If A and B are both invertible, then $[(A^{-1})B]^T$ is invertible. 
If $A$ and $B$ are both invertible, then $[A^{-1}B]^T$ is invertible.

Is this true or false? I have tried coming up with a proof but I don't really know where to start. 
 A: If $A$ and $B$ are both invertible, $(AB)^t=B^tA^t$, then the inverse of $(AB)^t$ is $(B^{-1}A^{-1})^t=(A^{-1})^t(B^{-1})^t$ since
\begin{align}
(AB)^t\cdot (B^{-1}A^{-1})^t&=(B^tA^t)\cdot((A^{-1})^t(B^{-1})^t)\\
&=B^t(A^t\cdot (A^{-1})^t)(B^{-1})^t\\
&=B^t(A^{-1}A)^t(B^{-1})^t\\
&=B^t I^t (B^{-1})^t\\
&=B^t I (B^{-1})^t\\
&=B^t \cdot (B^{-1})^t\\
&=(B^{-1}B)^t\\
&=I^t\\
&=I
\end{align}
Now for your question amend the above with $A$ replaced by $A^{-1}$. Also note a matrix $A$ is invertible iff its determinant, $\det(A)\neq0$, and then $\det(A^{-1})=(\det(A))^{-1}$. Plus $\det(A)=\det(A^t)$ and $\det(AB)=\det(A)\det(B)$.
A: $A, B, AB$ are invertable if their respective determinants don't equal 0.
Now what ruled do you know about determinants, products, inverses and transposes.
i.e. $det(AB) = det(A)det(B)$ and 
$(A^{-1} B)^T = B^T(A^{-1})^T$
etc.  
You should have enough to show that $det (A^{-1} B)^T \ne 0$
A: The product of invertible matrices is invertible; the transpose of an invertible matrix is invertible.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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Assume $\ds{C \equiv\pars{A^{-1}B}^{T}}$ is $\color{#f00}{not}$ invertible. It means $\ds{\det\pars{C} = 0}$ Then, you'll arrive to a contradiction because

\begin{align}
AC^{T} & = B\quad\implies\quad\det\pars{A}\det\pars{C} = \det\pars{B} = 0
\end{align}
A: Hint: Multiply by $[B^{-1}A]^T$, remembering that $[XY]^T=Y^TX^T$.
