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Thread: 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. $\displaystyle \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)$
    B, an enlargement by a factor 2. $\displaystyle \left(\begin{array}{cc}2&0\\0&2\end{array}\right)$
    C, a rotation by $\displaystyle \frac{\pi}{2}$ about O. $\displaystyle \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$
    D, a stretch by a factor of 3 in the direction Oy. $\displaystyle \left(\begin{array}{cc}1&0\\0&3\end{array}\right)$
    E, a reflection in the line y=x. $\displaystyle \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$
    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. $\displaystyle \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)$
    B, an enlargement by a factor 2. $\displaystyle \left(\begin{array}{cc}2&0\\0&2\end{array}\right)$
    C, a rotation by $\displaystyle \frac{\pi}{2}$ about O. $\displaystyle \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$
    D, a stretch by a factor of 3 in the direction Oy. $\displaystyle \left(\begin{array}{cc}1&0\\0&3\end{array}\right)$
    E, a reflection in the line y=x. $\displaystyle \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$
    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 $\displaystyle \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}0&-1\\-1&0\end{array}\right)$?
    and A following C would be $\displaystyle \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$
    The answers for these two are reversed.
    C carried out after E $\displaystyle \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\left(\begin{array}{cc}0&1 \\1&0\end{array}\right)=\left(\begin{array}{cc}1&0 \\0&-1\end{array}\right)$
    answer is $\displaystyle \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$
    and also A followed by B followed by C $\displaystyle \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}2&0\\0&2 \end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}2& 0\\0&-2\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}0&-2\\-2&0\end{array}\right)$
    answer is $\displaystyle \left(\begin{array}{cc}0&2\\2&0\end{array}\right)$
    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 $\displaystyle \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}0&-1\\-1&0\end{array}\right)$?
    and A following C would be $\displaystyle \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$
    The answers for these two are reversed.
    C carried out after E $\displaystyle \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\left(\begin{array}{cc}0&1 \\1&0\end{array}\right)=\left(\begin{array}{cc}1&0 \\0&-1\end{array}\right)$
    answer is $\displaystyle \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$
    and also A followed by B followed by C $\displaystyle \left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\left(\begin{array}{cc}2&0\\0&2 \end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}2& 0\\0&-2\end{array}\right)\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=\left(\begin{array}{cc}0&-2\\-2&0\end{array}\right)$
    answer is $\displaystyle \left(\begin{array}{cc}0&2\\2&0\end{array}\right)$
    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 $\displaystyle A$ followed by $\displaystyle C$, you need to work out the product $\displaystyle CA$.

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

    Can you see where you've gone wrong now?

    Grandad
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