Re: 4x4 matrices using Cramer's Rule

Virgil wrote:

In article <a1658$457d265b$82a1e228$31541@xxxxxxxxxxxxxxxx>,
Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> wrote:

Uinseann wrote:

Anyone got any ideas on how to solve 4x4 matrices by using Cramer's
Rule. I've looked in a multitude of math books and surfed the web for
hours to no avail. I can do a 3x3 no problem but unfortunately I dont
seem to be able to see how to do a 4X4. Any advice that anyone might
be able to offer regarding the problem below would be greatly
appreciated.

13 10 0 0 i1 6
-10 13 0 -3 X i2 = 10
0 0 18 -3 i3 0
0 -3 -3 6 i4 5

The following is in Delphi Pascal. I hope it's so much readable that you
can translate it into you own favorite programming language. The outcome
is, iff I've made no mistakes:

-2.38532110091743E-0001 = i1
9.10091743119266E-0001 = i2
2.34250764525994E-0001 = i3
1.40550458715596E+0000 = i4

Or det(M) = 24525 and
i1 = -5850/24525 = -26/109
i2 = 22320/24525 = 496/545
i3 = 5745/24525 = 383/1635
i4 = 34470/24525 = 766/545

Very good, Virgil.

Is there any particular reason why you need to use Cramer's rule?

There are much better methods.

Cramer's rule is only useful for very small matrices, BUT there are MANY
problems with such small matrices. Cramer's rule has the advantage that
it gives always an easy solution iff the matrix is non-singular. Integer
problems as the above can also be programmed in such a way that floating
point operations are avoided altogether, preserving an "exact" solution,
as you have demonstrated. An obvious drawback is that Cramer's rule very
quickly runs out of performance. Therefore use is delimited to matrices
with rank < 6 or 7 or some such. Quite a severe limitation, admittedly.

Han de Bruijn

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