Q.Given $A = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{bmatrix}$. Is $(AB)' = B'A'$?

Compute $(AB)'$ and $B'A'$ and compare.

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

So $(AB)'=\begin{bmatrix} 10 & 27 \\ 40 & 102 \end{bmatrix}$.

Now $B'=\begin{bmatrix} 1 & 2 & 1 \\ 4 & 8 & 3 \end{bmatrix}$, $A'=\begin{bmatrix} 2 & 3 \\ 4 & 9 \\ 0 & 6 \end{bmatrix}$: …

The property being checked

The transpose of a product reverses the order of the factors: $(AB)'=B'A'$ (not $A'B'$). We verify it for the given matrices.

Step 1 — compute $AB$

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

• $(1,1)$: $2(1)+4(2)+0(1)=2+8+0=10$
• $(1,2)$: $2(4)+4(8)+0(3)=8+32+0=40$
• $(2,1)$: $3(1)+9(2)+6(1)=3+18+6=27$
• $(2,2)$: $3(4)+9(8)+6(3)=12+72+18=102$

Step 2 — transpose it

Swap rows and columns:

Step 3 — form $B'$ and $A'$ and multiply

Then $B'A'$ is $2\times 2$:

• $(1,1)$: $1(2)+2(4)+1(0)=10$
• $(1,2)$: $1(3)+2(9)+1(6)=27$ …

Method: Verifying the transpose-of-a-product rule

Use this to check or apply $(AB)'=B'A'$.

Steps

Step 1: Compute the product, then transpose.

Find $AB$, then swap its rows and columns to get $(AB)'$.

Step 2: Transpose each factor and multiply in reversed order.

Form $B'$ and $A'$, then compute $B'A'$ — the order must reverse.

Step 3: Compare. …

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; $A'B'$ would be $(3\times2)(2\times3)=3\times3$, the wrong size to equal $(AB)'$. Correct approach: use $(AB)'=B'A'$.

Mistake 2: Transposing by negating or by moving only one row. …