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How to compute the permanent of a matrix
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Patrick D. Rockwell
Jan 26, 2013, 3:09:10 AM1/26/13
to
I've read how to calculate the permanent of a matrix, but I found the
notation hard to understand so I'd Like a practical demonstration. For
2 by 2 matrices It's easy. For example, the permanent of
2 4
9 6
Is 2*6+4*9=36+12=48
But what about
5 3 1
8 2 4
7 9 6
What is the permanent of the above matrix and how do you calculate it?
I know that there 30 million web pages out
There which give the answer, but I just don't
understand the notation so, I'd appreciate it
if some kind soul would give me the permanent
of the 3 by 3 matrix provided above, then I'll
try to use it understand the notation that I
mentioned. Please show me how you did the
work and thanks in advance.
notation hard to understand so I'd Like a practical demonstration. For
2 by 2 matrices It's easy. For example, the permanent of
2 4
9 6
Is 2*6+4*9=36+12=48
But what about
5 3 1
8 2 4
7 9 6
What is the permanent of the above matrix and how do you calculate it?
I know that there 30 million web pages out
There which give the answer, but I just don't
understand the notation so, I'd appreciate it
if some kind soul would give me the permanent
of the 3 by 3 matrix provided above, then I'll
try to use it understand the notation that I
mentioned. Please show me how you did the
work and thanks in advance.
Virgil
Jan 26, 2013, 3:51:04 AM1/26/13
to
In article
<25ecd314-f163-44e2...@vb8g2000pbb.googlegroups.com>,
Given matrix M = [[ a b c ]
[ d e f ]
[ g h i]]
The determinant is adi + bfg + cdh - afh - bdi - ceg
The permanent is adi + bfg + cdh + afh + bdi + ceg
The thing is that there are a lot of neat shortcuts to evaluating a
determinant, especially for larger square matrices, few if any of which
work for evaluating a permanent, so the number of terms of the permanent
of an n by n permanent is apparently n-factorial.
The determinant of a 5 by 5, or even larger, matrix of reasonably small
integers, can often be done by hand in a few minutes, using a variety of
simple shortcut procedures, whereas the permanent of a 5 by 5 would
require multiplying out 120 products of five integers each then adding
them up, and for a 6 by 6 there would be 720 products of 6 integers to
be added up.
--
<25ecd314-f163-44e2...@vb8g2000pbb.googlegroups.com>,
[ d e f ]
[ g h i]]
The determinant is adi + bfg + cdh - afh - bdi - ceg
The permanent is adi + bfg + cdh + afh + bdi + ceg
The thing is that there are a lot of neat shortcuts to evaluating a
determinant, especially for larger square matrices, few if any of which
work for evaluating a permanent, so the number of terms of the permanent
of an n by n permanent is apparently n-factorial.
The determinant of a 5 by 5, or even larger, matrix of reasonably small
integers, can often be done by hand in a few minutes, using a variety of
simple shortcut procedures, whereas the permanent of a 5 by 5 would
require multiplying out 120 products of five integers each then adding
them up, and for a 6 by 6 there would be 720 products of 6 integers to
be added up.
--
Patrick D. Rockwell
Jan 27, 2013, 11:18:11 PM1/27/13
to
On Jan 26, 12:51 am, Virgil <vir...@ligriv.com> wrote:
> In article
> <25ecd314-f163-44e2-867c-33f2d997e...@vb8g2000pbb.googlegroups.com>,
Thanks! Eh, if I'm not mistaken, adi in the above should be
aei, am I right?
> In article
> <25ecd314-f163-44e2-867c-33f2d997e...@vb8g2000pbb.googlegroups.com>,
aei, am I right?
Virgil
Jan 28, 2013, 12:05:59 AM1/28/13
to
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