| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I have a TI-68 calculator; it's a fairly standard 'scientific'
type unit with some interesting features, like complex math,
finding roots of polynomials, and solutions to simultaneous eqns.
It _doesn't_ support matrix operations, but this isn't something
I typically need...
Well, one of those rare needs has arisen, and I've been trying to
figure how to use the capabilities of the calculator (particularly
the simultaneous equation solver) to determine:
1) the inverse of a matrix
2) the determinant of a matrix.
I've figured out a relatively simple technique for (1); Given NxN
matrix A, I can extract A' by solving AX=B N times with B = [1,0,...],
then [0,1,...] etc, and each solution giving me a column of A'.
A little bit tedious, but hard to screw up...
I'm stumped on (2); does anyone have any ideas?
Dave
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 2044.1 | ACISS2::LENNIG | Dave (N8JCX), MIG, @CYO | Sun May 05 1996 23:08 | 29 | |
Bingo! I figured out a method for number (2)
Given NxN matrix A, solve AX=B with B=[1/aNN,0..] yeilding x1
Now solve AX=B for the minor of a11 with B=[x1,0...]
repeat for each successive minor (ie a22, a33, ...) feeding
x1 back in as the next b1. When done, 1/x1 = determinant.
Example:
given A solve
2 -1 1 -1 2x1 -1x2 +1x3 -1x4 = 1/a44 (1/2)
1 3 -4 2 1x1 +3x2 -4x3 +2x4 = 0
1 0 -2 1 1x1 +0x2 -2x3 +1x4 = 0
3 -1 5 2 3x1 -1x2 +5x3 +2x4 = 0
you'll get an error at this point if the determinent will be zero
else, feed resulting x1 into
+3x1 -4x2 +2x3 = prev x1
0x1 -2x2 +1x3 = 0
-1x1 +5x2 +2x3 = 0
feed resulting x1 back in again
-2x1 +1x2 = prev x1
+5x1 +2x2 = 0
invert the resulting x1, giving the determinent
Again, a bit tedious, but only data entry errors are possible.
BTW, if you know the determinant of the 2 by 2 minor, simply feed the
inverse of it into the first pass and stop after computing the 3 by 3.
Dave
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