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AQA FP4 25th May 2016 Discussion Thread watch

1. (Original post by Mm68)
I know this might be more fp1 but I've seen it come up in fp4, how are you supposed to work out the enlargement of a transformation matrix, for example if it's
2 -2
2 2
It would be a enlargement 2rt2 but how do you work that out.
s.f of enlargement is the square root of the determinant.
2. (Original post by EwanatTSR)
Does anyone know an easy method for finding eigenvalues in a 3x3 matrix? I use the Gaussian Elimination process but it is very confusing and a huge space for error.
What we learnt to do is set along the leading diagonal. For example if the top left part of the matrix was , you set it to . You do this for M1,1 , M2,2 and M3,3 (leading diagonal). Then you find the determinant in terms of and set it equal to 0. By factorising it, you can easily find the values of .
3. (Original post by milsy.afrochic)
anyone else getting slightly concerned about how few resources there are for FP4?? I mean the AQA textbook is just awful
I didn't see that the great man himself already suggested this website.
4. (Original post by AHRocks187)
How would you guys go about finding the shortest distance between a point and a plane, assuming you only know the point co-ordinates and the parametric equation of the plane? Just trying to brush up on geometry techniques as expecting a hard question on this this year
I always form an equation of a line that passes through the point and is perpendicular to the plane (using the normal vector to the plane as the direction vector of the line). so I would have something like , where is the position vector of the point you're trying to find the distance from.
Then solving for the scalar the shortest distance is .
5. Here's a question for you lot to try, it's about images of lines under matrix transformations.

The transformation, , is defined below as the following
.

The image of the line under is .
It is given that the equation of is
.

(a) Find the equation of the line .

(b) Find the image of the line under which has equation
.
6. (Original post by B_9710)
Here's a question for you lot to try, it's about images of lines under matrix transformations.

The transformation, , is defined below as the following
.

The image of the line under is .
It is given that the equation of is
.

(a) Find the equation of the line .

(b) Find the image of the line under which has equation
.
A. 9x^2+y^2+6xy+36x+10y=10

and B. 188x^2-216xy+62y^2=-3

Is that correct?
7. (Original post by Argylesocksrox)
A. 9x^2+y^2+6xy+36x+10y=10

and B. 188x^2-216xy+62y^2=-3

Is that correct?
I'll have to give the answers tomorrow as I'm in bed now. I think I had them written down somewhere.
8. I'm really struggling with knowing what I have to do at each question as if someone just told me what I have to do I'm normally alright! i suppose that just takes lots of practice of papers but I kind of left that too late :/ At least I'm helping lowering the grade boundaries for you all!

The one thing I really can't seem to do is invariant lines as the way my teacher taught us I've never been able to use in an exam...
9. (Original post by katie4202)
I'm really struggling with knowing what I have to do at each question as if someone just told me what I have to do I'm normally alright! i suppose that just takes lots of practice of papers but I kind of left that too late :/ At least I'm helping lowering the grade boundaries for you all!

The one thing I really can't seem to do is invariant lines as the way my teacher taught us I've never been able to use in an exam...
How have you been taught to find he invariant lines?
10. (Original post by B_9710)
How have you been taught to find he invariant lines?

To times the matrix by x and then mx+c and the equals the new x and y which is then used in the equation y=mx+c..

I'm not sure that makes sense...
11. (Original post by katie4202)
To times the matrix by x and then mx+c and the equals the new x and y which is then used in the equation y=mx+c..

I'm not sure that makes sense...
It dos make sense actually. I think you're saying this

?
This way works.
12. (Original post by B_9710)
It dos make sense actually. I think you're saying this

?
This way works.
We do almost the same thing, but let the last matrix be x' and y' and then say we require y' = mx' + c. Subbing in the values from multiplying out the left makes it all work out.
13. (Original post by student0042)
We do almost the same thing, but let the last matrix be x' and y' and then say we require y' = mx' + c. Subbing in the values from multiplying out the left makes it all work out.
I do normally put x' and y' but sometimes if it involves squaring then I find it more awkward to write it out.

14. Hey guys, does anyone know why for part C, you get an invariant line and an invariant plane?
I understand that that there is a line of invariants points for lambda =1 (Mx=1x). However I am just not getting the plane part for the other 2 repeated eigenvalues
15. (Original post by 01binary)

Hey guys, does anyone know why for part C, you get an invariant line and an invariant plane?
I understand that that there is a line of invariants points for lambda =1 (Mx=1x). However I am just not getting the plane part for the other 2 repeated eigenvalues
When you are trying to find the eigenvectors do you find that you form 3 identical equations?
Well there is not one answer to these 3 identical equations there are infinitely many answers (and I don't mean answers that are just multiples of each other). So from this you can actually find 2 eigenvectors that are not multiples of each other (they are linearly independent) and so if these eigenvectors are denoted and we can say that there are an invariant plane (if eigenvalues does not equal one) and we can give this equation of plane by .
16. (Original post by B_9710)
When you are trying to find the eigenvectors do you find that you form 3 identical equations?
Well there is not one answer to these 3 identical equations there are infinitely many answers (and I don't mean answers that are just multiples of each other). So from this you can actually find 2 eigenvectors that are not multiples of each other (they are linearly independent) and so if these eigenvectors are denoted and we can say that there are an invariant plane (if eigenvalues does not equal one) and we can give this equation of plane by .
Are those equations always identical, I find that sometimes they aren't e.g. for lambda =1?
Well for part b) I found that the eigenvector for lamda =4 as (1 1 2) (x y z) respectively. The equation I got 3 times was x+y-z=0
Why would I need to split the vector (1, 1, 2) this is the part I am having trouble with.
Also is this in the AQA FP4 book anywhere
17. (Original post by 01binary)
Are those equations always identical, I find that sometimes they aren't e.g. for lambda =1?
Well for part b) I found that the eigenvector for lamda =4 as (1 1 2) (x y z) respectively. The equation I got 3 times was x+y-z=0
Why would I need to split the vector (1, 1, 2) this is the part I am having trouble with.
Also is this in the AQA FP4 book anywhere
They aren't always identical. When there are 3 identical equations formed, that is the case when there is an invariant plane (or plane of invariant points if lambda =1).
For the bit you're having trouble with literally just find 2 sets of numbers that work. For x+y-z=0 I quickly find that (1,-1,0) and (2,1,3) work, but you could have (1,1,2) or (3,1,4) so I would have an invariant plane . But if you look carefully you will see that you can make any one of the solutions I gave above from linear combinations of the 2 vectors I gave for the plane. (Using i,j,k notation as it's quicker to do on here).
It's not explicitly stated in the FP4 book but there is an example where they from 3 identical equations and it does say that there is an invariant plane but does not explain why.
18. (Original post by B_9710)
They aren't always identical. When there are 3 identical equations formed, that is the case when there is an invariant plane (or plane of invariant points if lambda =1).
For the bit you're having trouble with literally just find 2 sets of numbers that work. For x+y-z=0 I quickly find that (1,-1,0) and (2,1,3) work, but you could have (1,1,2) or (3,1,4) so I would have an invariant plane . But if you look carefully you will see that you can make any one of the solutions I gave above from linear combinations of the 2 vectors I gave for the plane. (Using i,j,k notation as it's quicker to do on here).
It's not explicitly stated in the FP4 book but there is an example where they from 3 identical equations and it does say that there is an invariant plane but does not explain why.
Okay I think I am getting it now. So as there is a repeated eigenvalue, and that the equations you get have infinitely many solutions, you could many different eigenvectors for the same eigenvalue. But we only want 2 as eigenvalues each have a paired eigenvector. And what you're saying is that say e.g. (1, 1, 2) can be split into 2 separate direction vectors, which also satisfy the original eq. (x+y-z=0) so if we can split into 2 diff vectors this must is a plane as this is pretty much the definition of a plane (with the origin as the start point)?
19. (Original post by 01binary)
Okay I think I am getting it now. So as there is a repeated eigenvalue, and that the equations you get have infinitely many solutions, you could many different eigenvectors for the same eigenvalue. But we only want 2 as eigenvalues each have a paired eigenvector. And what you're saying is that say e.g. (1, 1, 2) can be split into 2 separate direction vectors, which also satisfy the original eq. (x+y-z=0) so if we can split into 2 diff vectors this must is a plane as this is pretty much the definition of a plane (with the origin as the start point)?
The fact that the eigenvalue is repeated is not important, it's just a coincidence in this case. It doesn't have to be repeated. Don't think of it as splitting a vector, think of it as just finding 2 'random' vectors that satisfy the equations (the 3 repeated ones).
20. 5D Jan 2011...

I don't fully get why the line where 2 planes intersect is perpendicular to both planes

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