By principle of mathematical induction we say that if a statement P(n) is true for n = 1 and if we assume P(k) to be true for some random natural number k and usnig it if we prove P(k+1) to be true we can say that P(n) is true for all natural numbers.

We are given to prove that (A’)^{n} = (A^{n})’.

Let P(n) be the statement : (A’)^{n} = (A^{n})’.

Clearly, P(1): (A’)^{1} = (A^{1})’

⇒ P(1) : A’ = A’

⇒ P(1) is true

Let P(k) be true.

∴ (A’)^{k} = (A^{k})’ …(1)

Let’s take P(k+1) now:

∵ (A^{k+1})’ = (A^{k}A)’

We know that by properties of transpose of a matrix:

(AB)^{T} = B^{T}A^{T}

∴ (A^{k}A)’ = A’(A^{k})’ = A’(A’)^{k} = (A’)^{k+1}

Thus, (A^{k+1})’ = (A’)^{k+1}

∴ P(k+1) is true.

Hence,

**We can say that: (A’)**^{n} = (A^{n})’ is true for all n ∈ N.