Q.If $A=\begin{pmatrix}1&0\\2&-1\end{pmatrix}$ and $B=\begin{pmatrix}3&1\\0&2\end{pmatrix}$, verify that $(AB)'=B'A'$.

Verifying this identity means computing $AB$ and its transpose on one side, and $B'A'$ on the other, then checking the two results are identical.

Step 1 — compute $AB$:

Row 1: $(1)(3)+(0)(0)=3$, $(1)(1)+(0)(2)=1$.

Row 2: $(2)(3)+(-1)(0)=6$, $(2)(1)+(-1)(2)=0$.

So $AB=\begin{pmatrix}3&1\\6&0\end{pmatrix}$, and transposing, $(AB)'=\begin{pmatrix}3&6\\1&0\end{pmatrix}$.

Step 2 — compute $B'A'$: first, $B'=\begin{pmatrix}3&0\\1&2\end{pmatrix}$ and $A'=\begin{pmatrix}1&2\\0&-1\end{pmatrix}$.

Row 1 of $B'A'$: $(3)(1)+(0)(0)=3$, $(3)(2)+(0)(-1)=6$.

Row 2: $(1)(1)+(2)(0)=1$, $(1)(2)+(2)(-1)=2-2=0$.

So $B'A'=\begin{pmatrix}3&6\\1&0\end{pmatrix}$.

Comparing, $(AB)'=\begin{pmatrix}3&6\\1&0\end{pmatrix}$ exactly equals $B'A'=\begin{pmatrix}3&6\\1&0\end{pmatrix}$ — the identity $(AB)'=B'A'$ is verified for these specific matrices.

Independent check: this identity is a general property of transposes (the transpose of a product reverses the order of the factors), so the numerical agreement here is exactly what theory predicts; as a further arithmetic cross-check, the $(2,1)$ entry of $(AB)'$ is the $(1,2)$ entry of $AB$, which was computed as $1$ above — and it matches the $(2,1)$ entry of $B'A'$, also $1$, confirming consistency across both computations.

Writing $A'B'$ instead of $B'A'$ on the right-hand side — the order must reverse when transposing a product; also forgetting to transpose $AB$ itself before comparing.