Question

If A and B are square matrices of 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. $n\left(A^{-1}BA\right)$

Answer 1

The correct option is C

$A^{-1}{B}^{n}A$

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

Similarly $(A^{-1}BA{)}^n=A^{-1}{B}^{n}A$