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Chapter 8 • Theory & Derivations

The Leontief Input-Output Economic Model

Inter-Industry Technological Matrices, The Open Leontief Equation, Hawkins-Simon Viability & Neumann Multipliers

§8.1 Economic Foundations of Inter-Industry Analysis

1. The Inter-Industry Economic Network

Wassily Leontief (1973 Nobel Laureate in Economics) developed input-output analysis to model the interdependence of industries in an economy. In an economy divided into $n$ sectors (e.g., Agriculture, Manufacturing, Energy, Transportation), the output of any one sector serves a dual role:

  • Intermediate Output: Consumed by other industries (and by itself) as inputs required for production.
  • Final Demand: Consumed by households, government, capital investment, or export.

2. The Flow Matrix of Transactions

Let $x_{ij}$ denote the dollar value of output from sector $i$ consumed as intermediate input by sector $j$ during a given production period.
Let $d_i$ be the external final consumer demand for sector $i$'s product.
Let $x_i$ be the total gross output produced by sector $i$. Conservation of economic output requires: $$\mathbf{x_i = \sum_{j=1}^n x_{ij} + d_i, \quad i = 1, 2, \dots, n}$$ Total Gross Output = Total Intermediate Inputs Consumed + Final Demand.

§8.2 The Consumption (Technological) Matrix $C$

1. The Technological Coefficients

Assuming constant returns to scale and fixed production recipes, the technological coefficient $c_{ij}$ is the dollar amount of sector $i$'s goods required to produce one dollar's worth of sector $j$'s output: $$\mathbf{c_{ij} \equiv \frac{x_{ij}}{x_j} \iff x_{ij} = c_{ij} x_j}$$ The $n \times n$ matrix $C = [c_{ij}]$ is called the consumption matrix (or technological matrix).

2. Properties of the Consumption Matrix

  • Every entry is non-negative: $c_{ij} \ge 0$.
  • Column $j$ represents the complete cost recipe per dollar produced by industry $j$: $$\mathbf{C_{*, j} = \begin{pmatrix} c_{1j} \\ c_{2j} \\ \vdots \\ c_{nj} \end{pmatrix}}$$
  • Economic Profitability Condition: In an economy where industries create value rather than destroying resources, the sum of intermediate material costs per dollar of output must be strictly less than one: $$\sum_{i=1}^n c_{ij} < 1, \quad \text{for all } j = 1, \dots, n$$ The remainder $v_j = 1 - \sum_{i=1}^n c_{ij} > 0$ represents the value added (wages, taxes, and operating profit) per dollar of production!

§8.3 The Open Leontief Production Equation

1. Derivation of the Matrix Equation

Substituting $x_{ij} = c_{ij} x_j$ into the economic conservation balance: $$x_i = \sum_{j=1}^n c_{ij} x_j + d_i \iff X = C X + D$$ where: $$X = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} \quad (\text{Gross Output Vector}), \qquad D = \begin{pmatrix} d_1 \\ d_2 \\ \vdots \\ d_n \end{pmatrix} \quad (\text{Final Demand Vector})$$ Rewriting into canonical linear system form: $$\mathbf{(I_n - C) X = D}$$ The matrix $I_n - C$ is called the Leontief Matrix.

2. The Equilibrium Production Solution

If the Leontief matrix $I_n - C$ is invertible, the gross output required across all sectors to satisfy final demand $D$ is uniquely determined by: $$\mathbf{X = (I_n - C)^{-1} D}$$ The inverse matrix $(I - C)^{-1}$ is termed the Leontief Inverse (or Total Requirements Matrix). Its $(i, j)$-th entry represents the total dollar amount that sector $i$ must produce directly and indirectly to supply one dollar of final demand to sector $j$!

§8.4 The Hawkins-Simon Economic Viability Conditions

1. The Economic Viability Problem

In real-world economics, negative production is impossible ($X \ge 0$). An economy is defined as economically viable if for every non-negative final demand vector $D \ge 0$, there exists a unique non-negative gross production vector $X \ge 0$ satisfying $(I - C)X = D$.

2. The Hawkins–Simon Theorem

Theorem (David Hawkins & Herbert Simon, 1949): An input-output system with consumption matrix $C \ge 0$ is economically viable if and only if all leading principal minors of the Leontief matrix $I - C$ are strictly positive:

$$\Delta_1 = 1 - c_{11} > 0$$ $$\Delta_2 = \begin{vmatrix} 1 - c_{11} & -c_{12} \\ -c_{21} & 1 - c_{22} \end{vmatrix} > 0$$ $$\dots$$ $$\Delta_n = \det(I - C) > 0$$
Economic Intuition:
  • $\Delta_1 > 0 \iff c_{11} < 1$: Sector 1 cannot consume more of its own product than it produces.
  • $\Delta_2 > 0 \iff (1 - c_{11})(1 - c_{22}) > c_{12} c_{21}$: The combined direct and indirect feedback loops between sectors 1 and 2 must not consume more than their collective net capacity.

§8.5 Neumann Series Multipliers & The Dual Leontief Price Model

1. The Neumann Power Series Expansion

If the spectral radius $\rho(C) < 1$, the Leontief inverse can be expanded as a convergent geometric matrix series (the Neumann Series): $$\mathbf{(I - C)^{-1} = I + C + C^2 + C^3 + \dots = \sum_{k=0}^\infty C^k}$$ Substituting into the output equation $X = (I - C)^{-1} D$: $$\mathbf{X = D + C D + C^2 D + C^3 D + \dots}$$ Economic Multiplier Breakdown:

  • $D$: Direct final consumer demand.
  • $CD$: First-round intermediate inputs required by industries to produce $D$.
  • $C^2 D$: Second-round inputs required to produce the intermediate inputs $CD$.
  • $C^k D$: $k$-th generation indirect supply chain requirements throughout the economy.

2. The Dual Leontief Price Model

Let $P = (p_1, p_2, \dots, p_n)$ be the unit price row vector across sectors, and let $V = (v_1, v_2, \dots, v_n)$ be the value-added row vector (wages + profits per unit). The equilibrium pricing relation states that price equals intermediate material costs plus value added: $$P = P C + V \iff P(I - C) = V$$ Multiplying by the Leontief inverse from the right yields the equilibrium price structure: $$\mathbf{P = V (I - C)^{-1}}$$ This allows governments and central banks to calculate how changes in wages or energy tax ($V$) propagate throughout the entire price level of the macroeconomy!

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Foundational Mechanics Example 8.1: Two-Sector Leontief Economy and Production Equilibrium

A two-sector economy consisting of Energy ($E$) and Manufacturing ($M$) has consumption matrix $C = \begin{pmatrix} 0.2 & 0.4 \\ 0.3 & 0.1 \end{pmatrix}$. (a) Verify the Hawkins-Simon conditions. (b) Compute the Leontief inverse $(I - C)^{-1}$. (c) Find the gross production vector $X$ required to satisfy final demand $D = \begin{pmatrix} 100 \\ 200 \end{pmatrix}$ million dollars.

Step 1: Verify Hawkins-Simon Viability Conditions
$$I - C = \begin{pmatrix} 1 - 0.2 & -0.4 \\ -0.3 & 1 - 0.1 \end{pmatrix} = \begin{pmatrix} 0.8 & -0.4 \\ -0.3 & 0.9 \end{pmatrix} \\ \Delta_1 = 0.8 > 0 \quad \checkmark \\ \Delta_2 = \det(I - C) = (0.8)(0.9) - (-0.4)(-0.3) = 0.72 - 0.12 = 0.60 > 0 \quad \checkmark$$

Both principal minors are strictly positive, guaranteeing that the economy is viable.

Step 2: Invert the 2x2 Leontief Matrix
$$(I - C)^{-1} = \frac{1}{\det(I - C)} \begin{pmatrix} 0.9 & 0.4 \\ 0.3 & 0.8 \end{pmatrix} = \frac{1}{0.6} \begin{pmatrix} 0.9 & 0.4 \\ 0.3 & 0.8 \end{pmatrix} = \begin{pmatrix} 1.5 & 0.667 \\ 0.5 & 1.333 \end{pmatrix}$$

Apply the $2 \times 2$ matrix inverse formula.

Step 3: Compute the Gross Output Vector X
$$X = (I - C)^{-1} D = \frac{1}{0.6} \begin{pmatrix} 0.9 & 0.4 \\ 0.3 & 0.8 \end{pmatrix} \begin{pmatrix} 100 \\ 200 \end{pmatrix} \\ = \frac{1}{0.6} \begin{pmatrix} 0.9(100) + 0.4(200) \\ 0.3(100) + 0.8(200) \end{pmatrix} = \frac{1}{0.6} \begin{pmatrix} 90 + 80 \\ 30 + 160 \end{pmatrix} = \frac{1}{0.6} \begin{pmatrix} 170 \\ 190 \end{pmatrix} = \begin{pmatrix} 283.33 \\ 316.67 \end{pmatrix}$$

Multiply the Leontief inverse by the external demand vector.

Final Answer & Physical Insight

\text{Hawkins-Simon conditions are satisfied; } \mathbf{(I - C)^{-1} = \begin{pmatrix} 1.5 & 0.667 \\ 0.5 & 1.333 \end{pmatrix}}; \quad \mathbf{X = \begin{pmatrix} 283.33 \\ 316.67 \end{pmatrix}} \text{ million dollars}.

Intermediate University Exam Example 8.2: Three-Sector Economic Shift and Output Reallocation

A 3-sector economy with technological matrix $C = \begin{pmatrix} 0.1 & 0.2 & 0.2 \\ 0.2 & 0.1 & 0.1 \\ 0.1 & 0.2 & 0.1 \end{pmatrix}$ has current final demand $D = \begin{pmatrix} 50 \\ 60 \\ 40 \end{pmatrix}$. If consumer demand in Sector 2 increases by 50% while others remain unchanged, compute the required change in gross output vector $\Delta X$.

Step 1: Form the Leontief Matrix I - C
$$I - C = \begin{pmatrix} 0.9 & -0.2 & -0.2 \\ -0.2 & 0.9 & -0.1 \\ -0.1 & -0.2 & 0.9 \end{pmatrix}$$

Subtract consumption matrix $C$ from identity matrix $I_3$.

Step 2: Determine Demand Shift ΔD
$$\Delta D = \begin{pmatrix} 0 \\ 0.50 \times 60 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 30 \\ 0 \end{pmatrix}$$

Only sector 2 experiences an external demand increase of 30 units.

Step 3: Solve (I - C) ΔX = ΔD
$$\begin{pmatrix} 0.9 & -0.2 & -0.2 \\ -0.2 & 0.9 & -0.1 \\ -0.1 & -0.2 & 0.9 \end{pmatrix} \begin{pmatrix} \Delta x_1 \\ \Delta x_2 \\ \Delta x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 30 \\ 0 \end{pmatrix} \\ \det(I - C) = 0.9(0.81 - 0.02) + 0.2(-0.18 - 0.01) - 0.2(0.04 + 0.09) \\ = 0.9(0.79) + 0.2(-0.19) - 0.2(0.13) = 0.711 - 0.038 - 0.026 = 0.647 \\ \text{Using Cramer's Rule for Column 2 of } (I - C)^{-1}: \\ \text{Cofactors: } C_{12} = -(-0.18 - 0.01) = 0.19, \quad C_{22} = 0.81 - 0.02 = 0.79, \quad C_{32} = -(-0.09 - 0.02) = 0.11 \\ \Delta X = \frac{30}{0.647} \begin{pmatrix} 0.19 \\ 0.79 \\ 0.11 \end{pmatrix} \approx \begin{pmatrix} 8.81 \\ 36.63 \\ 5.10 \end{pmatrix}$$

Compute the direct and indirect multiplier impact using cofactors.

Final Answer & Physical Insight

\mathbf{\Delta X \approx \begin{pmatrix} 8.81 \\ 36.63 \\ 5.10 \end{pmatrix}} \implies \text{All three sectors must expand output to support Sector 2's demand growth}.

Honors / Proof Challenge Example 8.3: Neumann Series Convergence & The Dual Price Equilibrium

Given an $n$-sector economy with non-negative consumption matrix $C$: (a) Prove that if the maximum column sum satisfies $\|C\|_1 = \max_j \sum_{i=1}^n c_{ij} < 1$, the spectral radius satisfies $\rho(C) < 1$, and the Neumann series $\sum_{k=0}^\infty C^k$ converges strictly to $(I - C)^{-1}$. (b) If $C = \begin{pmatrix} 0.3 & 0.2 \\ 0.1 & 0.4 \end{pmatrix}$ and the value-added vector per unit output is $V = (14, 21)$ dollars, determine the equilibrium price vector $P = (p_1, p_2)$.

Step 1: Prove Neumann Series Convergence
$$\text{For any induced matrix norm } \|\cdot\|, \quad \rho(C) \le \|C\|_1 < 1 \\ \text{Consider partial sum } S_m = \sum_{k=0}^m C^k. \quad (I - C) S_m = I - C^{m+1} \\ \text{Since } \rho(C) < 1, \quad \lim_{m \to \infty} C^{m+1} = 0 \\ \lim_{m \to \infty} (I - C) S_m = I \implies \sum_{k=0}^\infty C^k = (I - C)^{-1} \quad \blacksquare$$

Use operator norm and Gelfand's formula to prove absolute convergence of the matrix power series.

Step 2: Formulate the Dual Price Equation P = V(I - C)⁻¹
$$P(I - C) = V \iff \begin{pmatrix} p_1 & p_2 \end{pmatrix} \begin{pmatrix} 0.7 & -0.2 \\ -0.1 & 0.6 \end{pmatrix} = \begin{pmatrix} 14 & 21 \end{pmatrix}$$

Set up the horizontal row equation $P(I - C) = V$.

Step 3: Invert (I - C) and Compute Equilibrium Prices
$$\det(I - C) = (0.7)(0.6) - (-0.2)(-0.1) = 0.42 - 0.02 = 0.40 \\ (I - C)^{-1} = \frac{1}{0.40} \begin{pmatrix} 0.6 & 0.2 \\ 0.1 & 0.7 \end{pmatrix} = \begin{pmatrix} 1.5 & 0.5 \\ 0.25 & 1.75 \end{pmatrix} \\ P = \begin{pmatrix} 14 & 21 \end{pmatrix} \begin{pmatrix} 1.5 & 0.5 \\ 0.25 & 1.75 \end{pmatrix} \\ p_1 = 14(1.5) + 21(0.25) = 21 + 5.25 = 26.25 \\ p_2 = 14(0.5) + 21(1.75) = 7 + 36.75 = 43.75$$

Multiply the value-added row vector by the Leontief inverse.

Final Answer & Physical Insight

\mathbf{P = (26.25, \; 43.75)} \implies p_1 = \$26.25 \text{ and } p_2 = \$43.75 \text{ per unit}.