Question: Let A be a $3\times 3$ invertible matrix (with real coefficients) and let $B=A^TA^{-1}$. Prove that 

\begin{equation\*}
\det(I + B) = 2(1 + tr(B)).
\end{equation\*}

I know that 

\begin{equation\*}
\det (I+B)=\lambda\_1\lambda\_2\lambda\_3+\lambda\_1\lambda\_2 +\lambda\_1\lambda\_3+\lambda\_2\lambda\_3+\lambda\_1+\lambda\_2+\lambda\_3+1
\end{equation\*}

given $\lambda\_1,\lambda\_2$ and $\lambda\_3$ are three distinct eigenvalues of $B$. However, I don't know where to go on from here and how to utilise the fact that $B=A^TA^{-1}$. Any help or direction would be appreciated.

Answer: 
Hint: $tr(A^{-1}A^T)=tr(A^TA^{-1})=tr(A^{-T}A)=tr(B^{-1})$

$\sum \lambda\_i = tr(B)$

$\sum \lambda\_1\lambda\_2 = det(B)tr(B^{-1})$
It suffices to show that $1+\lambda\_1+\lambda\_2+\lambda\_3=\lambda\_1\lambda\_2\lambda\_3+\lambda\_1\lambda\_2+\lambda\_2\lambda\_3+\lambda\_3\lambda\_1$. But this is true because $\det(A^\top {A^{-1}})=1$, meaning $\lambda\_1\lambda\_2\lambda\_3=1$.