INconsistencies in a system of linear equations.

[nobahar]

Hello!
I was wondering about inconsistencies in linear equations.
If I have three equations and four unknowns, and I can make one a linear combination of the other two, then I can identify if there is an inconsistency. If I cannot make one from a linear combination of the other two, how do I identify if there is an inconsistency and therefore no solution, because I may end up with free variables and conclude that there is an infinite number of solutions? Do I have to plug in one of the solutions and see if it solves?
I hope this makes sense.
Any help appreciated.

 

[nobahar]

Apologies for belabouring the question, but I just cannot find any satisfactory answer to this. It primarily concerns whether or not it is possible to obtain the semblance of free variables in a set of linear equations which do not have any solutions. According to wiki, it is "tacitly" assumed that the equation can be solved. So if a system of equations can appear to have a solution but throws up some nonsensical result for a system with no free variables, can a system that yields free variables also have no solution?
Any input you may have would be appreciated, even if it's not a definitve answer; sorry I can't provide any examples, if I could, I wouldn't need to ask!
Many thanks,
Nobahar.

 

[AC130Nav]

Hello!
I was wondering about inconsistencies in linear equations.
If I have three equations and four unknowns, and I can make one a linear combination of the other two, then I can identify if there is an inconsistency. If I cannot make one from a linear combination of the other two, how do I identify if there is an inconsistency and therefore no solution, because I may end up with free variables and conclude that there is an infinite number of solutions? Do I have to plug in one of the solutions and see if it solves?
I hope this makes sense.
Any help appreciated.

It takes two unrelated equations to solve two unknowns. I would think it would take four to solve four unknowns. Not sure what you're getting at.

 

[Mark44]

Let's back up a bit and consider two equations in three unknowns, and the geometry of the system. These equations represent planes in R³. This situation is pretty simple, as the planes will either intersect everywhere, intersect in a line, or won't intersect at all because they are parallel.

In the first case, every point on either plane is a solution to the system; in the second case, every point on the line of intersection is a solution. In the third case, there are no solutions and the system is inconsistent.

In a system of three equations in four unknowns, the geometry is similar, although the system involves "hyperplanes" in four-dimensional space; i.e., "planes" that have dimension one less than the space they're embedded in.

If two or more of the hyperplanes are parallel, then the system will be inconsistent. The system will also be inconsistent if each pair of hyperplanes intersects in a separate line that is parallel to the line of intersection of any other pair of hyperplanes.

Hope this helps.

 

[nobahar]

Many thanks for the responses.
It was my error, I thought that perhpas being left with free variables didn't identofy whether there were solutions to the equation. I have since realized that that IS the solution to the equations, the equations are consistent, and has an infinite number of solutions.
Thanks again for the responses.

 

[Mark44]

Not necessarily. Here's a simple example of a system of two equations in three unknowns that is inconsistent.

x + 2y + z = 4
2x + 4y + 2z = 6

Geometrically, the system represents two planes in R³. The planes are parallel and don't intersect anywhere, so there are no solutions to this system.

 

[nobahar]

Thanks for the response Mark44.
My understanding of matrices is a little tenuous.
For example, I was concerned about rref(A), where A is some matrix, in which there is no 'zeroed out' row, but also has free variables. I was wondering, "How do I know the equation is consistent?".
That was my concern.
Many thanks once again.