Q.If $A = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 3 & -6 \end{bmatrix}$, verify that $(AB)' = B'A'$.

Concept: Matrix Transpose — the transpose of a product equals the product of transposes in reverse order.

First, compute $AB$:

Now take its transpose:

Next, compute $B'A'$:

$B' = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix}, \quad A' = \begin{bmatrix} -2 & 4 & 5 \end{bmatrix}$ …

Why this works: the logic of the transpose rule

The identity $(AB)' = B'A'$ is not just a notational trick — it comes from how matrix multiplication and transposition interact. When you multiply $A$ (a column vector) by $B$ (a row vector), you get a $3 \times 3$ matrix. Transposing that matrix swaps rows and columns. On the other side, $B'$ becomes a column, $A'$ becomes a row, and multiplying them in reverse order gives the same result. The reversal is essential because the inner dimensions must match.

Let’s verify it concretely.

1. Compute $AB$ $A$ is $3 \times 1$, $B$ is $1 \times 3$, so $AB$ is $3 \times 3$. Each entry $(AB)_{ij}$ is $A_{i1} \cdot B_{1j}$.

2. Find $(AB)'$ Transpose means swap rows and columns: row $i$ becomes column $i$.

3. Compute $B'$ and $A'$ separately $B$ is $1 \times 3$, so $B'$ is $3 \times 1$:

$A$ is $3 \times 1$, so $A'$ is $1 \times 3$:

4. Multiply $B'$ and $A'$ $B'$ is $3 \times 1$, $A'$ is $1 \times 3$, so $B'A'$ is $3 \times 3$. Each entry $(B'A')_{ij} = (B')_{i1} \cdot (A')_{1j}$. …

Method: Verifying the reversal law $(AB)' = B'A'$

Use this whenever you must confirm the transpose-of-a-product identity for given matrices.

Steps

Step 1: Compute the product $AB$, then transpose it.

Form $AB$ by the row-by-column rule and flip rows and columns to get $(AB)'$.

Step 2: Transpose each factor and multiply in reverse order. …

Common Mistakes

Mistake 1: Writing $(AB)' = A'B'$ instead of $B'A'$.

Why it's wrong: the transpose of a product reverses the factor order; keeping the order can even make the product undefined (here it would be $1\times1$, not $3\times3$). Correct approach: reverse the factors — $B'A'$.

Mistake 2: Transposing a factor incorrectly. …