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FP4 transformation question

IMG_7255.jpg Question 4b) I did T (x,mx) = (x,mx) ( (x,mx) is meant to be a 2x1 matrix)
Multiplied out LHS and simplified to get the equations 2x+mx=0 and 4x+2mx=0, both of which say that m=-2. On the mark scheme, positive 2 is also another correct solution, how would you get this?

Question 5b) I'm really not sure where to start

Any help would be really appreciated :biggrin:
Original post by Megan_101
Question 4b) I did T (x,mx) = (x,mx) ( (x,mx) is meant to be a 2x1 matrix)
Multiplied out LHS and simplified to get the equations 2x+mx=0 and 4x+2mx=0, both of which say that m=-2. On the mark scheme, positive 2 is also another correct solution, how would you get this?

Question 5b) I'm really not sure where to start

Any help would be really appreciated :biggrin:


4b) Can't be m=2m=2, so it's an error on the book. Run it through yourself to verify.

5b) If the lines are turned 90 degrees then the gradient of the transformed line is the negative reciprocal of the one you're transforming.
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Megan_101
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4
Original post by RDKGames
4b) Can't be m=2m=2, so it's an error on the book. Run it through yourself to verify.

5b) If the lines are turned 90 degrees then the gradient of the transformed line is the negative reciprocal of the one you're transforming.



Thank you :smile:
Original post by RDKGames
4b) Can't be m=2m=2, so it's an error on the book. Run it through yourself to verify.


:holmes: Unless I'm missing something, m=2 is valid:

Point on line is (x,2x)

Then T(x,2x) = (5x,10x) and y' = 2x'
(edited 6 years ago)
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B_9710
18
You have found a line of line of invariant points - which of course is an invariant line too - but you need to find the equations of any lines that map to themselves even if the points on the line are not mapped to themselves.

So you need to solve the equations
[2143][xmx]=[x′mx′] \displaystyle \begin{bmatrix} 2 & 1 \\ 4&3 \end{bmatrix} \begin{bmatrix} x \\ mx \end{bmatrix}= \begin{bmatrix} x' \\ mx' \end{bmatrix} .
Original post by ghostwalker
:holmes: Unless I'm missing something, m=2 is a valid

Point on line is (x,2x)

Then T(x,2x) = (5x,10x) and y' = 2x'


Oh, I lost track for a moment and proved that it's not a line of invariant points whereas y=−2xy=-2x is...

@Megan_101

A line which is mapped back onto itself is either the invariant line or the line of invariant points. What you've calculated is the line of invariant points.

For the invariant lines, look above.
Reply 6
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Megan_101
OP
4
Original post by ghostwalker
:holmes: Unless I'm missing something, m=2 is valid:

Point on line is (x,2x)

Then T(x,2x) = (5x,10x) and y' = 2x'


Original post by B_9710
You have found a line of line of invariant points - which of course is an invariant line too - but you need to find the equations of any lines that map to themselves even if the points on the line are not mapped to themselves.

So you need to solve the equations
[2143][xmx]=[x′mx′] \displaystyle \begin{bmatrix} 2 & 1 \\ 4&3 \end{bmatrix} \begin{bmatrix} x \\ mx \end{bmatrix}= \begin{bmatrix} x' \\ mx' \end{bmatrix} .


Thank you :biggrin:

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