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Prove by Mathematical Induction that (A′)n = (An)′, where n ∈ N for any square matrix A.

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Step 1: 

Let P(n) be (A′)n=(An)′ 

Let P(n) be true for all n∈ N 

For n=1 (A′)1=(A1)′ 

⇒(A)′=A′.

Hence LHS = RHS. 

Hence P(n) is true for n = 1

Step 2: 

Let P(n) be true for n=k. 

Put n = k 

(A′)k=(Ak)′ 

Multiply A′ on both the side 

(A′)k.A′=(Ak)′.A′ 

⇒ A′k.A′ = (Ak)′.A′ 

⇒ A′k.A′= (Ak.A1)′ 

⇒ (A′)k+1 = (Ak+1)′ 

Hence P(n) is true for n = k+1.

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