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Homogenous system

  1. Sep 27, 2009 #1
    If a homogenous system Ax = 0 has infinitely many solutions, then for a non-zero vector b, the associated system Ax = b ____ have _______


    In my assignment, the answer I wrote on this question is must have many solutions. However, what I got is wrong.

    Can anybody tell me why it is wrong?
     
    Last edited: Sep 27, 2009
  2. jcsd
  3. Sep 27, 2009 #2
    if Ay = b, then what can you say about any vector of the form x + y where Ax = 0?
     
  4. Sep 27, 2009 #3
    Well I am assuming there has to be many solutions since Ax=0 gives infinite solutions.

    Plus, no solution or/and one solution sounds less appropriate than many solutions at the moment I answer the question.
     
  5. Sep 27, 2009 #4
    [tex]\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}[/tex]

    Also, I might be being pedantic but I would say many is a broader term than infinite. I think that I've seen books using 'many' though so if your book or teacher does, just ignore me.
     
  6. Sep 27, 2009 #5
    Well I don't understand why many solutions is wrong since this tell me how many variables and how many unknowns.

    So, I ran out of options. Please help me.
     
  7. Sep 27, 2009 #6
    Must is surely wrong as the example shows. The homogeneous equation has infinite solutions. Can you find a solution to the non-homogeneous equation?

    Like I said, I was probably just being pedantic with the whole many/infinite thing.
     
  8. Sep 27, 2009 #7
    Teacher gave me these choices:

    A. may have exactly one solution
    B. must have many solutions
    C. must have either one solution or no solution
    D. may have no solution
    E. need not satisfy any of the above
     
  9. Sep 27, 2009 #8
    Well, you shouldn't have chosen B. Work my example to find out why.
     
  10. Sep 27, 2009 #9
    I would say many solutions and no solutions. B and D
     
  11. Sep 27, 2009 #10
    B is wrong in general. Do you see why? If you want, we can talk about when it's right.
     
  12. Sep 27, 2009 #11
    So B is wrong in any possible answers?
     
  13. Sep 27, 2009 #12
    It's wrong without conditions that weren't assumed in the question. Have you studied dimension yet?

    The key thing here is the use of the word "must". I've given you an example already that doesn't so think about it until you realize that it doesn't "must" have infinite solutions.
     
  14. Sep 27, 2009 #13
    So the correct answer is D only? May have no solutions?
     
  15. Sep 28, 2009 #14
    the answer is C? is it?
     
  16. Sep 28, 2009 #15
    [tex]\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}[/tex]


    [tex]\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}[/tex]
     
  17. Sep 28, 2009 #16
    I see x has a solution and y has no solution.

    Am I right?
     
  18. Sep 28, 2009 #17
    The second one has no solution, you are correct. How many solutions does the first one have again?
     
  19. Sep 28, 2009 #18
    the first one has one solution?
     
  20. Sep 28, 2009 #19
    No. (1 0) is a solution. (1 1) is a solution. (1 45/77) is also a solution.

    How many solutions does [tex]\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}[/tex] have? You can take any one of them, add it to (1 0) and get a solution to

    [tex]\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}[/tex]

    Right?
     
  21. Sep 28, 2009 #20
    so are you saying that there must have either one solution or no solution?
     
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