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Elemetary matrices proof problem PLS help

  1. Feb 13, 2008 #1
    If A is an n x n matrix and Ax = 0 has many solutions, then

    which ones are true??


    A. Ax = b has either no solutions or many solutions
    B. Ax = b has a unique solution for some b
    C. Ax = b has many solutions for any b
    D. Ax = b has no solutions for some b

    or are all of the above false??

    thanks
     
  2. jcsd
  3. Feb 13, 2008 #2

    Vid

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    Think about what it means for Ax = 0 to have many solutions. If the 0's in the augmented matrix of linear equations were changed to some constants, would that change how A reduces?
     
  4. Feb 14, 2008 #3

    HallsofIvy

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    Unfortunately, I don't know what you have to work with. If Ax= 0 has more than one solution, then the kernel of A is non-trivial and has non-zero dimension. That, in turn, means that the image of A is not all of Rn.

    Note also that is Ax= b and v is in the kernel of A, A(x+ v)= Ax+ Av= b+ 0= b.
     
  5. Feb 15, 2008 #4
    well this is what i kinda know

    Ax=0
    if you solve for x then x=0,
    which will make A invertible
    So, what does that say about Ax=B
     
  6. Feb 15, 2008 #5

    HallsofIvy

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    If x= 0 is the only solution to Ax= 0, then A is invertible. If A has an inverse, how would you solve Ax= B?

    However, that is completely irrelevant to your question since the question specifically says that A has MORE than one solution.

    I can only repeat: if v is any vector such that Av= 0, and x is a solution to Ax= b, then A(x+ v)= Ax+ Av= Ax+ 0= Ax= b. If there are many solutions to Ax= 0, what does that tell you about the number of solutions to Ax=b?
     
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