I need help with this. Please

Question

Assume this is a closed Leontief model.

Consider an international trade model consisting of three countries X, Y, and Z. Suppose X uses 1/4 of its production domestically, sends 1/2 of its production to Y, and sends 1/4 of its production to Z. Suppose Y uses 1/5 of its production domestically, sends 2/5 of its production to X, and sends 2/5 of its production to Z. Suppose Z uses 1/4 of its domestic goods, sends 1/4 of its production to X and sends 1/2 of its production to Y. If we know country Z produced 96 thousand units of goods, then country X produces ____thousand units of goods and country Y produces _____ thousand units of goods.

Answer 1

A closed Leontief model means that everything produced is also consumed. Thus, we can set up the following matrix A:

[ 0.25 0.4 0.25 ] [x] [x]

[ 0.5 0.2 0.5 ] [y] = [y]

[ 0.25 0.4 0.25 ] [z] [z]

Where x is the production of country X, y the production of country Y, and z the production of country Z.

How did I get this matrix? Since this is a closed model (production=consumption), I can set up systems of equations for the amount a country produces by the amount a country consumes (domestically and via trade).

For example, country X's equation would be:

0.25x + 0.4y + 0.25z = x

since country X uses 1/4 of its own production, country Y sends 2/5 of its production, and country Z sends 1/4 of its production. If you look at the first row of my matrix, it is exactly that equation.

Now we can rewrite the matrix equation as :

0 = (I - A)X

Where 0 is the zero vector, I is the 3x3 identity matrix, A is the matrix detailed above, and X is the vector of the three countries productions [x y Zz^T

This gives us the matrix system

[ 0.75 -0.4 -0.25 ]

[ -0.5 0.8 -0.5 ] X = 0

[ -0.25 -0.4 0.75 ]

Doing RREF gives us:

[ (3/4) -(2/5) -(1/4) ]

[ 0 (8/15) -(2/3) ] X = 0

[ 0 0 0 ]

This gives us a system of equations for x and y based on z:

(8/15)y -(2/3)z = 0 (equation from row 2 of the RREF matrix)

y = (5/4)z (simplified- call this equation ß)

(3/4)x -(2/5)y -(1/4)z = 0 (equation from row 1 of the RREF matrix)

(3/4)x = (2/5)(5/4)z + (1/4)z (substitute in equation ß)

x = z

Since we know z to be 96k, we have:

x = 96k, y=120k, z=96k