Q.If $A = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}$, then find $AB$, $BA$. Show that $AB \neq BA$.

Concept: Matrix Multiplication Compatibility — two matrices can be multiplied only when the number of columns in the first equals the number of rows in the second. The product $AB$ is defined if $A$ is $2 \times 3$ and $B$ is $3 \times 2$; $BA$ is also defined since $B$ is $3 \times 2$ and $A$ is $2 \times 3$, but the results are different sizes.

Step 1 — Compute $AB$:

$A$ is $2 \times 3$, $B$ is $3 \times 2$, so $AB$ is $2 \times 2$.

$AB = \begin{bmatrix} 1(2)+(-2)(4)+3(2) & 1(3)+(-2)(5)+3(1) \\ -4(2)+2(4)+5(2) & -4(3)+2(5)+5(1) \end{bmatrix}$

$= \begin{bmatrix} 2-8+6 & 3-10+3 \\ -8+8+10 & -12+10+5 \end{bmatrix} = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$.

Step 2 — Compute $BA$:

$B$ is $3 \times 2$, $A$ is $2 \times 3$, so $BA$ is $3 \times 3$.

$BA = \begin{bmatrix} 2(1)+3(-4) & 2(-2)+3(2) & 2(3)+3(5) \\ 4(1)+5(-4) & 4(-2)+5(2) & 4(3)+5(5) \\ 2(1)+1(-4) & 2(-2)+1(2) & 2(3)+1(5) \end{bmatrix}$

$= \begin{bmatrix} 2-12 & -4+6 & 6+15 \\ 4-20 & -8+10 & 12+25 \\ 2-4 & -4+2 & 6+5 \end{bmatrix} = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}$.

Step 3 — Compare:

$AB$ is $2 \times 2$, $BA$ is $3 \times 3$ — they are not even the same size, so $AB \neq BA$.

We start with the core idea: two matrices can be multiplied only when the number of columns in the first equals the number of rows in the second. This is the compatibility condition. For $A$ and $B$ given, $A$ is $2 \times 3$ (2 rows, 3 columns) and $B$ is $3 \times 2$ (3 rows, 2 columns). So $AB$ is defined and will be $2 \times 2$. Similarly, $BA$ is also defined (since $B$ has 2 columns and $A$ has 2 rows) and will be $3 \times 3$. Two matrices of different sizes can never be equal, so $AB \neq BA$ is immediate. But let’s compute both to see the actual numbers.

1. Compute $AB$

   $A$ is $2 \times 3$, $B$ is $3 \times 2$. The product $AB$ will be $2 \times 2$.

   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 $A$: $[1, -2, 3]$

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

     $(AB)_{11} = 1\cdot 2 + (-2)\cdot 4 + 3\cdot 2 = 2 - 8 + 6 = 0$

   • Row 1 of $A$ with column 2 of $B$: $\begin{bmatrix}3 \\ 5 \\ 1\end{bmatrix}$

     $(AB)_{12} = 1\cdot 3 + (-2)\cdot 5 + 3\cdot 1 = 3 - 10 + 3 = -4$

   • Row 2 of $A$: $[-4, 2, 5]$

     With column 1: $(-4)\cdot 2 + 2\cdot 4 + 5\cdot 2 = -8 + 8 + 10 = 10$

   • Row 2 with column 2: $(-4)\cdot 3 + 2\cdot 5 + 5\cdot 1 = -12 + 10 + 5 = 3$

   So

2. Compute $BA$

   $B$ is $3 \times 2$, $A$ is $2 \times 3$, so $BA$ is $3 \times 3$.

   Each entry is the dot product of a row of $B$ with a column of $A$.

   Rows of $B$:

   Row 1: $[2, 3]$

   Row 2: $[4, 5]$

   Row 3: $[2, 1]$

   Columns of $A$:

   Col 1: $\begin{bmatrix}1 \\ -4\end{bmatrix}$, Col 2: $\begin{bmatrix}-2 \\ 2\end{bmatrix}$, Col 3: $\begin{bmatrix}3 \\ 5\end{bmatrix}$

   Compute systematically:

   • $(BA)_{11}$: row 1 of $B$ with col 1 of $A$: $2\cdot 1 + 3\cdot (-4) = 2 - 12 = -10$

   • $(BA)_{12}$: row 1 with col 2: $2\cdot (-2) + 3\cdot 2 = -4 + 6 = 2$

   • $(BA)_{13}$: row 1 with col 3: $2\cdot 3 + 3\cdot 5 = 6 + 15 = 21$

   • $(BA)_{21}$: row 2 with col 1: $4\cdot 1 + 5\cdot (-4) = 4 - 20 = -16$

   • $(BA)_{22}$: row 2 with col 2: $4\cdot (-2) + 5\cdot 2 = -8 + 10 = 2$

   • $(BA)_{23}$: row 2 with col 3: $4\cdot 3 + 5\cdot 5 = 12 + 25 = 37$

   • $(BA)_{31}$: row 3 with col 1: $2\cdot 1 + 1\cdot (-4) = 2 - 4 = -2$

   • $(BA)_{32}$: row 3 with col 2: $2\cdot (-2) + 1\cdot 2 = -4 + 2 = -2$

   • $(BA)_{33}$: row 3 with col 3: $2\cdot 3 + 1\cdot 5 = 6 + 5 = 11$

   So

3. Compare $AB$ and $BA$ $AB$ is $2 \times 2$, $BA$ is $3 \times 3$. They don’t even have the same shape, so they cannot be equal. Even if we ignore size, the entries are completely different. This illustrates a fundamental fact: matrix multiplication is not commutative — in general, $AB \neq BA$, and often one product may not even be defined when the other is.

Method: Comparing $AB$ and $BA$ for non-commutativity

Use this when asked to compute both $AB$ and $BA$ and show they are unequal.

Steps

Step 1: Determine the order of each product first.

For $A_{m\times n}$ and $B_{n\times m}$, $AB$ is $m\times m$ and $BA$ is $n\times n$. If $m\neq n$ the products already differ in size, which alone proves $AB\neq BA$.

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

Even when both are defined, evaluate them fully to display the actual entries.

Step 3: Compare and conclude.

Different orders (or different entries) confirm matrix multiplication is not commutative.

Common Mistakes

Mistake 1: Assuming $AB = BA$ as with numbers.

Why it's wrong: matrix multiplication is generally non-commutative; here $AB$ is $2\times2$ while $BA$ is $3\times3$. Correct approach: compute both and compare orders/entries.

Mistake 2: Believing both products must have the same order.

Why it's wrong: $A_{2\times3}B_{3\times2}$ gives $2\times2$, but $B_{3\times2}A_{2\times3}$ gives $3\times3$. Correct approach: read the outer dimensions for each order separately.

Mistake 3: Sign errors accumulating in the dot products.

Why it's wrong: entries like $1(2)+(-2)(4)+3(2)$ need careful sign tracking. Correct approach: write each term with its sign before summing.