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If A and B are square matrices of the same order and A is non

Question

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

A^{- n} B^{n} A^{n}

A^{n} B^{n} A^{- n}

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

n (A^{- 1} B^{n} A)

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Liam Discussed

A

and

B

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

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

Consider,

(A^{- 1} B A)^{2} = (A^{- 1} B A) (A^{- 1} B A)

= A^{- 1} B (A A^{- 1}) B A

= A^{- 1} BB A = A^{- 1} B^{2} A

\Rightarrow (A^{- 1} B A)^{2} = A^{- 1} B^{2} A

Now,

(A^{- 1} B A)^{3} = (A^{- 1} B A)^{2} (A^{- 1} B A)

= (A^{- 1} B^{2} A) (A^{- 1} B A)

= A^{- 1} B^{2} (A A^{- 1}) B A

= A^{- 1} B^{3} A

\Rightarrow (A^{- 1} B A)^{3} = A^{- 1} B^{3} A

Hence,

(A^{- 1} B A)^{n} = A^{- 1} B^{n} A

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Question Text	If $A$ and $B$ 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
Answer Type	Text solution:1
Upvotes	150