Inverse of product of matrices Let $n>m$ and let $A$ and $B$ be $m\times n$ and $n\times n$ matrices. $B$ is invertible.
If $A$ was square and invertable, then obviously
$$
\left(ABA^T\right)^{-1} = A^{-T}B^{-1}A^{-1}
$$
But, if $A$ is not square, can we say something (assuming that $ABA^T$ is invertible)?
 A: I am going to give an answer only for a symmetric positive definite $B$.
In this case the answer is
$$
(ABA^{\rm T})^{-1} = A^{+ \rm T} B^{-1/2} XB^{-1/2}A^+,
$$
where
$$
X = I-B^{-1/2}(I-A^{+}A) \big( B^{-1/2}(I-A^{+}A)\big)^+
$$
and ${}^+$ stands for the Moore-Penrose inverse.

One way of derivation is as follows.
Denote
$$
F = B^{-1/2}, \quad \quad G = I-A^{+}A, \quad \quad H(\alpha) = \Big(A(\alpha I +FF)^{-1}A^{\rm T}\Big)^{-1}.
$$
Note that
$$
(ABA^{\rm T})^{-1} = \lim_{\alpha \to 0} \Big(A(\alpha I +FF)^{-1}A^{\rm T}\Big)^{-1} =  \lim_{\alpha \to 0} H(\alpha).
$$
Use Woodbury matrix identity to rewrite $H(\alpha)$ as
$$
H(\alpha) = \Big( \frac{1}{\alpha}AA^{\rm T}-\frac{1}{\alpha}AF(I+\frac{1}{\alpha}FF)^{-1}\frac{1}{\alpha}FA^{\rm T}\Big)^{-1} .
$$
Note that $AA^{\rm T}$ is invertible and use Woodbury matrix identity again to obtain
$$
H(\alpha) = 
\alpha (AA^{\rm T})^{-1} - (AA^{\rm T})^{-1} AF \Big( -(I+\frac{1}{\alpha}FF) + \frac{1}{\alpha}FA^{\rm T}(AA^{\rm T})^{-1}AF \Big)^{-1} FA^{\rm T} (AA^{\rm T})^{-1},
$$
$$
H(\alpha) = \alpha (AA^{\rm T})^{-1} + (AA^{\rm T})^{-1} AF \Big( I+\frac{1}{\alpha}F \big(I - A^{\rm T}(AA^{\rm T})^{-1}A \big) F \Big)^{-1} FA^{\rm T} (A A^{\rm T})^{-1}.
$$
Note that
$$
A^{\rm T} (A A^{\rm T})^{-1} = A^+, \quad \quad I - A^{\rm T}(AA^{\rm T})^{-1}A = G = G^2,
$$
and rewrite $H(\alpha)$ as
$$
H(\alpha) =\alpha (AA^{\rm T})^{-1} + A^{+ \rm T} F \Big( I + \frac{1}{\alpha}FGGF\Big)^{-1}FA^+.
$$
Use Woodbury matrix identity again to get
$$
H(\alpha) =\alpha (AA^{\rm T})^{-1} + A^{+ \rm T} F \Big(   
I-FG \big( \alpha I + GFFG \big)^{-1} GF
\Big)FA^+.
$$
Take the limit as $\alpha \to 0$ and apply the limit relation to obtain
$$
(ABA^{\rm T})^{-1} = A^{+ \rm T} F \Big(   
I-FG \big( FG \big)^+
\Big)FA^+.
$$
Substitute $F$ and $G$ to get the final answer.
A: If $ABA^T$ is invertible then $A(BA^T(ABA^T)^{-1})=I_m$, that is, $A$ is right-invertible. Similarly $A^T$ is left-invertible with left-inverse $(ABA^T)^{-1}AB$. 
Write, abusing notation, $A^{-1}$ and $A^{-T}$ for the right- and left inverses respectively, and $C$ for $ABA^T$. Then $C^{-1}=(C^{-1}AB)B^{-1}(BA^TC^{-1})=A^{-T}B^{-1}A^{-1}$: the same formula holds true.
