If `Aa n dB` are square matrices of the same order and `A` is non-singular, then for a positive integer `n ,(A^(-1)B A)^n` is equal to `A^(-n)B^n A^n`

Question

If `Aa n dB` are square matrices of the same order and `A` is non-singular, then for a positive integer `n ,(A^(-1)B A)^n` is equal to `A^(-n)B^n A^n`

If `Aa n dB` are square matrices of the same order and `A` is non-singular, then for a positive integer `n ,(A^(-1)B A)^n` is equal to `A^(-n)B^n A^n` b. `A^n B^n A^(-n)` c. `A^(-1)B^n A^` d. `n(A^(-1)B^A)^`
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)`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 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`
`(A^(-1) BA)^(3)=(A^(-1) B^(2)A) (A^(-1) BA)`
`=A^(-1) B^(2) (A A^(-1)) BA`
`=A^(-1) B^(3) A` and so on
`:. (A^(-1) BA)^(n)=A^(-1) B^(n)A`

5 views Answered 2023-02-05 15:29