If AB = BA for any two square matrices, prove by mathematical induction that (AB)ⁿ = Aⁿ Bⁿ.

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 (AB)ⁿ = AⁿBⁿ

Let P(n) be the statement : (AB)ⁿ = AⁿBⁿ

Clearly, P(1): (AB)¹ = A¹B¹

⇒ P(1) : AB = AB

⇒ P(1) is true

Let P(k) be true.

∴ (AB)^k = A^kB^k …(1)

Let’s take P(k+1) now:

∵ (AB)^k+1 = (AB)^k(AB)

⇒ (AB)^k+1 = A^kB^k(AB)

NOTE: As we know that Matrix multiplication is not commutative. So we can’t write directly that A^kB^k(AB) = A^k+1B^k+1

But we are given that AB = BA

∴ (AB)^k+1 = A^kB^k(AB)

⇒ (AB)^k+1 = A^kB^k-1(BAB)

As AB = BA

⇒ (AB)^k+1 = A^kB^k-1(ABB)

⇒ (AB)^k+1 = A^kB^k-1(AB²)

⇒ (AB)^k+1 = A^kB^k-2(BAB²)

⇒ (AB)^k+1 = A^kB^k-2(ABB²)

⇒ (AB)^k+1 = A^kB^k-2(AB³)

We observe that one power of B is decreasing while other is increasing. After certain repetitions decreasing power of B will become I

And at last step:

⇒ (AB)^k+1 = A^kI(AB^k+1)

⇒ (AB)^k+1 = A^kAB^k+1

⇒ (AB)^k+1 = A^k+1B^k+1

Thus P(k+1) is true when P(k) is true.

∴ (AB)ⁿ = Aⁿ Bⁿ ∀ n ∈ N when AB = BA.