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    Name:  DSCF5261[1].jpg
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Size:  509.1 KB Hi guys
    So I managed to complete 2a), find the 2 by 2 matrix (2 1, 2 3), then prove that it maps y+x=0 onto itself. However I can't work out the other line
    Any help would be greatly appreciated

    So for the first part you are showing that the line y=-x gets mapped to y=-x
    So we have  \displaystyle \begin{bmatrix} 2&1 \\ 2&3 \end{bmatrix} \begin{bmatrix} x \\ -x \end{bmatrix} = \begin{bmatrix} X \\ Y \end{bmatrix} .
    So just show that X+Y=0 and you will have shown that the line x+y=0 is invariant under T.
    So did invariant lines under T we have to do
     \displaystyle \begin{bmatrix} 2&1 \\ 2&3 \end{bmatrix} \begin{bmatrix} x \\ mx+c \end{bmatrix} = \begin{bmatrix} X \\ mX+c \end{bmatrix} .
    What this is basically saying is that any invariant line with equation y=mx+c Is transformed to Y=mX+c (the same equation). So now using this you have to try and find the equation of the line that under T is mapped to itself.
    So all you need to do is do what you did when you shown that x+y=0 is invariant under T but this time we are doing it for a general line with gradient m and passing through (0,c).
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Updated: March 23, 2016

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