TeachingLinear Algebra – Matrices – Applications
The open Leontief input-output model in economics
Suppose that the economy of a nation can be divided into \(n\) sectors that produce goods and services and let \(\mathbf{x}\) be a production vector in \(\mathbb{R}^n\) that lists the output of each sector for one year. Also, suppose that another part of the economy (the open sector) does not produce goods or services but only consumes them and let \(\mathbf{d}\) be a final demand vector that lists the values of the goods and services demanded from the various sectors by the nonproductive part of the economy. In order to meet this consumer demand, the producers themselves create additional intermediate demand for goods they need as inputs for their own production.
The interrelations between the sectors are very complex. The Russian economist Wassily Leontief asked if there is a production level \(\mathbf{x}\) such that the amounts produced will exactly balance the total demand for that production.
If \(\mathbf{c}_1,\mathbf{c}_2,\ldots,\mathbf{c}_n\) are so-called unit consumption vectors that lists the inputs per unit of output of the sector, then the open Leontief input-output model can be written as \(\mathbf{x}=C\mathbf{x}+\mathbf{d}\), where \(C=\Bigg(\mathbf{c}_1\;\mathbf{c}_2\;\ldots\;\mathbf{c}_n\Bigg)\) is the so-called consumption matrix. Note that this can be written as \((I-C)\mathbf{x}=\mathbf{d}\).
Theorem: Let \(C\) be the consumption matrix for an economy and let \(\mathbf{d}\) be the final demand vector. Then we have: if \(C\) and \(\mathbf{d}\) have nonnegative entries and if in each column of \(C\) the sum of the entries is less than \(1\), then \((I-C)^{-1}\) exists and the production vector \(\mathbf{x}=(I-C)^{-1}\mathbf{d}\) has nonnegative entries and is the unique solution of \(\mathbf{x}=C\mathbf{x}+\mathbf{d}\).
Example: If \(C=\begin{pmatrix}0.5&0.2&0.4\\0.2&0.2&0.1\\0.1&0.2&0.5\end{pmatrix}\) and \(\mathbf{d}=\begin{pmatrix}20\\30\\50\end{pmatrix}\). Then we have:
\begin{align*} (I-C)\mathbf{x}=\mathbf{d}:\quad&\left(\left.\begin{matrix}0.5&-0.2&-0.4\\-0.2&0.8&-0.1\\-0.1&-0.2&0.5\end{matrix}\,\right|\,\begin{matrix}20\\30\\50\end{matrix}\right) \sim\left(\left.\begin{matrix}1&2&-5\\-2&8&-1\\5&-2&-4\end{matrix}\,\right|\,\begin{matrix}-500\\300\\200\end{matrix}\right) \sim\left(\left.\begin{matrix}1&2&-5\\0&12&-11\\0&-12&21\end{matrix}\,\right|\,\begin{matrix}-500\\-700\\2700\end{matrix}\right)\\[2.5mm] &\quad\sim\left(\left.\begin{matrix}1&2&-5\\0&12&-11\\0&0&10\end{matrix}\,\right|\,\begin{matrix}-500\\-700\\2000\end{matrix}\right) \sim\left(\left.\begin{matrix}1&2&0\\0&12&0\\0&0&1\end{matrix}\,\right|\,\begin{matrix}500\\1500\\200\end{matrix}\right) \sim\left(\left.\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\,\right|\,\begin{matrix}250\\125\\200\end{matrix}\right) \quad\Longrightarrow\quad\mathbf{x}=\begin{pmatrix}250\\125\\200\end{pmatrix}. \end{align*}Last modified on March 29, 2021
