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Matrix problem

  1. May 10, 2008 #1
    Suppose A is a 3 x 4 matrix and there exists a 4 x 3 matrix C such that AC = I (the 3x3 identity matrix). Let b be an arbitrary vector in R3. Produce a solution of Ax=b.

    I'm not quite sure what the question is asking. I think I just need someone to point me in the right direction.

    Thanks
     
  2. jcsd
  3. May 11, 2008 #2

    tiny-tim

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    Hi flash! :smile:

    Hint: AC = I … so multiply something by I ! :wink:
     
  4. May 11, 2008 #3

    Defennder

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    Further hint if the one given above is too vague: Multiply it on the right side of I.
     
  5. May 11, 2008 #4

    HallsofIvy

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    If AC= I then CA= I.

    How would you solve Ax= b if A, x, and b were NUMBERS?
     
  6. May 11, 2008 #5
    Ax = b
    CAx = Cb
    Ix = Cb
    x = Cb

    Am I on the right track?
     
  7. May 11, 2008 #6

    tiny-tim

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    Not only on the right track … you've arrived at Grand Central! :biggrin:

    You have proved that A sends Cb to ACb = Ib = b.

    So A(Cb) = b.

    In other words, x = Cb is a solution to Ax = b. :smile:
     
  8. May 11, 2008 #7
    Ok, thanks. The other part of the question goes:
    A is a 4x3 matrix
    C is a 3x4 matrix such that CA = I
    Suppose, for some given b in R4 that Ax=b has at least one solution. Show that this solution is unique.

    Can I just say x = Cb which implies that there is only one solution for x? I'm thinking that I should say something along the lines of: if there exists a C such that CA = I then A must have no free variables.
     
  9. May 11, 2008 #8
    Hint: What does it mean for a matrix to be invertible? If you had an invertible matrix, what does this say about the number of solutions for every b in Ax = b?
     
  10. May 11, 2008 #9

    HallsofIvy

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    Yes, that's all you need to do. C is a given matrix , b is a given vector: since multiplication of a 3x4 matrix with a 4 dimensional vector (a 4x1 matrix) is "well defined", x= Cb must be a specific, unique vector.

    (I think BryanP's response is to your previous question- though then I don't know why he refers to " the number of solutions for every b in Ax = b". Here, A is not invertible. Only square matrices are invertible.)
     
    Last edited: May 11, 2008
  11. May 13, 2008 #10
    I apologize for that. I didn't notice my error about the invertibility.
     
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