If $A$ and $B$ are square matrices of the same order such that $(A + B) (A - B) = A^{2} - B^{2}$ then $(A B A^{- 1})^{2}$ is equal to

B^{2}

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I

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A^{2} B^{2}

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A^{2}

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Solution

The correct option is A $B^{2}$
Given, $(A + B) (A - B) = A^{2} - B^{2}$
$\Rightarrow A^{2} - A B + B A - B^{2} = A^{2} - B^{2}$
$\Rightarrow A B = B A$
Now, $(A B A^{- 1})^{2} = (B A A^{- 1})^{2} = B^{2}$ ...... $[∵ A A^{- 1} = I]$

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Similar questions

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

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

A

B

A^{2} = A, B^{2} = B, A B = B A = 0

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

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