If AB = BA for any two square matrices,

Solution:

Let P(n) : (AB)n = AnBn

So, P(1) : (AB)1 = A1B1

AB = AB

So, P(1) is true.

Let P(n) is true for some k ∈ N

Now,

$(A B)^{k+1}=(A B)^{k}(A B)$ (using (i))

$=A^{k} R^{k}(A B)$

$=A^{k} B^{k-1}(B A) B$

$=A^{k} B^{k-1}(A B) B \quad($ as given $A B=B A)$

$=A^{k} B^{k-1} A B^{2}$

$=A^{k} B^{k-2}(B A) B^{2}$

$=A^{k} B^{k-2} A B B^{2}$

$=A^{k} B^{k-2} A B^{3}$

..........

..........

$=A^{k+1} B^{k+1}$

Hence, P(1) is true and whenever P(k) is true P(k + 1) is true.

Thus, P(n) is true for all n ∈ N.