# How to solve the consumption matrix?

There are three industries interrelated so that their outputs are used as inputs by themselves.

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

Here

1. $$a_{jk}$$ is the fraction of the output of industry $$k$$ consumed by industry $$j$$.
2. $$p_j$$ 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. Show that this leads to $$Ap = p$$, where $$p = \left [ \begin{array}{ccc} p_1& p_2 & p_3\\ \end{array} \right ]$$, and find a solution $$p$$ with nonnegative $$p_1, p_2, p_3$$. Also, show that the consumption matrix $$A$$ must have columns which sum to $$1$$ and always has eigenvalue $$1$$. What approach should I follow to solve this question.?

Solving the equation $$Ap = p$$, you are looking for an eigenvector corresponding to the eigenvalue $$\lambda = 1$$, so the null space of $$A - I$$ should provide you with adequate vectors. To prove that $$1$$ is always an eigenvalue, first realize that $$\det A = \det A^T$$, which leads to the fact that $$A$$ and $$A^T$$ have the same eigenvalues. It is easily proven for a left stochastic matrix (note that these follow the exact same form as your consumption matrix) such as $$A$$ that the following vector $$\vec{u} =\begin{bmatrix}1 \\1 \\1 \end{bmatrix}$$ is always an eigenvector of $$A^T$$ corresponding to $$\lambda = 1$$ (see below). Consequently, $$1$$ must also be an eigenvalue of $$A$$. $$A^T\begin{bmatrix}1 \\1 \\1 \end{bmatrix}=\begin{bmatrix}0.1 & 0.8 &0.1 \\0.5 & 0 & 0.5 \\0.6 & 0.4 &0 \end{bmatrix}\begin{bmatrix}1 \\1 \\1 \end{bmatrix} = \begin{bmatrix}1 \\1 \\1 \end{bmatrix}$$