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OCR MEI FP2 (18th June) Exam Thread

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Original post by Peterphipps
I can't remember the full inverse matrix but i got the det(A)=13a-65 which works for x<>5.
Yeah I did but again really untidy I can't remember but i used x=7lambda to tidy up z i think but y was still a fraction. Wasn't happy with the answer but didn't have time to check.


what do u mean 13a - 65 isnt it just 1/k-5 ... because it said k cannot equal 5.. therefore determinant must be 1/k-5, then u sub whatevery they said was k in?
Original post by sreddy17
what do u mean 13a - 65 isnt it just 1/k-5 ... because it said k cannot equal 5.. therefore determinant must be 1/k-5, then u sub whatevery they said was k in?


It said prove that k<>5 so you had to work out the determinate which turned out to be 13a-65. When a=5 then det(A)=0 so k<>5 if the matrix has an inverse.
Original post by Peterphipps
It said prove that k<>5 so you had to work out the determinate which turned out to be 13a-65. When a=5 then det(A)=0 so k<>5 if the matrix has an inverse.


Yeah it said show that determinant =0 when k=5 therefore just evaluate the determinant with k=5? And u get determinant as 0 that's what I did , and then did the question say find the inverse matrix of A. What was the actual question altogether can u tell me for each part , the three parts of the matrix question thanks mate
Original post by Peterphipps
It said prove that k<>5 so you had to work out the determinate which turned out to be 13a-65. When a=5 then det(A)=0 so k<>5 if the matrix has an inverse.


Yeah well question said show that k cannot be 5 or something or determinant is zero so just simply evaluate determinat when it's 5 and u get 0 easy. Then work out the inverse
Reply 64
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NJam
2
Original post by sreddy17
Yeah well question said show that k cannot be 5 or something or determinant is zero so just simply evaluate determinat when it's 5 and u get 0 easy. Then work out the inverse


But by finding the determinant first, you cut out a stage of working for finding the inverse. Plus the 'Show that' generally indicates it wants you to actually work something out. If they just wanted you to sub in k=5, they would have said 'verify'. Provided you still got the inverse, you'll still get the marks.
Original post by NJam
But by finding the determinant first, you cut out a stage of working for finding the inverse. Plus the 'Show that' generally indicates it wants you to actually work something out. If they just wanted you to sub in k=5, they would have said 'verify'. Provided you still got the inverse, you'll still get the marks.


What did the actual question say mate show that k cannot be 5 , well just put k=5 And u get determinant as 0? Yeah I know where ur coming from, ur probably right , but yeah what was the next part of the question for that do u Remmeber? Was it just find inverse of A? But when what = what?
Reply 66
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Jimbobp
Original post by BrevityIs
Here's the two ways I could see of doing the last question.


That's a relief, 2nd way is how I did it :smile:
Original post by Jimbobp
That's a relief, 2nd way is how I did it :smile:


yep same! what did u get for all the matrcies questions? can u remember the questions?
Reply 68
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Original post by sreddy17
What did the actual question say mate show that k cannot be 5 , well just put k=5 And u get determinant as 0? Yeah I know where ur coming from, ur probably right , but yeah what was the next part of the question for that do u Remmeber? Was it just find inverse of A? But when what = what?


The first part said something along the lines of:

'Show that Matrix A has no inverse for k=5. Assuming k does not take this value, find the inverse of A in terms of k.'
Original post by NJam
The first part said something along the lines of:

'Show that Matrix A has no inverse for k=5. Assuming k does not take this value, find the inverse of A in terms of k.'

ahh yeah cool so u just do 1/(k-5) lots of the co factors all transposed am i right?
then what did the next question say something bout a simultaneous equation? and then the last one was where u work out p=4 and then put lambda as x and work out the rest?
Original post by NJam
The first part said something along the lines of:

'Show that Matrix A has no inverse for k=5. Assuming k does not take this value, find the inverse of A in terms of k.'


wait actually did it not make u use a value for k? im pretty sure it did.. then u had like 1/ something,, and then the matrix of co factors transposed?
has anyone got the paper?!
Still fairly confident that the det(A) was 13a-65 or 13(a-5) if you prefer
Then a=4 so the det(A)=-13 not -1
That's how I got my value of x=p-4 in the next part anyway.
Just realised that a lot of people are using the constant k rather than a which might be causing some confusion. Don't know which one it was, I seem to remember a but it is usually k but doesn't really matter.
Original post by Peterphipps
Still fairly confident that the det(A) was 13a-65 or 13(a-5) if you prefer
Then a=4 so the det(A)=-13 not -1
That's how I got my value of x=p-4 in the next part anyway.
Just realised that a lot of people are using the constant k rather than a which might be causing some confusion. Don't know which one it was, I seem to remember a but it is usually k but doesn't really matter.

yeahh i agree mate, should have done it like that really!! , and did u have to let a=4 to work out the inverse matrix? as in did u have to put a in then work out all the cofactors and transpose it?
No because in the first part you had to work out the inverse in terms of a.
So when it came to working out the inverse then a=4 you just subbed in the values. I mean you could do it that way but would take much longer.
Original post by Peterphipps
No because in the first part you had to work out the inverse in terms of a.
So when it came to working out the inverse then a=4 you just subbed in the values. I mean you could do it that way but would take much longer.


oh sh**, i accidently let a =4 in the first part and then worked out the inverse, cos i must have had read the question as let a=4 what is the inverse.. how many marks would i lose?
Reply 76
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maths42
13
Oh no! I did the last question using the chain rule easily but at the end I "simplified" 0.5ln^2(1+√2) to ln(1+√2) under time pressure - how many marks would I lose for this?
Original post by sreddy17
oh sh**, i accidently let a =4 in the first part and then worked out the inverse, cos i must have had read the question as let a=4 what is the inverse.. how many marks would i lose?


Well looking at past papers usually that question has 7 marks and if I'm honest if you subbed a=4 straight away then probably most of them. Seems to be split 3 marks are for working out the determinate then 4 for the matrix bit.
Original post by maths42
Oh no! I did the last question using the chain rule easily but at the end I "simplified" 0.5ln^2(1+√2) to ln(1+√2) under time pressure - how many marks would I lose for this?


I think if you got that far probably only 1. A lenient examiner might not even dock you that since you got the right answer then took it a step too far by completely ignoring that line of working.
Reply 79
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physicsnut
Original post by maths42
Oh no! I did the last question using the chain rule easily but at the end I "simplified" 0.5ln^2(1+√2) to ln(1+√2) under time pressure - how many marks would I lose for this?


It was only a three mark question so maybe 1 or 2 marks. I forgot that u integrates to 1/2u^2, total stupid moment. Only realised just as the examiner took my paper in :frown:, preparing to lose 2 or 3 marks for that mistake.

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