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Math Help - Affine Transformations

  1. #1
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    Post Affine Transformations

    In this question, f and g are both affine transformations. The transformations f is a reflection in the line y = -x +1, and g maps the points (0,0), (1,0) and (0,1) to the points (1,5), (1,-4) and (0,-5) respectively.

    a) Determine g in the form g(x) = Ax + a, where A is a 2 x 2 matrix and a is a vector with two components.

    b) Write down the matrix that represents reflection in an appropriate line through the origin, and find f (in the same form as for g in part (a)) by first translating an appropriate point to the origin.

    c) Find the affine transformation g o f (in the same form as for g and f in parts (a) and (b)).

    d) Hence or otherwise, find the images of the points (0,0), (0,-2) (2,-2) and (2,0) under g o f. Mark these points and images on the same diagram, making it clear which points maps to which. Describe g o f geometrically as a single transformation.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fair_lady0072002
    In this question, f and g are both affine transformations. The transformations f is a reflection in the line y = -x +1, and g maps the points (0,0), (1,0) and (0,1) to the points (1,5), (1,-4) and (0,-5) respectively.

    a) Determine g in the form g(x) = Ax + a, where A is a 2 x 2 matrix and a is a vector with two components.
    First we observe that:

    <br />
 g([0,0]')=A{0 \brack 0}+a={1 \brack 5}<br />

    which implies that:

    a = {1 \brack 5}\ \ \ \dots(1) .

    Now: g([1,0]')=A{1 \brack 0}+{1 \brack 5}={1 \brack -4}, so:

    <br />
A{1 \brack 0}={0 \brack -9}\ \ \ \dots(2)<br />
.

    Also: g([0,1]')=A{0 \brack 1}+{1 \brack 5}={0 \brack -5}, so:

    <br />
A{0 \brack 1}={-1 \brack -10}\ \ \ \dots(3)<br />
.

    Now (2) and (3) imply that:

    <br />
A=\left[ \begin{array}{cc}<br />
0&-1\\<br />
-9&-10<br />
\end{array}\right]<br />
,

    so:

    <br />
g(x)=\left[ \begin{array}{cc}<br />
0&-1\\<br />
-9&-10<br />
\end{array}\right]<br />
x+{1 \brack 5}<br />
.

    RonL
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