Economics at Zero Marginal Cost: Leontief Input-Output Model

1. Suppose that there are only three industries in the economy and one primary input and the input co-efficient matrix as follows:

$A=\begin{bmatrix} a_{11}\ \ \ \ \ a_{12} \ \ \ \ \ a_{13}\\ a_{21}\ \ \ \ \ a_{22} \ \ \ \ \ a_{23}\\ a_{31}\ \ \ \ \ a_{32} \ \ \ \ \ a_{33}\\ \end{bmatrix}= \begin{bmatrix} 0.2\ \ \ \ \ 0.3 \ \ \ \ \ 0.2\\ 0.4\ \ \ \ \ 0.1 \ \ \ \ \ 0.2\\ 0.1\ \ \ \ \ 0.3 \ \ \ \ \ 0.2\\ \end{bmatrix}$

If the specific final-demand vector say the final-output target of a development program happens to be $d=\begin{bmatrix} 10\\ 5\\ 6\\ \end{bmatrix}$ , what are the solution output levels for the three industries?

Solution:

Given:

Input co-efficient matrix, $A=\begin{bmatrix} 0.2\ \ \ \ \ 0.3 \ \ \ \ \ 0.2\\ 0.4\ \ \ \ \ 0.1 \ \ \ \ \ 0.2\\ 0.1\ \ \ \ \ 0.3 \ \ \ \ \ 0.2\\ \end{bmatrix}$

Final demand, $d=\begin{bmatrix} 10\\ 5\\ 6\\ \end{bmatrix}$

To find: Output levels for the three industries.

Using the Input-Output model equation, $X^*=(I-A)^{-1}.d$

Computing the Leontief matrix (I−A),

$I-A=\begin{bmatrix} 1\ \ \ \ \ 0 \ \ \ \ \ 0\\ 0\ \ \ \ \ 1\ \ \ \ \ 0\\ 0\ \ \ \ \ 0 \ \ \ \ \ 1\\ \end{bmatrix}-\begin{bmatrix} 0.2\ \ \ \ \ 0.3 \ \ \ \ \ 0.2\\ 0.4\ \ \ \ \ 0.1 \ \ \ \ \ 0.2\\ 0.1\ \ \ \ \ 0.3 \ \ \ \ \ 0.2\\ \end{bmatrix}$

$I-A=\begin{bmatrix} 1-0.2\ \ \ \ \ 0-0.3 \ \ \ \ \ 0-0.2\\ 0-0.4\ \ \ \ \ 1-0.1 \ \ \ \ \ 0-0.2\\ 0-0.1\ \ \ \ \ 0-0.3 \ \ \ \ \ 1-0.2\\ \end{bmatrix}$

$I-A=\begin{bmatrix} \ \ 0.8\ \ \ \ \ -0.3 \ \ \ \ \ -0.2\\ -0.4\ \ \ \ \ \ \ \ 0.9 \ \ \ \ \ -0.2\\ -0.1\ \ \ \ -0.3 \ \ \ \ \ \ \ \ \ 0.8\\ \end{bmatrix}$

Computing the inverse of I−A, i.e., $(I-A)^{-1}$ ,

$(I-A)^{-1}=\frac{1}{\left | I-A \right |}.Adj \left (I-A \right )$

$\\\left |I-A \right |=0.8\begin{vmatrix} 0.9\ \ \ \ -0.2\\ -0.3\ \ \ \ \0.8 \end{vmatrix}-(-0.3) \begin{vmatrix} -0.4\ \ -0.2\\ -0.1\ \ \ \ \0.8 \end{vmatrix}+(-0.2) \begin{vmatrix} -0.4\ \ \ \ \ \ 0.9\\ -0.1\ \ \ -0.3 \end{vmatrix}\\\ \\\left |I-A \right |=0.8(0.81-0.06)+0.3(-0.32-0.02)-0.2(0.12-(0.09))\\ \left |I-A \right |=0.8(0.66)+0.3(-0.34)-0.2(0.12+0.09)\\ \left |I-A \right |=0.528-0.102-0.2(0.21)\\ \left |I-A \right |=0.528-0.102-0.144\\ \left |I-A \right |=0.384\\$

$\\Co-factor\ of\ a_{11}, i.e.,\ 0.8=\begin{vmatrix} 0.9\ \ \ -0.2\\ -0.3\ \ \ \ \ 0.8\\ \end{vmatrix}=0.72-0.06=0.66\\ Co-factor\ of\ a_{12}, i.e.,\ -0.3=-\begin{vmatrix} -0.4\ \ \ -0.2\\ -0.1\ \ \ \ \ \ 0.8\\ \end{vmatrix}=-(-0.32-0.02)=-(-0.34)=0.34\\ Co-factor\ of\ a_{13},i.e.,\ -0.2=\begin{vmatrix} -0.4\ \ \ \0.9\\ -0.1\ -0.3\\ \end{vmatrix}=0.12-(-0.09)=0.12+0.09=0.21\\$

$\\Co-factor\ of\ a_{21}, i.e.,\ -0.4=-\begin{vmatrix} -0.3\ \ \ -0.2\\ -0.3\ \ \ \ \ 0.8\\ \end{vmatrix}=-(-0.24-0.06)=-(-0.30)=0.30\\ Co-factor\ of\ a_{22}, i.e.,\ 0.9=\begin{vmatrix} 0.8\ \ \ -0.2\\ -0.1\ \ \ \ \ \ 0.8\\ \end{vmatrix}=0.64-0.02=0.62\\ Co-factor\ of\ a_{23},i.e.,\ -0.2=-\begin{vmatrix} 0.8\ \ \ \ -0.3\\ -0.1\ \ -0.3\\ \end{vmatrix}=-(-0.24-0.03)=-(-0.27)=0.27\\$

$\\Co-factor\ of\ a_{31},i.e.,\ -0.1=\begin{vmatrix} -0.3\ \ \ \ -0.2\\ 0.9\ \ \ \ \ \ -0.2\\ \end{vmatrix}=0.06-(-0.18)=0.06+0.18=0.24\\ Co-factor\ of\ a_{32},i.e.,\ -0.3=-\begin{vmatrix} 0.8\ \ \ \ -0.2\\ -0.4\ \ -0.2\\ \end{vmatrix}=-(-0.16-0.08)=-(-0.24)=0.24\\ Co-factor\ of\ a_{33},i.e.,\ 0.8=\begin{vmatrix} 0.8\ \ \ \ -0.3\\ -0.4\ \ \ \ \ 0.9\\ \end{vmatrix}=0.72-0.12=0.60$

$Co-factor\ matrix\ (I-A)=\begin{bmatrix} 0.66\ \ \ \ \ 0.34 \ \ \ \ \ 0.21\\ 0.30\ \ \ \ \ 0.62 \ \ \ \ \ 0.27\\ 0.24\ \ \ \ \ 0.24\ \ \ \ \ 0.60\\ \end{bmatrix}$

$Adjoint\ matrix\ (I-A)=\begin{bmatrix} 0.66\ \ \ \ \ 0.30 \ \ \ \ \ 0.24\\ 0.34\ \ \ \ \ 0.62 \ \ \ \ \ 0.24\\ 0.21\ \ \ \ \ 0.27\ \ \ \ \ 0.60\\ \end{bmatrix}$

Now, $(I-A)^{-1}=\frac{1}{\left | I-A \right |}.Adj \left (I-A \right )$

$(I-A)^{-1}=\frac{1}{0.384}\begin{bmatrix} 0.66\ \ \ \ \ 0.30 \ \ \ \ \ 0.24\\ 0.34\ \ \ \ \ 0.62 \ \ \ \ \ 0.24\\ 0.21\ \ \ \ \ 0.27\ \ \ \ \ 0.60\\ \end{bmatrix}$

Next, $X^*=(I-A)^{-1}.d$

$X^*=\frac{1}{0.384}\begin{bmatrix} 0.66\ \ \ \ \ 0.30 \ \ \ \ \ 0.24\\ 0.34\ \ \ \ \ 0.62 \ \ \ \ \ 0.24\\ 0.21\ \ \ \ \ 0.27\ \ \ \ \ 0.60\\ \end{bmatrix}\begin{bmatrix} 10\\ 5\\ 6 \end{bmatrix}$

$X^*=\frac{1}{0.384}\begin{bmatrix} 0.66\times10 \ \ \ \ \ 0.30\times5 \ \ \ \ \ 0.24\times6\\ 0.34\times10\ \ \ \ \ 0.62\times5 \ \ \ \ \ 0.24\times6\\ 0.21\times10\ \ \ \ \ 0.27\times5\ \ \ \ \ 0.60\times6\\ \end{bmatrix}$