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 = (Aⁿ)’.

Let P(n) be the statement : (A’)ⁿ = (Aⁿ)’.

Clearly, P(1): (A’)¹ = (A¹)’

⇒ 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^kA)’

We know that by properties of transpose of a matrix:

(AB)^T = B^TA^T

∴ (A^kA)’ = 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 = (Aⁿ)’ is true for all n ∈ N.