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Proving in Linear Algebra

  1. Jul 5, 2009 #1
    1. The problem statement, all variables and given/known data
    I am studying linear algebra independently using a pdf book online. The question I have difficulty is this:
    Prove that, where a, b... e are real numbers and a does not equal 0, if
    (1) ax + by = c
    has the same solution set as
    (2) ax + dy = e
    then they are the same equation.

    2. Relevant equations

    3. The attempt at a solution

    Since they are a couple of linear equations, I performed a Gaussian operation on it
    (1) ax + by = c
    (2) ax + dy = e
    (1) - (2) y = c-e/b-d , x = c/a - (b/a)(c-e/b-d)

    So basically I subtracted (1) from the (2) to get the intercept point between the equations. That should be the solution set. Since I'm trying to prove that (1) and (2) are the same, then I should let b=d and c=e. However, if b = d and c = e, then y would be 0/0. So I'm not sure what I did wrong.
    Last edited: Jul 5, 2009
  2. jcsd
  3. Jul 5, 2009 #2


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    Gold Member

    Alright, if the two equations have the same solution set, that means that an ordered pair (x1,y1) must be a solution to each equation. Clearly (x2,y2) is a solution to each equation as well. Using this principle, subtract the two equations.

    [a(x1) + b(y1)] - [a(x1) + d(y1)] = c - e

    c - e = (b - d)y1

    Use a similar analysis for the pair (x2,y2) to finish the proof. I'd imagine there is a simple way to do this but I can't think of anything.

    Edit: Just thought of a much simpler way to do it but this should work.
    Last edited: Jul 5, 2009
  4. Jul 5, 2009 #3
    I've rethought the question (graphically this time on a x-y axis) and realize that the question is asking that if (1) and (2) intersect, they must be the same line.
  5. Jul 5, 2009 #4


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    Not quite. If they only intersect at one point, then they are not the same line. It's saying that if the lines intersect at infinitely many points (they have the same solution set) then they are the same line.

    Here, a simpler way to do it is to consider the solution (x1,0). Then we have, a(x1) = c and a(x1) = e. Can you take it from there?
  6. Jul 7, 2009 #5
    Actually, I have the answer key to the online textbook. But I just don't like the answer from the book.

    The answer to the book is the following:
    Let the first solution set to be x=(c-by)/a, let y=0 gives the solution set (c/a,0) Substitute that solution set to the second equations and you will get a(c/a) + d*0 = e so c = e. Let y =1 gives the solution ((c-b)/a,1) and substitute the solution to the second equation and get d = b.

    The problem I have with this method is that you have to pick 2 arbitrary points and then solve for the answer. How do you know that the 2 equations intersect at least two points? You have to make the assumption that the 2 equations does intersect at y= 1 and y = 2. I guess this is a reasonable assumption since we know that they are linear equations and that a does not equal zero, so the both equations span the entire number line. I feel that there should be a more general approach to this problem.
    The most important thing for me is that to identify the my flaw in understanding the problem.
    If ax + by = c and ax + dy = e both share the same solution set, then it means that the solution set is located at where the equations intersect (whether they intersect once or more times). The solution set they both share is ((c-e)/(b-d)), (b/a)((c-e)/(b-d)).This is the general expression for where the 2 equations intersect. The question tells us to prove that c = e, d = b.

    After that I have no idea what to do.
  7. Jul 8, 2009 #6


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    The book's solution is almost identical to the second solution I suggested. Since the equations have the same solution set - they intersect at infinitely many points - if (x1,0) is a solution to the first equation, it is also a solution to the second equation!
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