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Thread: 2x2 matrices - reflection & inverse

  1. #1
    Junior Member froodles01's Avatar
    Nov 2011
    Surrey UK

    2x2 matrices - reflection & inverse

    Problem is;
    f is reflection in the y-axis and
    g is the linear transformation represented by the matrix

    2 3
    1 1

    Find the matrices representing the following linear transformations
    i) f
    ii) gof
    iii) g^-1

    now iii) is fine & ii) would be OK if I knew i)

    Now, rule for a refletion has the form;
    qθ: R2 --> R2
    x--> qθ x

    where qθ = cos(2θ) sin(2θ)
    . . . . . . . sin(2θ) -cos(2θ)

    So please, how do I get f from his rule.
    Last edited by froodles01; Oct 17th 2012 at 03:33 AM. Reason: difficulty in formatting - probbly still is.
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  2. #2
    MHF Contributor

    Mar 2011

    Re: 2x2 matrices - reflection & inverse

    the reflection through the y-axis, preserves the y-axis, right? this means that the matrix for f sends (0,1) to (0,1) (since (0,1) lies on the y-axis). so all we need to know is what f does to (1,0), and we're done (almost).

    but clearly, f sends (1,0) to it's "reflection", (1,0) in the "opposite direction", which is -(1,0) = (-1,0).

    so let's say the matrix for f is:

    $\displaystyle \begin{bmatrix}a&b\\c&d \end{bmatrix}$.


    $\displaystyle \begin{bmatrix}a&b\\c&d \end{bmatrix} \begin{bmatrix}1\\0 \end{bmatrix} = \begin{bmatrix}-1\\0 \end{bmatrix}$

    and actually doing the matrix multiplication on the LHS gives us a = -1, c= 0.

    similarly, we see that:

    $\displaystyle \begin{bmatrix}a&b\\c&d \end{bmatrix} \begin{bmatrix}0\\1 \end{bmatrix} = \begin{bmatrix}0\\1 \end{bmatrix}$

    so b = 0, d = 1.

    thus the matrix for f is:

    $\displaystyle \begin{bmatrix}-1&0\\0&1 \end{bmatrix} $

    which clearly sends the point (x,y) to (-x,y).

    how does this relate to your qθ rule? note that the angle the y-axis forms with the x-axis is θ = π/2, so:

    cos(2θ) = cos(π) = -1
    sin(2θ) = sin(π) = 0

    so we get the same matrix as above.
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