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\centerline{Mathematics 244 -- Linear Algebra}
\centerline{Group Discussion 1 -- The Leontieff Input-Output Model}
\centerline{\it February 7 and 9, 2007}
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{\it The Leontieff Model}
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Recall that $x \in {\bf R}^n$ represents the production vector for 
a national economy, broken into some number $n$ of sectors.  The 
final consumer demand for outputs from each of the sectors is given
by the vector $d$.  But each sector uses inputs from other sectors
as the inputs to produce its outputs.  These inputs are represented
by the entries of the $n \times n$ consumption matrix $C$.  The basic
form of the model is the equation
$$x = Cx + d. \leqno(LM_1)$$
We usually think of the entries of $x$ and $d$ as expressed in 
monetary units (e.g. billions of dollars for a big economy like
the U.S. national economy), but we won't be specific about that
in the following.
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{\it Discussion Questions}
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\noindent
I.  Consider an economy with 3 sectors -- manufacturing ($m$), 
agriculture ($a$), and services ($s$).  For each unit of 
its output, $m$ uses $.1$ unit from other companies in its
sector, $.3$ units from $a$, and $.3$ units from $s$. 
For each unit of its output, $a$ uses $.2$ units of its
output and $.6$ units from $m$ and $.1$ units from $s$.
For each unit of its output, $s$ uses $.6$ units from $m$
and $.1$ units of its own output, but no output from $a$. 
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\item{A)}  What is the consumption matrix $C$ that matches
the given information?
\item{B)}  Determine the production levels needed to 
satisfy a final demand of $18$ units from $a$, but 
no demand for the other sectors.  
Hint:  When you set up your equations to solve, if 
you are doing this by hand, it might help to ``scale'' each 
row to start to get rid of the decimals. 
\item{C)}  Determine the production levels needed to 
satisfy a final demand of $18$ units from $m$, but 
no demand for the other sectors.  
\item{D)}  What are the production levels needed 
to satisfy a final demand of $18$ units from $m$, and
$18$ units from $a$, but no demand for $s$.
Hint:  You don't need any additional calculations here
if you have already done parts B and C.  Why not?
\item{E)}  Is $I_3 - C$ an invertible matrix in this
example?  If so, find $(I_3 - C)^{-1}$.
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\noindent
II.  As we have seen, if $(I_n - C)$ is invertible,
then $(LM_1)$ has a unique solution 
$$x = (I_n - C)^{-1}d$$
for all $d$.  Here is another way to think about 
what this means.  The meaning of the matrix $C$ is that
\item{$\bullet$} to produce $d$ directly, $Cd$ inputs are required (adding over all sectors), but
\item{$\bullet$} those inputs require $C(Cd) = C^2d$ ``second round'' inputs, and
\item{$\bullet$} the second round inputs require $C(C^2d) = C^3 d$ ``third round'' inputs, etc.
\item{$\bullet$} So if we follow this backward process {\it ad infinitum}, then the 
total inputs required will be 
$$x = d + Cd + C^2d + C^3d + \cdots = (I + C + C^2 + C^3 + \cdots\ )d$$
where the sum extends to infinity (that is, an {\it infinite series of matrices}!).
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\item{A)}  The infinite series 
$$I + C + C^2 + C^3 + \cdots$$
makes sense if and only if $\lim_{n\to\infty} C^n = 0$ (the zero matrix)
and 
$$\lim_{n\to\infty} I + C + C^2 + \cdots + C^n = D$$
for some matrix $D$.  Show that if these conditions are
satisfied, then the matrix $D$ {\it is} $(I - C)^{-1}$.  It is enough,
by the definition of matrix inverses, to show $D(I - C) = I = (I - C)D$. 
Start by looking at 
$$(I - C)(I + C + C^2 + \cdots + C^n).$$  
\item{B)}  The discussion in part A is closely related 
to a topic you have seen before, but in a different class.  What topic is this?
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IV.  In the ``full'' Leontieff model there is a second equation
$$p = C^tp + v\leqno(LM_2)$$
where $v,p$ are vectors in ${\bf R}^n$.  The vector $v$ represents
the ``value added'' -- its components give the value added per
unit input for each sector.  The vector $p$ is the price vector,
the components are the prices per unit output for all the sectors. 
The matrix $C^t$ is the {\it transpose} of $C$ above.  The transpose
of an $m \times n$ matrix $A$ is the $n\times m$ matrix $A^t$ whose rows
are the columns of $A$ and whose columns are the rows of $A$.  For
instance:
$$A = \pmatrix{1 & 2 & 3\cr 4& 5&6\cr} \Rightarrow A^t = \pmatrix{1 & 4\cr
                                                                  2 & 5\cr
                                                                  3 & 6\cr}$$
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\item{A)}  Show that if $A, B$ are $n\times n$ matrices, then 
$$(AB)^t = B^t A^t.$$
(Hint:  use the formula for the entries in a matrix product.)
\item{B)}  From part A, deduce that if $A$ is invertible, then
so is $A^t$ and 
$$(A^t)^{-1} = (A^{-1})^t.$$ 
\item{C)}  Show using matrix algebra that
$$ v^t x = p^t d .$$
You may assume that $I_n - C$ is invertible.  The products
on either side here are 
$$(1 \times n)(n \times 1) = (1\times 1)$$
matrices, so we can think of the result as a single number.  
The value we get from each these matrix products is
the {\it gross domestic product} of the economy.  
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{\it Assignment}
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One writeup from each group, due in class on Monday, February 12.
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