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Fp1 matrices

What am I doing wrong? I havent finished yet but do far it looks nothing like the ms
Q10ii.
http://m.imgur.com/G5pn2lC,BQWZ8HU,wEccZsU

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Reply 1

Zacken

Original post by Super199

What am I doing wrong? I havent finished yet but do far it looks nothing like the ms
Q10ii.
http://m.imgur.com/G5pn2lC,BQWZ8HU,wEccZsU

Look up finding the matrix of minors again, you shouldn't be inlcuding the coefficients of when computing the minor determinants (or whatever it's called).

So, like where you did 0(6a - 3) it should be just 6a - 3. Not including the coefficient thingy.

Reply 2

Super199

Original post by Zacken

Yh got it cheers. How do I work out 10iii?
(AB)^-1= B^-1A^-1.
Idk what to do after.

Reply 3

Zacken

Original post by Super199

Yh got it cheers. How do I work out 10iii?
(AB)^-1= B^-1A^-1.
Idk what to do after.

Yeah, so B^(-1) = (AB)^(-1) * A by multiply both sides by A. And you know (AB)^(-1).

Reply 4

Super199

Original post by Zacken

Yeah, so B^(-1) = (AB)^(-1) * A by multiply both sides by A. And you know (AB)^(-1).

Is A^-1A = I

or AA^-1= I?

Reply 5

Zacken

Original post by Super199

Is A^-1A = I

or AA^-1= I?

Both.

Reply 6

Super199

Original post by Zacken

Both.

is I = 1 0
0 1

or does it have a value of 1?

Reply 7

Zacken

Original post by Super199

is I = 1 0
0 1

Correct.

Reply 8

Super199

Original post by Zacken

Correct.

http://www.ocr.org.uk/Images/63184-question-paper-unit-4725-01-further-pure-mathematics-1.pdf

q5. What does it mean by using the determinant of an appropriate matrix?
No worries I think I got it actually

Reply 9

Zacken

Original post by Super199

Reply 10

Super199

Original post by Zacken

http://m.imgur.com/zHrzLmt,nxdr3vb

Q7ii. I dont think ive done the proof right

Reply 11

Zacken

Original post by Super199

http://m.imgur.com/zHrzLmt,nxdr3vb

Q7ii. I dont think ive done the proof right

Looks fine although you'd want to say -uk not +uk .

Reply 12

Super199

Original post by Zacken

Looks fine although you'd want to say -uk not +uk .

Yup my bad. Same paper 8iii.

Can I let x= alpha +1

So alpha= x-1 and then sub it in? or do I have to use the previous part some how?

Reply 13

Zacken

Original post by Super199

Yup my bad. Same paper 8iii.

Can I let x= alpha +1

So alpha= x-1 and then sub it in? or do I have to use the previous part some how?

Can't see the question.

Reply 14

Nish123

Original post by Super199

Yup my bad. Same paper 8iii.

Can I let x= alpha +1

So alpha= x-1 and then sub it in? or do I have to use the previous part some how?

As the question says, hence assume you should just continue.

Reply 15

Super199

Original post by Zacken

Can't see the question.

http://www.ocr.org.uk/Images/63184-question-paper-unit-4725-01-further-pure-mathematics-1.pdf

Reply 16

Super199

Original post by Nish123

As the question says, hence assume you should just continue.

I don't see how though.

Reply 17

Zacken

Original post by Super199

I don't see how though.

A quadratic equations with those roots are:

x^2 - x(\alpha + 1 + \beta - 1) + (\alpha +1)(\beta - 1)

Reply 18

Nish123

Original post by Nish123

As the question says, hence assume you should just continue.

Plus, the same thing is not happening to each root (symmetry) , so I am assuming that you cannot do the "slick method" ie substitution without making it longer.

(edited 7 years ago)

Reply 19

Nish123

Original post by Super199

I don't see how though.

Sorry, I have just joined Student Room so don't know how to use Latex however just sum the new roots and multiply them together, ie (a+1)(b-1) and (a+1)+(b-1) and then manipulate until you have that in terms of alpha and beta. I have used a instead of alpha and b instead of beta.