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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