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Math Help - another matrix transformation

  1. #1
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    another matrix transformation

    Hi folks,

    Given the following transformation:

    \overline{x'} = <br />
\left( \begin{array}{cc} -2 & 0 \\ 0 & 2 \end{array} \right) <br />
\left( \begin{array}{c} x \\ y \end{array} \right) +  <br />
\left( \begin{array}{c} 4 \\ 0 \end{array} \right)

    show that it may be expressed as a reflection in the line x = 4/3 followed by an enlargement and give the centre and scale factor of this enlargement.

    using:
    x' = -2x + 4
    y' =  2y

    I have plotted the results of the transformation as shown in the diagram. As far as I can see, the transform represents a reflection about x = 1 and an enlargement, scale factor 2 about the origin. The first part of my diagram shows the transformation, the second the reflection about x = 1 and the third the enlargement. Just to clarify this last bit: If the enlargement had a cenre at A (1,0) then A would stay where it is, O would move to (3,0) B would go to (3,2) and C to (1,2) but if we move the centre back to the origin, we should get the result in the top part of the diagram.

    I just don't see any way that the transform can be a reflection in x = 4/3 floowed by an enlargement of 2 centre (4/3, 0) which is the answer provided!

    Can anyone help? It's not so hard, is it?
    Attached Thumbnails Attached Thumbnails another matrix transformation-m6e17.jpg  
    Last edited by s_ingram; March 22nd 2010 at 11:19 AM. Reason: to try to get someone interested!
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  2. #2
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    Re: another matrix transformation

    Sorry to bump this old thread, but I'm having the same problem with this same question. I really don't know how I can break equation down to a reflection in the line x = 4/3 followed by an enlargement. Any suggestions?
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  3. #3
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    Re: another matrix transformation

    Okay, I think I might have it. To get a reflection in x = 4/3, it's like a reflection in the y-axis followed by a translation. Knowing that the coordinates (4/3, 0) will be invariant:

    (4/3 = (-1 0 (4/3 + (a
    0) 0 -1) 0) b)

    This results in a = 8/3 and b = 0.

    Now, the original equation written by the original poster can be changed to:

    (x' = (2 0 (-1 0 + (8/3 + (4/3
    y') 0 2) 0 1) 0) 0)

    We can see the two matrices that must be relevant to the enlargement (scale factor 2), and by

    (x = (2 0 (x + (4/3
    y) 0 2) y) 0)

    we can find the invariant point, which would be centre of the enlargement. Working the above out reveals this centre to be (4/3, 0).

    I'm not sure at how efficient this method was. Was there a better way to tackle this question?

    Edit: Ahhh, it won't accept the spaces for the matrices. I'll have to figure a way how to present the above properly, then edit again. I apologise.
    Last edited by SpriteAlpha; September 1st 2012 at 03:01 PM.
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  4. #4
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: another matrix transformation

    Quote Originally Posted by s_ingram View Post
    Given the following transformation:
    using:
    x' = -2x + 4
    y' =  2y
    I just don't see any way that the transform can be a reflection in x = 4/3 floowed by an enlargement of 2 centre (4/3, 0) which is the answer provided!
    Can anyone help? It's not so hard, is it?
    Any point (x,y) is transferred to (4-2x, 2y):

    \left(\begin{array}{cc} -2 & 0 \\ 0 & 2\end{array}\right).\left(\begin{array}{c} x \\ y\end{array}\right)+\left(\begin{array}{c} 4 \\ 0\end{array}\right) = \left(\begin{array}{c} 4-2 x \\ 2 y\end{array}\right)

    In vector form the resultant of this transformation is ( I use complex form):

    4-2x+2y i -(x+y i) = 4 - 3 x + y i

    Then: mirror line is: 4-3x=0 & y=0, or x=4/3...
    Attached Thumbnails Attached Thumbnails another matrix transformation-mirror.png  
    Last edited by MaxJasper; September 1st 2012 at 07:02 PM.
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