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Math Help - Finding the 2 x 2 matrix

  1. #1
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    Finding the 2 x 2 matrix

    Find the 2 by 2 matrix representing the transformation which maps triangle A onto triangle B.



    I need, not just the solution but the method to it, pretty quickly. Appreciate any help you offer me, thanks!

    As far as I can tell, it's a anti-clockwise 90 degree rotation. So would I work it out by...

    1. Select the points (1,0) and (0,1) on the graph, and then the matrix is
    (1 0
    (0 1)
    2. Find the images of these points under rotation
    3. Write the position vectors of these images.
    4. The position vector of the image of the first point becomes the first column.
    5. The position vector of the image of the second point is the second column of the required matrix.

    So then my answer is
    (1 0)
    (0-1) ?
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  2. #2
    Super Member Quacky's Avatar
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    I was always taught a general transformation for a rotation:

    \left(\begin{array}{cc}Cos\theta&-Sin\theta\\Sin\theta&Cos\theta\end{array}\right)

    Where \theta is the number of degrees anticlockwise that the shape is being rotated. I don't understand your method, admittedly, because I never had to use it, but I would suspect that there is a mistake within your working because it contradicts with my method, which is a general formula. I'll check your method more carefully and see if I can work out where the error lies - it may be that for whatever reason my formula doesn't apply here.
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  3. #3
    Behold, the power of SARDINES!
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    Not quite the matrix you give is a reflection across the x axis.

    Notice that the angle in triangle A across from the hypotenuse is represented by

    2e_1+e_2

    This gets mapped to

    T(2e_1+e+2)=2T(e_1)+T(e_2)=-e_1+2e_2

    Picking another point on the triangle we get

    T(3e_1+e+2)=3T(e_1)+T(e_2)=-e_1+3e_2

    Now if we subtract the 1st equation from the 2nd we get

    T(e_1)=e_2

    This gives

    T(e_2)=-e_1

     \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}
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  4. #4
    Super Member Quacky's Avatar
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    Aha! If you take the point (1,0), then the image under transformation becomes (0,1). So the first column is (0,1)

    If you take the point (0,1) and rotate it 90 degrees, then it becomes (-1,0) which is the second column.

    So this yields \left(\begin{array}{cc}0&-1\\1&0\end{array}\right) which agrees with my formula.

    Edit: was beaten by TES.
    Last edited by Quacky; May 8th 2011 at 10:44 AM. Reason: Grammatical Error.
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  5. #5
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    Ooookay, right! So the first column is (0,1), then the 2nd is (1,0), you rotate them anticlockwise 90degrees, then it becomes
    (0 -1)
    (1 0)

    Got it, thanks Quacky! By the way, I don't get your joke about Descartes... He's the one who made the Cartesian system right?
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  6. #6
    Super Member Quacky's Avatar
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    Quote Originally Posted by yorkey View Post
    Ooookay, right! So the first column is (0,1), then the 2nd is (1,0), you rotate them anticlockwise 90degrees, then it becomes
    (0 -1)
    (1 0)

    Got it, thanks Quacky! By the way, I don't get your joke about Descartes... He's the one who made the Cartesian system right?
    He also came up with the famous philosophical statement "I think, therefore I am".
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