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Homework Help: Matrices and transformations

  1. Feb 8, 2006 #1
    My question is:

    If P' is the image of P under a matrix D = (1, -4, 0, -1) as follows (top left, top right, bottom left, bottom right). If P is not on the x-axis, why is PP' bisected by the x-axis and is at a constant angle to the x-axis, for any choice of P? :confused:

    I can visually see what's happening but how can I show it? Please help
     
    Last edited: Feb 9, 2006
  2. jcsd
  3. Feb 8, 2006 #2

    0rthodontist

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    Your notation is exceedingly unclear.
     
  4. Feb 8, 2006 #3

    JasonRox

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    What is P'P?

    I'm completely lost myself.

    I see your matrices and all, but that's about it.
     
  5. Feb 9, 2006 #4
    Can anyone crack this one??:uhh:
     
  6. Feb 9, 2006 #5

    HallsofIvy

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    I started to do this yesterday but the Tex apparently wasn't working.

    The best I can say is do it!! Let P be (x,y) (with y non-zero) and multiply it by D: P'= DP so that you have P' in terms of x and y.

    What is the midpoint of the line segment PP'? what is the slope of that line?
     
  7. Feb 9, 2006 #6

    Well I get that the midpoint of the line segment PP' is the x-axis and the line is y = 1/4 x which means the slope is tan-1 (0.25) so theta = 14.04 degrees? Is this correct?
     
  8. Feb 9, 2006 #7

    HallsofIvy

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    The midpoint of PP' is on the x-axis. What does that tell you about "why is PP' bisected by the x-axis"?

    Technically, the slope is 1/4, slope is not the angle. In any case, the slope does not depend on the point P?? What does that tell you about "and is at a constant angle to the x-axis, for any choice of P"?
     
  9. Feb 9, 2006 #8
    I can't answer the why in the "why is PP' bisected by the x-axis"? I can only say that the x-axis will be the midpoint of any P chosen except if it's on the x-axis itself. But that of course is a consequence, can't answer why :-(

    Honestly I can't answer this question, I'm going to give it a miss I think. I can visualise see it by I can't explain why. The lines PP' will always be parallel but a part from that what can I say?
     
  10. Feb 9, 2006 #9

    HallsofIvy

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    "Bisect" means "divide into equal parts". If the midpoint of PP' is always on the x-axis then the x-axis bisects the PP'!

    And if the lines PP' are always parallel, no matter what P is, then they cross the x-axis at a constant angle, don't they?

    That's all this problem is asking you to say!
     
  11. Feb 9, 2006 #10
    Thanks Sir that will do :wink:
     
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