# another matrix transformation

• Mar 20th 2010, 10:41 AM
s_ingram
another matrix transformation
Hi folks,

Given the following transformation:

$\displaystyle \overline{x'} = \left( \begin{array}{cc} -2 & 0 \\ 0 & 2 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) + \left( \begin{array}{c} 4 \\ 0 \end{array} \right)$

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:
$\displaystyle x' = -2x + 4$
$\displaystyle y' = 2y$

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?
• Sep 1st 2012, 02:25 PM
SpriteAlpha
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?
• Sep 1st 2012, 02:59 PM
SpriteAlpha
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.
• Sep 1st 2012, 06:55 PM
MaxJasper
Re: another matrix transformation
Quote:

Originally Posted by s_ingram
Given the following transformation:
using:
$\displaystyle x' = -2x + 4$
$\displaystyle y' = 2y$
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):

$\displaystyle \left(\begin{array}{cc} -2 & 0 \\ 0 & 2\end{array}\right).\left(\begin{array}{c} x \\ y\end{array}\right)+\left(\begin{array}{c} 4 \\ 0\end{array}\right)$ = $\displaystyle \left(\begin{array}{c} 4-2 x \\ 2 y\end{array}\right)$

In vector form the resultant of this transformation is ( I use complex form):

$\displaystyle 4-2x+2y i -(x+y i)$ = $\displaystyle 4 - 3 x + y i$

Then: mirror line is: 4-3x=0 & y=0, or x=4/3...
http://mathhelpforum.com/attachment....1&d=1346554952