Question

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

Answer 1

As $A$ and $B$ are invertible matrices $A^{-1},B^{-1}$ both exist. Also, for every positive integer, $A"$ and $B"$ are invertible.
Suppose $(AB{)}^n=A^n{B}^n$ holds for three consecutive positive integers $m,m+1$ and $m+2$. We have
$(AB{)}^m=A^m{B}^m$ (1)
$(AB{)}^{m+1}=A^{m+1}{B}^{m+1}$ (2)
and $(AB{)}^{m+2}=A^{m+2}{B}^{m+2}$ (3)
From (2), we have
$A^{m+1}{B}^{m+1}=(AB{)}^{m+1}=(AB{)}^m\left(AB\right)$
$\Rightarrow A^{m}A{B}^{m}B=A^m{B}^{m}AB$ [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
$AB=BA$
A) $AB{A}^{-1}=BA{A}^{-1}=B$
B) $BA{B}^{-1}=AB{B}^{-1}=A$
C) $A{B}^2{A}^{-1}=B^{2}A{A}^{-1}$
D) $B{A}^2{B}^{-1}=A^{2}B{B}^{-1}$