# composite transformation

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• Apr 19th 2011, 12:15 AM
vidhi96
composite transformation
I am going through this past papers and I am utterly confused with the question- looked at all the notes and examples but simply can't understand what to do for part (b) Please explain. I have sucessfully done part (a) and answer is:

A= (-2 - -2, -1+2
3-2 2-2) = (0 1
1 0)

And a = (-2
2)

Hence g(x) = (0 1 + (-2
1 0)x 2)

this is the whole question:
In this question, f and g are both affine transformations. The
transformation f is reflection in the line y = 1, and the transformation g
maps the points (0, 0), (1, 0) and (0, 1) to the points (−2, 2), (−2, 3) and
(−1, 2), respectively.
(a) Determine g (in the form g(x) = Ax + a, where A is a 2×2 matrix
and a is a vector with two components).
(b) Express f as a composite of three transformations: a translation,
followed by reflection in a line through the origin, followed by a
translation. Hence determine f (in the same form as you found g in
part (a)).
(c) Use the expressions that you found for f and g in parts (a) and (b) to
calculate f(g(x)), and hence find the affine transformation f ◦ g in the
same form as you found g in part (a).

Thank you
• Apr 19th 2011, 05:03 AM
Deveno
for part (b) use the translation T(x,y) = (x,y) - (0,1), to move the the line y = 1 so that it now goes through the origin. then you may apply the reflection

R(x,y) = (x,-y), and finally apply the reverse translation (you should be able to easily write down what this is). can you see how to put f in the required form?
• Apr 19th 2011, 06:14 AM
vidhi96
what angle would I consoder
Hi
Thanks Deveno
What confuses me as I had been given a similar problem where x=2 - so after resolving part (a) as I did in this one, my tutor gave me follwoing method which we are expected to follow: can you help me make sense out of it:

this is what he wrote: Since x=2 makes an angle pi/2 with the positive x- axis, it follows that refelction in this line is qpi/2
so I need to start my part b like this- which I never understood at the time and I am still confused- can you please help to get this correct.(Worried)

• Apr 19th 2011, 09:18 AM
Deveno
Quote:

Originally Posted by vidhi96
Hi
Thanks Deveno
What confuses me as I had been given a similar problem where x=2 - so after resolving part (a) as I did in this one, my tutor gave me follwoing method which we are expected to follow: can you help me make sense out of it:

this is what he wrote: Since x=2 makes an angle pi/2 with the positive x- axis, it follows that refelction in this line is qpi/2
so I need to start my part b like this- which I never understood at the time and I am still confused- can you please help to get this correct.(Worried)

what do you mean by q? the line y = 1 makes an angle 0 with the x-axis. now, normally a reflection across a line of angle θ through the origin is:

[ cos(2θ) sin(2θ)]
[sin(2θ) -cos(2θ)], but when θ = 0, this matrix has a particularly simple form.

the form above is derived by seeing that a reflection about y = θx, can be viewed as a rotation of -θ, followed by a reflection across the x-axis, followed by a rotation of angle θ:

[cosθ -sinθ][ 1 0][ cosθ sinθ]
[ sinθ cosθ][0 -1][-sinθ cosθ]
• Apr 19th 2011, 10:06 AM
vidhi96
oh
Quote:

Originally Posted by Deveno
what do you mean by q? the line y = 1 makes an angle 0 with the x-axis. now, normally a reflection across a line of angle θ through the origin is:

[ cos(2θ) sin(2θ)]
[sin(2θ) -cos(2θ)], but when θ = 0, this matrix has a particularly simple form.

the form above is derived by seeing that a reflection about y = θx, can be viewed as a rotation of -θ, followed by a reflection across the x-axis, followed by a rotation of angle θ:

[cosθ -sinθ][ 1 0][ cosθ sinθ]
[ sinθ cosθ][0 -1][-sinθ cosθ]

sorry I should have mentioned- notation r= represents rotation and q is for reflection
• Apr 19th 2011, 10:12 AM
Deveno
are you still confused as to how to proceed?
• Apr 19th 2011, 11:47 PM
vidhi96
confused and frustrated
see attached file- that's what he is expecting and I am completely lost(Speechless)
• Apr 20th 2011, 12:43 AM
Deveno
what i have written as T(x,y) would be t(0,-1). and what i have written as R would be q0.

so the entire transformation is f = t(0,1) o q0 o t(0,-1).

t(0,-1) moves the line y = 1 "down 1" so that now we have a line going through the origin. q0 reflects the plane about the x-axis (y = 0),

which keeps all the x's the same, but exchanges y with -y. then t(0,1) moves the reflected image "back up 1". on an arbitrary point (x,y), we have:

f(x,y) = t(0,1)(q0(t(0,-1)(x,y))) =

[...([x]....[ 0])].....[0]
[q0([y] + [-1])] + [1] =

[1 .0][..x.].....[0]
[0 -1][y-1] + [1] =

[1 .0][x]...[1 .0][..0]....[0]
[0 -1][y] - [0 -1][-1] + [1] =

[1 .0][x]....[0]
[0 -1][y] + [2], so that our matrix A is just the matrix for q0 and the vector a is (0,2).

as a check, let's verify that f leaves points on the line y = 1 unchanged.

f(x,1) = (x,-1) + (0,2) = (x,1). see? now the point (x,1+k) should go to (x,1-k), we can verify this, too:

f(x,1+k) = (x,-1-k) + (0,2) = (x,1-k), as expected.
• Apr 20th 2011, 04:41 AM
vidhi96
thanks
Quote:

Originally Posted by Deveno
what i have written as T(x,y) would be t(0,-1). and what i have written as R would be q0.

so the entire transformation is f = t(0,1) o q0 o t(0,-1).

t(0,-1) moves the line y = 1 "down 1" so that now we have a line going through the origin. q0 reflects the plane about the x-axis (y = 0),

which keeps all the x's the same, but exchanges y with -y. then t(0,1) moves the reflected image "back up 1". on an arbitrary point (x,y), we have:

f(x,y) = t(0,1)(q0(t(0,-1)(x,y))) =

[...([x]....[ 0])].....[0]
[q0([y] + [-1])] + [1] =

[1 .0][..x.].....[0]
[0 -1][y-1] + [1] =

[1 .0][x]...[1 .0][..0]....[0]
[0 -1][y] - [0 -1][-1] + [1] =

[1 .0][x]....[0]
[0 -1][y] + [2], so that our matrix A is just the matrix for q0 and the vector a is (0,2).

as a check, let's verify that f leaves points on the line y = 1 unchanged.

f(x,1) = (x,-1) + (0,2) = (x,1). see? now the point (x,1+k) should go to (x,1-k), we can verify this, too:

f(x,1+k) = (x,-1-k) + (0,2) = (x,1-k), as expected.

Hey this is great it makes lot more sense. I will resolve the whole thing. Thanks a lot.
you are brilliant(Happy)
• Apr 20th 2011, 07:24 AM
vidhi96
for part c)
for the last bit
f o g i got
f o g ={1 0} * {0 1} X + {1 0} * {-2} + {0}
{0 -1} {1 0} {0 -1} {2} {2}

= {0 1} X + {-2}
{-1 0} {0}

so AM threw with part c. I will appriciate if you can guide me for part d

d) Use your answer to part (c) to determine any points (x, y) that are
left unchanged by the transformation f ◦ g, or to show that there are
no such points. (To do this, find any values of x and y for which
f(g(x, y)) = (x, y).)
• Apr 20th 2011, 10:48 AM
Deveno

if we explicity calculate f(g(x,y)), we get f(g(x,y)) = (y-2, -x).

so if (x,y) = (y-2,-x) we get 2 equations in x and y. can you solve these?
• Apr 21st 2011, 02:19 AM
vidhi96
hey
Quote:

Originally Posted by Deveno

if we explicity calculate f(g(x,y)), we get f(g(x,y)) = (y-2, -x).

so if (x,y) = (y-2,-x) we get 2 equations in x and y. can you solve these?

Hi follwoing this I get two equations:
x=y-2 and y=-x
solving them I get x=-1 and y=1

is this correct?
• Apr 21st 2011, 11:25 AM
Deveno
you can check this: calculate f(g(-1,1)). is it equal to (-1,1)?
• May 15th 2011, 02:21 PM
putney98
Hello Deveno,
I found this thread (and MHF) this weekend as I was searching for some help in the subject and this thread really helped me.
I've got to the conclusion that f(g(x)) = f o g.
Some places would write f(g(x)) = f o g (x).
Am I right or did I loose the plot?
Thank you
• May 15th 2011, 03:17 PM
Deveno
yes. to evaluate (fog) at the vector x, one evaluates g at the vector x, and then evaluates f at the vector g(x).

...g.....f.....
X-->Y-->Z (here X is the space where the x's live, Y is the "target space" of g (which could be the same as X, just "re-arranged"), and Z is the "target space" of f).

g(x) = y
f(y) = z
fog(x) = f(g(x)) = f(y) = z

in the above problem, X = Y = Z = R^2, the real 2-dimensional plane, and f and g are affine mappings.
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