Question

If $\left(AB\right){}^{\prime }=B^{{}^{\prime }}{A}^{{}^{\prime }}$, where $A$ and $B$ are not square matrices, then number of rows in $A$ is equal to number of column in $B$ and number of columns in $A$ is equal to number of rows in $B$.

• A. True
• B. False

Answer 1

The correct option is A True

It is true,

Because,

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

Since, $(AB{)}^{{}^{\prime }}=B^{{}^{\prime }}{A}^{{}^{\prime }}$

$\therefore A_{m\times n}{B}_{p\times q}$ is defined

$\Rightarrow n=p...\left(i\right)$

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

$\Rightarrow (AB{)}^{{}^{\prime }}$ is of order $q\times m...\left(ii\right)$

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 $q\times m...\left(iii\right)$

Also, equality of matrices $(AB{)}^{{}^{\prime }}=B^{{}^{\prime }}{A}^{{}^{\prime }}$, we get the given statement as true.

e.g. If $A$ is order $(3\times 1)$ and $B$ is of order $(1\times 3)$, we get

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