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cwrw238
 4 years ago
Please try this question from a math test. I am baffled by it.
4fccab66e4b0c6963ad793dfsmishra1338813325025copyofnewmicrosoftofficeworddocument.docx
cwrw238
 4 years ago
Please try this question from a math test. I am baffled by it. 4fccab66e4b0c6963ad793dfsmishra1338813325025copyofnewmicrosoftofficeworddocument.docx

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont think it uploaded properly try again and ill try and help

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1If 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 

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1i think its a determinant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what does GP stand for?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah nice. ok give me a minute

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1i assume that they are consecutive terms  can we do that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0unfortunately not, but we do know their positions, p,q,r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the nth term of a GP is A(r^(n1)) where A is the first term and r is the common ratio

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to avoid confusion lets use d instead of r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so lets call the first term of the progression "f" and the common ratio "d"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you see this: \[a= fd^{p1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cool have a go playing with that, and see how far you can get it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that was a lovely problem :) made me very happy. i haven't really come across logarithms intertwined with matrices before. thanks for showing me

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1i'm struggling  whats confusing me is that it says do not expand

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah ok well ill put up my solution and you can read through as far as you like

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\left\begin{matrix}log(a) & p & 1\\ log(b) & q & 1\\ log(c) & r & 1\end{matrix}\right \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\left\begin{matrix}\log(fd^{p1}) & p & 1\\ \log(fd^{q1}) & q & 1\\ \log(fd^{r1}) & r & 1\end{matrix}\right\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now just to check, are you familiar with how you can manipulate determinants, eg row operations?

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1yes  i got that far!  no more!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lets start by applying some log laws first we break apart the logarithms like this: \[\left\begin{matrix}\log(d^{p1}) + \log(f) & p & 1\\ \log(d^{q1}) + \log(f) & q & 1\\ \log(d^{r1}) + \log(f) & r & 1\end{matrix}\right\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then: \[\left\begin{matrix}\ (p1)\log(d) + \log(f) & p & 1\\ \ (q1)\log(d) + \log(f) & q & 1\\ \ (r1)\log(d) + \log(f) & r & 1\end{matrix}\right\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, 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: \[\left\begin{matrix}\ (pq)\log(d) & pq & 0\\ \ (q1)\log(d) + \log(f) & q & 1\\ \ (r1)\log(d) + \log(f) & r & 1\end{matrix}\right\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now we do R2 R3 : \[\left\begin{matrix}\ (pq)\log(d) & pq & 0\\ \ (qr)\log(d) & qr & 0\\ \ (r1)\log(d) + \log(f) & r & 1\end{matrix}\right\]

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1oh yes i remember doing something like this  to find inverse of matrix i think

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1i think i know whats coming  when you calculate the determinant the zeros will be significant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0probably 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
 4 years ago
Best ResponseYou've already chosen the best response.0because of the two existing zeros we can do this: \[\left\begin{matrix}\ (pq)\log(d) & pq & 0\\ \ (qr)\log(d) & qr & 0\\ \ 0 & 0 & 1\end{matrix}\right\] do you understand or should i explain this step more?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok no worries, ill explain :)

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1yes i think i see now  those elements in row 3 are multiplied by 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how about trying to do the operation (column 2)  r(column 3) here: \[\left\begin{matrix}\ (pq)\log(d) & pq & 0\\ \ (qr)\log(d) & qr & 0\\ \ (r1)\log(d) + \log(f) & r & 1\end{matrix}\right\] we would get: \[\left\begin{matrix}\ (pq)\log(d) & (pq)0 & 0\\ \ (qr)\log(d) & (qr)  0 & 0\\ \ (r1)\log(d) + \log(f) & (r) r & 1\end{matrix}\right\] to get: \[\left\begin{matrix}\ (pq)\log(d) & pq & 0\\ \ (qr)\log(d) & qr & 0\\ \ (r1)\log(d) + \log(f) & 0 & 1\end{matrix}\right\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there are two ways to think about it, by expanding or by operations, whichever you prefer :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if we were to expand it those two elements wouldn't count because, as you said, they just get multiplied by zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does this help or is it still confusing? i hope i haven't made it worse lol

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1i think i'm getting it  i remember finding this stuff difficult at alevel 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
 4 years ago
Best ResponseYou've already chosen the best response.0basically its all about using row operations. they through you a curveball with the logarithms and the GP

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here is a good site http://www.sosmath.com/matrix/determ1/determ1.html

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no problem :) thanks for the juicy problem
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