Question

If $A$ and $B$ are square matrices of 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) $AB=0 \Rightarrow A=0$ or $B=0$

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

Hint:

Matrix multiplication is generally \textbf{not commutative}. Always remember that \(AB \ne BA\) in most cases.

Answer 1

The Correct Option is C

Step 1: Check statement (i).
Consider the expression
\[ (A+B)(A-B) \] Expanding the product we obtain
\[ (A+B)(A-B) = A(A-B) + B(A-B) \] \[ = A^2 - AB + BA - B^2 \] Thus
\[ (A+B)(A-B) = A^2 - AB + BA - B^2 \] For this to become
\[ A^2 - B^2 \] we must have
\[ AB = BA \] But matrix multiplication is generally not commutative.
Therefore, statement (i) is not always true.
Step 2: Check statement (ii).
Statement (ii) says
\[ AB = BA \] However, in matrix algebra, multiplication is generally non-commutative.
Example:
\[ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \] Then
\[ AB \neq BA \] Thus, statement (ii) is not always true.
Step 3: Check statement (iii).
Consider
\[ (A+B)^2 \] \[ (A+B)^2 = (A+B)(A+B) \] Expanding the multiplication
\[ = A(A+B) + B(A+B) \] \[ = A^2 + AB + BA + B^2 \] This identity is always valid for matrices of the same order.
Thus, statement (iii) is always true.
Step 4: Check statement (iv).
Statement (iv) states
\[ AB = 0 \Rightarrow A = 0 \text{ or } B = 0 \] This property holds for real numbers but not necessarily for matrices.
Example:
\[ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \] Here
\[ AB = 0 \] but
\[ A \neq 0 \quad \text{and} \quad B \neq 0 \] Thus, statement (iv) is false.
Step 5: Final conclusion.
Only statement (iii) is always true for square matrices of the same order.
Final Answer: \( \boxed{\text{Only (iii)}} \)