<s>[INST] 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}}))$ ?
 [/INST] $B$ 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 $\det(A)=9$.
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