If $A$ and $B$ are square matrices of the same order such that $A^{2} = A, B^{2} = B, A B = B A = 0$ , then

A B^{2} = 0

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

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

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none of these.

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Solution

The correct options are
A $(A + B)^{2} = A + B$
C $A B^{2} = 0$
$A B^{2} = (A B) B = 0 B = 0$
$(A + B)^{2} = A^{2} + A B + B A + B^{2}$
$= A + 0 + 0 + B = A + B$
Also, $(A + B)^{2} = A + B \neq A - B$

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Similar questions

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

A

B

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

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

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