# Find a price vector p for various prices of industries.

( Leontief input-output model ) Suppose that three industries are interrelated so that their outputs are used as inputs by themselves, according to the $3 \times 3$ consumption matrix

A = [$a_{jk}$] = $\left[ \begin{array}{ccc} 0.1&0.5&0\\ 0.8&0&0.4\\ 0.1&0.5&0.6 \end{array} \right]$

where $a_{jk}$ is the fraction of the output of industry $k$ consumed (purchased) by industry $j$. Let $p_{j}$ be the price charged by industry $j$ for its total output. A problem is to find prices so that for each industry, total expenditures equal total income. Determine that there is a price vector such that $~~~~$ p = $[~~p_{1} ~~~ p_{2} ~~~ p_{3}~~]^{T}$ $~$ for this scenario.

Any ideas on how to go about solving this??

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Viktor has answered your question, but I thought I would add that you are looking for what's called an eigenvector for your matrix $A$. –  Mike Spivey Mar 18 '11 at 20:43
@Jasen: If you have a new question, you shouldn't just edit your previous one, you should open a new one. –  user5501 Mar 18 '11 at 22:15
@Jasen: I have reverted your question to its original form. If you want to ask a new question, start a new thread. –  Qiaochu Yuan Mar 18 '11 at 22:22
@Jasen: By the way, this question might resolve your difficulties with modulus notation. –  user5501 Mar 18 '11 at 22:24
@Jasen: If you like Viktor's answer, it would also be good form to upvote it (click on the up arrow by the answer) and then to formally accept it (click on the check mark by the answer). –  Mike Spivey Mar 18 '11 at 22:30

$Income = Expenditures$ condition would give you $A\,\vec p=\vec p$ or $(I-A)\vec p=0$. Such homogeneous SLE will have infinitely many solutions (a parametric family): $$\vec p=c\left[\begin{array}{c} p_1 \\ p_2 \\ p_3 \end{array}\right]$$
@Jasen SLE = system of linear equations. $A\vec p = \vec p \Leftrightarrow (I-A)\vec p=0$. Book ref: 10.8 Leontief Economic Models. –  Viktor Mar 15 '11 at 11:32
@Mike Spivey: I had been wondering, what does OP mean? hehe, thanks. And could you elaborate a bit more in a "answer" reply why does the transpose of the price vector $p$ need to be multiplied in such a manner? And how does this change the result, if any? –  Jasen Mar 18 '11 at 3:57
@Jasen: "OP" stands for "original poster" or "original post," depending on the context. So "OP" here means you. :) Also, Viktor is correct, and I was wrong; you do need to multiply $\vec{p}$ on the right and solve $A\vec{p} = \vec{p} \Leftrightarrow (I-A)\vec{p}=0$. This is important because you would get a different answer for $\vec{p}$ if you were to solve $\vec{p}^T A = \vec{p}^T$. Try it, and you'll see. –  Mike Spivey Mar 18 '11 at 20:42