Someone please answer this (Hahn? Erlich?):

I understand how to compute the Fokker

determinant for a 2-dimensional matrix:

| 4 -1 | = 12

| 0 3 |

Simple cross-multiplication: (4*3)-(-1*0) = 12.

But I don't understand how to calculate

the 3-dimensional determinant:

| 2 2 -1 |

| 4 -1 0 | = 31

| 1 0 3 |

How do you get this value from this matrix?

- Monzo

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> But I don't understand how to calculate

> the 3-dimensional determinant:

>

> | 2 2 -1 |

> | 4 -1 0 | = 31

> | 1 0 3 |

Determinants of larger matrices can be found in several ways. (I've

been doing mostly digital design for the past 15 years since I calculated

determinants all the time to get my EE degree, so it's been a while since

I've done this. Hopefully I'll remember it right.)

It's probably easiest to illustrate the pattern by example:

| a b c |

| d e f | = aei + bfg + cdh - afh - bdi - ceg

| g h i |

Similarly,

| a b c d |

| e f g h |

| i j k l | = afkp + bglm + chin + dejo - ahkn - belo - cfip - dgjm

| m n o p |

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.

> 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.

I should probably clarify that with an example of that. The determinant

of:

| a b c d |

| e f g h |

| i j k l |

| m n o p |

is "a" times the determinant of

| f g h |

| j k l |

| n o p |

plus "b" times the determinant of

| e g h |

| i k l |

| m o p |

plus "c" times the determinant of

| e f h |

| i j l |

| m n p |

plus "d" times the determinant of

| e f g |

| i j k |

| m n o |

On Sun, 7 Mar 1999, Gary Morrison wrote:

> | a b c |

>

> | d e f | = aei + bfg + cdh - afh - bdi - ceg

>

> | g h i |

This is correct. However,

> | a b c d |

>

> | e f g h |

>

> | i j k l | = afkp + bglm + chin + dejo - ahkn - belo - cfip - dgjm

>

> | m n o p |

This is not. There should be 24 terms, not just 8.

> 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]

On Mon, 8 Mar 1999, Joseph L Monzo wrote:

> I tried the formula for 3-D matrix

> determinant given by Gary Morrison,

> and was puzzled by a result I got.

[snip]

> The commas in question are:

> the 81/80 between 27/20 and 4/3

> the 64/63 between 4/3 and 7/4

> the 36/35 between 27/20 and 7/4

[snip]

> In matrix form, this is:

>

> | 4 -1 0 |

> |-2 0 -1 |

> | 2 -1 -1 |

>

> But when I used the formula:

> (4*0*-1)+(-1*-1*2)+(0*-2*-1)

> -(4*-1*-1)-(-1*-2*-1)-(0*0*2)

>

> the result my spreadsheet gave me

> was 0. What does this mean?

It means that the three unison vectors in question do not span a

three-dimensional space. If you think about the fact that the three

unison vectors are the three edges of a periodicity block, it means that

these three vectors are coplanar. This can be seen easily from the fact

that the first row of the matrix, plus the second row, sums to the third

row. To make this work you need to find a third unison vector, or

vanishing/bridge interval, which is independent of the other two.

--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 . . . "

NOTE: dehyphenate node to remove spamblock. <*>

> This is correct. However,

>

> > | a b c d |

> >

> > | e f g h |

> >

> > | i j k l | = afkp + bglm + chin + dejo - ahkn - belo - cfip - dgjm

> >

> > | m n o p |

>

> This is not. There should be 24 terms, not just 8.

Thanks for the refresher. I guess I've been in the world of nanoseconds,

microamps, and Karnaugh maps for too long!