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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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    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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