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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