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Fp1 matrices

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Reply 20
Original post by Zacken
A quadratic equations with those roots are:

x2−x(α+1+β−1)+(α+1)(β−1)x^2 - x(\alpha + 1 + \beta - 1) + (\alpha +1)(\beta - 1)


Yh got it.

For 9ii. Why is 1/(2n-3) included surely it cancels out with a term before.
Reply 21
Original post by Super199
Yh got it.

For 9ii. Why is 1/(2n-3) included surely it cancels out with a term before.


Nopes. What term would it cancel with?
Reply 22
Original post by Zacken
Nopes. What term would it cancel with?


Two terms before it. Since n=4, the 1/5 cancels with the 1/5 from n=2
Hey Can Someone Help Me With Matrix Transformations Please? How Do You Know Whether It Is A Reflection Etc?
Reply 24
Original post by Super199
Two terms before it. Since n=4, the 1/5 cancels with the 1/5 from n=2


From a quick glance, looks like you're right. (Can't do the question myself right now) does the markscheme say it's included? Have you looked at the direction of the cancelling? Probably best just to put your working up tbh.
(edited 7 years ago)
Reply 25
Original post by Zacken
From a quick glance, looks like you're right. (Can't do the question myself right now) does the markscheme say it's included? Have you looked at the direction of the cancelling? Probably best just to put your working up tbh.

Bit messy but this is what I tried to do.
http://m.imgur.com/0FB2bJb
Reply 26
Original post by Super199
Bit messy but this is what I tried to do.
http://m.imgur.com/0FB2bJb


Yeah, that's (almost) correct. The 12n−3\frac{1}{2n-3} does cancel but the 12n−1\frac{1}{2n-1} doesn't.
Reply 27
Original post by Zacken
Yeah, that's (almost) correct. The 12n−3\frac{1}{2n-3} does cancel but the 12n−1\frac{1}{2n-1} doesn't.


ah yes I see it now. How do I do q10ii same paper :/
Reply 28
Original post by Super199
ah yes I see it now. How do I do q10ii same paper :/


It's a disguised quadratic, make the sub u=z2u = z^2
Reply 29
Original post by Zacken
It's a disguised quadratic, make the sub u=z2u = z^2


So it's literally the same answers from the previous part?

u^2-4u+9=0

Completing the square gives: u= 2+- square root of 5i
Reply 30
Original post by Super199
So it's literally the same answers from the previous part?

u^2-4u+9=0

Completing the square gives: u= 2+- square root of 5i


Yep.
Reply 31
ffs dont worry
(edited 7 years ago)
Reply 32
Original post by Zacken
Yep.


Yo I think im going mad but can you explain how to work out the centre of the circle from loci's. e.g q6ii
https://3ed6c47c6c1ead854571a830aabac4d23871f1f4.googledrive.com/host/0B1ZiqBksUHNYQ0ZBTlpRSFI0eTA/June%202009%20QP%20-%20FP1%20OCR.pdf
Reply 33
Original post by Super199
Yo I think im going mad but can you explain how to work out the centre of the circle from loci's. e.g q6ii
https://3ed6c47c6c1ead854571a830aabac4d23871f1f4.googledrive.com/host/0B1ZiqBksUHNYQ0ZBTlpRSFI0eTA/June%202009%20QP%20-%20FP1%20OCR.pdf


Locus |z-a| = r has radius r and centre a.
Reply 34
Original post by Zacken
Locus |z-a| = r has radius r and centre a.


How do you work the matrix for a shear? q8ii this is?
Reply 35
Original post by Super199
How do you work the matrix for a shear? q8ii this is?


Set up a matrix with unknown components and make matrix * (1, 1) = (1,2) or whatever points they've given and use that to find the matrix components.
Reply 36
Original post by Zacken
Set up a matrix with unknown components and make matrix * (1, 1) = (1,2) or whatever points they've given and use that to find the matrix components.


How though

let the matrix be
a b
c d

a +b =1
c+d = 2
not sure what to do from then.
Reply 37
Original post by Super199
How though

let the matrix be
a b
c d

a +b =1
c+d = 2
not sure what to do from then.


Yeah but you also know any point on the y-axis is invariant

So (matrix) * (0 1) = (something - draw a diagram and think it through, what does a shear to to units vectors?)
(edited 7 years ago)
Reply 38
Original post by Zacken
Yeah but you also know any point on the y-axis is invariant

So (matrix) * (0 1) = (something - draw a diagram and think it through, what does a shear to to units vectors?)


a b
c d * (0 1) = 0 1 ?
Reply 39
Original post by Super199
a b
c d * (0 1) = 0 1 ?


I think so, I dunno. Try it.

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