Invertible Proof with transposed matrices Let A, B, C be square matrices that are invertible. Say I want to express X with no inverses
Say 
$$
(A^{T}A)^{-1}(X +B^ {T})(C^{-1}B^{-1})^{T} = I.
$$
I know that $A^{T}A$ = $I$, but where can I go from there?
 A: Use $(M^T)^{-1} = (M^{-1})^T$, $(M_1M_2)^T = M_2^TM_1^T$, $(M_1M_2)^{-1}= M_2^{-1}M_1^{-1}$ to see that $$(C^{-1}B^{-1})^{T}= ((BC)^{-1})^T = ((BC)^T)^{-1} = (C^TB^T)^{-1}. \quad (*)$$
Now note that
$$(A^{T}A)^{-1}(X +B^ {T})(C^{-1}B^{-1})^{T} = I \overset{(1)}{\iff} (X +B^ {T})(C^{-1}B^{-1})^{T} = A^TA \\ \overset{(*)}{\iff} (X+B^{T})(C^{T}B^{T})^{-1} = A^TA \overset{(2)}{\iff} X+B^{T}=A^TAC^{T}B^{T} \overset{(3)}{\iff} X =(A^TAC^{T}-I)B^{T}.$$
(1) Multiply on the left by $A^TA$
(2) Multiply on the right by $C^{T}B^{T}$
(3) Subtract $B^T$ and factorize
In particular if $A^TA=I,$ then $X = (C^T-I)B^T$.
A: Note that from $(B^{-1})^T = (B^T)^{-1}$ and $(C^{-1})^T = (C^T)^{-1}$ one have
\begin{eqnarray}
(A^TA)^{-1}(X+B^T)(C^{-1}B^{-1})^T 
&=&
(X+B^T)(C^{-1}B^{-1})^T
\\&=& 
X(C^{-1}B^{-1})^T+B^T(C^{-1}B^{-1})^T 
\\&=& 
X (B^{-1})^T(C^{-1})^T+B^T(B^{-1})^T(C^{-1})^T 
\\&=& 
X (B^{-1})^T(C^{-1})^T+B^T(B^T)^{-1}(C^{-1})^T 
\\&=& 
X (B^{T})^{-1}(C^{T})^{-1}+(C^{T})^{-1} 
\end{eqnarray}
Now from $X X (B^{T})^{-1}(C^{T})^{-1}+(C^{T})^{-1}  =I$, you have that
$$
X (B^{-1})^T+I = C^T
$$
and
$$
X = (C^T-I) B^T.
$$
