Understanding the Leontief inverse

Question

What I remember from economics about input/output analysis is that it basically analyses the interdependencies between business sectors and demand. If we use matrices we have $A$ as the input-output matrix, $I$ as an identity matrix and $d$ as final demand. In order to find the final input $x$ we may solve the Leontief Inverse:

$$ x = (I-A)^{-1}\cdot d $$

So here's my question: Is there a simple rationale behind this inverse? Especially when considering the form:

$$ (I-A)^{-1} = I+A + A^2 + A^3\ldots $$

What happens if we change an element $a_{i,j}$ in $A$? How is this transmitted within the system? And is there decent literature about this behaviour around? Thank you very much for your help!

Answer 1

The formulation $(I−A)^{−1} =I+A+A^2 +A^3 \cdots$ is the most interesting one. $I*d$ is the production of d itself, $A*d$ is supply of intermediate goods and services to the direct producers of d, $A^2d$ is supply of intermediate g&s to these, and so on and so on.

Answer 2

The equation you are concerned with relates total output $x$ to intermediate output $Ax$ plus final output $d$, $$ x = Ax + d $$.

If the inverse $(I - A)^{-1}$ exists, then a unique solution to the equation above exists. Note that some changes of $a_{ij}$ may cause a determinate system to become indeterminate, meaning there can be many feasible production plans.

Also, increasing $a_{ij}$ is equivalent to increasing the demand by sector $i$ for the good produced by sector $j$. Thus, as sector $i$ produces more, it will consume more of sector $j$'s goods in its production process.

Answer 3

This question has languished. At the level the question was asked, there is now a short, useful lecture available: https://www.youtube.com/watch?v=-1jT5NOk93w

If this information is insufficient, perhaps a followup question would be appropriate.

Answer 4

$ \def\p{\partial} \def\LR#1{\left(#1\right)} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\B{\LR{I-A}^{-1}} $The gradient of a matrix with respect to one of its components is given by $$\eqalign{ \grad{A}{A_{ij}} &= E_{ij} \\ }$$ where the components of the matrix $E_{ij}$ are all zero, except for the $(i,j)$ component which equals one. Applying this to the given expression yields $$\eqalign{ x &= \B d \\ dx &= -\B\LR{-dA}\B d \\ &= +\B\;dA\;x \\ \grad{x}{A_{ij}} &= \B E_{ij}\,x \\ }$$ which answers the question about how changes to $A$ will propagate through the solution.