cwrw238
  • cwrw238
Please try this question from a math test. I am baffled by it. 4fccab66e4b0c6963ad793df-smishra-1338813325025-copyofnewmicrosoftofficeworddocument.docx
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
i dont think it uploaded properly try again and ill try and help
cwrw238
  • cwrw238
i'll draw it
anonymous
  • anonymous
ok :)

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cwrw238
  • 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 |
anonymous
  • anonymous
sorry i didnt read
cwrw238
  • cwrw238
i think its a determinant
anonymous
  • anonymous
what does GP stand for?
cwrw238
  • cwrw238
geometric series
anonymous
  • anonymous
ah nice. ok give me a minute
cwrw238
  • cwrw238
i assume that they are consecutive terms - can we do that?
anonymous
  • anonymous
unfortunately not, but we do know their positions, p,q,r
anonymous
  • anonymous
the nth term of a GP is A(r^(n-1)) where A is the first term and r is the common ratio
anonymous
  • anonymous
to avoid confusion lets use d instead of r
anonymous
  • anonymous
so lets call the first term of the progression "f" and the common ratio "d"
anonymous
  • anonymous
can you see this: \[a= fd^{p-1}\]
cwrw238
  • cwrw238
yes
anonymous
  • anonymous
cool have a go playing with that, and see how far you can get it
cwrw238
  • cwrw238
b = fd^(q-1)
anonymous
  • anonymous
yes
anonymous
  • anonymous
that was a lovely problem :) made me very happy. i haven't really come across logarithms intertwined with matrices before. thanks for showing me
anonymous
  • anonymous
hows it going?
cwrw238
  • cwrw238
i'm struggling - whats confusing me is that it says do not expand
anonymous
  • anonymous
ah ok well ill put up my solution and you can read through as far as you like
cwrw238
  • cwrw238
ok
anonymous
  • anonymous
\[\left|\begin{matrix}log(a) & p & 1\\ log(b) & q & 1\\ log(c) & r & 1\end{matrix}\right| \]
anonymous
  • anonymous
\[\left|\begin{matrix}\log(fd^{p-1}) & p & 1\\ \log(fd^{q-1}) & q & 1\\ \log(fd^{r-1}) & r & 1\end{matrix}\right|\]
anonymous
  • anonymous
now just to check, are you familiar with how you can manipulate determinants, eg row operations?
cwrw238
  • cwrw238
yes - i got that far! - no more!
cwrw238
  • cwrw238
only vaguely
cwrw238
  • cwrw238
thats why I'm stuck
anonymous
  • anonymous
lets start by applying some log laws first we break apart the logarithms like this: \[\left|\begin{matrix}\log(d^{p-1}) + \log(f) & p & 1\\ \log(d^{q-1}) + \log(f) & q & 1\\ \log(d^{r-1}) + \log(f) & r & 1\end{matrix}\right|\]
cwrw238
  • cwrw238
ok
anonymous
  • anonymous
then: \[\left|\begin{matrix}\ (p-1)\log(d) + \log(f) & p & 1\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]
cwrw238
  • cwrw238
yea i follow that
anonymous
  • anonymous
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: \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]
anonymous
  • anonymous
now we do R2- R3 : \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]
cwrw238
  • cwrw238
oh yes i remember doing something like this - to find inverse of matrix i think
cwrw238
  • cwrw238
i think i know whats coming - when you calculate the determinant the zeros will be significant
anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
cwrw238
  • cwrw238
to be honest no
anonymous
  • anonymous
ok no worries, ill explain :)
cwrw238
  • cwrw238
yes i think i see now - those elements in row 3 are multiplied by 0
anonymous
  • anonymous
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: \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & 0 & 1\end{matrix}\right|\]
anonymous
  • anonymous
there are two ways to think about it, by expanding or by operations, whichever you prefer :)
anonymous
  • anonymous
if we were to expand it those two elements wouldn't count because, as you said, they just get multiplied by zero
anonymous
  • anonymous
does this help or is it still confusing? i hope i haven't made it worse lol
cwrw238
  • 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
anonymous
  • anonymous
basically its all about using row operations. they through you a curve-ball with the logarithms and the GP
anonymous
  • anonymous
here is a good site http://www.sosmath.com/matrix/determ1/determ1.html
cwrw238
  • cwrw238
thanks a lot
anonymous
  • anonymous
no problem :) thanks for the juicy problem

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