- cwrw238

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

- chestercat

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- anonymous

i dont think it uploaded properly try again and ill try and help

- cwrw238

i'll draw it

- anonymous

ok :)

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## More answers

- 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

sorry i didnt read

- cwrw238

i think its a determinant

- anonymous

what does GP stand for?

- cwrw238

geometric series

- anonymous

ah nice. ok give me a minute

- cwrw238

i assume that they are consecutive terms - can we do that?

- anonymous

unfortunately not, but we do know their positions, p,q,r

- 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

to avoid confusion lets use d instead of r

- anonymous

so lets call the first term of the progression "f" and the common ratio "d"

- anonymous

can you see this:
\[a= fd^{p-1}\]

- cwrw238

yes

- anonymous

cool have a go playing with that, and see how far you can get it

- cwrw238

b = fd^(q-1)

- anonymous

yes

- 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

hows it going?

- cwrw238

i'm struggling - whats confusing me is that it says do not expand

- anonymous

ah ok well ill put up my solution and you can read through as far as you like

- cwrw238

ok

- anonymous

\[\left|\begin{matrix}log(a) & p & 1\\ log(b) & q & 1\\ log(c) & r & 1\end{matrix}\right| \]

- 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

now just to check, are you familiar with how you can manipulate determinants, eg row operations?

- cwrw238

yes - i got that far! - no more!

- cwrw238

only vaguely

- cwrw238

thats why I'm stuck

- 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

ok

- 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

yea i follow that

- 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

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

oh yes i remember doing something like this - to find inverse of matrix i think

- cwrw238

i think i know whats coming - when you calculate the determinant the zeros will be significant

- 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

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

to be honest no

- anonymous

ok no worries, ill explain :)

- cwrw238

yes i think i see now - those elements in row 3 are multiplied by 0

- 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

there are two ways to think about it, by expanding or by operations, whichever you prefer :)

- anonymous

if we were to expand it those two elements wouldn't count because, as you said, they just get multiplied by zero

- anonymous

does this help or is it still confusing? i hope i haven't made it worse lol

- 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

basically its all about using row operations. they through you a curve-ball with the logarithms and the GP

- anonymous

here is a good site http://www.sosmath.com/matrix/determ1/determ1.html

- cwrw238

thanks a lot

- anonymous

no problem :) thanks for the juicy problem

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