Question

Please try this question from a math test. I am baffled by it.
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Answer 1

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

Answer 2

i'll draw it

Answer 3

ok :)

Answer 4

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 |

Answer 5

sorry i didnt read

Answer 6

i think its a determinant

Answer 7

what does GP stand for?

Answer 8

geometric series

Answer 9

ah nice. ok give me a minute

Answer 10

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

Answer 11

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

Answer 12

the nth term of a GP is A(r^(n-1)) where A is the first term and r is the common ratio

Answer 13

to avoid confusion lets use d instead of r

Answer 14

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

Answer 15

can you see this:

\[a= fd^{p-1}\]

Answer 16

yes

Answer 17

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

Answer 18

b = fd^(q-1)

Answer 19

yes

Answer 20

that was a lovely problem :) made me very happy. i haven't really come across logarithms intertwined with matrices before. thanks for showing me

Answer 21

hows it going?

Answer 22

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

Answer 23

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

Answer 24

ok

Answer 25

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

Answer 26

\[\left|\begin{matrix}\log(fd^{p-1}) & p & 1\\ \log(fd^{q-1}) & q & 1\\ \log(fd^{r-1}) & r & 1\end{matrix}\right|\]

Answer 27

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

Answer 28

yes - i got that far! - no more!

Answer 29

only vaguely

Answer 30

thats why I'm stuck

Answer 31

lets start by applying some log laws
first we break apart the logarithms like this:

\[\left|\begin{matrix}\log(d^{p-1}) + \log(f) & p & 1\\ \log(d^{q-1}) + \log(f) & q & 1\\ \log(d^{r-1}) + \log(f) & r & 1\end{matrix}\right|\]

Answer 32

ok

Answer 33

then:
\[\left|\begin{matrix}\ (p-1)\log(d) + \log(f) & p & 1\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]

Answer 34

yea i follow that

Answer 35

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:
\[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]

Answer 36

now we do R2- R3 :

\[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]

Answer 37

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

Answer 38

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

Answer 39

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.

Answer 40

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?

Answer 41

to be honest no

Answer 42

ok no worries, ill explain :)

Answer 43

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

Answer 44

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:
\[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & 0 & 1\end{matrix}\right|\]

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

thanks a lot

Answer 52

no problem :) thanks for the juicy problem