Solution Set for 4 Equations in 3 Unknowns

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Homework Help Overview

The discussion revolves around a system of 4 equations in 3 unknowns, exploring the nature of its solution set, which could include infinitely many solutions, a unique solution, or no solution at all.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of having more equations than unknowns, with some questioning the original poster's assumption about the nature of solutions. Others suggest considering the rank of the matrix to determine the number of free variables and potential solutions.

Discussion Status

There is an active exploration of different interpretations regarding the relationship between the number of equations and unknowns. Some participants provide insights into the conditions under which solutions may exist, while others emphasize the need for concrete examples to clarify vague reasoning.

Contextual Notes

Participants note the distinction between the augmented matrix and the coefficient matrix, highlighting the importance of understanding the system's structure in determining the solution set.

EV33
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Homework Statement


Determine all possibilities for the solution set (from among infinitely many solutions, a unique solution, or no solution) of the system of linear equations.

A system of 4 equations in 3 unknowns

Homework Equations


None.


The Attempt at a Solution


This is more of a word problem.

Under the assumption that "4 equations in 3 unknown's means a 4X3 Matrix then I would say that there are an infinite amound of solutions because 4X3 matrix could be filled with an infinite amount of options.

Is that correct?
 
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With n unknowns you need at least n equations to get a unique solution.
 
If the first equation is a=0 and the second equation is a=1, I doubt there are an infinite number of solutions. Just saying "a 4x3 matrix could be filled with an infinite number of options" means you are thinking about this really vaguely. Try and be more concrete. Any of those options is actually possible. Try and give an example of each.
 
rock.freak667 said:
With n unknowns you need at least n equations to get a unique solution.

"4" equations. "3" unknowns. There are plenty of equations.
 
4 equations in 3 unknown's means a 4X3 Matrix

Not necessairly, you didn't specify whether you were looking at the augmented matrix or the coefficient matrix. Consider the rank of the matrix(either augmented or coefficient), this should help you determine the number of parameters(free variables) and point you towards a conclusion.
 
Dick said:
"4" equations. "3" unknowns. There are plenty of equations.

well that was my point. Though I probably worded it badly.
 
A system of 4 equations in 3 unknowns


This is an overdetermind system of linear equations which if it happens to have a solution, it is unique. Otherwise it won't have any exact solution assuming no two eauations coincide. You can use methods like "square least" to get the nearest set of solutions that is not always guaranteed to be a good approximation, but still a very useful and reliable one.

AB
 

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