Re: Another problem with matrix inversion

[Jeroen Demeyer <jdemeyer@cage.ugent.be>]

Lorenz Minder wrote:
> So while A has an inverse in this case, it is not found.  The workaround
> using chinese remaindering works if the factorization is known, so this
> is going to be a problem if the modulus is an RSA modulus, for example.
> It would be better to try to run the algorithm as is, and as soon as a
> nonzero non-invertible remainder is found, to split into two instances
> with the now known partial factorization and continue.

This brings up another question regarding GP: is it possible to trap 
"impossible inverse modulo" errors and actually recover the element 
which caused the impossible inverse?  I think it would be nice to have 
that functionality in GP, but I don't really know what would be the 
correct way to do it.

Cheers,
Jeroen.