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Math Help - Equivalent transformations by matrices

  1. #1
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    Equivalent transformations by matrices

    I have done the first part. I can't think of a way to do the second part other than by making guesses. The answers are y=2x and y=-0.5x.
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  2. #2
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    I have done it now. The second part is done by starting similarly to the first part.
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  3. #3
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    Yes. To do the first part, look at \begin{pmatrix}2 & 1 \\ 2 & -2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}2 & 1 \\ 2 & -2\end{pmatrix}\begin{pmatrix}x \\ 3x\end{pmatrix} = \begin{pmatrix}2x+ 3x \\ 2x- 6x\end{pmatrix}= \begin{pmatrix}5x \\ -4x\end{pmatrix}.

    So x'= 5x and y'= -4x. x= x'/5 so y'= -4(x'/5)= (-4/5)x'.

    Any "line through the origin" is of the form y= ax with slope a. The line at right angles to that is y= (-1/a)x so you want to find a such that
    \begin{pmatrix}2 & 1 \\ 2 & -2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}2 & 1 \\ 2 & -2\end{pmatrix}\begin{pmatrix}x \\ ax\end{pmatrix} =\begin{pmatrix}2x+ ax \\ 2x- 2ax\end{pmatrix}= \begin{pmatrix}(2+a)x \\ (2- 2a)x\end{pmatrix}= \begin{pmatrix}x' \\ -(1/a)x'\end{pmatrix}

    That is, we must have x'= (2+a)x and y'= (-1/a)x'= (2- 2a)x. Dividing one equation by the other so that we eliminate x' on both sides of the equation, -\frac{1}{a}= \frac{2+ a}{2+ 2a}. Multiply on both sides by (2+ 2a)a to get rid of the fractions.

    2+ 2a= (2+ a)a= 2a+ a^2 which reduces to a^2= 2.

    Ouch, I divided the two equations incorrectly. Dividing x'= (2+ a)x by (-1/a)x'= (2- 2a)x gives \frac{x'}{-\frac{1}{a}x'}= \frac{(2+a)x}{(2- 2a)x} or -a= \frac{2+a}{2- 2a}. Multiplying both sides by 2- 2a gives -(2- 2a)a= -2a+ 2a^2= 2+ a. 2a^2- 3a2 2= (2a+ 1)(a- 2)= 0 which has roots a= 2 and a= -1/2. the lines, y= ax, are y= 2x and y= (-1/2)x as you said.
    Last edited by HallsofIvy; August 5th 2010 at 04:34 AM.
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  4. #4
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    I used a slightly different method, still observing that multiplied gradients of perpendicular lines equals -1. I think you might have a mistake. You haven't finished with the equations of lines required.
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  5. #5
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    Could you also fix the missing equals sign that is missing from the second row of matrix multiplications? Thanks.
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