The inverse of \(A^{t}\) is the transpose of \(A^{-1}\) :
$$(A^{t})^{-1}=(A^{-1})^{t}$$
Proof
We use the property of the transposed matrix (4) :
$$(AB)^{t}=B^{t}A^{t}$$
Then we have
\begin{eqnarray*}A^{t} (A^{-1})^{t}&=&(A^{-1}A)^{t}=I\\(A^{-1})^{t}A^{t}&=&(AA^{-1})^{t}=I\end{eqnarray*}
where \(I\) is an identity matrix.
This shows that the inverse of \(A^{t}\) is the transpose of \(A^{-1}\) :
$$(A^{t})^{-1}=(A^{-1})^{t}$$