Re: RREFMOD


"Adder" <david@xxxxxxxxxxxx> wrote in message news:1127677939.131793.149860@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 
Hi,

Browsing through the group I discovered RREFMOD which promises to do
exactly what I want, unfortunately it doesn't seem to work. I am
working mod 11 and have stored that in the CAS variable MODULO using
MODSTO, I enter my matrix (a Vandermonde matrix reduced mod 11):

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

and run RREFMOD and the HP49+ gives me back:

2 0 0 0 -2  3  2
0 1 0 0  4  4  3
0 0 2 0 -1  4 -2
0 0 0 1  4 -1 -2

Now the negative numbers I can cope with using a routine I wrote my
self (with help from the group), but the first four columns definitely
don't form an identity matrix. Or am I missing something?
I could 'fix up' the result with more code, but I started to ask myself
how confident I was in the rest of the result.

Using RREF correctly reduces the first part of the matix to the 4x4
identity, but the last part contains exact numbers like '67/76' which
MOD wont handle.

Adder


1/2 mod 11 = -5 , because -5*2= -10=1 mod 11.
This can be calculated by INVMOD.
In this case, multiply the first row by -5, multiply the third row by -5 and take mod 11.
then you can get a matrix with first four columns to form an identity matrix.

Takashi Matsubara 

.

Relevant Pages

  • Re: RREFMOD
    ... Browsing through the group I discovered RREFMOD which promises to do ... In this case, multiply the first row by -5, multiply the third row by -5 and take mod 11. ... then you can get a matrix with first four columns to form an identity matrix. ...
    (comp.sys.hp48)
  • RREFMOD
    ... Browsing through the group I discovered RREFMOD which promises to do ... Now the negative numbers I can cope with using a routine I wrote my ... don't form an identity matrix. ...
    (comp.sys.hp48)
  • Re: RREFMOD
    ... > Browsing through the group I discovered RREFMOD which promises to do ... > Now the negative numbers I can cope with using a routine I wrote my ... > don't form an identity matrix. ...
    (comp.sys.hp48)
  • Re: How do you find two orthogonal vectors (a plane) perpendicular to given vector?
    ... whichever makes the first entry bigger in magnitude, ... 3-by-3 identity matrix. ... Then the first row of H will be a unit vector parallel to your b, ...
    (sci.math)
  • Re: Stable finding of orthogonal vector to a normal in 3D?
    ... where I is the identity matrix and T denotes the transpose. ... The first row of H is a unit vector parallel to b. ... The other two rows are unit vectors perpendicular to b ...
    (sci.math)