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

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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 = AkBk …(1)

Let’s take P(k+1) now:

∵ (AB)k+1 = (AB)k(AB)

⇒ (AB)k+1 = AkBk(AB)

NOTE: As we know that Matrix multiplication is not commutative. So we can’t write directly that AkBk(AB) = Ak+1Bk+1

But we are given that AB = BA

∴ (AB)k+1 = AkBk(AB)

⇒ (AB)k+1 = AkBk-1(BAB)

As AB = BA

⇒ (AB)k+1 = AkBk-1(ABB)

⇒ (AB)k+1 = AkBk-1(AB2)

⇒ (AB)k+1 = AkBk-2(BAB2)

⇒ (AB)k+1 = AkBk-2(ABB2)

⇒ (AB)k+1 = AkBk-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 = AkI(ABk+1)

⇒ (AB)k+1 = AkABk+1

⇒ (AB)k+1 = Ak+1Bk+1

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

∴ (AB)n = An Bn ∀ n ∈ N when AB = BA.

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