# nt.quantity principle – Deciding unimodular versus a eccentric matrix retort

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## nt.quantity principle – Deciding unimodular versus a eccentric matrix

$$L$$ and $$R$$ are matrices in $${0,1}^{ntimes n}$$ given to you and one in every of $$L$$ and $$R$$ is eccentric and the opposite is unimodular on the identification as a biadjacency of a bipartite graph it has one consummate matching.

Is there a unostentatious path to establish which of $$L$$ or $$R$$ is unimodular with out computing determinants? By less complicated I meant to graze if there’s path to establish in $$NC^1$$ (https://complexityzoo.bag/Complexity_Zoo:N#nc1) or equally in $$5$$$$PBP$$ (https://complexityzoo.bag/Complexity_Zoo:P#kpbp). The drawback is in $$NC^2$$ (https://complexityzoo.bag/Complexity_Zoo:N#nc2).

If wanted I can present a foundation $$M$$ for a lattice in $$n$$ dimensions in $$mathbb R^n$$.

The drawback is contained in $$C_=L$$ (https://complexityzoo.bag/Complexity_Zoo:C#cequalsl) as it’s testing for singularity however it’s less complicated for the reason that non-eccentric matrix has determinant $$pm1$$.

It is in $$oplus L$$ (https://complexityzoo.bag/Complexity_Zoo:Symbols#parityl) however it’s less complicated for the reason that covenant is we’re constraining to eccentric matrices if determinant is $$0bmod 2$$ and to unimodular matrices if determinant is $$1bmod 2$$.

It is in $$UL$$ by developing a matrix whose determinant is a sq. of given matrix (https://complexityzoo.bag/Complexity_Zoo:U#ul) however it’s less complicated since $$UL$$ shouldn’t be recognized to amass full issues.

So wanting on the zoo https://www.math.ucdavis.edu/~greg/zoology/diagram.xml it seems the issue is rapid to $$L$$

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