Prove $\det(I + B) = 2(1 + tr(B)).$ 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.
 A: 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.
A: Hint: $tr(A^{-1}A^T)=tr(A^TA^{-1})=tr(A^{-T}A)=tr(B^{-1})$
$\sum \lambda_i = tr(B)$
$\sum \lambda_1\lambda_2 = det(B)tr(B^{-1})$
A: An alternative approach, which doesn't use the explicit formula in terms of eigenvalues.  Instead, find the characteristic polynomial.  See that, for any $\lambda$,
$$p(\lambda) = \det (B - \lambda I) = -\lambda^3 + \lambda^2 \mathrm{Tr} \,  B - \lambda x + \det B$$
for some number $x$.  Here we merely have $\lambda = -1$.
The number $x$ can be computed in a variety of ways.  Using exterior algebra is one method:  construct the $3 \times 3$ matrix $B_2$, which acts on $3 \times 1$ column vectors corresponding to elements of $\bigwedge^2 \mathbb R^3$.  Then $x = \mathrm{Tr} \, B_2$.
The relationship between $B_2$ and $B$ is explicitly
$$B_2 (a \wedge b) = B(a) \wedge B(b)$$
for any vectors $a, b$.
Now, use a common inversion identity:
$$B^{-1}(a) = \star B_2^T(\star a)/\det B$$
where $\star$ is the usual Hodge dual.  This means we can write $B_2$ as
$$B_2(a \wedge b) = \star (B^T)^{-1}(\star [a \wedge b]) \det B$$
You can verify now (e.g. by breaking into a basis) that $\mathrm{Tr} \, B_2 = \det B \, \mathrm{Tr} \, (B^T)^{-1}$.  By the arguments given in other answers, this is merely $\det B \, \mathrm{Tr} \, B$, and as a result, we have
$$p(\lambda) = \det(B - \lambda I) = -\lambda^3 + \lambda^2 \mathrm{Tr} \, B - \lambda [\det B \, \mathrm{Tr} \, B] + \det B$$
For $\lambda =-1$, and since $\det B = 1$, the result follows.
A: 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 $\lambda_1\lambda_2\lambda_3=1$. 
A: For my opinion, it seems more "comfortable" to use Newton's identities, especially when B is NOT invertible.
e.g.$\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3=\dfrac{1}{2} [(\lambda_1+\lambda_2+\lambda_3)^2-(\lambda^2_1+\lambda^2_2+\lambda^2_3)]=\dfrac{1}{2}(tr(B)^2-tr(B^2))$
