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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

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  1. eigenschmeigen Group Title
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    i dont think it uploaded properly try again and ill try and help

    • one year ago
  2. cwrw238 Group Title
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    i'll draw it

    • one year ago
  3. eigenschmeigen Group Title
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    ok :)

    • one year ago
  4. cwrw238 Group Title
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    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 |

    • one year ago
  5. eigenschmeigen Group Title
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    sorry i didnt read

    • one year ago
  6. cwrw238 Group Title
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    i think its a determinant

    • one year ago
  7. eigenschmeigen Group Title
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    what does GP stand for?

    • one year ago
  8. cwrw238 Group Title
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    geometric series

    • one year ago
  9. eigenschmeigen Group Title
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    ah nice. ok give me a minute

    • one year ago
  10. cwrw238 Group Title
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    i assume that they are consecutive terms - can we do that?

    • one year ago
  11. eigenschmeigen Group Title
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    unfortunately not, but we do know their positions, p,q,r

    • one year ago
  12. eigenschmeigen Group Title
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    the nth term of a GP is A(r^(n-1)) where A is the first term and r is the common ratio

    • one year ago
  13. eigenschmeigen Group Title
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    to avoid confusion lets use d instead of r

    • one year ago
  14. eigenschmeigen Group Title
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    so lets call the first term of the progression "f" and the common ratio "d"

    • one year ago
  15. eigenschmeigen Group Title
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    can you see this: \[a= fd^{p-1}\]

    • one year ago
  16. cwrw238 Group Title
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    yes

    • one year ago
  17. eigenschmeigen Group Title
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    cool have a go playing with that, and see how far you can get it

    • one year ago
  18. cwrw238 Group Title
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    b = fd^(q-1)

    • one year ago
  19. eigenschmeigen Group Title
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    yes

    • one year ago
  20. eigenschmeigen Group Title
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    that was a lovely problem :) made me very happy. i haven't really come across logarithms intertwined with matrices before. thanks for showing me

    • one year ago
  21. eigenschmeigen Group Title
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    hows it going?

    • one year ago
  22. cwrw238 Group Title
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    i'm struggling - whats confusing me is that it says do not expand

    • one year ago
  23. eigenschmeigen Group Title
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    ah ok well ill put up my solution and you can read through as far as you like

    • one year ago
  24. cwrw238 Group Title
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    ok

    • one year ago
  25. eigenschmeigen Group Title
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    \[\left|\begin{matrix}log(a) & p & 1\\ log(b) & q & 1\\ log(c) & r & 1\end{matrix}\right| \]

    • one year ago
  26. eigenschmeigen Group Title
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    \[\left|\begin{matrix}\log(fd^{p-1}) & p & 1\\ \log(fd^{q-1}) & q & 1\\ \log(fd^{r-1}) & r & 1\end{matrix}\right|\]

    • one year ago
  27. eigenschmeigen Group Title
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    now just to check, are you familiar with how you can manipulate determinants, eg row operations?

    • one year ago
  28. cwrw238 Group Title
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    yes - i got that far! - no more!

    • one year ago
  29. cwrw238 Group Title
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    only vaguely

    • one year ago
  30. cwrw238 Group Title
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    thats why I'm stuck

    • one year ago
  31. eigenschmeigen Group Title
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    lets start by applying some log laws first we break apart the logarithms like this: \[\left|\begin{matrix}\log(d^{p-1}) + \log(f) & p & 1\\ \log(d^{q-1}) + \log(f) & q & 1\\ \log(d^{r-1}) + \log(f) & r & 1\end{matrix}\right|\]

    • one year ago
  32. cwrw238 Group Title
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    ok

    • one year ago
  33. eigenschmeigen Group Title
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    then: \[\left|\begin{matrix}\ (p-1)\log(d) + \log(f) & p & 1\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]

    • one year ago
  34. cwrw238 Group Title
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    yea i follow that

    • one year ago
  35. eigenschmeigen Group Title
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    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: \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]

    • one year ago
  36. eigenschmeigen Group Title
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    now we do R2- R3 : \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]

    • one year ago
  37. cwrw238 Group Title
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    oh yes i remember doing something like this - to find inverse of matrix i think

    • one year ago
  38. cwrw238 Group Title
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    i think i know whats coming - when you calculate the determinant the zeros will be significant

    • one year ago
  39. eigenschmeigen Group Title
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    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.

    • one year ago
  40. eigenschmeigen Group Title
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    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?

    • one year ago
  41. cwrw238 Group Title
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    to be honest no

    • one year ago
  42. eigenschmeigen Group Title
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    ok no worries, ill explain :)

    • one year ago
  43. cwrw238 Group Title
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    yes i think i see now - those elements in row 3 are multiplied by 0

    • one year ago
  44. eigenschmeigen Group Title
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    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: \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & 0 & 1\end{matrix}\right|\]

    • one year ago
  45. eigenschmeigen Group Title
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    there are two ways to think about it, by expanding or by operations, whichever you prefer :)

    • one year ago
  46. eigenschmeigen Group Title
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    if we were to expand it those two elements wouldn't count because, as you said, they just get multiplied by zero

    • one year ago
  47. eigenschmeigen Group Title
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    does this help or is it still confusing? i hope i haven't made it worse lol

    • one year ago
  48. cwrw238 Group Title
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    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

    • one year ago
  49. eigenschmeigen Group Title
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    basically its all about using row operations. they through you a curve-ball with the logarithms and the GP

    • one year ago
  50. eigenschmeigen Group Title
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    here is a good site http://www.sosmath.com/matrix/determ1/determ1.html

    • one year ago
  51. cwrw238 Group Title
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    thanks a lot

    • one year ago
  52. eigenschmeigen Group Title
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    no problem :) thanks for the juicy problem

    • one year ago
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