Determinants Matrices Question 35
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
Options:
A) $ {A^{-n}}B^{n}A^{n} $
B) $ A^{n}B^{n}{A^{-n}} $
C) $ {A^{-1}}B^{n}A $
D) $ n({A^{-1}}BA) $
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ {{({A^{-1}}BA)}^{2}}=({A^{-1}}BA)({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)({A^{-1}}BA) $
$ ={A^{-1}}B^{2}(A{A^{-1}})BA $
$ ={A^{-1}}B^{2}IBA $
$ ={A^{-1}}B^{3}A $ and so on
$ \Rightarrow {{({A^{-1}}BA)}^{n}}={A^{-1}}B^{n}A $
BETA
