Question

If $A$ is a square matrix, $B$ is a singular matrix of same order, then for a positive integer $n,(A^{-1}BA{)}^n$ equals

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

Answer 1

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

Consider $n=2$

${\left(A^{-1}BA\right)}^2=\left(A^{-1}BA\right)\left(A^{-1}BA\right)=\left(A^{-1}B\right)\left(A^{-1}A\right)\left(BA\right)=A^{-1}{B}^{2}A$

Again for $n=3$ we have ${\left(A^{-1}BA\right)}^3=\left(A^{-1}{B}^{2}A\right)\left(A^{-1}BA\right)=A^{-1}{B}^{3}A$

Thus generalizing the case

${\left(A^{-1}BA\right)}^n=A^{-1}{B}^{n}A$