Question 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. 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 15 10 1 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. For any matrices A and B of suitable orders, we have (i) (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 23 3 2A and B 1 2 44 2 0   − = =      , verify that (i) (A′)′ = A, (ii) (A + B)′ = A′ + B′, (iii) (kB)′ = kB′, where k is any constant. Solution (i) We have A = ( ) 3 4 3 3 2 3 3 2A 3 2 A A 4 2 0 4 2 02 0        ′′ ′⇒ = ⇒ = =           Thus (A′)′ = A (ii) We have A = 3 3 2 , 4 2 0       B = 2 1 2 5 3 1 4A B1 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 3 1 4 4 4     −     Thus (A + B)′ = A′ + B′ (iii) We have kB = k 2 1 2 2 2 1 2 4 2 4

− −   =       Then (kB)′ = 2 2 1 2 1 2 B 2 4 2 4 k k

k k         ′− = − =           Thus (kB)′ = kB′ Example 21 If [ ] A 4 , B 1 3 6 −   = = −     , verify that (AB) ′ = B′A′. Solution We have A = [ ] 4 , B 1 3 6 −    = −     then AB = [ ] 4 1 3 6 −    −     = 2 6 12 4 12 24 5 15 30 − −   −   −  Now A′ = [-2 4 5] , 1    ′ =    −  B′A′ = [ ] 1 2 4 5 3 2 4 5 6 12 15 (AB) 6 12 24 30 −        ′− = − =       − − −    Clearly (AB)′ = B′A′