Q.For any two matrices $A$ and $B$, we have
(A) $AB = BA$
(B) $AB \neq BA$
(C) $AB = O$
(D) None of the above

The key idea is that matrix multiplication is not commutative in general — $AB$ and $BA$ are not always equal, and they may not even be defined if the dimensions don't match.

1. For two matrices $A$ and $B$, the product $AB$ is defined only if the number of columns in $A$ equals the number of rows in $B$. Similarly, $BA$ requires the number of columns in $B$ to equal the number of rows in $A$.
2. Even when both products are defined, they are rarely equal. A counterexample: let $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. Then $AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ but $BA = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. …

The question asks about a property that holds for any two matrices $A$ and $B$. That’s a very strong condition — it must be true no matter what $A$ and $B$ are, as long as they can be multiplied.

Let’s think about what matrix multiplication actually means. When you multiply $A$ (of size $m \times n$) by $B$ (of size $n \times p$), you get an $m \times p$ matrix. For $BA$ to even be defined, the number of columns of $B$ must equal the number of rows of $A$. That’s a separate condition — it doesn’t automatically hold.

So the first thing to notice is: $AB$ and $BA$ are not always both defined. If $A$ is $2 \times 3$ and $B$ is $3 \times 4$, then $AB$ is $2 \times 4$, but $BA$ is impossible because $B$ has 4 columns and $A$ has 2 rows. The question says “for any two matrices” — so even this basic case must be considered.

Now, even when both products are defined (which requires $A$ to be $m \times n$ and $B$ to be $n \times m$), they are rarely equal. Let’s test a simple case.

1. Take a concrete counterexample. Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. Compute $AB$:

Compute $BA$:

Clearly $AB \neq BA$. So option (A) is false.

2. Check option (B): “$AB \neq BA$ for any two matrices”. …

Method: Test a "for any two matrices" claim

Use this reasoning pattern for questions asserting a property holds for all matrices (e.g. "$AB = BA$ for any $A,B$").

Steps

Step 1: Read "for any" as "must hold in every case, with no exception".

A universal claim is false the moment a single valid counterexample exists.

Step 2: Recall the key facts about matrix multiplication. …

Common Mistakes

Mistake 1: Choosing "$AB \neq BA$" because multiplication is non-commutative.

Why it's wrong: "non-commutative in general" means not always equal, not always unequal — $A=I$ or $A=O$ commute with any $B$. Correct approach: reject a universal-inequality claim with a commuting example.

Mistake 2: Choosing "$AB = BA$" as a general truth.

Why it's wrong: a simple counterexample (e.g. $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&1\\0&0\end{bmatrix}$) gives $AB \neq BA$. Correct approach: one counterexample kills the claim. …