Question

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

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

Answer 1

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

$(A^{-1}BA{)}^2=(A^{-1}BA\left)\right(A^{-1}BA)=A^{1}B(A{A}^{-1})BA=A^{1}BIBA=A^{-1}{B}^{2}A$
$\Rightarrow (A^{-1}BA{)}^3=(A^{-1}{B}^{2}A\left)\right(A^{-1}BA)=A^{-1}{B}^2(A{A}^{-1})BA=A^{-1}{B}^{2}IBA=A^{-1}{B}^{3}Aandsoon$
$\Rightarrow (A^{-1}BA{)}^n=A^{-1}{B}^{n}A$