How to simplify $\langle A^{-1}b, Ac \rangle$? Assume $A$ is an n x n real invertible matrix, $b$ and $c$ are $n$-dimensional distinct vectors. I would like to rearrange / simplify the inner (dot) product
$$\langle A^{-1}b, Ac \rangle$$
into an expressions that resembles $\langle Xb,c \rangle$ for some n x n matrix $X$. Is this possible?
By inspection, for the $n=2$ case, I found  $$\langle A^{-1}b, Ac \rangle = \langle (A^{T})^{-1} Ac,b \rangle$$, which looks ok to me. Is that true for the general case?

Note that $\def\<#1>{\left<#1\right>}$we have for $x,y \in \def\R{\mathbf R}\R^n$, $B \in \def\M{\operatorname{Mat}}\M_n(\R)$: 
\begin{align*}
   \<Bx,y> &= \sum_i (Bx)_i y_i\\
           &= \sum_i \sum_j b_{ij}x_j y_i\\
           &= \sum_j \sum_i b_{ij}y_i x_j\\
           &= \sum_j x_j (B^t y)_j\\
           &= \<x, B^t y>
\end{align*}
that is, the transposed matrix represents (in the given bases) the euclidian adjoint. Hence, we have
 $$ \< A^{-1}b, Ac> = \<A^t A^{-1}b, c> $$
