Simultaneous Equation mini-challenge
Message #1 Posted by Rodger Rosenbaum on 31 Mar 2006, 1:52 a.m.
Quote:
In his submission "Calculator Precision: Sure" of December 21 which started all of this Valentin Albillo proposed an investigation of the set of linear equations
1.3 x1 + 7.2 x2 + 5.7 x3 + 9.4 x4 + 9.0 x5 + 9.2 x6 + 3.5 x7 = 45.3
4.0 x1 + 9.3 x2 + 9.0 x3 + 9.9 x4 + 0.1 x5 + 9.5 x6 + 6.6 x7 = 48.4
4.8 x1 + 9.1 x2 + 7.1 x3 + 4.8 x4 + 9.3 x5 + 3.2 x6 + 6.7 x7 = 45.0
0.7 x1 + 9.3 x2 + 2.9 x3 + 0.2 x4 + 2.4 x5 + 2.4 x6 + 0.7 x7 = 18.6
4.1 x1 + 8.4 x2 + 4.4 x3 + 4.0 x4 + 8.2 x5 + 2.7 x6 + 4.9 x7 = 36.7
0.3 x1 + 7.2 x2 + 0.6 x3 + 3.3 x4 + 9.7 x5 + 3.4 x6 + 0.4 x7 = 24.9
4.3 x1 + 8.2 x2 + 6.6 x3 + 4.3 x4 + 8.3 x5 + 2.9 x6 + 6.1 x7 = 40.7
which has the "unique" solution :
x1 = x2 = x3 = x4 = x5 = x6 = x7 = 1.0
Consider a linear system, Ax = B, where B is an all zeroes vector; in other words, the system Ax = 0. One obvious solution is x=0; this is called the trivial solution. We learn in linear algebra that the Ax = 0 system only has a solution if A is singular. Valentin's A matrix is almost singular, so maybe there is an approximate solution. Find a solution vector x with length 1 (use the ABS function on the HP49G+) that gives a result A*x that is the smallest possible.
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