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My acceleration is using the formula of minors of the inverse:

$$\det{A}=\frac{\det{A(\alpha)}}{\det{A^{-1}(\alpha')}}$$

In which, $\alpha\subset\{1,2,...,n\}$, $\alpha'$ is its complement and $A(\alpha)$ is a sub-matrix of $A$.

This way I need only to compute the determinants of two n/2-by-n/2 matrices.

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Using the answer here, $\det(A)$ is a multiple of $1/\text{Tr}(A^{-1})$. Not sure though how you can narrow it down further. –  user2468 Sep 4 '12 at 22:50
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@JenniferDylan: The determinant is a number, and so it is a multiple of every non-zero number. –  Fly by Night Sep 4 '12 at 23:09
    
@FlybyNight True, but what I meant is that $\det(A) \in \{ k\frac{1}{\text{Tr}(A^{-1})} : k \in \Bbb{R} \}$. If there's another relation (which I failed to find), then may be we can intersect it with this set to find $\det(A)$ easily, may be, may be not. –  user2468 Sep 5 '12 at 0:20
    
But $k$ is not a constant, it is a polynomial in the entries of $A$, homogeneous of degree $n-1$. –  Robert Israel Sep 5 '12 at 0:29
    
... namely $(-1)^{n-1}$ times the coefficient of $\lambda^1$ in the characteristic polynomial of $A$. –  Robert Israel Sep 5 '12 at 0:50
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