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

For, $(A^{-1}BA{)}^2=(A^{-1}BA\left)\right(A^{-1}BA)$
$=A^{-1}B\left(A{A}^{-1}\right)BA$
$=A^{-1}BIBA=A^{-1}{B}^{2}A$
Similarly, $(A^{-1}BA{)}^3=(A^{-1}BA{)}^2\left(A^{-1}BA\right)=\left(A^{-1}{B}^{2}A\right)\left(A^{-1}BA\right)$
$=A^{-1}{B}^2\left(A{A}^{-1}\right)BA$
$=A^{-1}{B}^{2}IBA$
$=A^{-1}{B}^{3}A$ and so on.
$\therefore (A^{-1}BA{)}^n=A^{-1}{B}^{n}A$