If $A$ and $B$ are square matrices of same order such that $A B = A$ and $B A = B$ , then

A^{2} = A

and

B^{2} = B

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A^{2} = B

and

B^{2} = A

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A B = A B

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A

and

B

are periodic matrices

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Solution

The correct option is A $A^{2} = A$ and $B^{2} = B$
$A B = A \to (1)$
$B A = B \to (2)$
Post multiply (1) with $A$
$A B A = A^{2}$
$⟹ A (B A) = A^{2}$
$⟹ A B = A^{2} = A$

Similarly, $B^{2} = B$

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Similar questions

A

B

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

Show that if A and B are square matrices such that AB = BA, then $(A + B)^{2} = A^{2} + 2 A B + B^{2}$ .

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