Example 2

Consider an economy with three industries with the following consumption matrix

$$C = \left[ \begin{array}{rrr} k & k&k \\ k & k&k \\ k & k&k \\ \end{array} \right]$$

where $0 < k < 1$

a) For what values of $k$ will $( I-C)^{-1}$ exist ?

b) For what values of $k$ will $( I-C)^{-1}$ be nonnegative ?

c) What will be the production vector if the demand vector is $d= \left[ \begin{array}{c} 1000 \\ 1000 \\ 1000 \\ \end{array} \right]$ ?

Solution:

a) Eigenvalues of $C$ are $\lambda _1 =0$ and $\lambda _2 =3k$. Multiplicity of $\lambda _1 =0$ is $2$ and multiplicity of $\lambda _2 =3k$ is 1. So the eigenvalues of $( I-C)$ are 1-0= 1 and $1-3k$. Therefore, for $k \ne \frac{ 1}{3}$ zero is not an eigenvalue of $( I-C)$, hence $( I-C)$ is invertible.

b) Note that $( I-C)^{-1}$ is nonnegative if and only if $C$ is profitable. Since the sum of the rows of $C$ are equal to $3k$, $C$ is profitable if $3k<1$ or $k<\frac {1} {3}$.

c) We want to find a production matrix $p$ such that $d= ( I- C )p$. Note that $v= \left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ \end{array} \right]$ is an eigenvector of $( I-C)$ with the eigenvalue of $1-3k$. Hence

$$( I-C) \left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ \end{ar... ... \begin{array}{c} 1 \\ 1 \\ 1 \\ \end{array} \right],$$

and for $p = \frac{ 1} { 1-3k}\left[ \begin{array}{c} 1000 \\ 1000 \\ 1000 \\ \end{array} \right]$ we will get :

$d = \left[ \begin{array}{c} 1000 \\ 1000 \\ 1000 \\ \end{array} \right] = (I-C) p$.