Question

If A is $3\times 4$ matrix and B is a matrix such that A’B and B’A are both defined, then order of B is
A. $4\times 4$
B. $3\times 3$
C. $3\times 4$
D. $4\times 3$

Answer 1

Hint: Let there be matrix $X$ and $Y$.
$\begin{aligned} & X={{\left[ {{x}_{ij}} \right]}_{m\times n}},\text{ 1}\le \text{i}\le m,\text{ 1}\le \text{j}\le n \\\ & Y={{\left[ {{y}_{ij}} \right]}_{p\times q}},\text{ 1}\le \text{i}\le \text{p, 1}\le \text{j}\le \text{q} \\\ \end{aligned}$

Complete step-by-step answer:
Here, $m$ is the number of rows and $n$ is the number of columns in matrix $X$ respectively . Similarly, $p$ is the number of rows and $q$ is the number of columns in the matrix $Y$. So, two matrices $X$ and $Y$ can be multiplied as $XY$ if the number of columns of $X$ is equal to the number of rows of $Y$. That is $n=p$.
Mathematically,
Number of columns of$X$ $=$ Number of rows of $Y$
that is $n=p$
So, $XY=\left( \begin{matrix} {{x}_{11}} & \ldots & {{x}_{1n}} \\\ \vdots & \ddots & \vdots \\\ {{x}_{m1}} & \cdots & {{x}_{mn}} \\\ \end{matrix} \right)\times \left( \begin{matrix} {{y}_{11}} & \ldots & {{y}_{1q}} \\\ \vdots & \ddots & \vdots \\\ {{y}_{n1}} & \cdots & {{y}_{nq}} \\\ \end{matrix} \right)$
So, here, $n=p$
$X'$ of a matrix is the transpose of matrix$X$
So, if $X=\left( \begin{matrix} {{x}_{11}} & \ldots & {{x}_{1n}} \\\ \vdots & \ddots & \vdots \\\ {{x}_{m1}} & \cdots & {{x}_{mn}} \\\ \end{matrix} \right)$
Then, $X'={{X}^{T}}=X={{\left( \begin{matrix} {{x}_{11}} & \ldots & {{x}_{1n}} \\\ \vdots & \ddots & \vdots \\\ {{x}_{m1}} & \cdots & {{x}_{mn}} \\\ \end{matrix} \right)}^{T}}_{m\times n}={{\left( \begin{matrix} {{x}_{11}} & \ldots & {{x}_{m1}} \\\ \vdots & \ddots & \vdots \\\ {{x}_{1n}} & \cdots & {{x}_{nm}} \\\ \end{matrix} \right)}_{n\times m}}$

So, clearly we can see the order reverses after the transpose of a matrix.
So, for $A={{\left[ {{a}_{ij}} \right]}_{3\times 4}},\text{ 1}\le \text{i}\le \text{3,}\ \text{1}\le \text{j}\le \text{4}$
$A'={{\left[ {{a}_{ji}} \right]}_{4\times 3}},\text{ 1}\le \text{j}\le \text{4,}\ \text{1}\le \text{i}\le \text{3}$
So, order of$A'$is$4\times 3$ and for$B={{\left[ {{b}_{ij}} \right]}_{m\times n}},\ \ \text{1}\le \text{i}\le \text{m, 1}\le \text{j}\le \text{n}$
$B'={{\left[ {{b}_{ji}} \right]}_{n\times m}},\ \text{1}\le \text{j}\le \text{n, 1}\le \text{i}\le \text{m}$
Now, for $A'B$, number of columns of $A'$ $=$ number of rows of$B$ $\Rightarrow 3=m$
Also, for $B'A$ to be defined number of columns of $B'=$number of rows of$A$ $\Rightarrow m=3$
For both cases to be defined, $m$ equal number of rows of $B$ should be $3$ and $n$ can be any number. So, it can be $3$ and $4$ both. So, following that the order of $B$ could be $3\times 3$ or $3\times 4$ respectively.
So, the correct option is option B and option C.

Note: $A'$ is the transpose of matrix $A$. One may not consider it inverse of $A$ unless specified so. Also, if $A'B$ exists then $B'A$ also exists automatically because,
$(A'B)'=B'\left( A' \right)'=B'A$ ( General property of matrices)
So, in simple terms if matrix $A$ exists then its transpose also exists.