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

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eigenschmeigenBest ResponseYou've already chosen the best response.1
i dont think it uploaded properly try again and ill try and help
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
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

eigenschmeigenBest ResponseYou've already chosen the best response.1
sorry i didnt read
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
i think its a determinant
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
what does GP stand for?
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
ah nice. ok give me a minute
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
i assume that they are consecutive terms  can we do that?
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
unfortunately not, but we do know their positions, p,q,r
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
the nth term of a GP is A(r^(n1)) where A is the first term and r is the common ratio
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
to avoid confusion lets use d instead of r
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
so lets call the first term of the progression "f" and the common ratio "d"
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
can you see this: \[a= fd^{p1}\]
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
cool have a go playing with that, and see how far you can get it
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
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

cwrw238Best ResponseYou've already chosen the best response.1
i'm struggling  whats confusing me is that it says do not expand
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
ah ok well ill put up my solution and you can read through as far as you like
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
\[\left\begin{matrix}log(a) & p & 1\\ log(b) & q & 1\\ log(c) & r & 1\end{matrix}\right \]
 one year ago

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

eigenschmeigenBest ResponseYou've already chosen the best response.1
now just to check, are you familiar with how you can manipulate determinants, eg row operations?
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
yes  i got that far!  no more!
 one year ago

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

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

eigenschmeigenBest ResponseYou've already chosen the best response.1
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: \[\left\begin{matrix}\ (pq)\log(d) & pq & 0\\ \ (q1)\log(d) + \log(f) & q & 1\\ \ (r1)\log(d) + \log(f) & r & 1\end{matrix}\right\]
 one year ago

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

cwrw238Best ResponseYou've already chosen the best response.1
oh yes i remember doing something like this  to find inverse of matrix i think
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
i think i know whats coming  when you calculate the determinant the zeros will be significant
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
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

eigenschmeigenBest ResponseYou've already chosen the best response.1
because 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?
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
ok no worries, ill explain :)
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
yes i think i see now  those elements in row 3 are multiplied by 0
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
how 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: \[\left\begin{matrix}\ (pq)\log(d) & pq & 0\\ \ (qr)\log(d) & qr & 0\\ \ (r1)\log(d) + \log(f) & 0 & 1\end{matrix}\right\]
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
there are two ways to think about it, by expanding or by operations, whichever you prefer :)
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
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

eigenschmeigenBest ResponseYou've already chosen the best response.1
does this help or is it still confusing? i hope i haven't made it worse lol
 one year ago

cwrw238Best ResponseYou've already chosen the best response.1
i 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
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
basically its all about using row operations. they through you a curveball with the logarithms and the GP
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
here is a good site http://www.sosmath.com/matrix/determ1/determ1.html
 one year ago

eigenschmeigenBest ResponseYou've already chosen the best response.1
no problem :) thanks for the juicy problem
 one year ago
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