Understanding the Leontief inverse 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!
 A: 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.
A: 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.
A: 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.
A: $
\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.
