Question

If $A$ and $B$ are square matrices of the same order and $A$ is non singular, then for a positive integer $n$, $({A^{-1}BA)}^n$ is equal to

• A. $A^{-n}{B}^n{A}^n$
• B. $A^n{B}^n{A}^{-n}$
• C. $A^{-1}{B}^{n}A$
• D. $n\left(A^{-1}{B}^{n}A\right)$

Answer 1

The correct option is C $A^{-1}{B}^{n}A$

Consider, $({A^{-1}BA)}^2=(A^{-1}BA\left)\right(A^{-1}BA)$
$=A^{-1}B\left(A{A}^{-1}\right)BA$
$=A^{-1}BBA=A^{-1}{B}^{2}A$
$\Rightarrow ({A^{-1}BA)}^2=A^{-1}{B}^{2}A$
Now, $({A^{-1}BA)}^3=({A^{-1}BA)}^2\left(A^{-1}BA\right)$
$=\left(A^{-1}{B}^{2}A\right)\left(A^{-1}BA\right)$
$=A^{-1}{B}^2\left(A{A}^{-1}\right)BA$
$=A^{-1}{B}^{3}A$
$\Rightarrow ({A^{-1}BA)}^3=A^{-1}{B}^{3}A$
Hence, $({A^{-1}BA)}^n=A^{-1}{B}^{n}A$