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Matrix Manipulation/Algebra

  • #1

Homework Statement



A multi-product firm has total cost function C(q) = qtAq and faces inter-related but linear demand schedules for the n goods it produces: q = Bp + c. Both A and B are symmetric and B is invertible. Obtain an expression for total profit π(q) in the form π(q) = qtDq - etq where D and e are appropriate matrices.


Homework Equations



We are given that Total Cost is, C(q) = qtAq.
We also are given a production function: q = Bp + c

Total profits is just Total Revenue, p*q, minus Total Costs.
Hence:
π = p*q - C(q).
And we want to get it into a form like this:
π(q) = qtDq - etq


The Attempt at a Solution



So far I have done some manipulation of the production function:
q = Bp+c
B-1(q-c)=p
and then substituted into the equation: π(q) = p*q-C(q) = (B-1(q-c))q - qtAq

The remaining dificulities I'm having is figuring out how to get this resulting equation to look something like the requested, π(q) = qtDq - etq. I'm also not sure if the algebra is fully legal.

While I know this isn't Physics, it's really just math. The only econ part of it is in the word problem and equation definitions. If you could help at all I would be most appreciative. If you need any more info please let me know.

Thanks!
 

Answers and Replies

  • #2
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0
π = p*q - C(q).
Try thinking about this equation right here. Remember that both [itex]\mathbf{p}[/itex] and [itex]\mathbf{q}[/itex] are vectors, while [itex]n[/itex] is a scalar, so the "[itex]*[/itex]" is scalar multiplication. But scalar multiplication can be rewritten using matrix multiplication, right? And once you write it in that form, try looking at its transpose.
 
  • #3
I'm not sure if I completely understand, but this is what I've managed to come up with.
If we substitute the solved p value into the equation as well as the total cost we come up with something like:

n=p*q - c(q) = [B-1(q-c)]q - qAqt

If I take the transpose of the substituted p, pt, I get:

pt = [B-1(q-c)]t = (B-1)t(q-c)t = (B-1)t(qt-ct).

Am I then able to try and carry some terms through, such as:

qpt = q*(B-1)t(qt-ct) = q(B-1)tqt-B-1ctq

And then combine with the original, n = qp - c(q) into:

n = q(B-1)tqt - B-1ctq - qAqt.

In this case the B-1ct could be equal to the et but I would still have to find a way to combine the q(B-1)tqt - qAqt.

Thank you for your help. It's been quite some time since I've done anything with matrices and my brain seems to be full of rust.
 
  • #4
I've reworked this now, trying to pay a little more attention to the way the distributive property works with matrices as well as how transpose carries through. If I start from defining the profit, n.

n = TR - TC.
TR = pq
TC = C(q) = qtAq

We know that q = Bp+c which means that B-1(q-c) = p.
Now if I take the transpose of that, I get:
pt=(B-1(q-c))t=(q-c)t(B-1)t=(qt-ct)(B-1)t=(qt-ct)(Bt)-1=(qt-ct)B-1=qtB-1-ctB-1 since we knew B was symmetrical, and therefore B=Bt

Now when I substitute that back into the equation, I get:
ptq=(qtB-1-ctB-1)q=qtB-1q-ctB-1q

If I do the same with taking C(q)t, I get:
C(q)t = (qAqt)t=qtAtq=qtAq because A is symmetrical as well.

Now, putting that all into the equation we arrive at:
n = ptq-C(q)t
n = qtB-1q-ctB-1q-qtAq=qt(B-1-A)q-ctB-1q
and that falls into the n = qtDq - etq kind of format.

*I guess the issue is this: if I let n = pq - c = ptq-ct it works. Is this possible or do I have to start from the beginning and say nt = (pq)t-ct?
 
Last edited:
  • #5
I figured it out. I don't know what I was possibly thinking to begin with as it was actually a simple task. Thanks for bearing with my multiple posts and bumps.
 

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