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why does a system of linear equations really have no solution or one unique solution, or infinite solutions?

What forbids a system to a finite number of solutions?

thanks

fisico30

- Thread starter fisico30
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- #1

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why does a system of linear equations really have no solution or one unique solution, or infinite solutions?

What forbids a system to a finite number of solutions?

thanks

fisico30

- #2

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- #3

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Consider the following three systems of linear equations

1) ax+by+cz=0; dx+ey+fz=0

2) ax+by+cz=0; dx+ey+fz=0; gx+hy+iz=0

3) ax+by+cz=0; dx+ey+fz=0; gx+hy+iz=0; jx+ky+lz=0

What then?

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- #5

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thanks everyone.

So if the system of composed of linear equations, then I can see how, 1,0, infinity would the the solutions....

The linear equations can be algebraic or differential, correct? The same solutions (1,0, infinity) would work...

If the system was made of nonlinear equations, then there could be a finite number of solutions, correct?

Is it possible to have a mixed, hybrid, system, composed of linear equations and nonlinear equations?

thanks

fisico30

- #6

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Yes you can have a mixture of linear and non linear equations, but the system is automatically non linear if it includes even one non linear equation.

Please note that my examples are different from Murphrid ( who is not wrong) because they show a different situation.

Case (1) is an under-determined system because there are more unknowns than equations.

Case (2) is fully determined since the number of equations matches the number of unknowns.

Case (3) is overdetermined since there are more equations than unknowns.

- #7

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I guess if the system is nonlinear (at least one nonlinear equation), then we cannot even think about matrices and linear algebra to find the solution(s), correct?

Matrices are only useful for linear systems...

In nonlinear systems, a consistent system can have more than 1 unique solution...

thanks

fisico30

- #8

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Normally a great deal of effort goes in to find linear approximations or substitutions or restricted ranges over which linearity can be assumed, in order to use matrix algorithms. It all depends upon the equations.

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