Q. If A and B are square matrices of the same order and A is non-singular, then for a positive integer $n,\ \left( A^{- 1}BA \right)^{n}$ is equal to
A) $A^{- n}B^{n}A^{n}$
B) $A^{n}B^{n}{A -}^{n}$
C) $A^{- 1}B^{n}A$
D) $n(A^{- 1}BA)$
Explanation
(c)
$\left( A^{- 1}BA \right)^{2} = \left( A^{- 1}BA \right)\left( A^{- 1}BA \right)$
$= A^{- 1}B\left( AA^{- 1} \right)BA$
$= A^{- 1}BIBA = A^{- 1}B^{2}A$
$\left( A^{- 1}BA \right)^{3} = \left( A^{- 1}B^{2}A \right)\left( A^{- 1}BA \right)$
$= A^{- 1}B^{2}\left( AA^{- 1} \right)BA$
$= A^{- 1}B^{3}A$ and so on
$\therefore\left( A^{- 1}BA \right)^{n} = A^{- 1}B^{n}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.