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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](http://mathoverflow.net/questions/46553/fast-trace-of-inverse-of-a-square-matrix), $\det(A)$ is a multiple of $1/\text{Tr}(A^{-1})$. Not sure though how you can narrow it down further.2012-09-04
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    @JenniferDylan: The determinant is a number, and so it is a multiple of every non-zero number.2012-09-04
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    @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.2012-09-05
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    But $k$ is not a constant, it is a polynomial in the entries of $A$, homogeneous of degree $n-1$.2012-09-05
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    ... namely $(-1)^{n-1}$ times the coefficient of $\lambda^1$ in the characteristic polynomial of $A$.2012-09-05
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    I suppose you could apply this iteratively to reduce it to 4 determinants of order $n/4$, 8 of order $n/8$, and so on.2012-09-05

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