Simultaneous system... with a twist! Your help would be good...

Studying matricies at the moment and we were shown a problem that cannot be solved using the methods taught to us. We were asked to think about it and I've done some research but are unable to get the correct answers.

It's more of a "see what you can find out" than getting it right but I'm now curious...

Q: A group of drinkers keep a record of what they drink over the course of an evening in a bar. The following matrix shows the simultaneous system after four rounds when they drink 4 different drinks w, x, y & z...

$\displaystyle \left| \begin{matrix} 1.5 & 2 & 2.5 & 0.5 \\ 2.5 & 0.5 & 2 & 2.5 \\ 2 & 2.5 & 3 & 1\\ 1 & 2 & 0 & 2.5 \end{matrix} \right| * \left| \begin{matrix} w\\x\\y\\z\end{matrix} \right| = \left| \begin{matrix} 8.99\\10.46\\11.78\\7.99 \end{matrix} \right|$

The problems arises that it appears the drinks cost:

w = - 3.86

x= 0.60

y= 4.58

z= 4.26

Which seems that they were paid to drink drink w. What is the true price of the drinks?

So , I did some research and came accross ill-conditioned systems which carry errors. These errors can be maginifed when finding solutions. I guess that the method is beyond my course but I'd still like to know how to solve it....

Any takers?

Thanks, felix

Re: Simultaneous system... with a twist! Your help would be good...

well, there is a "hidden assumption" in the problem, which is:

that the bar charged accurately for the drinks. to amplify, suppose the cost of drink w is odd (in cents). buying 0.5 of a drink would result in a cost of a fractional amount of cents. now, the error introduced by "rounding up" to the nearest penny, could be 0.5 cents. this makes little difference to the total a customer pays, but it drastically affects the solution space of the problem.

by playing around with the values of the fourth column of the augmented matrix, i discovered (largely by trial-and-error, i am sure there is a more precise numerical method) the following solution:

w = $1.23

x = $1.44

y = $1.39

z = $1.55

assuming the bar rounds up for 0.5 a drink, this makes the "true value" of the fourth column of the matrix:

8.975 (drinker 1 paid 1.5 cents extra for his 3 0.5 drinks)

10.45 (drinker 2 paid 1 cent extra for his 3 0.5 drinks, one of which is actually "fair price", since the cost per drink is an even number of cents)

11.78 (drinker 3 didn't pay extra because his one 0.5 drink was at a "fair price")

7.985 (drinker 4 paid 0.5 extra for his one 0.5 drink).

Re: Simultaneous system... with a twist! Your help would be good...

Ah-ha! I see. Roundings affect the result massiviely don't they? Thanks for pointing this out....