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**Hartlw** A 4th order determinant (aij) is given by

e(ijkl) a1i a2j a3k a4l

and the summation convention over repeated indices. e(ijkl) is the general kronecker delta (for ex e(1234)=1, e(2134)=-1, etc)

The a23 term in the expansion is

e(i3kl) a1i a3k a4l,

which is the cofactor of a23 by def. i,k,l is the column subscript and varies over 1,2,4 so that cofactor is the minor except for sign. How do you extract the sign from above formula? The answer is (-1)^(2+3), but how do you get it?