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Math Help - Matrices

  1. #1
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    Matrices

    I need to check that my work is correct for these transformation matrices.
    A, a reflection in Ox. \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)
    B, an enlargement by a factor 2. \left(\begin{array}{cc}2&0\\0&2\end{array}\right)
    C, a rotation by \frac{\pi}{2} about O. \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)
    D, a stretch by a factor of 3 in the direction Oy. \left(\begin{array}{cc}1&0\\0&3\end{array}\right)
    E, a reflection in the line y=x. \left(\begin{array}{cc}0&1\\1&0\end{array}\right)
    Can someone help me see of they are correct?
    Thanks
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  2. #2
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    Hello arze
    Quote Originally Posted by arze View Post
    I need to check that my work is correct for these transformation matrices.
    A, a reflection in Ox. \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)
    B, an enlargement by a factor 2. \left(\begin{array}{cc}2&0\\0&2\end{array}\right)
    C, a rotation by \frac{\pi}{2} about O. \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)
    D, a stretch by a factor of 3 in the direction Oy. \left(\begin{array}{cc}1&0\\0&3\end{array}\right)
    E, a reflection in the line y=x. \left(\begin{array}{cc}0&1\\1&0\end{array}\right)
    Can someone help me see of they are correct?
    Thanks
    Yep! I reckon they're spot on. Good work!

    Grandad
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  3. #3
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    so to find the matrices of a series of transformations:
    A followed by C I'll get \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}0&-1\\-1&0\end{array}\right)?
    and A following C would be \left(\begin{array}{cc}0&1\\1&0\end{array}\right)
    The answers for these two are reversed.
    C carried out after E \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\left(\begin{array}{cc}0&1  \\1&0\end{array}\right)=\left(\begin{array}{cc}1&0  \\0&-1\end{array}\right)
    answer is \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)
    and also A followed by B followed by C \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}2&0\\0&2  \end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}2&  0\\0&-2\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}0&-2\\-2&0\end{array}\right)
    answer is \left(\begin{array}{cc}0&2\\2&0\end{array}\right)
    thanks
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  4. #4
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    Hello arze
    Quote Originally Posted by arze View Post
    so to find the matrices of a series of transformations:
    A followed by C I'll get \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}0&-1\\-1&0\end{array}\right)?
    and A following C would be \left(\begin{array}{cc}0&1\\1&0\end{array}\right)
    The answers for these two are reversed.
    C carried out after E \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\left(\begin{array}{cc}0&1  \\1&0\end{array}\right)=\left(\begin{array}{cc}1&0  \\0&-1\end{array}\right)
    answer is \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)
    and also A followed by B followed by C \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}2&0\\0&2  \end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}2&  0\\0&-2\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}0&-2\\-2&0\end{array}\right)
    answer is \left(\begin{array}{cc}0&2\\2&0\end{array}\right)
    thanks
    You need to remember that you work from right to left when you're combining transformation matrices. So to find the matrix that represents A followed by C, you need to work out the product CA.

    This is because the transformation matrix is always written on the left of the object that it's working on. So if I transform (p,q) with matrix A, I'll get:
    A\left(\begin{array}{c}p\\q\end{array}\right)
    =\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{c}p\\q\end{  array}\right)
    Then if we do C to this, we get:
    CA\left(\begin{array}{c}p\\q\end{array}\right)
    =\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\left(\begin{array}{cc}1&0  \\0&-1\end{array}\right)\left(\begin{array}{c}p\\q\end{  array}\right)
    So A followed by C is CA, not AC.

    Can you see where you've gone wrong now?

    Grandad
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