[?] Solve equation system with more variables than equations?

Questions
,
1

I have this problem that driving me crazy. Is it possible to solve a equation system with more variables than equations?

Solve[
 {x/72 + y/63 + z/56 == 12/3,
  z/72 + y/63 + x/56 == 14/3}]

The unknown variables are distance and then divided by speed, so I am looking for the total length the object will travel. First equation is from point a to b and then return the same distance but in different speeds. Note that the distance “y” is the same and in the same speed.

Any ideas?

2

What is the problem with the result obtained from that system?

3

Remove[“Global`*”]

Solve[
{x/72 + y/63 + z/56 == 12/3,
z/72 + y/63 + x/56 == 14/3}, {x, y, z}]

Solve::svars: Equations may not give solutions for all “solve” variables.

Out[45]= {{y → 441 - 2 x, z → -168 + x}}

4

This is a degenerate case!

A = {{1/72, 1/63, 1/56}, {1/56, 1/63, 1/72}};
MatrixForm@A
A.{x, y, z}
MatrixRank[A]

Consider first the kernel of A: a vector V

V = First@NullSpace[A]
A.(k V) // Simplify

Complete V by other sandard vectors, and change of basis:

B = Transpose@{{1, 0, 0}, {0, 1, 0}, V};
MatrixForm@B
Det[B]

AI = A.B;
MatrixForm@AI
MatrixForm[AI.Inverse[B]]

The tranformed matrix AI is better-suited! Now, we can restrict ouselves to a 2-dimensional non-degenerated problem:

TA = Transpose@Drop[Transpose@AI, -1];
MatrixForm@TA
{X, Y} = (Inverse[TA]).{12/3, 14/3}
TA.{X, Y}
AI.{X, Y, z}

Come back in the original basis, and verify:

B.{X, Y, z}
A.B.{X, Y, z}
5

Thx! But this is way beyond my Mathematica skills :slight_smile: Sinse the system is build up of two equations maybe I could make a third that define the total length it will travel and then have three equations and three unknown variabels.

…