Let A and B be two non-singular matrices such that ${\left(AB\right)}^k=A^k{B}^k$ for three consecutive positive integral value’s of k.

	Column_I		Column_II
A)	$AB{A}^{-1}$	P)	$A^2$
B)	$BA{B}^{-1}$	Q)	B
C)	$A{B}^2{A}^{-1}$	R)	A
D)	$B{A}^2{B}^{-1}$	S)	$B^2$
		T)	AB

As A and B are invertible matrices $A^{-1},B^{-1}$ both exist, Also, for every positive integer, $A^{n}and{B}^n$ are invertible .
Suppose ${\left(AB\right)}^n=A^n{B}^n$ holds three consecutive positive integer $m, m + 1 and m + 2$. We have ${\left(AB\right)}^m=A^m{B}^m$  ………………..(1)
 ${\left(AB\right)}^{m+1}=A^{m+1}{B}^{m+1}\left(2\right)$
And  ${\left(AB\right)}^{m+2}=A^{m+2}{B}^{m+2}\left(3\right)$
From (2), we have
$$\begin{gathered} A^{m+1}{B}^{m+1}={\left(AB\right)}^{{ }^{m+1}}={\left(AB\right)}^m\left(AB\right) \\ \Rightarrow A^{m}A{B}^{m}B=A^m{B}^{m}AB \end{gathered}$$

[using (1)]
Since $A^m$ and B are invertible matrices
$A{B}^m=B^{m}A$…………………(4)
Similarly, using (2) and(3) we can show that
$A{B}^{m+1}=B^{m+1}A$………………………..(5)
We have  $\left(AB\right)B^m=A{B}^{m+1}=B^{m+1}A$
[using (5)]
$=B\left(B^{m}A\right)=B\left(A{B}^m\right)=\left(BA\right)B^m$
[Using (4)]
Thus,  $\left(AB\right)B^m=\left(BA\right)B^m$
As $B^m$ is an invertible matrix, we can cancel $B^m$ from both the sides to obtain $AB – BA$$\Rightarrow A^{-1}BA=Band{B}^{-1}AB=Aetc.,$