<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] 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 $det(adj(A^{-1}))$.
</s>