Rearranging Matrix Equation I have an equation of matrices, where all are positive symmetric, and $D$ matrices are also diagonal and $D \neq kI$:
$\alpha = (D^{-1}_{0}AD^{-1}_{0})^{-1}(D^{-1}_{1}BD^{-1}_{1})$
$\alpha = D_{0}A^{-1}D_{0}D^{-1}_{1}BD^{-1}_{1}$
How do I solve for $A^{-1}B$?
My attempt at a solution starts as follows:
$ D^{-1}_{0}\alpha D_{1} =A^{-1}D_{0}D_{1}^{-1}B$
But due to my limited linear algebra, I get stuck here as to how to get the $D$s on the outside on the right hand side.
 A: I tried but the answers might not be satisfactory.
Attempt 1: Let $M = D_0^{-1} \alpha D_1$ and $D = D_0 D_1^{-1} = \text{diag}(d_1,\dots,d_n) > 0$. Let
$$
A^{-1} = \left[\begin{array}{ccc}
a_1 & \cdots & a_n
\end{array}\right] \quad \text{and} \quad 
B = \left[\begin{array}{c}
b_1^T \\ \vdots \\ b_n^T
\end{array}\right].
$$
From the equation $M = A^{-1} D B$, we have $M = d_1 a_1 b_1^T + \cdots + d_n a_n b_n^T$. By dividing by $d_i$ on both sides, for $i=1,2,\dots,n$, we obtain
$$
\frac{1}{d_i} M = \sum_{j=1}^n \frac{d_j}{d_i} a_j b_j^T.
$$
Let $A^{-1} B = W$, then $W = X_1 + \cdots + X_n$, where $X_i = a_i b_i^T$.
Finally, let $V_i = W - \frac{1}{d_i} M$. Then, we have the following equation to solve
$$
\underbrace{\left[\begin{array}{c}
V_1 \\ V_2 \\ \vdots \\ V_n
\end{array}\right]}_{V} = \underbrace{\left[\begin{array}{cccc}
0 & c_{12} & \cdots & c_{1n} \\
c_{21} & 0 & \cdots & c_{2n} \\
\vdots & &\ddots & \vdots \\
c_{n1} & c_{n2} & \cdots & 0
\end{array}\right]}_{C} \underbrace{\left[\begin{array}{c}
X_1 \\ X_2 \\ \vdots \\ X_n
\end{array}\right]}_{X}
$$
where $c_{ij} = \frac{d_i - d_j}{d_i}$ for $i\neq j$. For $n>2$, this is a dead end because $\text{rank}(C) = 2$ for all $n\geq 2$, and thus $C$ is not invertible for $n>2$.

Conclusion from Attempt 1: If you want the expression of $A^{-1}B$ to not contain either $A$ and/or $B$, then the solution is not feasible. Because, in general, it is impossible to infer the unweighted sum $W = X_1 + \dots + X_n$ from the weighted sum $M = d_1 X_1 + \dots + d_n X_n$.

The following attempt gives a solution in terms of $A$ or $B$.
Attempt 2: Let $M = D_0^{-1} \alpha D_1$ and $D = D_0 D_1^{-1} = \text{diag}(d_1,\dots,d_n)$. Notice that $D$ is invertible since $d_i > 0$ for all $i\in\{1,\dots,n\}$. Then, the equation is written as $M = A^{-1} D B = A^{-1} D A A^{-1} B$ and $A^{-1} B = A^{-1} D^{-1} A M$. Or, $M = A^{-1} B B^{-1} D B$ and $A^{-1} B = MB^{-1} D^{-1} B$.
