If A and B are square matrices of the same order, explain, why in gene

1.	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
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.

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

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.