Problem 13 Solution: "Characteristic Characters"

Consider the system with figured characteristics. Suppose there are N total characteristics. We will make the following definitions:
Let T be an N-dimensional row vector representing the cost per point for each characteristic.
Let P be an N-dimensional column vector representing the number of points of each characteristic that is purchased.
Let R be an N-dimensional column vector representing the number of points of each characteristic that the player actually ends up with.
Let X be the total number of points the player spends on characteristics.
We will now find relationships between these variables.
First, observe that X is just the dot product of T and P - or in other words TP.
Now consider the relationship between P and R. R is equal to P plus the vector V representing the number of "free points" of figured characteristics obtained. But since each element of V is a linear combination of elements of P (of course zero in the case of base characteristics), we can represent this as AP where A is a matrix representing the set of linear equations. So we have R = P+V = P+AP = (I+A)P where I is the N-by-N identity matrix.

In other words, the player gets a final characteristic vector of (I+A)P for a cost of TP points.
Now observe that I+A has ones all along its main diagonal, and is upper triangular (if you list the base characteristics before the figured ones), which is sufficient to show that it is nonsingular, and thus has an inverse, which we will designate as Z.

Now consider a system without figured characteristics, and with a cost vector of TZ. Then in order to get the same characteristic vector as before - (I+A)P - the player he simply buys those characteristics directly, for a total cost of TZ(I+A)P. But since Z and I+A are inverses, they cancel each other out, so the total cost is TP, just as before. This shows that the two systems have the same cost, regardless of what characteristics the player wants to buy.