OK, I am having a problem with this economic theory by Piero Sraffa. He says that production affects value, that's not the hard part. Here is his first example of a hypothetical economy:

280 qr. wheat + 12 t. iron --> 400 qr. Wheat

120 qr. wheat + 8 t. iron --> 20 t. iron

 

I figured, hey what the heck, just set the value of a ton of iron to x and a qr. of wheat to y.

280y + 12x = 400y

120y + 8x = 20x

Then subtract and we get:

12x = 120y

120y = 12x

Thus 10 qr. wheat is exchangeable for 1 t. iron.

 

But now he introduces the production with a surplus of 175 qr. wheat. There is a "rate of profit" scalar "r" that is [math]0\geq r\geq1[/math]and then we multiply both (1+r)(inputs).

 

He then says OK given

(1+r)(280y + 12x) = 575y

(1+r)(120y + 8x) = 20x

then magically comes to the conclusion r=0.25 and that 15y=x. I don't understand that little minor part.

This isn't exactly a matrix problem. What you have there is a non-linear equation in three variables; no amount of playing with matrices is going to give you the answer you're looking for. You should probably note that there is another trivial solution, x = y = 0; for the rest of this post I'll assume that we've already found this result and hence dividing by x/y won't be a problem.

 

Finding the 15y = x part is fairly easy. You can simply do the following:

 

[math]\frac{(1+r)(280y + 12x)}{(1+r)(120y+8x)} = \frac{575y}{20x}[/math]

 

Then simplify to [math]20x(280x + 12x) = 575y(120y+8x)[/math]. Expand and simplify to give you [math]24x^2+100xy-6900y^2 = 0[/math]. Now, this may look nasty and horrible but assuming you fix y, you can find x - it's just a quadratic equation.

 

So from this we get [math]x = \frac{-100y \pm \sqrt{10000y^2 + 662400y^2}}{48} = \frac{-100y \pm 820y}{48}[/math]

 

Now this is obviously going to give us two answers, and we want the positive one. So ignore the negative result (which would give x = -115y/6).

 

Hence we get [math]x = \frac{720y}{48}[/math] and the result follows.

 

Finding r is fairly trivial from this. Simply expand the two equations and add them together to get something along the lines of:

 

[math]-175y + r(400y + 20x) = 0[/math]

 

Now use your previous result to get: [math]r(400y+300y) = 175y[/math]. Divide through by y and simplify to get your answer.