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A system of linear equations can't have exactly two solutions. Why?

  1. Feb 1, 2013 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Problem:
    A system of linear equations can't have exactly two solutions. Why?

    (a) If (x, y, z) and (X, Y, Z) are two solutions, what is another solution?
    (b) If 25 planes meet at two points, where else do they meet?

    Solution:
    (a) Another solution is 1/2 * (x + X, y + Y, z + Z).
    (b) If 25 planes meet at two points, they meet along the whole line through those two points.

    2. Relevant equations
    N/A

    3. The attempt at a solution
    Am I correct in thinking that a system of linear equations can't have exactly two solutions because a line in three-dimensional space must intersect at zero points or one point or a full line's worth of points and a plane in three-dimensional space must intersect at zero points, a line's worth of points or a plane's worth of points when attempting to answer the question asked before part (a)?

    Assuming I'm correct, I know that, for example, when I say “a line's worth” it means an infinite amount of points and that that is the same thing as “a plane's worth” etc but, I'm just expressing myself like that to clarify what I am saying.

    For part (a), I was thinking (x + X, y + Y, z + Z) would be another solution. Am I right? Are all answers with scalars multiplying each component and then operators summing, dividing, multiplying or subtracting those respective (transformed) components solutions as well?

    For part (b), is it correct to add that not only could they meet at an entire (infinite) line's worth of points but that another case is that if the planes overlap, they can meet at an entire (infnite) plane's worth of points?

    If something is unclear, tell me and, I will attempt to clarify it.

    Any input would be greatly appreciated!
     
  2. jcsd
  3. Feb 1, 2013 #2

    Dick

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    You are on the right track. But suppose the equation is x-y=1. Then A=(3,2) is a solution and B=(2,1) is a solution. I'd agree that (A+B)/2 is a solution. I would not agree that A+B is also a solution. What wrong with A+B? Can you figure out how to write down an infinite number of solutions in terms of A and B?
     
  4. Feb 1, 2013 #3

    HallsofIvy

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    Another way if looking at it is this: Any system of linear equations can be written as a matrix equation, Ax= b, where A is the matrix of coefficients, x is a column matrix having the unkowns as entries and b is the column matrix having the 'right side' of the equations as entries.

    So if X and x are two solutions. That is if Ax= b and Ax= b, then A((x+ X)/2)= (1/2)Ax+ (1/2)AX= (1/2)b+ (1/2)b= b so that (x+ X)/2 is also a solution. In fact, it is easy to see that if x and X are solutions the so it [itex]\alpha x+ \beta X[/itex] any [itex]\alpha[/itex], [itex]\beta[/itex], between 0 and 1 such that [itex]\alpha+ \beta= 1[/itex].

    That is, I believe, what you are saying.
     
  5. Feb 1, 2013 #4

    Dick

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    Since the OP said, "For part (a), I was thinking (x + X, y + Y, z + Z) would be another solution. Am I right?", I don't think that was what was being said. I think with a few hints the OP might have gotten around to saying what you said.
     
  6. Feb 2, 2013 #5

    s3a

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    What you/HallsofIvy said was not what I was saying but was what I needed to hear.

    Thank you both!
     
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