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cwrw238

  • 3 years ago

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

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  1. eigenschmeigen
    • 3 years ago
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    i dont think it uploaded properly try again and ill try and help

  2. cwrw238
    • 3 years ago
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    i'll draw it

  3. eigenschmeigen
    • 3 years ago
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    ok :)

  4. cwrw238
    • 3 years ago
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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 |

  5. eigenschmeigen
    • 3 years ago
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    sorry i didnt read

  6. cwrw238
    • 3 years ago
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    i think its a determinant

  7. eigenschmeigen
    • 3 years ago
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    what does GP stand for?

  8. cwrw238
    • 3 years ago
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    geometric series

  9. eigenschmeigen
    • 3 years ago
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    ah nice. ok give me a minute

  10. cwrw238
    • 3 years ago
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    i assume that they are consecutive terms - can we do that?

  11. eigenschmeigen
    • 3 years ago
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    unfortunately not, but we do know their positions, p,q,r

  12. eigenschmeigen
    • 3 years ago
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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

  13. eigenschmeigen
    • 3 years ago
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    to avoid confusion lets use d instead of r

  14. eigenschmeigen
    • 3 years ago
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    so lets call the first term of the progression "f" and the common ratio "d"

  15. eigenschmeigen
    • 3 years ago
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    can you see this: \[a= fd^{p-1}\]

  16. cwrw238
    • 3 years ago
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    yes

  17. eigenschmeigen
    • 3 years ago
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    cool have a go playing with that, and see how far you can get it

  18. cwrw238
    • 3 years ago
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    b = fd^(q-1)

  19. eigenschmeigen
    • 3 years ago
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    yes

  20. eigenschmeigen
    • 3 years ago
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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

  21. eigenschmeigen
    • 3 years ago
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    hows it going?

  22. cwrw238
    • 3 years ago
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    i'm struggling - whats confusing me is that it says do not expand

  23. eigenschmeigen
    • 3 years ago
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    ah ok well ill put up my solution and you can read through as far as you like

  24. cwrw238
    • 3 years ago
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    ok

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

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

  27. eigenschmeigen
    • 3 years ago
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    now just to check, are you familiar with how you can manipulate determinants, eg row operations?

  28. cwrw238
    • 3 years ago
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    yes - i got that far! - no more!

  29. cwrw238
    • 3 years ago
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    only vaguely

  30. cwrw238
    • 3 years ago
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    thats why I'm stuck

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

  32. cwrw238
    • 3 years ago
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    ok

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

  34. cwrw238
    • 3 years ago
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    yea i follow that

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

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

  37. cwrw238
    • 3 years ago
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    oh yes i remember doing something like this - to find inverse of matrix i think

  38. cwrw238
    • 3 years ago
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    i think i know whats coming - when you calculate the determinant the zeros will be significant

  39. eigenschmeigen
    • 3 years ago
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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.

  40. eigenschmeigen
    • 3 years ago
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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?

  41. cwrw238
    • 3 years ago
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    to be honest no

  42. eigenschmeigen
    • 3 years ago
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    ok no worries, ill explain :)

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

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

  45. eigenschmeigen
    • 3 years ago
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    there are two ways to think about it, by expanding or by operations, whichever you prefer :)

  46. eigenschmeigen
    • 3 years ago
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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

  47. eigenschmeigen
    • 3 years ago
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    does this help or is it still confusing? i hope i haven't made it worse lol

  48. cwrw238
    • 3 years ago
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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

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

  50. eigenschmeigen
    • 3 years ago
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    here is a good site http://www.sosmath.com/matrix/determ1/determ1.html

  51. cwrw238
    • 3 years ago
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    thanks a lot

  52. eigenschmeigen
    • 3 years ago
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    no problem :) thanks for the juicy problem

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