Question

If A and B are square matrices of the same order and A is nonsingular, 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 $A^{-1}{B}^{n}A$

Given, theme=atrix $A$ is non singular. Therefore, $A^{-1}$ exists.

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

$=\left[A^{-1}BA{A}^{-1}BA\right](A^{-1}BA{)}^{n-2}$

` $=\left(A^{-1}{B}^{2}A\right)\left(A^{-1}BA\right)(A^{-1}BA{)}^{n-3}$

$=\left(A^{-1}{B}^{3}A\right)\left(A^{-1}BA\right)(A^{-1}BA{)}^{n-4}$

$=\left(A^{-1}{B}^{4}A\right)(A^{-1}BA{)}^{n-4}$ and so on ......

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

Hence, the correct option is $\left(C\right)$