Question

If $A$ and $B$ are square matrices of the same order, then which of the following statements is/are always true?
(i) $(A + B)(A - B) = A^2 - B^2$
(ii) $AB = BA$
(iii) $(A + B)^2 = A^2 + AB + BA + B^2$
(iv) $AB = 0 \implies A = 0$ or $B = 0$

• A. Only (i) and (iii)
• B. Only (ii) and (iii)
• C. Only (iii)
• D. Only (iii) and (iv)

Hint:

Never treat matrix algebra like scalar algebra! Always keep the matrix order intact: $(A+B)^2 = A^2 + AB + BA + B^2$. It only becomes $A^2 + 2AB + B^2$ if you are explicitly told that $AB = BA$.

Answer 1

The Correct Option is C

Concept: Matrix multiplication is fundamentally different from real number algebra because matrix multiplication is generally non-commutative, meaning $AB \neq BA$ in most cases. We must preserve order when expanding algebraic terms.

Step 1: Analyze statement (i).
Let us expand the left-hand side expansion keeping multiplication order strict: \[ (A + B)(A - B) = A(A - B) + B(A - B) = A^2 - AB + BA - B^2 \] For this to equal $A^2 - B^2$, the terms $-AB + BA$ must cancel out, which requires $AB = BA$. Since matrix multiplication is non-commutative, this statement is not always true.

Step 2: Analyze statement (ii).
As established, matrix multiplication is non-commutative in general ($AB \neq BA$). There are specific matrices that commute, but it is not always true for any arbitrary square matrices.

Step 3: Analyze statement (iii).
Let us expand $(A + B)^2$: \[ (A + B)^2 = (A + B)(A + B) = A(A + B) + B(A + B) = A^2 + AB + BA + B^2 \] This algebraic expansion perfectly preserves the multiplication order without assuming commutativity. Therefore, this statement is always true.

Step 4: Analyze statement (iv).
In matrix algebra, it is possible to have two non-zero matrices whose product is a zero matrix (these are called divisors of zero). For example: Let $A = \begin{bmatrix} 0 & 1
0 & 0 \end{bmatrix} \neq 0$ and $B = \begin{bmatrix} 1 & 0
0 & 0 \end{bmatrix} \neq 0$. Their product is $AB = \begin{bmatrix} 0 & 0
0 & 0 \end{bmatrix} = 0$. Hence, $AB = 0$ does not necessarily imply $A=0$ or $B=0$. This statement is not always true. Thus, only statement (iii) is true, matching option (C).