Question 10.3.7: Using the Leontief Input–Output Model In the preceding discu...
Using the Leontief Input–Output Model
In the preceding discussion, suppose the consumer demand for energy is 1000 units and for food is 3000 units. Find the level of production (X) that will meet interindustry and consumer demand.
For the matrix A, we have
I-A=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]-\left[\begin{array}{ll}\frac{1}{4} & \frac{1}{2} \\\frac{1}{3} & \frac{1}{4}\end{array}\right]=\left[\begin{array}{rr}\frac{3}{4} & -\frac{1}{2} \\-\frac{1}{3} & \frac{3}{4}\end{array}\right].
Use the formula for the inverse of a 2×2 matrix.
(I-A)^{-1}=\frac{1}{\frac{3}{4} \cdot \frac{3}{4}-\left(-\frac{1}{2}\right)\left(-\frac{1}{3}\right)}\left[\begin{array}{cc}\frac{3}{4} & \frac{1}{2} \\\frac{1}{3} & \frac{3}{4}\end{array}\right]
=\frac{48}{19}\left[\begin{array}{ll}\frac{3}{4} & \frac{1}{2} \\\frac{1}{3} & \frac{3}{4}\end{array}\right]=\frac{1}{19}\left[\begin{array}{ll}36 & 24 \\16 & 36\end{array}\right] Simplify.
The matrix D=\left[\begin{array}{l}1000 \\3000\end{array}\right]. Therefore, the gross production matrix X is
X=(I-A)^{-1} D=\frac{1}{19}\left[\begin{array}{ll}36 & 24 \\16 & 36\end{array}\right]\left[\begin{array}{l}1000 \\3000\end{array}\right] \quad \begin{aligned}&\text { Substitute for } \\&(I-A)^{-1} \text { and } D\end{aligned}
=\frac{1}{19}\left[\begin{array}{l}108,000 \\124,000\end{array}\right]=\left[\begin{array}{c}\frac{108,000}{19} \\\frac{124,000}{19}\end{array}\right] .
To meet the consumer demand for 1000 units of energy and 3000 units of food, the energy produced must be \frac{108,000}{19} units and food production must be \frac{124,000}{19} units.
RECALL
If ad – bc ≠ 0, then for A=\left[\begin{array}{ll}a & b \\c & d\end{array}\right] we have