =
[
]
2
4
1
3
6
5
−
−
=
2
6
12
4
12
24
5
15
30
−
−
−
−
Now
A′ = [–2 4 5] ,
1
B
3
6
′ =
−
B′A′ =
[
]
1
2
4
5
3
2
4
5
6
12
15
(AB)
6
12
24
30
−
′
−
= −
=
−
−
−
Clearly
(AB)′ = B′A′
3 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 For example
3
2
3
A
2
1 |
1 | 1297-1300 | Solution We have
A =
[
]
2
4 , B
1
3
6
5
−
=
−
then
AB =
[
]
2
4
1
3
6
5
−
−
=
2
6
12
4
12
24
5
15
30
−
−
−
−
Now
A′ = [–2 4 5] ,
1
B
3
6
′ =
−
B′A′ =
[
]
1
2
4
5
3
2
4
5
6
12
15
(AB)
6
12
24
30
−
′
−
= −
=
−
−
−
Clearly
(AB)′ = B′A′
3 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 For example
3
2
3
A
2
1 5
1
3
1
1
=
−
−
−
is a symmetric matrix as A′ = A
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 |
1 | 1298-1301 | 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 For example
3
2
3
A
2
1 5
1
3
1
1
=
−
−
−
is a symmetric matrix as A′ = A
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 Now, if we put i = j, we
have aii = – aii |
1 | 1299-1302 | For example
3
2
3
A
2
1 5
1
3
1
1
=
−
−
−
is a symmetric matrix as A′ = A
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 Now, if we put i = j, we
have aii = – aii Therefore 2aii = 0 or aii = 0 for all i’s |
1 | 1300-1303 | 5
1
3
1
1
=
−
−
−
is a symmetric matrix as A′ = A
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 Now, if we put i = j, we
have aii = – aii Therefore 2aii = 0 or aii = 0 for all i’s This means that all the diagonal elements of a skew symmetric matrix are zero |
1 | 1301-1304 | Now, if we put i = j, we
have aii = – aii Therefore 2aii = 0 or aii = 0 for all i’s This means that all the diagonal elements of a skew symmetric matrix are zero Rationalised 2023-24
64
MATHEMATICS
For example, the matrix
0
B
0
0
e
f
e
g
f
g
=
−
−
−
is a skew symmetric matrix as B′= –B
Now, we are going to prove some results of symmetric and skew-symmetric
matrices |
1 | 1302-1305 | Therefore 2aii = 0 or aii = 0 for all i’s This means that all the diagonal elements of a skew symmetric matrix are zero Rationalised 2023-24
64
MATHEMATICS
For example, the matrix
0
B
0
0
e
f
e
g
f
g
=
−
−
−
is a skew symmetric matrix as B′= –B
Now, we are going to prove some results of symmetric and skew-symmetric
matrices Theorem 1 For any square matrix A with real number entries, A + A′ is a symmetric
matrix and A – A′ is a skew symmetric matrix |
1 | 1303-1306 | This means that all the diagonal elements of a skew symmetric matrix are zero Rationalised 2023-24
64
MATHEMATICS
For example, the matrix
0
B
0
0
e
f
e
g
f
g
=
−
−
−
is a skew symmetric matrix as B′= –B
Now, we are going to prove some results of symmetric and skew-symmetric
matrices Theorem 1 For any square matrix A with real number entries, A + A′ is a symmetric
matrix and A – A′ is a skew symmetric matrix Proof Let B = A + A′, then
B′ = (A + A′)′
= A′ + (A′)′ (as (A + B)′ = A′ + B′)
= A′ + A (as (A′)′ = A)
= 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′)′ (Why |
1 | 1304-1307 | Rationalised 2023-24
64
MATHEMATICS
For example, the matrix
0
B
0
0
e
f
e
g
f
g
=
−
−
−
is a skew symmetric matrix as B′= –B
Now, we are going to prove some results of symmetric and skew-symmetric
matrices Theorem 1 For any square matrix A with real number entries, A + A′ is a symmetric
matrix and A – A′ is a skew symmetric matrix Proof Let B = A + A′, then
B′ = (A + A′)′
= A′ + (A′)′ (as (A + B)′ = A′ + B′)
= A′ + A (as (A′)′ = A)
= 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′)′ (Why )
= A′ – A (Why |
1 | 1305-1308 | Theorem 1 For any square matrix A with real number entries, A + A′ is a symmetric
matrix and A – A′ is a skew symmetric matrix Proof Let B = A + A′, then
B′ = (A + A′)′
= A′ + (A′)′ (as (A + B)′ = A′ + B′)
= A′ + A (as (A′)′ = A)
= 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′)′ (Why )
= A′ – A (Why )
= – (A – A′) = – C
Therefore
C = A – A′ is a skew symmetric matrix |
1 | 1306-1309 | Proof Let B = A + A′, then
B′ = (A + A′)′
= A′ + (A′)′ (as (A + B)′ = A′ + B′)
= A′ + A (as (A′)′ = A)
= 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′)′ (Why )
= A′ – A (Why )
= – (A – A′) = – C
Therefore
C = A – A′ is a skew symmetric matrix Theorem 2 Any square matrix can be expressed as the sum of a symmetric and a
skew symmetric matrix |
1 | 1307-1310 | )
= A′ – A (Why )
= – (A – A′) = – C
Therefore
C = A – A′ is a skew symmetric matrix Theorem 2 Any square matrix can be expressed as the sum of a symmetric and 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 |
1 | 1308-1311 | )
= – (A – A′) = – C
Therefore
C = A – A′ is a skew symmetric matrix Theorem 2 Any square matrix can be expressed as the sum of a symmetric and 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 Since for any matrix A, (kA)′ = kA′, it follows that 1 (A
A )
2
′
+
Rationalised 2023-24
MATRICES 65
is symmetric matrix and 1 (A
A )
2
′
−
is skew symmetric matrix |
1 | 1309-1312 | Theorem 2 Any square matrix can be expressed as the sum of a symmetric and 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 Since for any matrix A, (kA)′ = kA′, it follows that 1 (A
A )
2
′
+
Rationalised 2023-24
MATRICES 65
is symmetric matrix and 1 (A
A )
2
′
−
is skew symmetric matrix Thus, any square
matrix can be expressed as the sum of a symmetric and a skew symmetric matrix |
1 | 1310-1313 | 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 Since for any matrix A, (kA)′ = kA′, it follows that 1 (A
A )
2
′
+
Rationalised 2023-24
MATRICES 65
is symmetric matrix and 1 (A
A )
2
′
−
is skew symmetric matrix Thus, any square
matrix can be expressed as the sum of a symmetric and a skew symmetric matrix Example 22 Express the matrix
2
2
4
B
1
3
4
1
2
3
−
−
= −
−
−
as the sum of a symmetric and a
skew symmetric matrix |
1 | 1311-1314 | Since for any matrix A, (kA)′ = kA′, it follows that 1 (A
A )
2
′
+
Rationalised 2023-24
MATRICES 65
is symmetric matrix and 1 (A
A )
2
′
−
is skew symmetric matrix Thus, any square
matrix can be expressed as the sum of a symmetric and a skew symmetric matrix Example 22 Express the matrix
2
2
4
B
1
3
4
1
2
3
−
−
= −
−
−
as the sum of a symmetric and a
skew symmetric matrix Solution Here
B′ =
2
1
1
2
3
2
4
4
3
−
−
−
−
−
Let
P =
4
3
3
1
1
(B + B )
3
6
2
2
2
3
2
6
−
−
′ =
−
−
−
=
3
3
2
2
2
3
3
1
32
1
3
2
−
−
−
−
−
,
Now
P′ =
3
3
2
2
2
3
3
1
2
3
1
3
2
−
−
−
−
−
= P
Thus
P = 1 (B + B )
2
′ is a symmetric matrix |
1 | 1312-1315 | Thus, any square
matrix can be expressed as the sum of a symmetric and a skew symmetric matrix Example 22 Express the matrix
2
2
4
B
1
3
4
1
2
3
−
−
= −
−
−
as the sum of a symmetric and a
skew symmetric matrix Solution Here
B′ =
2
1
1
2
3
2
4
4
3
−
−
−
−
−
Let
P =
4
3
3
1
1
(B + B )
3
6
2
2
2
3
2
6
−
−
′ =
−
−
−
=
3
3
2
2
2
3
3
1
32
1
3
2
−
−
−
−
−
,
Now
P′ =
3
3
2
2
2
3
3
1
2
3
1
3
2
−
−
−
−
−
= P
Thus
P = 1 (B + B )
2
′ is a symmetric matrix Also, let
Q =
1
5
0
2
2
0
1
5
1
1
1
(B – B )
1
0
6
0
3
2
2
2
5
6
0
5
3
0
2
−
−
−
−
′ =
=
−
−
Rationalised 2023-24
66
MATHEMATICS
Then
Q′ =
1
5
0
2
3
1
0
3
Q
52
3
0
2
−
−
= −
−
Thus
Q = 1 (B – B )
2
′ is a skew symmetric matrix |
1 | 1313-1316 | Example 22 Express the matrix
2
2
4
B
1
3
4
1
2
3
−
−
= −
−
−
as the sum of a symmetric and a
skew symmetric matrix Solution Here
B′ =
2
1
1
2
3
2
4
4
3
−
−
−
−
−
Let
P =
4
3
3
1
1
(B + B )
3
6
2
2
2
3
2
6
−
−
′ =
−
−
−
=
3
3
2
2
2
3
3
1
32
1
3
2
−
−
−
−
−
,
Now
P′ =
3
3
2
2
2
3
3
1
2
3
1
3
2
−
−
−
−
−
= P
Thus
P = 1 (B + B )
2
′ is a symmetric matrix Also, let
Q =
1
5
0
2
2
0
1
5
1
1
1
(B – B )
1
0
6
0
3
2
2
2
5
6
0
5
3
0
2
−
−
−
−
′ =
=
−
−
Rationalised 2023-24
66
MATHEMATICS
Then
Q′ =
1
5
0
2
3
1
0
3
Q
52
3
0
2
−
−
= −
−
Thus
Q = 1 (B – B )
2
′ is a skew symmetric matrix Now
3
3
1
5
2
0
2
2
2
2
2
2
4
3
1
P + Q
3
1
0
3
1
3
4
B
2
2
1
2
3
3
5
1
3
3
0
2
2
−
−
−
−
−
−
−
=
+
= −
=
−
−
−
−
−
Thus, B is represented as the sum of a symmetric and a skew symmetric matrix |
1 | 1314-1317 | Solution Here
B′ =
2
1
1
2
3
2
4
4
3
−
−
−
−
−
Let
P =
4
3
3
1
1
(B + B )
3
6
2
2
2
3
2
6
−
−
′ =
−
−
−
=
3
3
2
2
2
3
3
1
32
1
3
2
−
−
−
−
−
,
Now
P′ =
3
3
2
2
2
3
3
1
2
3
1
3
2
−
−
−
−
−
= P
Thus
P = 1 (B + B )
2
′ is a symmetric matrix Also, let
Q =
1
5
0
2
2
0
1
5
1
1
1
(B – B )
1
0
6
0
3
2
2
2
5
6
0
5
3
0
2
−
−
−
−
′ =
=
−
−
Rationalised 2023-24
66
MATHEMATICS
Then
Q′ =
1
5
0
2
3
1
0
3
Q
52
3
0
2
−
−
= −
−
Thus
Q = 1 (B – B )
2
′ is a skew symmetric matrix Now
3
3
1
5
2
0
2
2
2
2
2
2
4
3
1
P + Q
3
1
0
3
1
3
4
B
2
2
1
2
3
3
5
1
3
3
0
2
2
−
−
−
−
−
−
−
=
+
= −
=
−
−
−
−
−
Thus, B is represented as the sum of a symmetric and a skew symmetric matrix EXERCISE 3 |
1 | 1315-1318 | Also, let
Q =
1
5
0
2
2
0
1
5
1
1
1
(B – B )
1
0
6
0
3
2
2
2
5
6
0
5
3
0
2
−
−
−
−
′ =
=
−
−
Rationalised 2023-24
66
MATHEMATICS
Then
Q′ =
1
5
0
2
3
1
0
3
Q
52
3
0
2
−
−
= −
−
Thus
Q = 1 (B – B )
2
′ is a skew symmetric matrix Now
3
3
1
5
2
0
2
2
2
2
2
2
4
3
1
P + Q
3
1
0
3
1
3
4
B
2
2
1
2
3
3
5
1
3
3
0
2
2
−
−
−
−
−
−
−
=
+
= −
=
−
−
−
−
−
Thus, B is represented as the sum of a symmetric and a skew symmetric matrix EXERCISE 3 3
1 |
1 | 1316-1319 | Now
3
3
1
5
2
0
2
2
2
2
2
2
4
3
1
P + Q
3
1
0
3
1
3
4
B
2
2
1
2
3
3
5
1
3
3
0
2
2
−
−
−
−
−
−
−
=
+
= −
=
−
−
−
−
−
Thus, B is represented as the sum of a symmetric and a skew symmetric matrix EXERCISE 3 3
1 Find the transpose of each of the following matrices:
(i)
5
21
1
−
(ii)
1
1
2
3
−
(iii)
1
5
6
3
5
6
2
3
1
−
−
2 |
1 | 1317-1320 | EXERCISE 3 3
1 Find the transpose of each of the following matrices:
(i)
5
21
1
−
(ii)
1
1
2
3
−
(iii)
1
5
6
3
5
6
2
3
1
−
−
2 If
1
2
3
4
1
5
A
5
7
9
and B
1
2
0
2
1
1
1
3
1
−
−
−
=
=
−
, then verify that
(i) (A + B)′ = A′ + B′,
(ii) (A – B)′ = A′ – B′
3 |
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