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If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n ∈ N.

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(i) Let P(n) be the statement 

P(n) ⇒ Abn = Bn

P(1) ⇒ AB = BA 

∴ P( 1) is true 

Let P(k) be true 

P(k) ⇒ ABk = Bk

P(k +1) ⇒ ABk+1 

= A(BkB) = (ABk)B 

∵ matrix multiplication is associative. 

⇒ (BkA) B ⇒ Bk (AB) = Bk (BA) 

= (BkB) A = Bk+1

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

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