affine transformation

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gurnet

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Jan 20, 2012
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In this question, f and g are both affine transformations. The
transformation f is reflection in the line y = 2, and the transformation g
maps the points (0, 0), (1, 0) and (0, 1) to the points (1, 1), (2, 2) and
(3,−1), 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). [4]

(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)). [6]

(c) Use the expressions that you found for f(x) and g(x) 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). [5]

(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).)



I have already done part a but am quite confused from b onwards, any help would be greatly appreciated.
 
For part b:
Find a transformation \(\displaystyle f_1\) that sends (x,y) to (x,y-2).
Then, find a transformation \(\displaystyle f_2\) that sends (x,y) to (x,-y).
Then find a transformation \(\displaystyle f_3\) that sends (x,y) to (x,y+2).
Let \(\displaystyle f=f_3\circ f_2 \circ f_1\).

For parts c and d, just do what the problem says.
 
im still really confused, how do i translate from (x,y) to (x,y-2)? more detail would be great.
 
\(\displaystyle f_1(v) = Av + a\)
\(\displaystyle f_1(v) = Iv + (0,-2)\)
where \(\displaystyle v = (x,y)\)
Does that help?
 
not really,

sorry, i feel really stupid now. Im not sure what Iv is also how do i get (X,Y) when all i know is y=2

im having real difficulties with this, could you break it down in more detail for me,

thanks so much for your help.
 
i have tried and come up with

matrix
0 1. x + 0
0 -1 .....4
am i anywhere near the answer?
 
Last edited:
I apologize. I was using shorthand notation. In this case, I meant for \(\displaystyle I\) to be the 2x2 identity matrix.

\(\displaystyle f_1\left(\begin{pmatrix}x\\y\end{pmatrix}\right) = \begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} + \begin{pmatrix}0\\-2\end{pmatrix}\)

What you want overall:
You want a function that will take a point \(\displaystyle \begin{pmatrix}x\\y\end{pmatrix}\) and return a point that is its reflection across the line \(\displaystyle y=2\). So, you want:
\(\displaystyle \begin{pmatrix}x\\2+y\end{pmatrix} \mapsto \begin{pmatrix}x\\2-y\end{pmatrix}\) (this is the reflection across the line \(\displaystyle y=2\)).
So, my suggestion was to find three functions. The first would take \(\displaystyle \begin{pmatrix}x\\y\end{pmatrix} \mapsto \begin{pmatrix}x\\y-2\end{pmatrix}\). Let's see what happens if we apply this function to \(\displaystyle f_1\left(\begin{pmatrix}x\\2+y\end{pmatrix}\right) = \begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}x\\2+y\end{pmatrix}+\begin{pmatrix}0\\-2\end{pmatrix} = \begin{pmatrix}x\\y\end{pmatrix}\)

Next, I suggested we want a function that will take a point \(\displaystyle \begin{pmatrix}x\\y\end{pmatrix} \mapsto \begin{pmatrix}x\\-y\end{pmatrix}\)
Finally, I suggested we want a function that will take a point \(\displaystyle \begin{pmatrix}x\\y\end{pmatrix} \mapsto \begin{pmatrix}x\\2+y\end{pmatrix}\)

So, if \(\displaystyle f_1\) takes \(\displaystyle \begin{pmatrix}x\\2+y\end{pmatrix} \mapsto \begin{pmatrix}x\\y\end{pmatrix}\), \(\displaystyle f_2\) takes \(\displaystyle \begin{pmatrix}x\\y\end{pmatrix} \mapsto \begin{pmatrix}x\\-y\end{pmatrix}\) and \(\displaystyle f_3\) takes \(\displaystyle \begin{pmatrix}x\\-y\end{pmatrix} \mapsto \begin{pmatrix}x\\2+(-y)\end{pmatrix}\) then the composition of those three functions does exactly what we want.

Can you come up with the functions \(\displaystyle f_2, f_3\)?
 
Thank you so much!!!!!

i think i have it now, my final answer after combining all three is

1 0.....x....+...0
0 -1.....y...+....4

That was a really good explanation.
 
for part c i have combined both f(x) and g(x) and come up with

1 2.....x...+.....9
-1 2....y....+....-9

is this correct?

sorry for the way it is written but im not sure how to write a matrix on this site.
 
I don't believe that is correct. To write matrices, use this:

Code: 
[ tex]\begin{pmatrix}a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2}\end{pmatrix}[/tex]

But remove the space before the first "tex" in [ tex]

What did you get for the function \(\displaystyle g\)?
 
g(x) =
\(\displaystyle \begin{pmatrix}{1} & 2} \\ 1} & -2}\end{pmatrix}\) X +\(\displaystyle \begin{pmatrix}{1} \\ 1}\end{pmatrix}\)​




f(x) = \(\displaystyle \begin{pmatrix}{1} & 0} \\ 0} & -1}\end{pmatrix}\) \(\displaystyle \begin{pmatrix}{X} \\ Y}\end{pmatrix}\)+\(\displaystyle \begin{pmatrix}{0 \\ 4}\end{pmatrix}\)


So fog(x) = \(\displaystyle \begin{pmatrix}{1} & 2} \\ -1} & 2}\end{pmatrix}\) \(\displaystyle \begin{pmatrix}{X} \\Y}\end{pmatrix}\) + \(\displaystyle \begin{pmatrix}{1} \\3}\end{pmatrix}\)

Am i getting closer?​
 
Last edited:
so for part d i used a simultaneous equation but it does not work... any other suggestions to show if we have or do not have points????
 
bennyd2012 said:
so for part d i used a simultaneous equation but it does not work... any other suggestions to show if we have or do not have points????
Click to expand...
That depends on the simultaneous equation you are using. gurnet said above, \(\displaystyle f\circ g\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}1 & 2\\-1 & 2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix} \)

So, let \(\displaystyle f\circ g\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}x\\y\end{pmatrix}\).

\(\displaystyle \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}1 & 2\\-1 & 2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix} = \begin{pmatrix}x+2y+1\\-x+2y+3\end{pmatrix}\)

Thus, \(\displaystyle x = x+2y+1\) and \(\displaystyle y=-x+2y+3\). Solving, we get: \(\displaystyle y=-\frac{1}{2}, x=\frac{5}{2}\). Plugging that in, we see:

\(\displaystyle f\circ g\begin{pmatrix}\frac{5}{2}\\-\frac{1}{2}\end{pmatrix} = \begin{pmatrix}1 & 2\\-1 & 2\end{pmatrix}\begin{pmatrix}\frac{5}{2}\\-\frac{1}{2}\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix}\) \(\displaystyle = \begin{pmatrix}\frac{3}{2}\\-\frac{7}{2}\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix} = \begin{pmatrix}\frac{5}{2}\\-\frac{1}{2}\end{pmatrix}\)
 
Last edited:
SlipEternal said:
That depends on the simultaneous equation you are using. gurnet said above, \(\displaystyle g\circ f\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}1 & 2\\-1 & 2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix} \)

So, let \(\displaystyle g\circ f\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}x\\y\end{pmatrix}\).

\(\displaystyle \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}1 & 2\\-1 & 2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix} = \begin{pmatrix}x+2y+1\\-x+2y+3\end{pmatrix}\)

Thus, \(\displaystyle x = x+2y+1\) and \(\displaystyle y=-x+2y+3\). Solving, we get: \(\displaystyle y=-\frac{1}{2}, x=\frac{5}{2}\). Plugging that in, we see:

\(\displaystyle g\circ f\begin{pmatrix}\frac{5}{2}\\-\frac{1}{2}\end{pmatrix} = \begin{pmatrix}1 & 2\\-1 & 2\end{pmatrix}\begin{pmatrix}\frac{5}{2}\\-\frac{1}{2}\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix}\) \(\displaystyle = \begin{pmatrix}\frac{3}{2}\\-\frac{7}{2}\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix} = \begin{pmatrix}\frac{5}{2}\\-\frac{1}{2}\end{pmatrix}\)
Click to expand...

Thats exactly waht i got but i didnt think it was correct! as it said f(g(x,y)) = (x,y)
so should what we sub in for f of g not = our answer. as in if we sub in 5/2 for x in the equation and y =1/2 and work it out we should get back the original value we subbed in??
 
ignore the last post have it now... sorry had a wrong sign when worked it out... thanks :p
 
FoG

Hi I was just wondering about FoG (part c) and how the vector becomes (1,3) when f is (0,4) and g is (1,1)? Just curious :)
 
Zeuxis said:
Hi I was just wondering about FoG (part c) and how the vector becomes (1,3) when f is (0,4) and g is (1,1)? Just curious :)
Click to expand...

\(\displaystyle f\circ g(x) = f(g(x))\)
\(\displaystyle g\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}1 & 2 \\ 1 & -2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} + \begin{pmatrix}1 \\ 1\end{pmatrix}\)
\(\displaystyle g\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}x+2y+1\\x-2y+1\end{pmatrix}\)
\(\displaystyle f\begin{pmatrix}x+2y+1\\x-2y+1\end{pmatrix} = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}x+2y+1\\x-2y+1\end{pmatrix}+\begin{pmatrix}0\\4\end{pmatrix}\)
\(\displaystyle f\circ g\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}x+2y+1\\-x+2y+3\end{pmatrix} = \begin{pmatrix}1&2\\-1&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix}\)
 
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