|P*1 - P*1(T) = 0||P y1 - x1 P*1 = 0||P*2 - P*2(T) = 0||P y2 - x2 P*2 = 0||x1 + x2 - 1 = 0||y1+ y2 - 1 = 0|
|P*1 - P*1(T) = 0||(1)||p1 - x1 P*1 = 0||(2)||P*2 - P*2(T) = 0||(3)||p2 - x2 P*2 = 0||(4)||x1 + x2 - 1 = 0||(5)||p1+ p2 - P = 0||(6)|
The remaining unknowns in the equations are then:
P , P*1 , p1 , P*2 , x2 , p2
We can draw up an incidence or occurrence table for the set of equations as shown below, with a row for each equation and a column for each unknown, where an `x' indicates that the specified variable occurs in the equation and a blank that it does not.
|Equation (2)||x||x||Equation (3)||x||Equation (4)||x||x||x||Equation (5)||x||Equation (6)||x||x||x|
We might want to write this in `mathematical' notation we would replace the x's by ones, leaving the blanks, which are conventionally taken to imply zeros. The matrix would then look like this:
We explain the use of this type of matrix or table for ordering equations in section 220.127.116.11. To summarise briefly, the matrix shows that equations (1), (3) and (5) contain only a single unknown each, and so can be solved for each of these, i.e. respectively for P*1 , P*2 and x2. It can then be seen that equation (2) now contains only one remaining unknown, p1 as P*1 is no longer unknown, and so it can be solved. In fact all the remaining unknowns can now be determined by solving equation (4) and then (6) each for a single unknown.
Use of the matrix has shown how a problem which might have been thought to involve the simultaneous solution of 6 equations in six unknowns actually involves solving only one equation at a time.