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another matrix transformation
Hi folks,
Given the following transformation:
 <br />
\left( \begin{array}{c} x \\ y \end{array} \right) + <br />
\left( \begin{array}{c} 4 \\ 0 \end{array} \right))
show that it may be expressed as a reflection in the line x = 4/3 followed by an enlargement and give the centre and scale factor of this enlargement.
using:


I have plotted the results of the transformation as shown in the diagram. As far as I can see, the transform represents a reflection about x = 1 and an enlargement, scale factor 2 about the origin. The first part of my diagram shows the transformation, the second the reflection about x = 1 and the third the enlargement. Just to clarify this last bit: If the enlargement had a cenre at A (1,0) then A would stay where it is, O would move to (3,0) B would go to (3,2) and C to (1,2) but if we move the centre back to the origin, we should get the result in the top part of the diagram.
I just don't see any way that the transform can be a reflection in x = 4/3 floowed by an enlargement of 2 centre (4/3, 0) which is the answer provided!
Can anyone help? It's not so hard, is it?
Re: another matrix transformation
Sorry to bump this old thread, but I'm having the same problem with this same question. I really don't know how I can break equation down to a reflection in the line x = 4/3 followed by an enlargement. Any suggestions?
Re: another matrix transformation
Okay, I think I might have it. To get a reflection in x = 4/3, it's like a reflection in the y-axis followed by a translation. Knowing that the coordinates (4/3, 0) will be invariant:
(4/3 = (-1 0 (4/3 + (a
0) 0 -1) 0) b)
This results in a = 8/3 and b = 0.
Now, the original equation written by the original poster can be changed to:
(x' = (2 0 (-1 0 + (8/3 + (4/3
y') 0 2) 0 1) 0) 0)
We can see the two matrices that must be relevant to the enlargement (scale factor 2), and by
(x = (2 0 (x + (4/3
y) 0 2) y) 0)
we can find the invariant point, which would be centre of the enlargement. Working the above out reveals this centre to be (4/3, 0).
I'm not sure at how efficient this method was. Was there a better way to tackle this question?
Edit: Ahhh, it won't accept the spaces for the matrices. I'll have to figure a way how to present the above properly, then edit again. I apologise.
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Re: another matrix transformation
Quote:
Originally Posted by
s_ingram
Given the following transformation:
using:

I just don't see any way that the transform can be a reflection in x = 4/3 floowed by an enlargement of 2 centre (4/3, 0) which is the answer provided!
Can anyone help? It's not so hard, is it?
Any point (x,y) is transferred to (4-2x, 2y):
= )
In vector form the resultant of this transformation is ( I use complex form):
=
Then: mirror line is: 4-3x=0 & y=0, or x=4/3...
http://mathhelpforum.com/attachment....1&d=1346554952