Q. 673.5( 4 Votes )

If A and B are sq

Answer :

(i) Given that A and B are square matrices of the same order.


We know (A + B)2 = (A + B)(A + B)


(A + B)2 = A(A + B) + B(A + B)


(A + B)2 = A2 + AB + BA + B2


For the equation (A + B)2 = A2 + 2AB + B2 to be valid, we need AB = BA.


As the multiplication of two matrices does not satisfy the commutative property in general, AB ≠ BA.


Thus, (A + B)2 ≠ A2 + 2AB + B2.


(ii) Given that A and B are square matrices of the same order.


We know (A – B)2 = (A – B)(A – B)


(A – B)2 = A(A – B) – B(A – B)


(A – B)2 = A2 – AB – BA + B2


For the equation (A – B)2 = A2 – 2AB + B2 to be valid, we need AB = BA.


As the multiplication of two matrices does not satisfy the commutative property in general, AB ≠ BA.


Thus, (A – B)2 ≠ A2 – 2AB + B2.


(iii) Given that A and B are square matrices of the same order.


We have (A + B)(A – B) = A(A – B) + B(A – B)


(A + B)(A – B) = A2 – AB + BA – B2


For the equation (A + B)(A – B) = A2 – B2 to be valid, we need AB = BA.


As the multiplication of two matrices does not satisfy the commutative property in general, AB ≠ BA.


Thus, (A + B)(A – B) ≠ A2 – B2.


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