If A and B are matrices of the same order, then ${(A + B)}^{2} = A^{2} + 2 A B + B^{2}$ is possible, iff

AB= I

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BA= I

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AB= BA

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none of these

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Solution

The correct option is C AB= BA
${(A + B)}^{2} = (A + B) (A + B)$
$= A^{2} + A B + B A + B^{2}$
If $A B = B A$ then
${(A + B)}^{2} = A^{2} + 2 A B + B^{2}$

Suggest Corrections

Similar questions

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

B = - A^{- 1} B A

(A + B)^{2}

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