Q.If $A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & 4 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}$, then $AB$ and $BA$ are defined and equal.

Concept: Matrix Multiplication Compatibility — for $AB$ to be defined, the number of columns in $A$ must equal the number of rows in $B$; the product $AB$ and $BA$ are generally not equal even when both are defined.

Step 1: Check dimensions. $A$ is $2 \times 3$, $B$ is $3 \times 2$. So $AB$ is $2 \times 2$, and $BA$ is $3 \times 3$. Both products are defined, but they are not the same size — hence they cannot be equal.

Step 2: Compute $AB$:

Step 3: Compute $BA$: …

Why this question is a trap

Many students see that both products are defined and assume they must be equal. That’s a natural guess — but matrix multiplication is not commutative. Even when both products exist, they rarely give the same result. Here, the dimensions alone tell you they can’t be equal: $AB$ is a $2 \times 2$ matrix, while $BA$ is $3 \times 3$. Two matrices of different sizes cannot be equal.

Let’s work through the multiplication to confirm.

Step-by-step computation

1. Check compatibility for $AB$

$A$ is $2 \times 3$ (2 rows, 3 columns). $B$ is $3 \times 2$ (3 rows, 2 columns). The inner dimensions match (3 = 3), so $AB$ is defined and will be $2 \times 2$.

2. Compute $AB$

The entry in row $i$, column $j$ of $AB$ is the dot product of row $i$ of $A$ with column $j$ of $B$.

• Row 1 of $A$: $[2, 3, -1]$

• Row 2 of $A$: $[1, 4, 2]$

• Column 1 of $B$: $\begin{bmatrix}2 \\ 4 \\ 2\end{bmatrix}$

• Column 2 of $B$: $\begin{bmatrix}3 \\ 5 \\ 1\end{bmatrix}$

Now:

$(AB)_{11} = 2 \cdot 2 + 3 \cdot 4 + (-1) \cdot 2 = 4 + 12 - 2 = 14$

$(AB)_{12} = 2 \cdot 3 + 3 \cdot 5 + (-1) \cdot 1 = 6 + 15 - 1 = 20$

$(AB)_{21} = 1 \cdot 2 + 4 \cdot 4 + 2 \cdot 2 = 2 + 16 + 4 = 22$

$(AB)_{22} = 1 \cdot 3 + 4 \cdot 5 + 2 \cdot 1 = 3 + 20 + 2 = 25$

So

3. Check compatibility for $BA$

$B$ is $3 \times 2$, $A$ is $2 \times 3$. Inner dimensions match (2 = 2), so $BA$ is defined and will be $3 \times 3$.

4. Compute $BA$

Row $i$ of $B$ with column $j$ of $A$.

Rows of $B$:

Row 1: $[2, 3]$

Row 2: $[4, 5]$

Row 3: $[2, 1]$

Columns of $A$:

Col 1: $\begin{bmatrix}2 \\ 1\end{bmatrix}$, Col 2: $\begin{bmatrix}3 \\ 4\end{bmatrix}$, Col 3: $\begin{bmatrix}-1 \\ 2\end{bmatrix}$

Now:

$(BA)_{11} = 2 \cdot 2 + 3 \cdot 1 = 4 + 3 = 7$

$(BA)_{12} = 2 \cdot 3 + 3 \cdot 4 = 6 + 12 = 18$

$(BA)_{13} = 2 \cdot (-1) + 3 \cdot 2 = -2 + 6 = 4$

$(BA)_{21} = 4 \cdot 2 + 5 \cdot 1 = 8 + 5 = 13$ …

Method: Deciding Whether $AB$ and $BA$ Can Be Equal (Shape-First)

Use this for any claim that two products of the same two matrices are equal. Check orders before computing anything — mismatched sizes settle it instantly.

Steps

Step 1: Write the orders and the product shapes.

For $A$ of order $m\times n$ and $B$ of order $n\times m$, $AB$ is $m\times m$ and $BA$ is $n\times n$. Both are defined, but they are equal-sized only if $m=n$.

Step 2: If the sizes differ, stop — they cannot be equal. …

Common Mistakes

Mistake 1: Assuming that because both $AB$ and $BA$ exist, they are equal.

Why it's wrong: for a $2\times3$ and a $3\times2$ matrix, $AB$ is $2\times2$ while $BA$ is $3\times3$ — different orders, so never equal. Correct approach: compare orders first.

Mistake 2: Rushing into multiplication before checking shapes.

Why it's wrong: the size mismatch already answers the question and saves all the arithmetic. Correct approach: read off the product orders before computing entries. …