Mathematics Matrices 1 Mark
Appeared in: 2026 (2 sets)
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 (iii) and (iv)
✓ Verified CBSE Marking Scheme Official Evaluation Steps & EL10 Insight

Step 1 (Governing Law & Definitions): [0.25 Mark]

Matrix multiplication is not commutative in general, i.e., $AB \neq BA$. Consequently, algebraic identities involving binomial expansion must be expanded using the distributive property of matrices: $A(B+C) = AB + AC$ and $(A+B)(C+D) = AC + AD + BC + BD$.

Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]

  • Analysis of (i): $(A + B)(A - B) = A(A) - A(B) + B(A) - B(B) = A^2 - AB + BA - B^2$. Since $AB$ is not necessarily equal to $BA$, this does not equal $A^2 - B^2$. (False)
  • Analysis of (ii): Matrix multiplication is not commutative for all square matrices. (False)
  • Analysis of (iii): $(A + B)^2 = (A + B)(A + B) = A(A + B) + B(A + B) = A^2 + AB + BA + B^2$. This follows directly from the distributive law. (True)
  • Analysis of (iv): In matrix algebra, the product of two non-zero matrices can be a zero matrix (e.g., if $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, then $AB = 0$ but $A \neq 0$ and $B \neq 0$). (False)

Step 3 (Final Concluded Answer): [0.25 Mark]

Based on the analysis above, only statement (iii) is always true. Therefore, the correct option is (C).

💡 EL10 Concept Insight: Think of matrix multiplication like a specific set of instructions where the order matters—like putting on your socks and then your shoes; if you swap the order, the result is completely different, which is why standard algebra rules don't always apply!