Question
If AB = BA for any two sqaure matrices, prove by mathematical induction that (AB)n = AnBn.

Answer

Let P(n) : (AB)n = AnBn

$\therefore$ P(1) : (AB)1 = A1B1 ⇒ AB = AB

So, P(1) is true.

Now, P(k) : (AB)k = AkBk $\text{k}\in\text{N}$

So, P(k) is true, whenever P(k + 1) is true.

$\therefore$ P(k + 1) ⇒ (AB)k+1 = Ak+1.Bk+1

$\therefore$ P(k + 1 : AB)k+1 = Ak+1Bk+1

⇒ Ak.Bk.AB

⇒ Ak.BkBA ⇒ AkBk+1A

⇒ AkA.Bk+1 ⇒ Ak+1Bk+1

⇒ (A.B)k+1 = Ak+1Bk+1

So, P(k+1) is true for all $\text{n}\in\text{N},$ whenever P(k) is true.

By mathematical induction (AB) = AnBn is true for all $\text{n}\in\text{N}.$

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