matrix determinant

[Piers Barber]

i'm using Thomas's 42s on iPhone and Free42s as packaged for Fedora
16.

i get a non-zero result for:

| 1 2 3 |
| 4 5 6 |
| 7 8 9 |

i.e. det [[1,2,3],[4,5,6],[7,8,9]]

the determinant should be zero as this matrix is singular. it follows
that there should also be no inverse - yet one is calculated.

the determinant is very small (and the matrix inverse is very large as
a result of 1/det)

but i don't understand why it's non-zero - HP48GX is correct, and
complains with INV Error: Infinite result

what's the explanation?

thanks

[Byron Foster]

Here's a little discussion about it:

http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv014.cgi?read=65551

I also tried this with a real 42S and the determinant is also a very small non-zero number. So the behavior is consistent. Thomas added a singular matrix error option, but I'm not sure why it's not triggered in this case when the option is on.

If you are still curious I would ask Thomas Okken.

http://thomasokken.com/free42/

[Piers Barber]

Thanks Byron, that's great - answers my question especially after a
quick look at the source and a glance at my 30 year old college math
book

My 3x3 A is being LU decomposed into:

A = LU

| 1 2 3 | | 1 0 0 | | p q r |
| 4 5 6 | = | a 1 0 | | 0 s t |
| 7 8 9 | | b c 1 | | 0 0 u |

det(A) = det(L)det(U)
= det(U)
= p * s * u

That's going to be very fast on big matrices, but the LU decomp
scaling factors will result in errors and these show up in my final
result.

[Piers Barber]

Nah - still confused because I've now done an LU decomposition
manually and i get:

LU=A

| 1 0 0 || 1 2 3 | | 1 2 3 |
| 4 1 0 || 0 -3 -6 | = | 4 5 6 |
| 7 2 1 || 0 0 0 | | 7 8 9 |

det(U) = 1 * -3 * 0 = 0

Looks as though I'm really going to have to go through what actually
happens in the code - time to learn ddd debugger...

[Byron Foster]

Hmm, curious... I'll take a closer look this weekend.

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