# Linear Transformations

• Mar 30th 2012, 12:22 PM
GrigOrig99
Linear Transformations
The question statement says find f(L), which I have highlighted in bold. However, the book's answer seems to continue on and calcualte for y from the f(L) equation, which I proceed to work out, although my final answer is slightly off. I'm slightly confused as to whether or not the highlihted value for f(L) is in fact f(L)! Can anyone help?

Many thanks.

Q.
L is the line x - y + 1 = 0. f is the transformation f: (x, y) ---> (x', y') where: x' = 2x - y & y' = y. Find f(L).

Attempt: If y = y', then x' = 2x - y => 2x = x' - y => 2x = x' - y' => x = $\displaystyle \frac{x' - y'}{2}$

If L = x - y + 1 = 0, then f(L) = $\displaystyle \frac{x' - y'}{2}$ - y' + 1 = 0 => x' - y' - 2y' + 2 = 0 => x' - 3y' + 2 = 0

Now apply x/ y values to f(L) = x' - 3y' + 2 = 0 => 2x - y - 3y + 2 = 0 => 2x - 4y + 2 = 0 => x - 2y + 1 = 0 => 2y = x + 1 => y = $\displaystyle \frac{x + 1}{2}$

Ans:
(From text book): y = x + 2
• Mar 31st 2012, 12:09 AM
biffboy
Re: Linear Transformations
You have a sign wrong in first line of your attempt x'=2x-y gives 2x=x'+y
• Mar 31st 2012, 07:05 AM
HallsofIvy
Re: Linear Transformations
Quote:

Originally Posted by GrigOrig99
The question statement says find f(L), which I have highlighted in bold. However, the book's answer seems to continue on and calcualte for y from the f(L) equation

I don't know what you mean by this. From the equation of the line, x- y+ 1= 0, y= x+ 1 so the set is {(x, x+1)}. Applying L to that we have x'= 2x- y= 2x- (x+ 1)= x- 1 and y'= y= x+ 1 so that (x, x+1) is mapped to (x- 1, x+ 1). If x'= x- 1 then x= x'+ 1 and x+ 1= x'+ 2, so we can also write this set as {(x', x'+ 2)}.
• Mar 31st 2012, 07:36 AM
biffboy
Re: Linear Transformations
I just noticed that in the first line it said x'=2x-y => 2x=x'-y. I'm obviously not sure what f(L) means (?). I thought they were asking for the equation which the book gives.
• Mar 31st 2012, 12:21 PM
GrigOrig99
Re: Linear Transformations
Thanks for the help. I'll give it a 2nd look.