# If A and B are sq

(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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