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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Nomvalt on April 05, 2010, 12:37:06 pm

Title: Matrix transformations
Post by: Nomvalt on April 05, 2010, 12:37:06 pm: For the following transformations, where T: R²--> R², state what transformation T represents and determine the image of the equation $f(x) = \frac{1}{x}$ .

$T\left (\begin{bmatrix} x\\y \end{bmatrix}\right )= \begin{bmatrix} 2 &0 \\ 0 & \frac{-1}{2} \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}+ \begin{bmatrix} 3\\-1 \end{bmatrix}$

$=\begin{bmatrix} 2x + 3 \\ \frac{-1}{2}y -1 \end{bmatrix}$

Dilation by a factor of $\frac{1}{2}$ from the x-axis
Reflection in the x-axis
Dilation by a factor of 2 from the y axis
Translation by 3 units to the right and 1 unit down

The answer for the image is $f(x) = \frac{-1}{x-3} -1$

For some strange reason I got $f(x) = \frac{-1}{2(2x-3)^{2}} - 1$ which is obviously wrong.
I don't understand how you get from the image described by the matrix to the image as the equation above. Anybody out there willing to help? :(
Title: Re: Matrix transformations
Post by: Blakhitman on April 05, 2010, 01:09:17 pm: ${x}'=2x + 3$

$\therefore x= \frac{{x}'-3}{2}$

${y}'=\frac{-1}{2}\cdot y - 1$

${y}' + 1 = \frac{-1}{2}\cdot y$

multiply both sides by 2 to get:

$2{y}' + 2= -1y$

multiply both sides by negative one to get:

$y=-2y - 2$

Then sub them back into original equation

$-2y - 2= \frac{1}{\frac{x-3}{2}}$

then solve for y to get:

$-2y - 2= \frac{2}{x-3}$

$-2y= \frac{2}{x-3} + 2$

Divide both sides by negative 2:

$y= -\frac{1}{x-3} - 1$

$\therefore$ the rule of the image is: $y= -\frac{1}{x-3} - 1$

hope it makes sense
Title: Re: Matrix transformations
Post by: the.watchman on April 06, 2010, 07:24:43 am: Shortcut: You have $x = \frac{x'-3}{2}$ AND $y'=-\frac{1}{2}y-1$

Sub $y=\frac{1}{x}$ into the second eqn instead of solving for y:

$y'=-\frac{1}{2x}-1$

Then sub the first eqn in to get your answer, much easier than solving both eqns :)
Title: Re: Matrix transformations
Post by: Juddinator on April 06, 2010, 10:26:16 am: Quote from: the.watchman on April 06, 2010, 07:24:43 am
Shortcut: You have $x = \frac{x'-3}{2}$ AND $y'=-\frac{1}{2}y-1$

Sub $y=\frac{1}{x}$ into the second eqn instead of solving for y:

$y'=-\frac{1}{2x}-1$

Then sub the first eqn in to get your answer, much easier than solving both eqns :)
yeah, that's awesomeness for ya in methods right there :P
good work watchman! :D
Title: Re: Matrix transformations
Post by: Blakhitman on April 06, 2010, 03:31:42 pm: Quote from: the.watchman on April 06, 2010, 07:24:43 am
Shortcut: You have $x = \frac{x'-3}{2}$ AND $y'=-\frac{1}{2}y-1$

Sub $y=\frac{1}{x}$ into the second eqn instead of solving for y:

$y'=-\frac{1}{2x}-1$

Then sub the first eqn in to get your answer, much easier than solving both eqns :)

What do you mean by "much easier than solving both equations"?
Title: Re: Matrix transformations
Post by: the.watchman on April 06, 2010, 03:48:31 pm: Quote from: Blakhitman on April 06, 2010, 03:31:42 pm
Quote from: the.watchman on April 06, 2010, 07:24:43 am
Shortcut: You have $x = \frac{x'-3}{2}$ AND $y'=-\frac{1}{2}y-1$

Sub $y=\frac{1}{x}$ into the second eqn instead of solving for y:

$y'=-\frac{1}{2x}-1$

Then sub the first eqn in to get your answer, much easier than solving both eqns :)

What do you mean by "much easier than solving both equations"?

I meant that with the two x'=... AND y'=... you wouldn't have to solve both of them for x and y and sub them in, just solve one instead. It shortens it by at least four or five steps :)
Title: Re: Matrix transformations
Post by: Blakhitman on April 06, 2010, 03:52:21 pm: Oh lol, didn't even realise you didn't solve both. Anyway the way I did it may seem long because I modified the post to show each step, but really you don't need those "four" steps when actually working out :P.

I'm not saying yours isn't faster...it's just not "that" much faster :P
Title: Re: Matrix transformations
Post by: the.watchman on April 06, 2010, 03:53:54 pm: Quote from: Blakhitman on April 06, 2010, 03:52:21 pm
Oh lol, didn't even realise you didn't solve both. Anyway the way I did it may seem long because I modified the post to show each step, but really you don't need those "four" steps when actually working out :P.

I'm not saying yours isn't faster...it's just not "that" much faster :P

Lol, not "that" much, but still is better in my opinion :P
Anyway, it doesn't matter, fairly minor thing :D