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: 
Note that $\det B = 1$. 

Note additionally that $B^{-1} = A(A^T)^{-1}$, so that
$$
\operatorname{trace}(B^{-1}) = 
\operatorname{trace}(A(A^T)^{-1}) = 
\operatorname{trace}((A^T)^{-1}A) = 
\operatorname{trace}(B^T) = \operatorname{trace}(B)
$$
Now, your polynomial can be written as
$$
\det(I + B) = \det(B) + \det(B)\operatorname{trace}(B^{-1})
+ \operatorname{trace}(B) + 1
$$
the conclusion follows.
Hint: $tr(A^{-1}A^T)=tr(A^TA^{-1})=tr(A^{-T}A)=tr(B^{-1})$

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

$\sum \lambda\_1\lambda\_2 = det(B)tr(B^{-1})$