STEP Maths I, II, III 1993 Solutions

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Reply 20

STEP III Question 3

$\begin{pmatrix} a&0&0 \\ b&c&0 \\ d&e&f \end{pmatrix} \begin{pmatrix} 1&p&q \\ 0&1&r \\ 0&0&r \end{pmatrix} = \begin{pmatrix} 1&3&2 \\ 4&13&5 \\ 3&8&7 \end{pmatrix}$

$\begin{pmatrix} a&ap&aq \\ b& bp+ c & bq+rc \\ d & dp+e & dq+f+er \end{pmatrix} = \begin{pmatrix} 1&3&2 \\ 4&13&5 \\ 3&8&7 \end{pmatrix}$

Comparing elements

a = 1

b = 4

d = 3

ap = 3

p = 3

aq = 2

q = 2

bp + c = 13

12 + c = 13

c = -1

dp + e = 8

9 + e = 8

e = -1

bq + rc = 5

8 + r = 5

r = - 3

dq + f + er = 7

6 + f + 3 = 7

f = -2

So

A = 1, B = 4 , C = -1 , D = 3 , E = -1, F = -2, P = 3, Q = 2, R = -3

So

$A \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 3 & -1 & -2 \end{pmatrix} \hspace{10} B \begin{pmatrix} 1 & 3 & 2 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix}$

Inverse of A

$A \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 3 & -1 & -2 \end{pmatrix}$

Matrix of Minors

$\begin{pmatrix} -2 & -8 & -7 \\ 0 & -2 & -1 \\ 0 & 0 & 1 \end{pmatrix}$

Matrix of Cofactors

$\begin{pmatrix} -2 & 8 & -7 \\ & 0 & -2 & 1 \\ 0 & 0 & 1 \end{pmatrix}$

Transpose of Matrix of Cofactors

$\begin{pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0 \\ -7 & 1 & 1 \end{pmatrix}$

Determinant = -2

$A^{-1} = -\frac{1}{2} \begin {pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0 \\ -7 & 1 & 1 \end{pmatrix}$

Inverse of B

Matrix of Minors

$\begin{pmatrix} -1 & 0 & 0 \\ 3 & 1 & 0 & \\ -11 & -3 1 \end{pmatrix}$

Matrix of Cofactors

$\begin{pmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ -11 & 3 & 1 \end{pmatrix}$

Transpose of Matrix of Cofactors

$\begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}$

Determinant = 1

$B^{-1} = \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}$

Looking at the simultaneous equations

$\begin{pmatrix} 1&3&2 \\ 4&13&5 \\ 3&8&7 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}$

$(AB)X = \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}$

Pre multiply by $(AB)^{-1}$

$(B^{-1}A^{-1})(AB) X = (B^{-1}A^{-1})\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}$

$X = (B^{-1}A^{-1})\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}$

$X = -\frac{1}{2} \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} \begin {pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0 \\ -7 & 1 & 1\end{pmatrix} \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}$

$X = -\frac{1}{2} \begin{pmatrix} 51 & -5 & -11 \\ -13 & 1 & 3 \\ -7 & 1 & 1 \end{pmatrix} \begin{pmatrix} 7 \\ 8 \\ 25 \end{pmatrix}$

$X = -\frac{1}{2} \begin{pmatrix} -8 \\ 2 \\ -6 \end{pmatrix}$

$X = \begin{pmatrix} 4 \\ -1 \\ 3 \end{pmatrix}$

x = 4, y = -1 , z =3

Just to check it works

x + 3y + 2z = 7

4 - 3 + 6 = 7

Hoar!

Reply 21

*bobo*

mechanics 11 part 1( i think STEPII):

mechanics11.doc501.2KB

Reply 22

*bobo*

q11 part2:

mechanics11part2.doc517.1KB

Reply 23

*bobo*

q11 part3:

mechanics11part3.doc449.0KB

Reply 24

Speleo

Is that the entire question? (if not, please say when it is)

Reply 25

*bobo*

sorry, yes its the entire question

Reply 26

insparato

I think i can get Questions 1,2 out on STEP III too. But im too tired atm.

Reply 27

insparato

Put it this way. I take my hat off to people chasing STEP offers. I couldnt get a 1 in either STEP I,II,III.

Reply 28

nota bene

I think I'll have a crack at STEP I Q 2, have solved some of it already but I'm trying to get the last parts to work out...

$S(n)=a+ar^2+ar^4+...+ar^{2n} \newline[br]r^2S(n)=ar^2+ar^4+...+ar^{2n}+ar^{2n+2}$
First line - second line (keeping in mind a=1 here):
$(1-r^2)S(n)=1-r^{2n+2} \newline[br]\frac{1-r^{2n+2}}{1-r^2}$ as required.

$S(r)=r+r^2+r^4+r^5+...+r^{3n-2}+r^{3n-1} \newline[br]r^3S(r)=r^4+r^5+...+r^{3n-2}+r^{3n-1}+r^{3n+1}+r^{3n+2}$
First line - second line:
$(1-r^3)S(r)=$

Reply 29

DFranklin

nota bene

Whatever xD (I proved they were the only ones as well lol...) Actually I'm very surprised if I just had to verify that I did the question in <10 minutes :confused:

.I would say "confirm" is pretty explicit "exam-speak" for "this is the answer, you just need to show it's right". Note also that there's no mention of "the winning strategy" or anything like that that might imply you need to show this is the only solution.

Now the bad news. I don't think you'd get score that many marks for what you've written. To my mind there's too much "it is quite convincing that ..." and not enough actual proof. And when you list the combinations, you don't actually say what you're doing with them. I was left thinking "I don't know why you've done this" when I saw the long list at the end.

For example, for the first part of the question, I think the answer should go something like:

[indent]Since Betty knows the numbers a1,a2,a3, their order makes no difference to the analysis, so WLOG, a1<=a2<=a3. We enumerate the possibilities and show in each case a choice Betty can make that beats it:

Adam... Betty
(1,1,7) (4,4,1)
(1,2,6) (4,4,1)
(1,3,5) (4,4,1)
(1,4,4) (3,5,1)
(2,2,5) (3,5,1)
(2,3,4) (3,5,1)
(3,3,3) (4,4,1)

Therefore we see that in all cases, Betty can find a choice that beats Adam.[/indent]

Your answer to the 2nd part is probably OK (though again, you've just said "(1,1,7) works" without actually proving it. But it's pretty obvious, so you'd probably get away with it).

But with the 3rd part, I think you'd really get hammered by the examiners; particularly if my interpretation of the question is correct, because in that case you've basically been asked "show that X wins" and your answer has been "there are no counterstrategies, therefore X wins". (Again, without proving there are no counterstrategies).

Of course, I might be completely wrong about what the examiners were expecting, and I suspect this kind of question won't pop up again anyhow. But for what it's worth, that's my opinion.

Reply 30

Swayum

nota bene, I've been working on question 2 as well (I didn't know you were). I'm not sure if I understand what they what, but does this seem right for the first 2 parts?

STEP I q2:

$1 + r^2 + r^4 + ... + r^{2n} = 1 + \frac{r^2(1-r^{2n})}{1-r^2}$
$= 1 + \frac{r^2 - r^{2n + 2}}{1-r^2}$
$= \frac{1 - r^2 + r^2 - r^{2n+2}}{1-r^2} = \frac{1 - r^{2n+2}}{1-r^2}$

WWWWW

For the next part, Snr seems to be the sum of r^(3n-1) + the sum of r^(3n+1).

$S_nr^{3n-1} = \frac{r^2(1-r^{3n})}{1-r^3}$
$S_nr^{3n+1} = \frac{r-r^{3n+1}}{1-r^3}$

Adding those together gives:

$S_nr = \frac{r + r^2 - r^{3n+1} - r^{3n+2}}{1-r^3}$

$T_n(r) = 1 + r^2 + r^3 + r^4 + r^6 + r^8 + r^9 + r^{10} + r^{12} + r^{14} + r^{15} + r^{16} + ... + r^{6n}$

That looks to me like T_n(r) is 1 + sum of r^2 numbers (starting from r^2) + r^6 numbers (starting at r^3)

Looking at the r^2 numbers:

n = 1:

r^2 + r^4 + r^6

n = 2

r^2 + r^4 + r^6 + r^8 + r^10 + r^12

It seems like we have 3n r^2 numbers (3 r^2 numbers when n = 1 and 6 r^2 numbers when n = 2)

So the sum of the r^2 numbers, starting at r^2, in the T_n(r) series is:

$S_{3n} = r^2(1-r^{2n * 3})/(1-r^2)$
$S_{3n} = \frac{r^2 - r^{6n+2}}{1-r^2}$

Looking at the r^6 numbers:

when n = 1

r^3

when n = 2

r^3 + r^9

when n = 3

r^3 + r^9 + r^15

So their sum would be:

$S_n = \frac{r^3(1-r^{6n})}{(1-r^6)}$

And so T_nr:

$T_n(r) = 1 + \frac{r^2 - r^{6n+2}}{1-r^2} + \frac{r^3(1-r^{6n})}{(1-r^6)}$
$T_n(r) = 1 + \frac{r^2 - r^{6n+2}}{1-r^2} + \frac{r^3 - r^{6n + 3}}{1-r^6}$

As T_n(r) tends to infinity;

$T_n(r) = 1 + \frac{r^2}{1-r^2} + \frac{r^3}{1-r^6}$

When r = 1, T_n(r) is undefined (or does it diverge?) because of 1-r^2 as the denominator.
When r = -1, T_n(r) is undefined (or does it diverge?) because of 1-r^2 as the denominator.

I don't know about what happens when r > 1.

Reply 31

nota bene

DFranklin

But with the 3rd part, I think you'd really get hammered by the examiners; particularly if my interpretation of the question is correct, because in that case you've basically been asked "show that X wins" and your answer has been "there are no counterstrategies, therefore X wins". (Again, without proving there are no counterstrategies).

For part i) I agree I need to show the possibilities explicitly...

For part iii) I'll say in my defence that I probably was lasy typing that out because I have 3 pages of different possibilities of different cases (some quite irrelevant) scribbled by hand where I test all possible combinations of counter the 531 315 choice. Really that list of the 28 possibilities just has to be matched with the statement and then the proof should be formally complete - or?

edit: DeathAwaitsU, S_n(r) is just a silly way of writing $\displaystyle\sum_1^n r^{3n-1}$ (like f(x) for a function of x this is a sum of r) as far as I know...
You can finish it off I've not done the last part yet, and I'm going to bed soon so just keep working :smile:

Reply 32

DFranklin

nota bene

For part iii) I'll say in my defence that I probably was lasy typing that out because I have 3 pages of different possibilities of different cases (some quite irrelevant) scribbled by hand where I test all possible combinations of counter the 531 315 choice. Really that list of the 28 possibilities just has to be matched with the statement and then the proof should be formally complete - or?

I think you'd need to list the 28 and in each case say "doesn't beat 531" or "doesn't beat by 315" as appropriate (with whatever shorthand to save writing you prefer).

(I just spent 10 minutes trying to do a "use logic to do better than listing 28 cases" and decided my first judgement was right. Straight brute force is better!).

Not to bang on, but I see similar issues in your answer to the dice question. I'd let you get away with this line:

Only one combination and as we know there is at least one 1 there must be at least one 3(not on the same die as the one/s) OR there must be a two on one die and a two on the other, however the latter leads up the wrong creek pretty soon after testing it.

if I could see within 30 seconds or so why you can't have a 2 on both dice, but I couldn't. Possibly I'm just being stupid here, but the problem is, this question isn't "really" about finding the numbers on the dice. It's about presenting a well argued and easy to follow proof. So I think you'd lose a lot of marks for each "you can see why this is true if you think about it enough" step in your solution.

Although again, I appreciate that maybe the issue is the difficulty of writing these solutions up on the forum as opposed to pencil and paper.

Reply 33

DFranklin

DeathAwaitsU

For the next part, Snr seems to be the sum of r^(3n-1) + the sum of r^(3n+1).Huh? Why do you think there's an r^(3n+1) term in S_n(r)?

Reply 34

ad absurdum

DeathAwaitsU

.

I just want to confirm that this is right before doing the next parts.

I'm not sure you've considered $S_n(r)$ correctly (or maybe you have, and I'm just not seeing it :s-smilie:

). A different approach that would lead you nicely into the deduction based on S_n(r) would be can you write S_n(r) as (one series)-(another, related, series)?

Reply 35

nota bene

DFranklin

Yeah I tried to think of something before I started the question; but definitely ended with the same conclusion.

Reply 36

DFranklin

insparato

STEP III Question 3I'm sure this answer would get full marks, but (from the lofty position of having been taught LU decomposition), I'm fairly sure the examiners hoped you'd be a little cleverer about inverting A and B. The idea is to invert using an augmented matrix and row operations - which is particularly easy because of the form of A and B.

i.e. to invert A form the augmented matrix (where LH = A, RH = identity).

1 0 0 | 1 0 0
4 1 0 | 0 1 0
3 -1 -2 | 0 0 1

Subtract 4 x row1 from row2, 3 x row1 from row3:

1 0 0 | 1 0 0
0 1 0 | -4 1 0
0 -1 -2 | -3 0 1

Add 1 x row2 to row3:

1 0 0 | 1 0 0
0 1 0 | -4 1 0
0 0 -2 | -7 1 1

Multiply row3 by -1/2:

1 0 0 | 1 0 0
0 1 0 | -4 1 0
0 0 1 | 3/2 -1/2 -1/2

Having made the LH = the identiy, the RH is now A^{-1}.

(Although I'm sure LU decomposition was never on any syllabus, my impression is that this general method of inverting a matrix by row operations is no longer taught, even though it was taught in my day and so would have been the natural approach here).

Reply 37

Kolya

DFranklin

Although again, I appreciate that maybe the issue is the difficulty of writing these solutions up on the forum as opposed to pencil and paper.

Feel free to scan in your pencil/paper and attach to a post, if necessary :smile:

Reply 38

insparato

Ive never head of augmented matrix and row operations. I only did it the way ive been taught to do the inverse of a 3x3. But thanks ill definitely look that up.

Reply 39

ad absurdum

DFranklin

(Although I'm sure LU decomposition was never on any syllabus, my impression is that this general method of inverting a matrix by row operations is no longer taught, even though it was taught in my day and so would have been the natural approach here).

We're still taught that method in Scotland :smile:

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