If A and B are square matrices of the same order, explain, why in general
(i) (A + B)2 ≠ A2 + 2AB + B2
(ii) (A – B)2 ≠ A2 – 2AB + B2
(iii) (A + B)(A – B) = A2 – B2
(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.