Alternative calculation of determinant.

The alternative calculation is very efficient.  I cannot prove that it delivers the goods without recourse to algebraic topology.  See  Lickorish  Corollary , 9.5 for the proof.

Choose a chess boarding, of the knot diagram.  Define a (symmetric) matrix as follows. Both rows and columns are labeled by the white regions.  Attach to each crossing  an index , as shown.



The matrix entry for row A and column B. With  A not equal to B is the sum of indices where A meets B. The diagonal entries are chosen so that row and column sums are zero.
This is the Goeritz matrix (except that I changed all signs) of the diagram.
Delete a row and column and compute the  determinant.

Example      The pretzel link P(p,q,r)

There are only three white regions  labelled A B C  and p + q - 1 black regions.

Now show that the determinant of  10₁₂₄ is 1 and hence cannot be coloured mod any number. Here  is another diagram for 10₁₂₄

9₄₆ is another pretzel knot with determinant 9 which comes up in lecture 9