Question

I'm analyzing systems of linear equations for my linear algebra class. We are being asked to determine whether it is possible to have an inconsistent underdetermined system or an overdetermined system with infinite solutions. My guess is that They both aren't possible.

Answer 1

If a system is underdetermined, it will have to have one of its variables parameterized. This would prevent a system from having no solution because the system really depends on what we make the parameter equal to. As for the overdetermined system, for it to be infinite solutions, the bottom row would need all zeroes. But if there are more eqns. than variables, we wouldn't be able to make them all zeros because adding the rows further would cause us to lose a zero I would think.

Answer 2

Could you have 3 variables with 2 equations, but the 2 equations contradict each other? Or would that not qualify as underdetermined?

Answer 3

That would be inconsistent and underdetermined, but I can't think of how they would contradict each other when one the variables is a free variable. Underdetermined systems usually end with one variable being parameterized. Any inconsistencies with the other variables could be remedied by the free variable I think.

Answer 4

OK, so if you had three variables x,y,z and the two equations were x+y+z=0 and x+y+z=1 that wouldn't be considered underdetermined? Because there are fewer equations than variables?

Answer 5

Yes, I looked at the matrix for that and it appears that we do find an inconsistency. Thanks for the help =)

Answer 6

Sure thing. You don't really need a matrix, since it says x+y+z is both 0 and 1, which would mean 0=1

Answer 7

transitive property... I must not be thinking straight today

Answer 8

And I think if we have an overdetermined system with all of the eqns. varying only by a constant multiple, then those will have infinite solutions...

Answer 9

Yes, if all the equations essentially say the same thing.