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1 | 1318-1321 | 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 If
3 4
1
2
1
A
1 2
and B
1
2 3
0 1
−
′ = −
=
, then verify that
(i) (A + B)′ = A′ + B′
(ii) (A – B)′ = A′ – B′
Rationalised 2023-24
MATRICES 67
4 |
1 | 1319-1322 | 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 If
3 4
1
2
1
A
1 2
and B
1
2 3
0 1
−
′ = −
=
, then verify that
(i) (A + B)′ = A′ + B′
(ii) (A – B)′ = A′ – B′
Rationalised 2023-24
MATRICES 67
4 If
2
3
1
0
A
and B
1
2
1
2
−
−
′ =
=
, then find (A + 2B)′
5 |
1 | 1320-1323 | 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 If
3 4
1
2
1
A
1 2
and B
1
2 3
0 1
−
′ = −
=
, then verify that
(i) (A + B)′ = A′ + B′
(ii) (A – B)′ = A′ – B′
Rationalised 2023-24
MATRICES 67
4 If
2
3
1
0
A
and B
1
2
1
2
−
−
′ =
=
, then find (A + 2B)′
5 For the matrices A and B, verify that (AB)′ = B′A′, where
(i)
[
]
1
A
4
, B
1
2
1
3
= −
= −
(ii)
[
]
0
A
1
, B
1
5
7
2
=
=
6 |
1 | 1321-1324 | If
3 4
1
2
1
A
1 2
and B
1
2 3
0 1
−
′ = −
=
, then verify that
(i) (A + B)′ = A′ + B′
(ii) (A – B)′ = A′ – B′
Rationalised 2023-24
MATRICES 67
4 If
2
3
1
0
A
and B
1
2
1
2
−
−
′ =
=
, then find (A + 2B)′
5 For the matrices A and B, verify that (AB)′ = B′A′, where
(i)
[
]
1
A
4
, B
1
2
1
3
= −
= −
(ii)
[
]
0
A
1
, B
1
5
7
2
=
=
6 If (i)
cos
sin
A
sin
cos
α
α
=
−
α
α
, then verify that A′ A = I
(ii) If
sin
cos
A
cos
sin
α
α
=
−
α
α
, then verify that A′ A = I
7 |
1 | 1322-1325 | If
2
3
1
0
A
and B
1
2
1
2
−
−
′ =
=
, then find (A + 2B)′
5 For the matrices A and B, verify that (AB)′ = B′A′, where
(i)
[
]
1
A
4
, B
1
2
1
3
= −
= −
(ii)
[
]
0
A
1
, B
1
5
7
2
=
=
6 If (i)
cos
sin
A
sin
cos
α
α
=
−
α
α
, then verify that A′ A = I
(ii) If
sin
cos
A
cos
sin
α
α
=
−
α
α
, then verify that A′ A = I
7 (i) Show that the matrix
1
1
5
A
1
2
1
5
1
3
−
= −
is a symmetric matrix |
1 | 1323-1326 | For the matrices A and B, verify that (AB)′ = B′A′, where
(i)
[
]
1
A
4
, B
1
2
1
3
= −
= −
(ii)
[
]
0
A
1
, B
1
5
7
2
=
=
6 If (i)
cos
sin
A
sin
cos
α
α
=
−
α
α
, then verify that A′ A = I
(ii) If
sin
cos
A
cos
sin
α
α
=
−
α
α
, then verify that A′ A = I
7 (i) Show that the matrix
1
1
5
A
1
2
1
5
1
3
−
= −
is a symmetric matrix (ii) Show that the matrix
0
1
1
A
1
0
1
1
1
0
−
= −
−
is a skew symmetric matrix |
1 | 1324-1327 | If (i)
cos
sin
A
sin
cos
α
α
=
−
α
α
, then verify that A′ A = I
(ii) If
sin
cos
A
cos
sin
α
α
=
−
α
α
, then verify that A′ A = I
7 (i) Show that the matrix
1
1
5
A
1
2
1
5
1
3
−
= −
is a symmetric matrix (ii) Show that the matrix
0
1
1
A
1
0
1
1
1
0
−
= −
−
is a skew symmetric matrix 8 |
1 | 1325-1328 | (i) Show that the matrix
1
1
5
A
1
2
1
5
1
3
−
= −
is a symmetric matrix (ii) Show that the matrix
0
1
1
A
1
0
1
1
1
0
−
= −
−
is a skew symmetric matrix 8 For the matrix
1
5
A
6
7
=
, verify that
(i) (A + A′) is a symmetric matrix
(ii) (A – A′) is a skew symmetric matrix
9 |
1 | 1326-1329 | (ii) Show that the matrix
0
1
1
A
1
0
1
1
1
0
−
= −
−
is a skew symmetric matrix 8 For the matrix
1
5
A
6
7
=
, verify that
(i) (A + A′) is a symmetric matrix
(ii) (A – A′) is a skew symmetric matrix
9 Find
(
)
1 A
A
2
′
+
and
(
)
1 A
A
2
′
−
, when
0
A
0
0
a
b
a
c
b
c
= −
−
−
10 |
1 | 1327-1330 | 8 For the matrix
1
5
A
6
7
=
, verify that
(i) (A + A′) is a symmetric matrix
(ii) (A – A′) is a skew symmetric matrix
9 Find
(
)
1 A
A
2
′
+
and
(
)
1 A
A
2
′
−
, when
0
A
0
0
a
b
a
c
b
c
= −
−
−
10 Express the following matrices as the sum of a symmetric and a skew symmetric
matrix:
Rationalised 2023-24
68
MATHEMATICS
(i)
3
5
1
1
−
(ii)
6
2
2
2
3
1
2
1
3
−
−
−
−
(iii)
3
3
1
2
2
1
4
5
2
−
−
−
−
−
(iv)
1
5
1
2
−
Choose the correct answer in the Exercises 11 and 12 |
1 | 1328-1331 | For the matrix
1
5
A
6
7
=
, verify that
(i) (A + A′) is a symmetric matrix
(ii) (A – A′) is a skew symmetric matrix
9 Find
(
)
1 A
A
2
′
+
and
(
)
1 A
A
2
′
−
, when
0
A
0
0
a
b
a
c
b
c
= −
−
−
10 Express the following matrices as the sum of a symmetric and a skew symmetric
matrix:
Rationalised 2023-24
68
MATHEMATICS
(i)
3
5
1
1
−
(ii)
6
2
2
2
3
1
2
1
3
−
−
−
−
(iii)
3
3
1
2
2
1
4
5
2
−
−
−
−
−
(iv)
1
5
1
2
−
Choose the correct answer in the Exercises 11 and 12 11 |
1 | 1329-1332 | Find
(
)
1 A
A
2
′
+
and
(
)
1 A
A
2
′
−
, when
0
A
0
0
a
b
a
c
b
c
= −
−
−
10 Express the following matrices as the sum of a symmetric and a skew symmetric
matrix:
Rationalised 2023-24
68
MATHEMATICS
(i)
3
5
1
1
−
(ii)
6
2
2
2
3
1
2
1
3
−
−
−
−
(iii)
3
3
1
2
2
1
4
5
2
−
−
−
−
−
(iv)
1
5
1
2
−
Choose the correct answer in the Exercises 11 and 12 11 If A, B are symmetric matrices of same order, then AB – BA is a
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
12 |
1 | 1330-1333 | Express the following matrices as the sum of a symmetric and a skew symmetric
matrix:
Rationalised 2023-24
68
MATHEMATICS
(i)
3
5
1
1
−
(ii)
6
2
2
2
3
1
2
1
3
−
−
−
−
(iii)
3
3
1
2
2
1
4
5
2
−
−
−
−
−
(iv)
1
5
1
2
−
Choose the correct answer in the Exercises 11 and 12 11 If A, B are symmetric matrices of same order, then AB – BA is a
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
12 If
cos
sin
A
,
sin
cos
α
−
α
=
α
α
and A + A′ = I, then the value of α is
(A) 6
π
(B) 3
π
(C) π
(D) 3
2
π
3 |
1 | 1331-1334 | 11 If A, B are symmetric matrices of same order, then AB – BA is a
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
12 If
cos
sin
A
,
sin
cos
α
−
α
=
α
α
and A + A′ = I, then the value of α is
(A) 6
π
(B) 3
π
(C) π
(D) 3
2
π
3 7 Invertible Matrices
Definition 6 If A is a square matrix of order m, and if there exists another square
matrix B of the same order m, such that AB = BA = I, then B is called the inverse
matrix of A and it is denoted by A– 1 |
1 | 1332-1335 | If A, B are symmetric matrices of same order, then AB – BA is a
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
12 If
cos
sin
A
,
sin
cos
α
−
α
=
α
α
and A + A′ = I, then the value of α is
(A) 6
π
(B) 3
π
(C) π
(D) 3
2
π
3 7 Invertible Matrices
Definition 6 If A is a square matrix of order m, and if there exists another square
matrix B of the same order m, such that AB = BA = I, then B is called the inverse
matrix of A and it is denoted by A– 1 In that case A is said to be invertible |
1 | 1333-1336 | If
cos
sin
A
,
sin
cos
α
−
α
=
α
α
and A + A′ = I, then the value of α is
(A) 6
π
(B) 3
π
(C) π
(D) 3
2
π
3 7 Invertible Matrices
Definition 6 If A is a square matrix of order m, and if there exists another square
matrix B of the same order m, such that AB = BA = I, then B is called the inverse
matrix of A and it is denoted by A– 1 In that case A is said to be invertible For example, let
A =
2
3
1
2
and B =
2
3
1
2
−
−
be two matrices |
1 | 1334-1337 | 7 Invertible Matrices
Definition 6 If A is a square matrix of order m, and if there exists another square
matrix B of the same order m, such that AB = BA = I, then B is called the inverse
matrix of A and it is denoted by A– 1 In that case A is said to be invertible For example, let
A =
2
3
1
2
and B =
2
3
1
2
−
−
be two matrices Now
AB =
2
3
2
3
1
2
1
2
−
−
=
4
3
6
6
1
0
I
2
2
3
4
0
1
−
− +
=
=
−
− +
Also
BA =
1
0
I
0
1
=
|
1 | 1335-1338 | In that case A is said to be invertible For example, let
A =
2
3
1
2
and B =
2
3
1
2
−
−
be two matrices Now
AB =
2
3
2
3
1
2
1
2
−
−
=
4
3
6
6
1
0
I
2
2
3
4
0
1
−
− +
=
=
−
− +
Also
BA =
1
0
I
0
1
=
Thus B is the inverse of A, in other
words B = A– 1 and A is inverse of B, i |
1 | 1336-1339 | For example, let
A =
2
3
1
2
and B =
2
3
1
2
−
−
be two matrices Now
AB =
2
3
2
3
1
2
1
2
−
−
=
4
3
6
6
1
0
I
2
2
3
4
0
1
−
− +
=
=
−
− +
Also
BA =
1
0
I
0
1
=
Thus B is the inverse of A, in other
words B = A– 1 and A is inverse of B, i e |
1 | 1337-1340 | Now
AB =
2
3
2
3
1
2
1
2
−
−
=
4
3
6
6
1
0
I
2
2
3
4
0
1
−
− +
=
=
−
− +
Also
BA =
1
0
I
0
1
=
Thus B is the inverse of A, in other
words B = A– 1 and A is inverse of B, i e , A = B–1
Rationalised 2023-24
MATRICES 69
ANote
1 |
1 | 1338-1341 | Thus B is the inverse of A, in other
words B = A– 1 and A is inverse of B, i e , A = B–1
Rationalised 2023-24
MATRICES 69
ANote
1 A rectangular matrix does not possess inverse matrix, since for products BA
and AB to be defined and to be equal, it is necessary that matrices A and B
should be square matrices of the same order |
1 | 1339-1342 | e , A = B–1
Rationalised 2023-24
MATRICES 69
ANote
1 A rectangular matrix does not possess inverse matrix, since for products BA
and AB to be defined and to be equal, it is necessary that matrices A and B
should be square matrices of the same order 2 |
1 | 1340-1343 | , A = B–1
Rationalised 2023-24
MATRICES 69
ANote
1 A rectangular matrix does not possess inverse matrix, since for products BA
and AB to be defined and to be equal, it is necessary that matrices A and B
should be square matrices of the same order 2 If B is the inverse of A, then A is also the inverse of B |
1 | 1341-1344 | A rectangular matrix does not possess inverse matrix, since for products BA
and AB to be defined and to be equal, it is necessary that matrices A and B
should be square matrices of the same order 2 If B is the inverse of A, then A is also the inverse of B Theorem 3 (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique |
1 | 1342-1345 | 2 If B is the inverse of A, then A is also the inverse of B Theorem 3 (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique Proof Let A = [aij] be a square matrix of order m |
1 | 1343-1346 | If B is the inverse of A, then A is also the inverse of B Theorem 3 (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique Proof Let A = [aij] be a square matrix of order m If possible, let B and C be two
inverses of A |
1 | 1344-1347 | Theorem 3 (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique Proof Let A = [aij] be a square matrix of order m If possible, let B and C be two
inverses of A We shall show that B = C |
1 | 1345-1348 | Proof Let A = [aij] be a square matrix of order m If possible, let B and C be two
inverses of A We shall show that B = C Since B is the inverse of A
AB = BA = I |
1 | 1346-1349 | If possible, let B and C be two
inverses of A We shall show that B = C Since B is the inverse of A
AB = BA = I (1)
Since C is also the inverse of A
AC = CA = I |
1 | 1347-1350 | We shall show that B = C Since B is the inverse of A
AB = BA = I (1)
Since C is also the inverse of A
AC = CA = I (2)
Thus
B = BI = B (AC) = (BA) C = IC = C
Theorem 4 If A and B are invertible matrices of the same order, then (AB)–1 = B–1 A–1 |
1 | 1348-1351 | Since B is the inverse of A
AB = BA = I (1)
Since C is also the inverse of A
AC = CA = I (2)
Thus
B = BI = B (AC) = (BA) C = IC = C
Theorem 4 If A and B are invertible matrices of the same order, then (AB)–1 = B–1 A–1 Proof From the definition of inverse of a matrix, we have
(AB) (AB)–1 = 1
or
A–1 (AB) (AB)–1 = A–1I
(Pre multiplying both sides by A–1)
or
(A–1A) B (AB)–1 = A–1
(Since A–1 I = A–1)
or
IB (AB)–1 = A–1
or
B (AB)–1 = A–1
or
B–1 B (AB)–1 = B–1 A–1
or
I (AB)–1 = B–1 A–1
Hence
(AB)–1 = B–1 A–1
1 |
1 | 1349-1352 | (1)
Since C is also the inverse of A
AC = CA = I (2)
Thus
B = BI = B (AC) = (BA) C = IC = C
Theorem 4 If A and B are invertible matrices of the same order, then (AB)–1 = B–1 A–1 Proof From the definition of inverse of a matrix, we have
(AB) (AB)–1 = 1
or
A–1 (AB) (AB)–1 = A–1I
(Pre multiplying both sides by A–1)
or
(A–1A) B (AB)–1 = A–1
(Since A–1 I = A–1)
or
IB (AB)–1 = A–1
or
B (AB)–1 = A–1
or
B–1 B (AB)–1 = B–1 A–1
or
I (AB)–1 = B–1 A–1
Hence
(AB)–1 = B–1 A–1
1 Matrices A and B will be inverse of each other only if
(A) AB = BA (B) AB = BA = 0
(C) AB = 0, BA = I (D) AB = BA = I
EXERCISE 3 |
1 | 1350-1353 | (2)
Thus
B = BI = B (AC) = (BA) C = IC = C
Theorem 4 If A and B are invertible matrices of the same order, then (AB)–1 = B–1 A–1 Proof From the definition of inverse of a matrix, we have
(AB) (AB)–1 = 1
or
A–1 (AB) (AB)–1 = A–1I
(Pre multiplying both sides by A–1)
or
(A–1A) B (AB)–1 = A–1
(Since A–1 I = A–1)
or
IB (AB)–1 = A–1
or
B (AB)–1 = A–1
or
B–1 B (AB)–1 = B–1 A–1
or
I (AB)–1 = B–1 A–1
Hence
(AB)–1 = B–1 A–1
1 Matrices A and B will be inverse of each other only if
(A) AB = BA (B) AB = BA = 0
(C) AB = 0, BA = I (D) AB = BA = I
EXERCISE 3 4
Rationalised 2023-24
70
MATHEMATICS
Miscellaneous Examples
Example 23 If
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then prove that
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, n ∈ N |
1 | 1351-1354 | Proof From the definition of inverse of a matrix, we have
(AB) (AB)–1 = 1
or
A–1 (AB) (AB)–1 = A–1I
(Pre multiplying both sides by A–1)
or
(A–1A) B (AB)–1 = A–1
(Since A–1 I = A–1)
or
IB (AB)–1 = A–1
or
B (AB)–1 = A–1
or
B–1 B (AB)–1 = B–1 A–1
or
I (AB)–1 = B–1 A–1
Hence
(AB)–1 = B–1 A–1
1 Matrices A and B will be inverse of each other only if
(A) AB = BA (B) AB = BA = 0
(C) AB = 0, BA = I (D) AB = BA = I
EXERCISE 3 4
Rationalised 2023-24
70
MATHEMATICS
Miscellaneous Examples
Example 23 If
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then prove that
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, n ∈ N Solution We shall prove the result by using principle of mathematical induction |
1 | 1352-1355 | Matrices A and B will be inverse of each other only if
(A) AB = BA (B) AB = BA = 0
(C) AB = 0, BA = I (D) AB = BA = I
EXERCISE 3 4
Rationalised 2023-24
70
MATHEMATICS
Miscellaneous Examples
Example 23 If
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then prove that
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, n ∈ N Solution We shall prove the result by using principle of mathematical induction We have
P(n) : If
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, n ∈ N
P(1) :
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, so
1
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
Therefore,
the result is true for n = 1 |
1 | 1353-1356 | 4
Rationalised 2023-24
70
MATHEMATICS
Miscellaneous Examples
Example 23 If
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then prove that
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, n ∈ N Solution We shall prove the result by using principle of mathematical induction We have
P(n) : If
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, n ∈ N
P(1) :
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, so
1
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
Therefore,
the result is true for n = 1 Let the result be true for n = k |
1 | 1354-1357 | Solution We shall prove the result by using principle of mathematical induction We have
P(n) : If
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, n ∈ N
P(1) :
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, so
1
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
Therefore,
the result is true for n = 1 Let the result be true for n = k So
P(k) :
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then
cos
sin
A
sin
cos
k
k
k
k
k
θ
θ
=
−
θ
θ
Now, we prove that the result holds for n = k +1
Now
Ak + 1 =
cos
sin
cos
sin
A A
sin
cos
sin
cos
k
k
k
k
k
θ
θ
θ
θ
⋅
=
−
θ
θ
−
θ
θ
=
cos cos
– sin sin
cos sin
sin cos
sin cos
cos sin
sin sin
cos cos
k
k
k
k
k
k
k
k
θ
θ
θ
θ
θ
θ +
θ
θ
−
θ
θ +
θ
θ
−
θ
θ +
θ
θ
=
cos(
)
sin(
)
cos(
1)
sin (
1)
sin(
)
cos(
)
sin (
1)
cos(
1)
k
k
k
k
k
k
k
k
θ + θ
θ + θ
+
θ
+
θ
=
−
θ + θ
θ + θ
−
+
θ
+
θ
Therefore, the result is true for n = k + 1 |
1 | 1355-1358 | We have
P(n) : If
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, n ∈ N
P(1) :
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, so
1
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
Therefore,
the result is true for n = 1 Let the result be true for n = k So
P(k) :
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then
cos
sin
A
sin
cos
k
k
k
k
k
θ
θ
=
−
θ
θ
Now, we prove that the result holds for n = k +1
Now
Ak + 1 =
cos
sin
cos
sin
A A
sin
cos
sin
cos
k
k
k
k
k
θ
θ
θ
θ
⋅
=
−
θ
θ
−
θ
θ
=
cos cos
– sin sin
cos sin
sin cos
sin cos
cos sin
sin sin
cos cos
k
k
k
k
k
k
k
k
θ
θ
θ
θ
θ
θ +
θ
θ
−
θ
θ +
θ
θ
−
θ
θ +
θ
θ
=
cos(
)
sin(
)
cos(
1)
sin (
1)
sin(
)
cos(
)
sin (
1)
cos(
1)
k
k
k
k
k
k
k
k
θ + θ
θ + θ
+
θ
+
θ
=
−
θ + θ
θ + θ
−
+
θ
+
θ
Therefore, the result is true for n = k + 1 Thus by principle of mathematical induction,
we have
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, holds for all natural numbers |
1 | 1356-1359 | Let the result be true for n = k So
P(k) :
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then
cos
sin
A
sin
cos
k
k
k
k
k
θ
θ
=
−
θ
θ
Now, we prove that the result holds for n = k +1
Now
Ak + 1 =
cos
sin
cos
sin
A A
sin
cos
sin
cos
k
k
k
k
k
θ
θ
θ
θ
⋅
=
−
θ
θ
−
θ
θ
=
cos cos
– sin sin
cos sin
sin cos
sin cos
cos sin
sin sin
cos cos
k
k
k
k
k
k
k
k
θ
θ
θ
θ
θ
θ +
θ
θ
−
θ
θ +
θ
θ
−
θ
θ +
θ
θ
=
cos(
)
sin(
)
cos(
1)
sin (
1)
sin(
)
cos(
)
sin (
1)
cos(
1)
k
k
k
k
k
k
k
k
θ + θ
θ + θ
+
θ
+
θ
=
−
θ + θ
θ + θ
−
+
θ
+
θ
Therefore, the result is true for n = k + 1 Thus by principle of mathematical induction,
we have
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, holds for all natural numbers Example 24 If A and B are symmetric matrices of the same order, then show that AB
is symmetric if and only if A and B commute, that is AB = BA |
1 | 1357-1360 | So
P(k) :
cos
sin
A
sin
cos
θ
θ
=
−
θ
θ
, then
cos
sin
A
sin
cos
k
k
k
k
k
θ
θ
=
−
θ
θ
Now, we prove that the result holds for n = k +1
Now
Ak + 1 =
cos
sin
cos
sin
A A
sin
cos
sin
cos
k
k
k
k
k
θ
θ
θ
θ
⋅
=
−
θ
θ
−
θ
θ
=
cos cos
– sin sin
cos sin
sin cos
sin cos
cos sin
sin sin
cos cos
k
k
k
k
k
k
k
k
θ
θ
θ
θ
θ
θ +
θ
θ
−
θ
θ +
θ
θ
−
θ
θ +
θ
θ
=
cos(
)
sin(
)
cos(
1)
sin (
1)
sin(
)
cos(
)
sin (
1)
cos(
1)
k
k
k
k
k
k
k
k
θ + θ
θ + θ
+
θ
+
θ
=
−
θ + θ
θ + θ
−
+
θ
+
θ
Therefore, the result is true for n = k + 1 Thus by principle of mathematical induction,
we have
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, holds for all natural numbers Example 24 If A and B are symmetric matrices of the same order, then show that AB
is symmetric if and only if A and B commute, that is AB = BA Solution Since A and B are both symmetric matrices, therefore A′ = A and B′ = B |
1 | 1358-1361 | Thus by principle of mathematical induction,
we have
cos
sin
A
sin
cos
n
n
n
n
n
θ
θ
=
−
θ
θ
, holds for all natural numbers Example 24 If A and B are symmetric matrices of the same order, then show that AB
is symmetric if and only if A and B commute, that is AB = BA Solution Since A and B are both symmetric matrices, therefore A′ = A and B′ = B Rationalised 2023-24
MATRICES 71
Let
AB be symmetric, then (AB)′ = AB
But
(AB)′ = B′A′= BA (Why |
1 | 1359-1362 | Example 24 If A and B are symmetric matrices of the same order, then show that AB
is symmetric if and only if A and B commute, that is AB = BA Solution Since A and B are both symmetric matrices, therefore A′ = A and B′ = B Rationalised 2023-24
MATRICES 71
Let
AB be symmetric, then (AB)′ = AB
But
(AB)′ = B′A′= BA (Why )
Therefore
BA = AB
Conversely, if AB = BA, then we shall show that AB is symmetric |
1 | 1360-1363 | Solution Since A and B are both symmetric matrices, therefore A′ = A and B′ = B Rationalised 2023-24
MATRICES 71
Let
AB be symmetric, then (AB)′ = AB
But
(AB)′ = B′A′= BA (Why )
Therefore
BA = AB
Conversely, if AB = BA, then we shall show that AB is symmetric Now
(AB)′ = B′A′
= B A (as A and B are symmetric)
= AB
Hence AB is symmetric |
1 | 1361-1364 | Rationalised 2023-24
MATRICES 71
Let
AB be symmetric, then (AB)′ = AB
But
(AB)′ = B′A′= BA (Why )
Therefore
BA = AB
Conversely, if AB = BA, then we shall show that AB is symmetric Now
(AB)′ = B′A′
= B A (as A and B are symmetric)
= AB
Hence AB is symmetric Example 25 Let
2
1
5
2
2
5
A
, B
, C
3
4
7
4
3
8
−
=
=
=
|
1 | 1362-1365 | )
Therefore
BA = AB
Conversely, if AB = BA, then we shall show that AB is symmetric Now
(AB)′ = B′A′
= B A (as A and B are symmetric)
= AB
Hence AB is symmetric Example 25 Let
2
1
5
2
2
5
A
, B
, C
3
4
7
4
3
8
−
=
=
=
Find a matrix D such that
CD – AB = O |
1 | 1363-1366 | Now
(AB)′ = B′A′
= B A (as A and B are symmetric)
= AB
Hence AB is symmetric Example 25 Let
2
1
5
2
2
5
A
, B
, C
3
4
7
4
3
8
−
=
=
=
Find a matrix D such that
CD – AB = O Solution Since A, B, C are all square matrices of order 2, and CD – AB is well
defined, D must be a square matrix of order 2 |
1 | 1364-1367 | Example 25 Let
2
1
5
2
2
5
A
, B
, C
3
4
7
4
3
8
−
=
=
=
Find a matrix D such that
CD – AB = O Solution Since A, B, C are all square matrices of order 2, and CD – AB is well
defined, D must be a square matrix of order 2 Let
D = a
b
c
d
|
1 | 1365-1368 | Find a matrix D such that
CD – AB = O Solution Since A, B, C are all square matrices of order 2, and CD – AB is well
defined, D must be a square matrix of order 2 Let
D = a
b
c
d
Then CD – AB = 0 gives
2
5
2
1
5
2
3
8
3
4
7
4
a
b
c
d
−
−
= O
or
2
5
2
5
3
0
3
8
3
8
43
22
a
c
b
d
a
c
b
d
+
+
−
+
+
= 0
0
0
0
or
2
5
3
2
5
3
8
43
3
8
22
a
c
b
d
a
c
b
d
+
−
+
+
−
+
−
= 0
0
0
0
By equality of matrices, we get
2a + 5c – 3 = 0 |
1 | 1366-1369 | Solution Since A, B, C are all square matrices of order 2, and CD – AB is well
defined, D must be a square matrix of order 2 Let
D = a
b
c
d
Then CD – AB = 0 gives
2
5
2
1
5
2
3
8
3
4
7
4
a
b
c
d
−
−
= O
or
2
5
2
5
3
0
3
8
3
8
43
22
a
c
b
d
a
c
b
d
+
+
−
+
+
= 0
0
0
0
or
2
5
3
2
5
3
8
43
3
8
22
a
c
b
d
a
c
b
d
+
−
+
+
−
+
−
= 0
0
0
0
By equality of matrices, we get
2a + 5c – 3 = 0 (1)
3a + 8c – 43 = 0 |
1 | 1367-1370 | Let
D = a
b
c
d
Then CD – AB = 0 gives
2
5
2
1
5
2
3
8
3
4
7
4
a
b
c
d
−
−
= O
or
2
5
2
5
3
0
3
8
3
8
43
22
a
c
b
d
a
c
b
d
+
+
−
+
+
= 0
0
0
0
or
2
5
3
2
5
3
8
43
3
8
22
a
c
b
d
a
c
b
d
+
−
+
+
−
+
−
= 0
0
0
0
By equality of matrices, we get
2a + 5c – 3 = 0 (1)
3a + 8c – 43 = 0 (2)
2b + 5d = 0 |
1 | 1368-1371 | Then CD – AB = 0 gives
2
5
2
1
5
2
3
8
3
4
7
4
a
b
c
d
−
−
= O
or
2
5
2
5
3
0
3
8
3
8
43
22
a
c
b
d
a
c
b
d
+
+
−
+
+
= 0
0
0
0
or
2
5
3
2
5
3
8
43
3
8
22
a
c
b
d
a
c
b
d
+
−
+
+
−
+
−
= 0
0
0
0
By equality of matrices, we get
2a + 5c – 3 = 0 (1)
3a + 8c – 43 = 0 (2)
2b + 5d = 0 (3)
and
3b + 8d – 22 = 0 |
1 | 1369-1372 | (1)
3a + 8c – 43 = 0 (2)
2b + 5d = 0 (3)
and
3b + 8d – 22 = 0 (4)
Solving (1) and (2), we get a = –191, c = 77 |
1 | 1370-1373 | (2)
2b + 5d = 0 (3)
and
3b + 8d – 22 = 0 (4)
Solving (1) and (2), we get a = –191, c = 77 Solving (3) and (4), we get b = – 110,
d = 44 |
1 | 1371-1374 | (3)
and
3b + 8d – 22 = 0 (4)
Solving (1) and (2), we get a = –191, c = 77 Solving (3) and (4), we get b = – 110,
d = 44 Rationalised 2023-24
72
MATHEMATICS
Therefore
D =
191
110
77
44
a
b
c
d
−
−
=
Miscellaneous Exercise on Chapter 3
1 |
1 | 1372-1375 | (4)
Solving (1) and (2), we get a = –191, c = 77 Solving (3) and (4), we get b = – 110,
d = 44 Rationalised 2023-24
72
MATHEMATICS
Therefore
D =
191
110
77
44
a
b
c
d
−
−
=
Miscellaneous Exercise on Chapter 3
1 If A and B are symmetric matrices, prove that AB – BA is a skew symmetric
matrix |
1 | 1373-1376 | Solving (3) and (4), we get b = – 110,
d = 44 Rationalised 2023-24
72
MATHEMATICS
Therefore
D =
191
110
77
44
a
b
c
d
−
−
=
Miscellaneous Exercise on Chapter 3
1 If A and B are symmetric matrices, prove that AB – BA is a skew symmetric
matrix 2 |
1 | 1374-1377 | Rationalised 2023-24
72
MATHEMATICS
Therefore
D =
191
110
77
44
a
b
c
d
−
−
=
Miscellaneous Exercise on Chapter 3
1 If A and B are symmetric matrices, prove that AB – BA is a skew symmetric
matrix 2 Show that the matrix B′AB is symmetric or skew symmetric according as A is
symmetric or skew symmetric |
1 | 1375-1378 | If A and B are symmetric matrices, prove that AB – BA is a skew symmetric
matrix 2 Show that the matrix B′AB is symmetric or skew symmetric according as A is
symmetric or skew symmetric 3 |
1 | 1376-1379 | 2 Show that the matrix B′AB is symmetric or skew symmetric according as A is
symmetric or skew symmetric 3 Find the values of x, y, z if the matrix
0
2
A
y
z
x
y
z
x
y
z
=
−
−
satisfy the equation
A′A = I |
1 | 1377-1380 | Show that the matrix B′AB is symmetric or skew symmetric according as A is
symmetric or skew symmetric 3 Find the values of x, y, z if the matrix
0
2
A
y
z
x
y
z
x
y
z
=
−
−
satisfy the equation
A′A = I 4 |
1 | 1378-1381 | 3 Find the values of x, y, z if the matrix
0
2
A
y
z
x
y
z
x
y
z
=
−
−
satisfy the equation
A′A = I 4 For what values of x : [
]
1
2
0
0
1
2
1
2
0
1
2
1
0
2
x
= O |
1 | 1379-1382 | Find the values of x, y, z if the matrix
0
2
A
y
z
x
y
z
x
y
z
=
−
−
satisfy the equation
A′A = I 4 For what values of x : [
]
1
2
0
0
1
2
1
2
0
1
2
1
0
2
x
= O 5 |
1 | 1380-1383 | 4 For what values of x : [
]
1
2
0
0
1
2
1
2
0
1
2
1
0
2
x
= O 5 If
3
1
A
1
2
=
−
, show that A2 – 5A + 7I = 0 |
1 | 1381-1384 | For what values of x : [
]
1
2
0
0
1
2
1
2
0
1
2
1
0
2
x
= O 5 If
3
1
A
1
2
=
−
, show that A2 – 5A + 7I = 0 6 |
1 | 1382-1385 | 5 If
3
1
A
1
2
=
−
, show that A2 – 5A + 7I = 0 6 Find x, if [
]
1
0
2
5
1
0
2
1
4
O
2
0
3
1
x
x
−
−
=
7 |
1 | 1383-1386 | If
3
1
A
1
2
=
−
, show that A2 – 5A + 7I = 0 6 Find x, if [
]
1
0
2
5
1
0
2
1
4
O
2
0
3
1
x
x
−
−
=
7 A manufacturer produces three products x, y, z which he sells in two markets |
1 | 1384-1387 | 6 Find x, if [
]
1
0
2
5
1
0
2
1
4
O
2
0
3
1
x
x
−
−
=
7 A manufacturer produces three products x, y, z which he sells in two markets Annual sales are indicated below:
Market
Products
I
10,000
2,000
18,000
II
6,000
20,000
8,000
Rationalised 2023-24
MATRICES 73
(a) If unit sale prices of x, y and z are ` 2 |
1 | 1385-1388 | Find x, if [
]
1
0
2
5
1
0
2
1
4
O
2
0
3
1
x
x
−
−
=
7 A manufacturer produces three products x, y, z which he sells in two markets Annual sales are indicated below:
Market
Products
I
10,000
2,000
18,000
II
6,000
20,000
8,000
Rationalised 2023-24
MATRICES 73
(a) If unit sale prices of x, y and z are ` 2 50, ` 1 |
1 | 1386-1389 | A manufacturer produces three products x, y, z which he sells in two markets Annual sales are indicated below:
Market
Products
I
10,000
2,000
18,000
II
6,000
20,000
8,000
Rationalised 2023-24
MATRICES 73
(a) If unit sale prices of x, y and z are ` 2 50, ` 1 50 and ` 1 |
1 | 1387-1390 | Annual sales are indicated below:
Market
Products
I
10,000
2,000
18,000
II
6,000
20,000
8,000
Rationalised 2023-24
MATRICES 73
(a) If unit sale prices of x, y and z are ` 2 50, ` 1 50 and ` 1 00, respectively,
find the total revenue in each market with the help of matrix algebra |
1 | 1388-1391 | 50, ` 1 50 and ` 1 00, respectively,
find the total revenue in each market with the help of matrix algebra (b) If the unit costs of the above three commodities are ` 2 |
1 | 1389-1392 | 50 and ` 1 00, respectively,
find the total revenue in each market with the help of matrix algebra (b) If the unit costs of the above three commodities are ` 2 00, ` 1 |
1 | 1390-1393 | 00, respectively,
find the total revenue in each market with the help of matrix algebra (b) If the unit costs of the above three commodities are ` 2 00, ` 1 00 and
50 paise respectively |
1 | 1391-1394 | (b) If the unit costs of the above three commodities are ` 2 00, ` 1 00 and
50 paise respectively Find the gross profit |
1 | 1392-1395 | 00, ` 1 00 and
50 paise respectively Find the gross profit 8 |
1 | 1393-1396 | 00 and
50 paise respectively Find the gross profit 8 Find the matrix X so that
1
2
3
7
8
9
X 4
5
6
2
4
6
−
−
−
=
Choose the correct answer in the following questions:
9 |
1 | 1394-1397 | Find the gross profit 8 Find the matrix X so that
1
2
3
7
8
9
X 4
5
6
2
4
6
−
−
−
=
Choose the correct answer in the following questions:
9 If A = is such that A² = I, then
(A) 1 + α² + βγ = 0
(B) 1 – α² + βγ = 0
(C) 1 – α² – βγ = 0
(D) 1 + α² – βγ = 0
10 |
1 | 1395-1398 | 8 Find the matrix X so that
1
2
3
7
8
9
X 4
5
6
2
4
6
−
−
−
=
Choose the correct answer in the following questions:
9 If A = is such that A² = I, then
(A) 1 + α² + βγ = 0
(B) 1 – α² + βγ = 0
(C) 1 – α² – βγ = 0
(D) 1 + α² – βγ = 0
10 If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
11 |
1 | 1396-1399 | Find the matrix X so that
1
2
3
7
8
9
X 4
5
6
2
4
6
−
−
−
=
Choose the correct answer in the following questions:
9 If A = is such that A² = I, then
(A) 1 + α² + βγ = 0
(B) 1 – α² + βγ = 0
(C) 1 – α² – βγ = 0
(D) 1 + α² – βγ = 0
10 If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
11 If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to
(A) A
(B) I – A
(C) I
(D) 3A
Summary
® A matrix is an ordered rectangular array of numbers or functions |
1 | 1397-1400 | If A = is such that A² = I, then
(A) 1 + α² + βγ = 0
(B) 1 – α² + βγ = 0
(C) 1 – α² – βγ = 0
(D) 1 + α² – βγ = 0
10 If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
11 If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to
(A) A
(B) I – A
(C) I
(D) 3A
Summary
® A matrix is an ordered rectangular array of numbers or functions ® A matrix having m rows and n columns is called a matrix of order m × n |
1 | 1398-1401 | If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
11 If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to
(A) A
(B) I – A
(C) I
(D) 3A
Summary
® A matrix is an ordered rectangular array of numbers or functions ® A matrix having m rows and n columns is called a matrix of order m × n ® [aij]m × 1 is a column matrix |
1 | 1399-1402 | If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to
(A) A
(B) I – A
(C) I
(D) 3A
Summary
® A matrix is an ordered rectangular array of numbers or functions ® A matrix having m rows and n columns is called a matrix of order m × n ® [aij]m × 1 is a column matrix ® [aij]1 × n is a row matrix |
1 | 1400-1403 | ® A matrix having m rows and n columns is called a matrix of order m × n ® [aij]m × 1 is a column matrix ® [aij]1 × n is a row matrix ® An m × n matrix is a square matrix if m = n |
1 | 1401-1404 | ® [aij]m × 1 is a column matrix ® [aij]1 × n is a row matrix ® An m × n matrix is a square matrix if m = n ® A = [aij]m × m is a diagonal matrix if aij = 0, when i ≠ j |
1 | 1402-1405 | ® [aij]1 × n is a row matrix ® An m × n matrix is a square matrix if m = n ® A = [aij]m × m is a diagonal matrix if aij = 0, when i ≠ j ® A = [aij]n × n is a scalar matrix if aij = 0, when i ≠ j, aij = k, (k is some
constant), when i = j |
1 | 1403-1406 | ® An m × n matrix is a square matrix if m = n ® A = [aij]m × m is a diagonal matrix if aij = 0, when i ≠ j ® A = [aij]n × n is a scalar matrix if aij = 0, when i ≠ j, aij = k, (k is some
constant), when i = j ® A = [aij]n × n is an identity matrix, if aij = 1, when i = j, aij = 0, when i ≠ j |
1 | 1404-1407 | ® A = [aij]m × m is a diagonal matrix if aij = 0, when i ≠ j ® A = [aij]n × n is a scalar matrix if aij = 0, when i ≠ j, aij = k, (k is some
constant), when i = j ® A = [aij]n × n is an identity matrix, if aij = 1, when i = j, aij = 0, when i ≠ j ® A zero matrix has all its elements as zero |
1 | 1405-1408 | ® A = [aij]n × n is a scalar matrix if aij = 0, when i ≠ j, aij = k, (k is some
constant), when i = j ® A = [aij]n × n is an identity matrix, if aij = 1, when i = j, aij = 0, when i ≠ j ® A zero matrix has all its elements as zero ® A = [aij] = [bij] = B if (i) A and B are of same order, (ii) aij = bij for all
possible values of i and j |
1 | 1406-1409 | ® A = [aij]n × n is an identity matrix, if aij = 1, when i = j, aij = 0, when i ≠ j ® A zero matrix has all its elements as zero ® A = [aij] = [bij] = B if (i) A and B are of same order, (ii) aij = bij for all
possible values of i and j α
β
γ
−α
Rationalised 2023-24
74
MATHEMATICS
® kA = k[aij]m × n = [k(aij)]m × n
® – A = (–1)A
® A – B = A + (–1) B
® A + B = B + A
® (A + B) + C = A + (B + C), where A, B and C are of same order |
1 | 1407-1410 | ® A zero matrix has all its elements as zero ® A = [aij] = [bij] = B if (i) A and B are of same order, (ii) aij = bij for all
possible values of i and j α
β
γ
−α
Rationalised 2023-24
74
MATHEMATICS
® kA = k[aij]m × n = [k(aij)]m × n
® – A = (–1)A
® A – B = A + (–1) B
® A + B = B + A
® (A + B) + C = A + (B + C), where A, B and C are of same order ® k(A + B) = kA + kB, where A and B are of same order, k is constant |
1 | 1408-1411 | ® A = [aij] = [bij] = B if (i) A and B are of same order, (ii) aij = bij for all
possible values of i and j α
β
γ
−α
Rationalised 2023-24
74
MATHEMATICS
® kA = k[aij]m × n = [k(aij)]m × n
® – A = (–1)A
® A – B = A + (–1) B
® A + B = B + A
® (A + B) + C = A + (B + C), where A, B and C are of same order ® k(A + B) = kA + kB, where A and B are of same order, k is constant ® (k + l ) A = kA + lA, where k and l are constant |
1 | 1409-1412 | α
β
γ
−α
Rationalised 2023-24
74
MATHEMATICS
® kA = k[aij]m × n = [k(aij)]m × n
® – A = (–1)A
® A – B = A + (–1) B
® A + B = B + A
® (A + B) + C = A + (B + C), where A, B and C are of same order ® k(A + B) = kA + kB, where A and B are of same order, k is constant ® (k + l ) A = kA + lA, where k and l are constant ®® If A = [aij]m × n and B = [bjk]n × p, then AB = C = [cik]m × p, where =
(i) A(BC) = (AB)C, (ii) A(B + C) = AB + AC, (iii) (A + B)C = AC + BC
®® If A = [aij]m × n, then A′ or AT = [aji]n × m
(i) (A′)′ = A, (ii) (kA)′ = kA′, (iii) (A + B)′ = A′ + B′, (iv) (AB)′ = B′A′
® A is a symmetric matrix if A′ = A |
1 | 1410-1413 | ® k(A + B) = kA + kB, where A and B are of same order, k is constant ® (k + l ) A = kA + lA, where k and l are constant ®® If A = [aij]m × n and B = [bjk]n × p, then AB = C = [cik]m × p, where =
(i) A(BC) = (AB)C, (ii) A(B + C) = AB + AC, (iii) (A + B)C = AC + BC
®® If A = [aij]m × n, then A′ or AT = [aji]n × m
(i) (A′)′ = A, (ii) (kA)′ = kA′, (iii) (A + B)′ = A′ + B′, (iv) (AB)′ = B′A′
® A is a symmetric matrix if A′ = A ® A is a skew symmetric matrix if A′ = –A |
1 | 1411-1414 | ® (k + l ) A = kA + lA, where k and l are constant ®® If A = [aij]m × n and B = [bjk]n × p, then AB = C = [cik]m × p, where =
(i) A(BC) = (AB)C, (ii) A(B + C) = AB + AC, (iii) (A + B)C = AC + BC
®® If A = [aij]m × n, then A′ or AT = [aji]n × m
(i) (A′)′ = A, (ii) (kA)′ = kA′, (iii) (A + B)′ = A′ + B′, (iv) (AB)′ = B′A′
® A is a symmetric matrix if A′ = A ® A is a skew symmetric matrix if A′ = –A ® Any square matrix can be represented as the sum of a symmetric and a
skew symmetric matrix |
1 | 1412-1415 | ®® If A = [aij]m × n and B = [bjk]n × p, then AB = C = [cik]m × p, where =
(i) A(BC) = (AB)C, (ii) A(B + C) = AB + AC, (iii) (A + B)C = AC + BC
®® If A = [aij]m × n, then A′ or AT = [aji]n × m
(i) (A′)′ = A, (ii) (kA)′ = kA′, (iii) (A + B)′ = A′ + B′, (iv) (AB)′ = B′A′
® A is a symmetric matrix if A′ = A ® A is a skew symmetric matrix if A′ = –A ® Any square matrix can be represented as the sum of a symmetric and a
skew symmetric matrix ® If A and B are two square matrices such that AB = BA = I, then B is the
inverse matrix of A and is denoted by A–1 and A is the inverse of B |
1 | 1413-1416 | ® A is a skew symmetric matrix if A′ = –A ® Any square matrix can be represented as the sum of a symmetric and a
skew symmetric matrix ® If A and B are two square matrices such that AB = BA = I, then B is the
inverse matrix of A and is denoted by A–1 and A is the inverse of B ® Inverse of a square matrix, if it exists, is unique |
1 | 1414-1417 | ® Any square matrix can be represented as the sum of a symmetric and a
skew symmetric matrix ® If A and B are two square matrices such that AB = BA = I, then B is the
inverse matrix of A and is denoted by A–1 and A is the inverse of B ® Inverse of a square matrix, if it exists, is unique —v
v
v
v
v—
1
=
n
ik
ij
jk
j
c
a b
∑
Rationalised 2023-24
MATRICES 75
NOTES
Rationalised 2023-24
76
MATHEMATICS
v All Mathematical truths are relative and conditional |
1 | 1415-1418 | ® If A and B are two square matrices such that AB = BA = I, then B is the
inverse matrix of A and is denoted by A–1 and A is the inverse of B ® Inverse of a square matrix, if it exists, is unique —v
v
v
v
v—
1
=
n
ik
ij
jk
j
c
a b
∑
Rationalised 2023-24
MATRICES 75
NOTES
Rationalised 2023-24
76
MATHEMATICS
v All Mathematical truths are relative and conditional — C |
1 | 1416-1419 | ® Inverse of a square matrix, if it exists, is unique —v
v
v
v
v—
1
=
n
ik
ij
jk
j
c
a b
∑
Rationalised 2023-24
MATRICES 75
NOTES
Rationalised 2023-24
76
MATHEMATICS
v All Mathematical truths are relative and conditional — C P |
1 | 1417-1420 | —v
v
v
v
v—
1
=
n
ik
ij
jk
j
c
a b
∑
Rationalised 2023-24
MATRICES 75
NOTES
Rationalised 2023-24
76
MATHEMATICS
v All Mathematical truths are relative and conditional — C P STEINMETZ v
4 |
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