equitable matrix for money exchange
This example shows how equitable matrices arise in money exchange. Consider $n$ currencies ${C}_{1},{C}_{2},\mathrm{\dots},{C}_{n}$, where ${C}_{i}$ stays for some name like “dollar”, “euro”, “pound”, etc. Denote the exchange rate between currencies ${C}_{i}$ and ${C}_{j}$ as ${a}_{ij}>0$, i.e.
$$1{C}_{i}\u27f6{a}_{ij}{C}_{j}.$$  (1) 
We will call $A={({a}_{ij})}_{i,j=1}^{n}$ an exchange rates matrix. Suppose, $A$ is an equitable matrix, i.e.
$${a}_{ij}={a}_{ik}\cdot {a}_{kj},i,j,k=1,\mathrm{\dots},n.$$  (2) 
Let us discuss consequences of this. First of all, there is no loss when exchanging between two currencies: if one exchanges $u$ units of ${C}_{i}$ to ${C}_{j}$ and then back to ${C}_{i}$, one will have again $u$ units. Indeed,
$$u{C}_{i}\u27f6u\cdot {a}_{ij}{C}_{j}\u27f6u\cdot ({a}_{ij}\cdot {a}_{ji}){C}_{i}$$ 
and desired conjecture follows from the fact that
$${a}_{ij}\cdot {a}_{ji}=1,i,j=1,\mathrm{\dots},n,$$  (3) 
which can be proven by putting $j=i$ in (2) and using diagonal property (${a}_{ii}=1$, for all $i=1,\mathrm{\dots},n$). But equitable property (2) suggests in fact more than just (3), i.e. more than just no loss by changing from one currency to another and back. For illustration consider an example.
Example 1.
Let us take three currencies ${C}_{1}$, ${C}_{2}$, ${C}_{3}$ with the following exchange rates
$1{C}_{1}$  $\u27f6$  $2{C}_{2},$  
$1{C}_{1}$  $\u27f6$  $3{C}_{3},$  
$1{C}_{2}$  $\u27f6$  $2{C}_{3}.$ 
The above relations^{} define ${a}_{12},{a}_{13},{a}_{23}$ in the exchange rates matrix $A$. Let us define other elements by (3). This gives us
$$A=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1/2\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 1/3\hfill & \hfill 1/2\hfill & \hfill 1\hfill \end{array}\right).$$ 
Now assume one has 100 ${C}_{3}$ units and wants to exchange them to ${C}_{1}$. If one does this directly, one will obtain 300 ${C}_{1}$ units. But if one exchanges first to ${C}_{2}$ and then to ${C}_{1}$, one will obtain 400 ${C}_{1}$ units.
For an equitable exchange rates matrix the above described situation is impossible: exchanging $u{C}_{i}$ units to ${C}_{j}$ is the same as first exchanging to ${C}_{k}$ and then to ${C}_{j}$. Indeed,
$$\begin{array}{ccccc}& & \hfill u{C}_{i}\hfill & \hfill \u27f6\hfill & \hfill u\cdot {a}_{ij}{C}_{j}\hfill \\ \hfill u{C}_{i}\hfill & \hfill \u27f6\hfill & \hfill u\cdot {a}_{ik}{C}_{k}\hfill & \hfill \u27f6\hfill & \hfill u\cdot ({a}_{ik}\cdot {a}_{kj}){C}_{j}\hfill \end{array}$$ 
and the final result is the same due to the equitable property (2). Note, that the matrix from the example is not equitable
$${a}_{23}\ne {a}_{21}\cdot {a}_{13}.$$ 
The above consideration shows that equitable property does not allow making money just by exchanging currencies. If (2) does not hold for some indexes, for example
$$ 
then having $u{C}_{i}$ units one can make money just by exchanging them to ${C}_{j}$ through ${C}_{k}$ and back
$$u{C}_{i}\u27f6u\cdot {a}_{ik}{C}_{k}\u27f6u\cdot ({a}_{ik}\cdot {a}_{kj}){C}_{j}\u27f6u\cdot \left(\frac{{a}_{ik}\cdot {a}_{kj}}{{a}_{ij}}\right){C}_{i}.$$ 
If we denote $q:=\frac{{a}_{ik}\cdot {a}_{kj}}{{a}_{ij}}>1$, then after making $N$ such exchanges instead of $u$ one would have $u\cdot {q}^{N}$ units — the capital would increase like geometric progression! If there is an opposite inequality
$${a}_{ij}>{a}_{ik}\cdot {a}_{kj},$$ 
then such advantage have those with ${C}_{j}$ currency. So, condition (2) guarantees that no one can speculate on currency exchange, thus motivating the name “equitable” — “to be fair”.
Let us give an interpretation^{} of the matrixvector multiplication for exchange rates matrices. This interpretation is not connected with the equitable property (2) and shows why exchange rates matrices are useful in general.
Consider a company which operates on the international market (e.g., a company which makes furniture/cars/household equipment/etc, and sells their products to more than one country) and, thus, obtains money in different currencies ${C}_{1},{C}_{2},\mathrm{\dots},{C}_{n}$. From time to time, for such a company the natural question arises: what is the total amount of money we have? Specifically, at a given time the company obtained the following money: ${u}_{1}{C}_{1},{u}_{2}{C}_{2},\mathrm{\dots},{u}_{n}{C}_{n}$. With given exchange rates ${a}_{ij}$, the total amount of money in the currency ${C}_{i}$ is ${\sum}_{j=1}^{n}{u}_{j}\cdot {a}_{ji}$. That’s why we have
if matrix $A$ is the exchange rates matrix for currencies ${C}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{C}_{n}$, $u\mathrm{=}\mathrm{(}{u}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{u}_{n}\mathrm{)}$ is a rowvector with components^{} expressing amount of units in each currency, then components of the vector $u\mathrm{\cdot}A$ express the total amount of money in each currency.
The following example gives illustration to this.
Example 2.
Imagine a company located in Germany, let’s call it “Peaut”, which makes auto “Leoptera”. It sells this auto to five different countries: Germany (its home country), USA, United Kingdom, Japan, and Switzerland. Thus, it needs to operate with the following currencies:
${C}_{1}$  $=$  $\text{\u201ceuro\u201d=\u201cEUR\u201d},$  
${C}_{2}$  $=$  $\text{\u201cUSA dollar\u201d=\u201cUSD\u201d},$  
${C}_{3}$  $=$  $\text{\u201cBritish pound\u201d=\u201cGBP\u201d},$  
${C}_{4}$  $=$  $\text{\u201cYapanese yen\u201d=\u201cJPY\u201d},$  
${C}_{5}$  $=$  $\text{\u201cSwiss frank\u201d=\u201cCHF\u201d}.$ 
The corresponding exchange rates matrix $A$ is presented in Table 1. The values were first collected from \htmladdnormallinkWikipediahttp://en.wikipedia.org/ at the middle of 2005, and then they were modified such that the resulting matrix is equitable.
EUR 
USD 
GBP 
JPY 
CHF 

EUR 
1 
1.25 
0.625 
125 
1.5626 
USD 
0.8 
1 
0.5 
100 
1.25 
GBP 
1.6 
2 
1 
200 
2.5 
JPY 
0.008 
0.01 
0.005 
1 
0.0125 
CHF 
0.64 
0.8 
0.4 
80 
1 
The price of “Leoptera” in each country, number of sold cars during a year, and corresponding amount of money are gathered in Table 2. The last column gives components of the rowvector $u$ in our example. To answer the question what is the total amount of money are obtained by “Peaut”, one needs to compute $u\cdot A$. The result is presented in Table 3.
Country where the cars were sold 
Price for one car 
Amount of sold cars (in thousands) 
Obtained money (in milliards) 

Germany 
21.000 EUR 
100 
2.1 EUR 
USA 
32.000 USD 
35 
1.12 USD 
UK 
14.000 GBP 
40 
0.56 GBP 
Japan 
3.1 mln JPY 
8 
24.8 JPY 
Switzerland 
34.000 CHF 
5 
0.17 CHF 
Country’s currency 
The total amount of money in different currencies (in milliards) 
EUR 
4.1992 
USD 
5.249 
GBP 
2.6245 
JPY 
524.9 
CHF 
6.56146 
Title  equitable matrix for money exchange 

Canonical name  EquitableMatrixForMoneyExchange 
Date of creation  20130322 14:59:25 
Last modified on  20130322 14:59:25 
Owner  mathforever (4370) 
Last modified by  mathforever (4370) 
Numerical id  9 
Author  mathforever (4370) 
Entry type  Example 
Classification  msc 9100 
Classification  msc 91B28 
Classification  msc 1500 