Example 1.  Consider the closed three sector economy  [Graphics:Images/LeontiefModelMod_gr_19.gif] consisting of say:  Energy, Manufacturing, and Services where the input-output matrix is given by

[Graphics:Images/LeontiefModelMod_gr_20.gif].   Find the production vector  [Graphics:Images/LeontiefModelMod_gr_21.gif].   

Solution 1.

Enter the matrix  [Graphics:../Images/LeontiefModelMod_gr_22.gif].  

[Graphics:../Images/LeontiefModelMod_gr_23.gif]


[Graphics:../Images/LeontiefModelMod_gr_24.gif]

If  [Graphics:../Images/LeontiefModelMod_gr_25.gif] is an eigenvalue of  [Graphics:../Images/LeontiefModelMod_gr_26.gif]  and  [Graphics:../Images/LeontiefModelMod_gr_27.gif]  an eigen-pair, then the solution to  [Graphics:../Images/LeontiefModelMod_gr_28.gif] will be a multiple of the eigenvector  [Graphics:../Images/LeontiefModelMod_gr_29.gif].
Let us verify that
  [Graphics:../Images/LeontiefModelMod_gr_30.gif] is an eigenvalue of  [Graphics:../Images/LeontiefModelMod_gr_31.gif].  

[Graphics:../Images/LeontiefModelMod_gr_32.gif]


[Graphics:../Images/LeontiefModelMod_gr_33.gif]

The eigen-pair   [Graphics:../Images/LeontiefModelMod_gr_34.gif]  is easily computed using Mathematica's subroutines  Eigenvalues and  Eigenvectors.  

[Graphics:../Images/LeontiefModelMod_gr_35.gif]



[Graphics:../Images/LeontiefModelMod_gr_36.gif]

Since  [Graphics:../Images/LeontiefModelMod_gr_37.gif]  is a solution for any constant  c, we are permitted to choose any multiple of  [Graphics:../Images/LeontiefModelMod_gr_38.gif] we desire for the solution.  For illustration:

[Graphics:../Images/LeontiefModelMod_gr_39.gif]



[Graphics:../Images/LeontiefModelMod_gr_40.gif]

Aside.  Any value of  c  is permitted, it is your choice.  When [Graphics:../Images/LeontiefModelMod_gr_41.gif] we get

[Graphics:../Images/LeontiefModelMod_gr_42.gif]



[Graphics:../Images/LeontiefModelMod_gr_43.gif]

We are done.

How did we get that "nice" vector ?
By using exact arithmetic and the following calculations.

Form the homogeneous system [Graphics:../Images/LeontiefModelMod_gr_44.gif].   
and the augmented matrix   [Graphics:../Images/LeontiefModelMod_gr_45.gif]  and find its reduced row echelon form.

[Graphics:../Images/LeontiefModelMod_gr_46.gif]


[Graphics:../Images/LeontiefModelMod_gr_47.gif]

 

 

The step used in finding the row reduced echelon form are familiar row operations.

[Graphics:../Images/LeontiefModelMod_gr_48.gif]

[Graphics:../Images/LeontiefModelMod_gr_49.gif]


[Graphics:../Images/LeontiefModelMod_gr_50.gif]

[Graphics:../Images/LeontiefModelMod_gr_51.gif]


[Graphics:../Images/LeontiefModelMod_gr_52.gif]

[Graphics:../Images/LeontiefModelMod_gr_53.gif]


[Graphics:../Images/LeontiefModelMod_gr_54.gif]

[Graphics:../Images/LeontiefModelMod_gr_55.gif]


[Graphics:../Images/LeontiefModelMod_gr_56.gif]

[Graphics:../Images/LeontiefModelMod_gr_57.gif]


[Graphics:../Images/LeontiefModelMod_gr_58.gif]

[Graphics:../Images/LeontiefModelMod_gr_59.gif]

The equations for this augmented matrix are  

    [Graphics:../Images/LeontiefModelMod_gr_60.gif]  

There is one free variables which we choose to be  [Graphics:../Images/LeontiefModelMod_gr_61.gif].   It is used in computing  [Graphics:../Images/LeontiefModelMod_gr_62.gif]  and   [Graphics:../Images/LeontiefModelMod_gr_63.gif].  
Use [Graphics:../Images/LeontiefModelMod_gr_64.gif] and solve the previous equations for  [Graphics:../Images/LeontiefModelMod_gr_65.gif]  and   [Graphics:../Images/LeontiefModelMod_gr_66.gif]

    [Graphics:../Images/LeontiefModelMod_gr_67.gif]  

Get    
    [Graphics:../Images/LeontiefModelMod_gr_68.gif]

The solution vector  [Graphics:../Images/LeontiefModelMod_gr_69.gif]  is

 

[Graphics:../Images/LeontiefModelMod_gr_70.gif]

We are done.

Aside.  We can verify that this is the solution by direct multiplication A X.  This is just for fun !

[Graphics:../Images/LeontiefModelMod_gr_71.gif]



[Graphics:../Images/LeontiefModelMod_gr_72.gif]

We are really done.

Aside.  Iteration can be used to solve for the solution of the closed Leontief model.  If the sum of the production levels is known and a starting vector is given  [Graphics:../Images/LeontiefModelMod_gr_73.gif] then the simple iteration  [Graphics:../Images/LeontiefModelMod_gr_74.gif]  will converge to  [Graphics:../Images/LeontiefModelMod_gr_75.gif].  Because the largest eigenvalue is  [Graphics:../Images/LeontiefModelMod_gr_76.gif],  this is a simplified version of the "power method" for  finding the dominant eigen-pair.  For the example above  [Graphics:../Images/LeontiefModelMod_gr_77.gif] and the following iteration will converge.

[Graphics:../Images/LeontiefModelMod_gr_78.gif]



[Graphics:../Images/LeontiefModelMod_gr_79.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004