If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that ${(AB)^n} = {A^n}{B^n}$for all $n \in N$.

.: Given AB = BA,

To prove

(1) $A{B^n} = {B^n}A$ and (2) ${(AB)^n} = {A^n}{B^n}$ $\forall n \in N$ ...(i)
We will prove it by mathematical induction.

(1) Given that AB = BA ...(ii)

We have to prove $A{B^n} = {B^n}A$
For n = 1, $A{B^1} = {B^1}A$
$\Rightarrow$ AB = BA, which is true [from (ii)]

Let it be true for n = ABm = BmA ...(iii)

Then, for n = m + 1,

$A{B^{m + 1}} = A({B^m}B) = (A{B^m})B = ({B^m}A)B$

[using (iii)]
$= {B^m}(AB) = {B^m}(BA)$

[using (ii)]
$= ({B^m}B)A = {B^{m + 1}}A.$So, it is true for n = m + 1

$\therefore$ $A{B^n} = {B^n}A.$

(2) For $n = 1,{(AB)^1} = {A^1}{B^1}$
$\Rightarrow$ $AB = BA$ which is true for n = 1
Let (i) be true for a positive integer n = m.
i.e., $(AB)^m = A^mB^m$

....(iv)

then for n = m + 1, $(AB)^m + 1 = (AB)^m (AB)$

$= (A^mB^m) (AB)$ (from (iv))

$= A^m(B^mA)B$
$= A^m(AB^m)B [AB^n = B^nA$

$\forall n,$whenever AB = BA]

= $(A^mA)(B^mB) = A^m + 1B^m + 1$
So, it holds for n = m + 1
Hence. (AB$)^n - A^nB^n­­\forall n \in N.$