Matrix complicated equation Let $$A = \begin{bmatrix}
    1 & 3 & 4\\
    3 & 6 & 9\\
    1 & 6 & 4
  \end{bmatrix},$$
$B$ be a $3\times 3$ matrix and $$A \cdot A^{T} \cdot A +3B^{-1} =0$$ 
What would be the value of 
$ \det( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}}))$ ?
 A: Let $$X =  |( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}})|
 = |(\operatorname{adj}(2B^T) \operatorname{adj}(B^{-1})\operatorname{adj}(A^{-1}))| $$
$$ = |2B^T|^2 |B^{-1}|^2 |A^{-1}|^2 = 2^6 |B|^2 \cfrac{1}{(|A||B|)^2} = \cfrac{2^6}{|A|^2}$$ 
I hope now you can figure out $|A|$.
A: $B$ is a red herring here and might be replaced by any invertible $3\times3$-matrix.  
Since $\operatorname{adj}(C)=\det(C)C^{-1}$ for invertible $C$ we have
$$\det(\operatorname{adj}(C)=\det(C^{-1}\cdot\det(C))=(\det(C))^3\det(C^{-1})=(\det(C))^2$$
if $C$ is of type $3\times3$.
Now happily compute 
\begin{align}\det( \operatorname{adj} (A^{-1}B^{-1}{2B^{T}})
&=\bigl(\det(A^{-1}B^{-1}2B^{T})\bigr)^2\\
&=\bigl(2^3\cdot\det(A^{-1})
\underbrace{\det(B^{-1})\det(B^{T})}_{=1}\bigr)^2\\
&=\left(2^3\cdot\frac{1}{9}\right)^2
=\frac{64}{81}
\end{align}
as $\det(A)=9$.
A: Since $A$ is invertible, it follows that $B = -3A^{-1}A^{-T}A^{-1}$ and therefore that 
\begin{align}
\det( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}})) &=\det\left( \operatorname{adj} \left(A^{-1}\left(-\frac{1}{3}A A^{T}  A \right){(-6A^{-T}A^{-1}A^{-T}})\right)\right). 
\end{align}
Since you already know $A$, this is just a computation.
A: No need to find the value of matrix $B$ since
\begin{equation}
B^{-1}=-\frac{1}{3} A A^T A
\end{equation}
then
\begin{equation}
B=-\frac{1}{3}(A A^T A)^{-1} = 3 A^{-1} A^{-T} A^{-1}
\end{equation}
put into the desired system
\begin{align}
A^{-1}(B^{-1})2B^T &=A^{-1} (-\frac{1}{3} A A^T A) 2 (3 A^{-1} A^{-T} A^{-1})
\notag \\
&= 2(A^{-1}A)A^T (A A^{-1}) A^{-T} A^{-1}
\notag \\
&= 2A^T  A^{-T} A^{-1}
\notag \\
&= 2A^{-1} 
\end{align}
where
\begin{equation}
A^{-1}=-\frac{1}{3}
\begin{pmatrix}-10 & 4 & 1\\ -1 & 0 & 1 \\ 4 & -1 & -1
\end{pmatrix}
\end{equation}
then find the result of $det(adj(A^{-1}))$.
