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Math Help - Transformations Involving Matrices

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    2

    Transformations Involving Matrices

    Hello, I am stuck on a question in my FP1 textbook, in the practice exam section. I'm doing the OCR board.

    It tells me that there are two 2 x 2 matrices R and S represent the following transformations in the x-y place.

    R: a rotation through angle θ anticlockwise about origin
    S: a stretch with scale factor 2 parallel to the x-axis (with the y-axis invariant)

    RSR- is denoted by M

    I have found that M is:

    \left(<br />
  \begin{array}{cc}<br />
    1 + cos^2 \theta & sin \theta cos \theta \\<br />
    sin \theta cos \theta & 1 + sin^2 \theta<br />
  \end{array}<br />
\right)

    The next part says the the point P has the column matrix:

    \left(<br />
  \begin{array}{c}<br />
    x \\<br />
    y<br />
  \end{array}<br />
\right)

    And that the point Q is the image of P under the transformation represented by M.

    I have found that the column matrix of Q would be:

    <br />
\left(<br />
  \begin{array}{c}<br />
    x + xcos^2 \theta + ysin \theta cos \theta \\<br />
    ysin \theta cos \theta + y+xsin^2 \theta<br />
  \end{array}<br />
\right)

    But the next part asks me to show that the line joining P to Q makes an angle θ with the x-axis.

    How would I show this last part? O____O Please help me and thank you very much in advance.
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    You have made a mistake on Q which is actually

    <br />
\left(<br />
  \begin{array}{c}<br />
    x + xcos^2 \theta + ysin \theta cos \theta \\<br />
    xsin \theta cos \theta + y+ysin^2 \theta<br />
  \end{array}<br />
\right)

    Then the column matrix of PQ is

    <br />
 \left(<br />
   \begin{array}{c}<br />
     xcos^2 \theta + ysin \theta cos \theta \\<br />
     xsin \theta cos \theta +ysin^2 \theta<br />
   \end{array}<br />
 \right) = (x \cos \theta + y \sin \theta) \left(<br />
    \begin{array}{c}<br />
      cos \theta \\<br />
      sin \theta<br />
    \end{array}<br />
  \right)
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  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    2
    Oh yes, sorry about that. >_<

    But I am unsure how the column matrix of PQ proves that it makes an angle of θ? Please could you explain?
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  4. #4
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    The column matrix of PQ is

    <br />
 \left(<br />
   \begin{array}{c}<br />
     xcos^2 \theta + ysin \theta cos \theta \\<br />
     xsin \theta cos \theta +ysin^2 \theta<br />
   \end{array}<br />
 \right) = (x \cos \theta + y \sin \theta) \left(<br />
    \begin{array}{c}<br />
      cos \theta \\<br />
      sin \theta<br />
    \end{array}<br />
  \right)

    PQ^2 = (x \cos \theta + y \sin \theta)^2 (\cos^2\theta + \sin^2\theta) = (x \cos \theta + y \sin \theta)^2

    ||\overrightarrow{PQ}|| = PQ = |x \cos \theta + y \sin \theta|

    \frac{\overrightarrow{PQ}}{||\overrightarrow{PQ}||  } = \pm \left(<br />
    \begin{array}{c}<br />
      cos \theta \\<br />
      sin \theta<br />
    \end{array}<br />
  \right)

    This shows that the angle between PQ and x-axis is θ or θ+pi
    Therefore the angle between the line (PQ) and x-axis is θ
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