My colleague K.-J. Wirths came up with another Rule of Sarrus for matrices. His suggestion is somehow closeto the original (at least graphically) and is easier to memorize. One has to use the “original” Rule of Sarrus for the
case but now three times. For the first case use the original matrix and for the next two case one has to permute two columns. Graphically this gives the following the pictures:
In principle this generalizes to larger matrices. But beware:
is large! For the
case one has a sum of 120 products but each “standard Sarrus” only gives 10 of them. Hence, one has to figure out 12 different permutations. In the
case one even needs to memorize
permutation, let alone all the computations…
I am sure that somebody with stronger background in algebra and more knowledge about permutation groups could easily figure out what is going on here, and to visualize the determinants better.
Update: Indeed! Somebody with more background in algebra already explored how to generalize the Sarrus rule to larger matrices. Again it was my colleague K.-J. Wirths who found the reference and here it is:
- Обобщенное правило Саррюса, by С. Аршон, Матем. сб., 42:1 (1935), 121–128
and it is from 1935 already! If you don’t speak Russian, in German it is
- “Verallgemeinerte Sarrussche Regel”, S. Arschon, Mat. Sb., 42:1 (1935), 121–128
and if you don’t speak German either, you can visit the page in mathnet.ru or to the page in the Zentralblatt (but it seems that there is no English version of the paper or the abstract available…) Anyway, you need permutations of the columns and apply the plain rule of Sarrus to all these (and end up, of course, with
summands, each of which has
factors – way more than using LU of QR factorization.)
September 7, 2011 at 2:00 pm
[…] a follow-up post, I have show a simpler visualization. Share this:TwitterFacebookLike this:LikeBe the first to like […]
August 16, 2012 at 6:16 am
[…] rules don’t originate with me, of course; you can see the same rule here. I’m sure I’m the seven millionth person to have done […]
November 18, 2013 at 6:34 pm
Very interesting! I’ve done a diagrammatic version of the 4×4 rule, based on an octagon: http://www.theoremoftheday.org/GeometryAndTrigonometry/Sarrus/Sarrus4x4.pdf
August 13, 2014 at 4:39 pm
The simplest method was found in October in the year 200, by the Mexican mathematical Gustavo Villalobos Hernandez of the University of Guadalajara. It is in Spanish in the following wikipedia page:
http://es.wikipedia.org/wiki/Regla_de_Villalobos
August 13, 2014 at 5:18 pm
Yeah, you can also permute the rows… Seems a bit simpler to memorize since one uses the same sign pattern that way.
August 13, 2014 at 7:20 pm
Thanks for your comment. Actually I do not speak English. I speak Spanish and Russian. I think it would be appropriate wikipedia page
https://es.wikipedia.org/wiki/Regla_de_Villalobos
I could be in English. I could use a virtual translator, but are not very accurate.
Greetings
August 13, 2014 at 7:26 pm
Thanks for your comment. Actually I do not speak English. I speak Spanish and Russian. I think it would be appropriate wikipedia page
https://es.wikipedia.org/wiki/Regla_de_Villalobos
I could be in English. I could use a virtual translator, but are not very accurate.
Greetings.
August 17, 2014 at 9:39 am
thanks/ i have also tried similar – but this yours is better/easier please send some actual numerical solved showing actions /few things not clear
September 11, 2018 at 10:11 pm
Hello chimpintrin,
It’s been a long time since Your post. Hope You’ll read this.
Can You explain this rule de Villalobos or give a link where to find it? The Wiki link ist empty as far as I can tell.
Thank You.
Cheers Tobias
PS: Or anyone Else? Thanks.
February 23, 2016 at 3:42 am
[…] rules don’t originate with me, of course; you can see the same rule here. I’m sure I’m the seven millionth person to have done this. However, the pictures here […]