## cwrw238 Group Title Please try this question from a math test. I am baffled by it. 4fccab66e4b0c6963ad793df-smishra-1338813325025-copyofnewmicrosoftofficeworddocument.docx one year ago one year ago

1. eigenschmeigen Group Title

i dont think it uploaded properly try again and ill try and help

2. cwrw238 Group Title

i'll draw it

3. eigenschmeigen Group Title

ok :)

4. cwrw238 Group Title

If a, b, c are positive and are pth, qth and rth terms of a GP; then show without expanding that | loga p 1 | | logb q 1 | = 0 | logc r 1 |

5. eigenschmeigen Group Title

6. cwrw238 Group Title

i think its a determinant

7. eigenschmeigen Group Title

what does GP stand for?

8. cwrw238 Group Title

geometric series

9. eigenschmeigen Group Title

ah nice. ok give me a minute

10. cwrw238 Group Title

i assume that they are consecutive terms - can we do that?

11. eigenschmeigen Group Title

unfortunately not, but we do know their positions, p,q,r

12. eigenschmeigen Group Title

the nth term of a GP is A(r^(n-1)) where A is the first term and r is the common ratio

13. eigenschmeigen Group Title

to avoid confusion lets use d instead of r

14. eigenschmeigen Group Title

so lets call the first term of the progression "f" and the common ratio "d"

15. eigenschmeigen Group Title

can you see this: $a= fd^{p-1}$

16. cwrw238 Group Title

yes

17. eigenschmeigen Group Title

cool have a go playing with that, and see how far you can get it

18. cwrw238 Group Title

b = fd^(q-1)

19. eigenschmeigen Group Title

yes

20. eigenschmeigen Group Title

that was a lovely problem :) made me very happy. i haven't really come across logarithms intertwined with matrices before. thanks for showing me

21. eigenschmeigen Group Title

hows it going?

22. cwrw238 Group Title

i'm struggling - whats confusing me is that it says do not expand

23. eigenschmeigen Group Title

ah ok well ill put up my solution and you can read through as far as you like

24. cwrw238 Group Title

ok

25. eigenschmeigen Group Title

$\left|\begin{matrix}log(a) & p & 1\\ log(b) & q & 1\\ log(c) & r & 1\end{matrix}\right|$

26. eigenschmeigen Group Title

$\left|\begin{matrix}\log(fd^{p-1}) & p & 1\\ \log(fd^{q-1}) & q & 1\\ \log(fd^{r-1}) & r & 1\end{matrix}\right|$

27. eigenschmeigen Group Title

now just to check, are you familiar with how you can manipulate determinants, eg row operations?

28. cwrw238 Group Title

yes - i got that far! - no more!

29. cwrw238 Group Title

only vaguely

30. cwrw238 Group Title

thats why I'm stuck

31. eigenschmeigen Group Title

lets start by applying some log laws first we break apart the logarithms like this: $\left|\begin{matrix}\log(d^{p-1}) + \log(f) & p & 1\\ \log(d^{q-1}) + \log(f) & q & 1\\ \log(d^{r-1}) + \log(f) & r & 1\end{matrix}\right|$

32. cwrw238 Group Title

ok

33. eigenschmeigen Group Title

then: $\left|\begin{matrix}\ (p-1)\log(d) + \log(f) & p & 1\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|$

34. cwrw238 Group Title

35. eigenschmeigen Group Title

ok, now we do some row operations. remember row operations do not change the value of the determinant so we can use them here. first we do R1 - R2 ( that's row 1 - row 2) you should get this: $\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|$

36. eigenschmeigen Group Title

now we do R2- R3 : $\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|$

37. cwrw238 Group Title

oh yes i remember doing something like this - to find inverse of matrix i think

38. cwrw238 Group Title

i think i know whats coming - when you calculate the determinant the zeros will be significant

39. eigenschmeigen Group Title

probably because you need determinants to find inverses. anyway back to the determinant :D now can you see IF we were to expand the determinant using column 3 to expand, the two left elements of R3 would not contribute to the determinant? yeah you guessed it, but we are not allowed to expand, we'll have to keep manipulating.

40. eigenschmeigen Group Title

because of the two existing zeros we can do this: $\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ 0 & 0 & 1\end{matrix}\right|$ do you understand or should i explain this step more?

41. cwrw238 Group Title

to be honest no

42. eigenschmeigen Group Title

ok no worries, ill explain :)

43. cwrw238 Group Title

yes i think i see now - those elements in row 3 are multiplied by 0

44. eigenschmeigen Group Title

how about trying to do the operation (column 2) - r(column 3) here: $\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|$ we would get: $\left|\begin{matrix}\ (p-q)\log(d) & (p-q)-0 & 0\\ \ (q-r)\log(d) & (q-r) - 0 & 0\\ \ (r-1)\log(d) + \log(f) & (r)- r & 1\end{matrix}\right|$ to get: $\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & 0 & 1\end{matrix}\right|$

45. eigenschmeigen Group Title

there are two ways to think about it, by expanding or by operations, whichever you prefer :)

46. eigenschmeigen Group Title

if we were to expand it those two elements wouldn't count because, as you said, they just get multiplied by zero

47. eigenschmeigen Group Title

does this help or is it still confusing? i hope i haven't made it worse lol

48. cwrw238 Group Title

i think i'm getting it - i remember finding this stuff difficult at a-level and it never 'stuck' in my brain very well . i'll keep at it though - thanks very much - i'm glad you enjoyed doing the problem

49. eigenschmeigen Group Title

basically its all about using row operations. they through you a curve-ball with the logarithms and the GP

50. eigenschmeigen Group Title

here is a good site http://www.sosmath.com/matrix/determ1/determ1.html

51. cwrw238 Group Title

thanks a lot

52. eigenschmeigen Group Title

no problem :) thanks for the juicy problem