(i) Let P(n) be the statement
P(n) ⇒ Abn = BnA
P(1) ⇒ AB = BA
∴ P( 1) is true
Let P(k) be true
P(k) ⇒ ABk = BkA
P(k +1) ⇒ ABk+1
= A(BkB) = (ABk)B
∵ matrix multiplication is associative.
⇒ (BkA) B ⇒ Bk (AB) = Bk (BA)
= (BkB) A = Bk+1 A
Hence P(k + 1) is true.
∴ P(n) is true for all the values of n ∈ N
(ii) Let P(n) be the statement
P(n) ⇒ (AB)n = AnBn
P(1) = (AB)1 = AB
∴ P(1) is true
Let P(k) be true
P(k) ⇒ (AB)k = AkBkP(k+1)
⇒ (AB)k+1 = (AB)kAB = (AkBk) AB
= Ak (BkA)B = Ak (ABk) B
(∵ ABn = BnA whenever Ab = BA)
= Ak+1 Bk+1
∴ P (k+1) is true
Hence P(n) is true for all the natures of n ∈ N