If ${(AB)^\prime } = {B^\prime }{A^\prime }$, where A and B are not square matrices,

then number of rows in A is equal to number of columns in B and number of columns in A is equal to number of rows in B.

Correct Answer True

Let A is of order $m \times n$ and B is of order $p \times q$.

Since, ${(AB)^\prime } = {B^\prime }{A^\prime }$ …..(i)
$\therefore$ ${A_{(m \times n)}}{B_{(p \times q)}}$ is defined $\Rightarrow$ $n = p$

and AB is of order $m \times q$.

$\Rightarrow$ ${(AB)^\prime }$ is of order $q \times m$ ….(ii)
Also, ${B^\prime }$ is of order $q \times p$ and ${A^\prime }$ is of order $n \times m$

$\therefore$ ${B^\prime }{A^\prime }$ is defined $\Rightarrow$ $p = n$

and ${B^\prime }{A^\prime }$ is of order $q \times m$.

Also, equality of matrices ${(AB)^\prime } = {B^\prime }{A^\prime }$,

Therefore we get the given statement as true.
e.g., If A is of order $(3 \times 1)$ and B is of order $(1 \times 3)$,

we get
Order of ${(AB)^\prime } =$ Order of $\left( {{B^\prime }{A^\prime }} \right) = 3 \times 3$