meeting the demand, profitability

If $(I-C) p =d$ we say the demand exactly met with out any surplus or shortage.

So given a consumption matrix $p$, and a demand vector $d$ , we are interested in finding a production vector $p \ge 0$ so that $(I-C) p =d$ .

Note that the notation $A > 0$ where $A$ is a matrix means that all entries of $A$ are positive. Similar definition holds for $A \ge 0$.

If $( I-C)$ is invertible, then $p = ( I-C)^{-1} d$. Since the demand vector is positive, we want $( I-C)^{-1}$ to be positive. A consumption matric $C$ is called productive if $( I-C)^{-1}$ exists and $( I-C)^{-1} > ge 0$.

It can be shown that a consumption matrix $C$ is productive if and only if there is a vector $x$ such that $x> Cx$.

As a result we can show that, a consumption matrix $C$ is productive if the sum of each of its rows is less than 1. And, also a consumption matrix $C$ is productive if the sum of each of its columns is less than 1.

This has an application about the profitability of each industry:

An industry $i$ is called profitable if the sum of the $i^{th}$ column of the consumption matrix $C$ is less than 1.