## cwrw238 3 years ago Please try this question from a math test. I am baffled by it. 4fccab66e4b0c6963ad793df-smishra-1338813325025-copyofnewmicrosoftofficeworddocument.docx

1. eigenschmeigen

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

2. cwrw238

i'll draw it

3. eigenschmeigen

ok :)

4. cwrw238

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

6. cwrw238

i think its a determinant

7. eigenschmeigen

what does GP stand for?

8. cwrw238

geometric series

9. eigenschmeigen

ah nice. ok give me a minute

10. cwrw238

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

11. eigenschmeigen

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

12. eigenschmeigen

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

to avoid confusion lets use d instead of r

14. eigenschmeigen

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

15. eigenschmeigen

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

16. cwrw238

yes

17. eigenschmeigen

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

18. cwrw238

b = fd^(q-1)

19. eigenschmeigen

yes

20. eigenschmeigen

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

hows it going?

22. cwrw238

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

23. eigenschmeigen

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

24. cwrw238

ok

25. eigenschmeigen

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

26. eigenschmeigen

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

27. eigenschmeigen

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

28. cwrw238

yes - i got that far! - no more!

29. cwrw238

only vaguely

30. cwrw238

thats why I'm stuck

31. eigenschmeigen

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

ok

33. eigenschmeigen

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

35. eigenschmeigen

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

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

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

38. cwrw238

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

39. eigenschmeigen

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

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

to be honest no

42. eigenschmeigen

ok no worries, ill explain :)

43. cwrw238

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

44. eigenschmeigen

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

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

46. eigenschmeigen

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

47. eigenschmeigen

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

48. cwrw238

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

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

50. eigenschmeigen

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

51. cwrw238

thanks a lot

52. eigenschmeigen

no problem :) thanks for the juicy problem