The input - output matrix for a two sector economy is given by $$A = \begin{pmatrix} 0.25 & 0.40 \\ 0.10 & 0.10 \end{pmatrix}$$​. If the external demand for the outputs of the two sectors is $$D = \begin{pmatrix} 10 \\ 20 \end{pmatrix}$$, what will be the optimal output levels of the two commodities?

Correct option is C

Correct Option: C. 26.77 and 25.20

Explanation:

The problem is based on the Leontief Input-Output Model. To find the gross output vector X required to satisfy a specific final demand D, we use the formula: $$X = (I - A)^{-1} D$$, where I is the identity matrix and A is the technology coefficient matrix.

Information Booster:

• Step 1: Calculate (I - A)

  ​$$I - A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 0.25 & 0.40 \\ 0.10 & 0.10 \end{pmatrix} = \begin{pmatrix} 0.75 & -0.40 \\ -0.10 & 0.90 \end{pmatrix}$$​​
  

• Step 2: Find the Inverse $$(I - A)^{-1}$$​

  • Determinant $$|I - A| = (0.75)(0.90) - (-0.40)(-0.10) = 0.675 - 0.04 = 0.635$$​​
  • Adjoint Matrix = $$\begin{pmatrix} 0.90 & 0.40 \\ 0.10 & 0.75 \end{pmatrix}$$ 
  • ​Inverse =  $$\frac{1}{0.635} \begin{pmatrix} 0.90 & 0.40 \\ 0.10 & 0.75 \end{pmatrix}$$​​

• Step 3: Solve for X

  ​$$X = \frac{1}{0.635} \begin{pmatrix} 0.90 & 0.40 \\ 0.10 & 0.75 \end{pmatrix} \begin{pmatrix} 10 \\ 20 \end{pmatrix}$$
  
  
  $$X_1 = \frac{1}{0.635} (0.90 \times 10 + 0.40 \times 20) = \frac{9 + 8}{0.635} = \frac{17}{0.635} \approx 26.77$$
  ​​​
  
  ​$$X_2 = \frac{1}{0.635} (0.10 \times 10 + 0.75 \times 20) = \frac{1 + 15}{0.635} = \frac{16}{0.635} \approx 25.196$$​​
  

Additional Knowledge:
1. Hawkins-Simon Conditions: For the system to be viable (produce positive outputs), the determinant |I-A| must be positive and the diagonal elements of (I-A) must be positive. Here, 0.635 > 0, 0.75 > 0, and 0.90 > 0, so the system is viable.
2. Leontief Inverse: The matrix $$(I-A)^{-1}$$​ is often called the Leontief Inverse Matrix or the Multiplier Matrix, showing the total (direct and indirect) requirements.