Q.If possible, find $BA$ and $AB$, where $A = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$, $B = \begin{bmatrix} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{bmatrix}$.

Check compatibility first: $A$ is $2\times 3$ and $B$ is $3\times 2$, so both products exist.

$BA$ ($3\times 2$ times $2\times 3$ $\Rightarrow 3\times 3$):

$AB$ ($2\times 3$ times $3\times 2$ $\Rightarrow 2\times 2$): …

Are the products defined?

A product $XY$ exists only when the columns of $X$ equal the rows of $Y$. Here $A$ is $2\times 3$ and $B$ is $3\times 2$:

• $BA$: $B$'s columns ($2$) $=$ $A$'s rows ($2$) ✓, result $3\times 3$.
• $AB$: $A$'s columns ($3$) $=$ $B$'s rows ($3$) ✓, result $2\times 2$.

Both exist, but they are different sizes — a clear reminder that $AB\neq BA$ in general.

Computing $BA$

Each entry is (a row of $B$) $\cdot$ (a column of $A$). Rows of $B$: $[4,1],[2,3],[1,2]$; columns of $A$: $\begin{bmatrix}2\\1\end{bmatrix},\begin{bmatrix}1\\2\end{bmatrix},\begin{bmatrix}2\\4\end{bmatrix}$.

Row 1: $4(2)+1(1)=9$, $4(1)+1(2)=6$, $4(2)+1(4)=12$.

Row 2: $2(2)+3(1)=7$, $2(1)+3(2)=8$, $2(2)+3(4)=16$.

Row 3: $1(2)+2(1)=4$, $1(1)+2(2)=5$, $1(2)+2(4)=10$.

Computing $AB$ …

Method: Computing both $AB$ and $BA$

Use this when asked to find both products of two rectangular matrices.

Steps

Step 1: Check conformability each way.

For $A_{2\times3}$ and $B_{3\times2}$: $AB$ needs $3=3$ (result $2\times2$); $BA$ needs $2=2$ (result $3\times3$). Both exist but have different sizes.

Step 2: Compute each product by the row-times-column rule. …

Common Mistakes

Mistake 1: Assuming $AB=BA$.

Why it's wrong: here $AB$ is $2\times2$ while $BA$ is $3\times3$ — not even the same size, let alone equal. Correct approach: compute each product separately.

Mistake 2: Mixing up the result's order.

Why it's wrong: the product takes the outer dimensions, so $A_{2\times3}B_{3\times2}$ is $2\times2$ and $B_{3\times2}A_{2\times3}$ is $3\times3$. Correct approach: read the result size from the outer numbers. …