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Uniqueness of a system of equations

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Its number four on this link:

    http://www.math.pitt.edu/~dwang/math0280/math0280-r1.pdf" [Broken]


    3. The attempt at a solution

    Well I reduced it to echelon form, and thats not really what I have the question on.
    But I have three equations now, but I am not sure what values of k would imply what type of solutions.

    My equations are:

    0 = k2-2
    3y-2z+2=k
    x-2y+3z=2

    I am thinking there is a unique solution at k equals plus or minus the square root of 2.
    Because the first equation is really the only thing that would limit the solutions. Is that right?
    SO would there be no solutions when k is anything BUT plus or minus the square root of 2?

    But then I am not sure what k value would yield infinite results?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 8, 2010 #2

    Mark44

    Staff: Mentor

    Your 2nd and 3rd equations are correct, but your first is not, so check your work.

    Just for the sake of argument, however, let's suppose that your work is correct. If k = +/-sqrt(2), then k2 - 2 = 0, so you have two other equations in three unknowns. This system is underdetermined, so there will be an infinite number of solutions.

    If k is any value other than sqrt(2) or -sqrt(2), then the first equation is saying that 0 = <some nonzero number>, so the system of equation is inconsistent, and therefore has no solution.

    There are no other possibilities.
     
  4. Feb 8, 2010 #3
    OH yeah! Sorry, you are right.
    I forgot to do the row operation on the right side.
    So the first equation should be:

    k2-3k+2=0 ?

    This is only true when k= 1 or 2.

    So, by what you said (which makes sense), if k equals anything besides this it is inconsistent and so there are no solutions?

    I guess I am just confused by, what would a UNIQUE solution look like?
    There has to be a value for k that x,y, and z had only 1 possible solution.
    So is it if you can explicitly find a value for x, y and x?

    But then for the first part, you said there were infinite solutions, do you have to solve for x y and z to know that or could you just tell somehow?

    Thanks for bearing with me, this helps a lot.
     
  5. Feb 8, 2010 #4

    Mark44

    Staff: Mentor

    No, that's not it either. Try it once more, but this time be more careful.

    You'll end up with an equation k2 + <stuff> = 0. It turns out that there are two solutions for k, one positive and one negative.

    If k is either of those two values, the system boils down to two equations in three unknowns. Since there is a free variable, such a system has an infinite number of solutions. Geometrically what the system represents is two planes in space that intersect in a line. Each point on the line is one of an infinite number of solutions.

    If k is any value other than the two numbers described in the previous paragraph, the system is inconsistent, meaning it has no solutions. A geometric interpretation might be a situation where there are three planes in space, and two of them are parallel. Although some pairs of planes have solutions (lines of intersection), there is no point that lies on all three planes.

    Since every real value of k is either a solution of the quadratic or it isn't, the system is either underdetermined or inconsistent. There are no values of k that can lead to the system being consistent, which means that there are no values of k for which the system has a unique solution.
     
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