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) $AB=0\Rightarrow A=0\text{ or }B=0$}

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

Hint:

Never assume \( AB = BA \) unless the question explicitly states that the matrices commute.
Standard algebraic identities like \( (a+b)^2 = a^2 + 2ab + b^2 \) only apply to matrices if \( AB = BA \).
If you suspect a statement is false, try to think of a simple \( 2 \times 2 \) matrix counterexample involving zeros.

Answer 1

The Correct Option is C

Concept:
• Matrix multiplication is generally non-commutative, meaning \( AB \neq BA \) in most cases.
• Algebraic identities for matrices must respect the order of multiplication.
• A statement is "always true" only if it holds for all possible square matrices \( A \) and \( B \) of the same order.

Step 1: Analyze statement (i): \( (A + B)(A - B) = A^2 - B^2 \)
Let's expand the left-hand side using the distributive property:
\[ (A + B)(A - B) = A(A - B) + B(A - B) \]
\[ = A^2 - AB + BA - B^2 \]
This equals \( A^2 - B^2 \) only if \( -AB + BA = 0 \), which implies \( AB = BA \).
Since matrices are not always commutative, this is not always true.

Step 2: Analyze statement (ii): \( AB = BA \)
This is the commutative property of multiplication.
For matrices, this is a special case and does not hold for all matrices.
Therefore, it is not "always" true.

Step 3: Analyze statement (iii): \( (A + B)^2 = A^2 + AB + BA + B^2 \)
Expanding the square:
\[ (A + B)^2 = (A + B)(A + B) \]
\[ = A(A + B) + B(A + B) \]
\[ = A^2 + AB + BA + B^2 \]
This expansion is derived directly from the distributive laws of matrix addition and multiplication.
It does not assume commutativity, so it is always true.

Step 4: Analyze statement (iv): \( AB = 0 \Rightarrow A = 0 \text{ or } B = 0 \)
In matrix algebra, the product of two non-zero matrices can be a zero matrix (these are called zero-divisors).
For example, if \( A = \begin{bmatrix} 0 & 1 0 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \), then \( AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \), but neither \( A \) nor \( B \) is zero.
So, this statement is not always true.

Step 5: Conclusion
Only statement (iii) is always true.