Linear system: Ux + y = 3, 7x + y = V

[Clifford]

Ux + y = 3
7x + y = V

Given these two equations find the conditions on U and V to produce a system that has no solutions, a unique solution and infinite solutions if possible.

In order to get an infinite number of solutions you need N equations with N+1 unknowns, giving you a free variable. The only other way to have infinite solutions is if one solutions is a linear combination of the other. In this case, the coefficient on each y is 1, so you cannot construct a linear combination of these two. Thus, there is conditions on U and V to produce an infinite solution.

To get a unique solution I think that and conditions on U and V will give you a unique solution.

Since any conditions will provide you with a unique solution, there is no condition that will give us no solutions.

I'm not so sure about the unique solution part, but from trying multiple values for each that is what I obtained. Is there a proof behind this to prove this?

 

[tkhunny]

You are starting to get it. A little more thought.

No Solution: Try U = 7 and V = 0

Infinitely Many Solutions: Try U = 7 and V = 3

Read the definitions carefully. Things are not always as they seem.

 

[stapel]

Ux + y = 3
7x + y = V

Given these two equations find the conditions on U and V to produce a system that has no solutions, a unique solution and infinite solutions if possible.

Think about the form of systems solutions that you've seen. :idea:

i) If a system, such as the above, has "no solution", then, at some point in your computations, you've ended up with some sort of nonsense, such as "0 = 3".

ii) If the system had infinitely-many solutions, you ended up with something like "0 = 0" or "7 = 7". That is, you ended up with something that was trivially true, and the values of x and y didn't matter.

iii) If the system had one solution, you ended up with "x = (something), y = (something else)".

Now look at your system. You can easily reduce this, using row operations. For instance, you could multiply the first equation by -7/U, and add to the second equation; in the new (but equivalent) system, you could multiply the (new) second equation by -U/(U - 7), and add to the first equation. After simplification, you would then end up with:

. . . . .x = (3 - V) / (U - 7)

. . . . .y = (UV - 21) / (U - 7)

Since the above simplification assumed that U was not zero, you'll need to consider that case separately, of course.

Obviously U cannot be 7, or you'll be dividing by zero. So what happens if you plug "7" in for "U" in the original system? Can you arrive at a "no solutions" value for V? How about an "infinitely-many solutions" solution? :wink:

For a unique solution, if U is not zero or seven, can you solve for V in terms of U, to find a formula for all the values which give single point solutions?

Eliz.