If A and B are any two different square matrices of order n with A – B is non-singular $A^{3} = B^{3}$ and A(AB) = B(BA) , then

A^{2} + B^{2} = O

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A^{2} + B^{2} = I

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A^{2} + B^{3} = I

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A^{3} + B^{3} = O

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Solution

The correct option is A $A^{2} + B^{2} = O$
$A^{3} = B^{3} . . . . . (i)$
$A^{2} B = B^{2} A . . . . . . . . . (i i)$
$(A^{2} + B^{2}) (A - B) = 0$
$∵ | A - B | \neq 0$
$A^{2} + B^{2} = 0$

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Similar questions

A^{3} = B^{3}

A^{3} = B^{3}

a^{3} b^{2}, a^{2} b^{3}, b^{2} a^{3}, b^{3} a^{2} .

A

B

2

A^{3} = B^{3}

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

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

(i) (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)

(i i) (a + b)^{3} = a^{3} + b^{3} + 3 a b (a - b)

(i i i) (a - b)^{3} = a^{3} - b^{3} - 3 a b (a + b)

(i v) a^{2} - b^{2} = (a + b) (a - b)

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