high-order determinants

I wrote:

> On Sun, 7 Mar 1999, Gary Morrison wrote:
>> You can also define them recursively: The determinant of a n-by-n
>> matrix M is given by the sum for i = 1 to n of M[i,1] times the determinant
>> of the (n-1)-by-(n-1) matrix formed by leaving out the top row and column
>> i.
>
> This is correct. [example snipped]

And then John Starrett wrote:
> Don't forget to aternate signs when computing the determinant of a
> matrix by minors!

D'oh! Sorry. Listen to John, not me.

--pH <manynote@lib-rary.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "Well, so far, every time I break he runs out.
-\-\-- o But he's gotta slip up sometime . . . "

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