Answer
$(A^{T})_{ij}=A_{ji}$
(the ij-th element of $A^{T}$ is the ji-th element of $A)$
Work Step by Step
If $A$ is an $m\times n$ matrix, then its transpose is the $n\times m$ matrix obtained by writing its rows as columns,
so that the $i\mathrm{t}h$ row of $A$ is the ith column of the transpose, $A^{T}$.
So if $A= \left[\begin{array}{llll}
a_{11} & a_{12} & \ldots & a_{1m}\\
a_{21} & a_{22} & \ldots & a_{2m}\\
\vdots & & & \\
a_{n1} & a_{n2} & \ldots & a_{mn}
\end{array}\right]$, an m$\times$n matrix
Then $A^{T}=\left[\begin{array}{llll}
a_{11} & a_{21} & \ldots & a_{n1}\\
a_{12} & a_{22} & \ldots & a_{n2}\\
\vdots & & & \\
a_{1m} & a_{2m} & \ldots & a_{nm}
\end{array}\right]$, an n$\times$m matrix
we see that $(A^{T})_{ij}=A_{ji}$
(the ij-th element of $A^{T}$ is the ji-th element of $A)$
