# [OPE-L:5034] Re: Re: RE: numerical example!!!

From: Allin Cottrell (cottrell@wfu.edu)
Date: Thu Feb 22 2001 - 09:21:38 EST

```On Wed, 21 Feb 2001, Fred B. Moseley wrote:

> Assuming to begin with that the rate of profit = 0.4, the equations are:
>
> 	(1.4) [28 p1+ 56 (5/80 p3)]   =   56 p1
>
> 	(1.4) [16 p1+ 16 (5/80 p3)]   =   48 p2
>
> 	(1.4) [12 p1+  8 (5/80 p3)]    =   8 p3
>
> Please solve for the three prices.

Andrew's right, so far as I can see.

Call the equations (1), (2) and (3) respectively.  They simplify
thus:

(1) => (1.4) [0.5 p1 + 5/80 p3] =   p1
(2) => (1.4) [    p1 + 5/80 p3] = 3 p2
(3) => (1.4) [1.5 p1 + 5/80 p3] =   p3

So:

(1) => 0.7 p1 + 7/80 p3 =   p1
(2) => 1.4 p1 + 7/80 p3 = 3 p2
(3) => 2.1 p1 + 7/80 p3 =   p3

Write P for the vector [p1, p2, p3]', and M for the coefficient
matrix, such that MP = 0.  Then M is

-0.3   0     7/80
1.4  -3     7/80
2.1   0   -73/80

The system MP = 0 has a non-trival solution for P iff M is singular,
but in this case |M| = -0.27, so there exists no non-trivial solution.

Allin.
```

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