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} = A^{n}B^{n}

**Let P(n) be the statement : (AB)**^{n} = A^{n}B^{n}

Clearly, P(1): (AB)^{1} = A^{1}B^{1}

⇒ P(1) : AB = AB

⇒ P(1) is true

Let P(k) be true.

∴ (AB)^{k} = A^{k}B^{k} …(1)

Let’s take P(k+1) now:

∵ (AB)^{k+1} = (AB)^{k}(AB)

⇒ (AB)^{k+1} = A^{k}B^{k}(AB)

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

But we are given that AB = BA

∴ (AB)^{k+1} = A^{k}B^{k}(AB)

⇒ (AB)^{k+1} = A^{k}B^{k-1}(BAB)

As AB = BA

⇒ (AB)^{k+1} = A^{k}B^{k-1}(ABB)

⇒ (AB)^{k+1} = A^{k}B^{k-1}(AB^{2})

⇒ (AB)^{k+1} = A^{k}B^{k-2}(BAB^{2})

⇒ (AB)^{k+1} = A^{k}B^{k-2}(ABB^{2})

⇒ (AB)^{k+1} = A^{k}B^{k-2}(AB^{3})

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^{k}I(AB^{k+1})

⇒ (AB)^{k+1} = A^{k}AB^{k+1}

⇒ (AB)^{k+1} = A^{k+1}B^{k+1}

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

**∴ (AB)**^{n} = A^{n} B^{n} ∀ n ∈ N when AB = BA.