Applications of Linear Systems

Homogeneous Systems in Economics

Linear algebra provides a framework for modeling complex economic relationships through homogeneous systems of equations. One of the primary applications is the Leontief Exchange Model.

• Concept of Economic Sectors: Suppose a nation's economy is divided into many distinct sectors. For each sector, we can track:

  • Total Output: The total production for one year.
  • Distribution/Exchange: Exactly how this output is divided or "exchanged" among the other sectors of the economy.
  • Price: The total dollar value of a sector's output.

• The Leontief Result: Wassily Leontief proved that there exist equilibrium prices that can be assigned to the total outputs of various sectors. These equilibrium prices ensure that the income of each sector exactly balances its expenses.

• An Example of a Simple Economy:     Consider an economy consisting of three sectors: Coal ($p_1$), Electric ($p_2$), and Steel ($p_3$). The distribution of output is summarized in an exchange table.

    Table 1: Distribution of Output

    | Purchased by | Output from Coal | Output from Electric | Output from Steel |     | :--- | :---: | :---: | :---: |     | Coal | $0$ | $0.4$ | $0.6$ |     | Electric | $0.6$ | $0.1$ | $0.2$ |     | Steel | $0.4$ | $0.5$ | $0.2$ |

• Understanding the Exchange Table:

  • Columns: A sector looks down a column to see where its output goes. For example, Coal ($p_1$) sends $0.6$ ($60\%$) of its output to Electric and $0.4$ ($40\%$) to Steel.
  • Rows: A sector looks across a row to see what inputs it requires (expenses). For instance, Row 1 indicates that Coal receives (and must pay for) $0.4$ of the Electric output and $0.6$ of the Steel output.

• Formulating Equilibrium Equations:     The equilibrium condition states: $\text{Income} = \text{Expenses}$.

  • For Coal: $p_1 = 0 \times p_1 + 0.4 \times p_2 + 0.6 \times p_3$
  • For Electric: $p_2 = 0.6 \times p_1 + 0.1 \times p_2 + 0.2 \times p_3$
  • For Steel: $p_3 = 0.4 \times p_1 + 0.5 \times p_2 + 0.2 \times p_3$

• Solving the Economic System:     To solve, move all variables to the left side and combine like terms to form a homogeneous system:

  • $1.0p_1 - 0.4p_2 - 0.6p_3 = 0$
  • $-0.6p_1 + 0.9p_2 - 0.2p_3 = 0$
  • $-0.4p_1 - 0.5p_2 + 0.8p_3 = 0$

    Row reducing the augmented matrix for this system yields:     $p_1 = 0.94p_3$$p_2 = 0.85p_3$$p_3 \text{ is a free variable}$

• Application of the Solution: Any positive choice for the price of steel ($p_3$) results in a set of equilibrium prices.
  • Case Example: If we set $p_3 = 100 \text{ million}$, equilibrium is reached if:
    • Coal output is priced at $94 \text{ million}$.
    • Electric output is priced at $85 \text{ million}$.
    • Steel output is priced at $100 \text{ million}$.

Balancing Chemical Equations

Chemical equations describe the quantities of substances consumed and produced in a reaction. Linear equations are used to ensure the law of conservation of mass is upheld.

• Propane Combustion Example: When propane gas ($C_3H_8$) burns and combines with oxygen ($O_2$), it produces carbon dioxide ($CO_2$) and water ($H_2O$).     The reaction: $x_1(C_3H_8) + x_2(O_2) \rightarrow x_3(CO_2) + x_4(H_2O)$

• Vector Representation of Molecules: Each molecule can be represented as a vector containing the number of atoms for Carbon (C), Hydrogen (H), and Oxygen (O):

  • $C_3H_8 = \begin{pmatrix} 3 \\ 8 \\ 0 \end{pmatrix}$, $O_2 = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix}$, $CO_2 = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$, $H_2O = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$

• The Vector Equation: To balance the equation, we solve:     $x_1 \begin{pmatrix} 3 \\ 8 \\ 0 \end{pmatrix} + x_2 \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} = x_3 \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + x_4 \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$

    Moving all terms to the left side:     $x_1 \begin{pmatrix} 3 \\ 8 \\ 0 \end{pmatrix} + x_2 \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} - x_3 \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} - x_4 \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

• Solving the Chemical System: Solving the augmented system leads to the general solution:
  • $x_1 = \frac{1}{4}x_4$
  • $x_2 = \frac{5}{4}x_4$
  • $x_3 = \frac{3}{4}x_4$
  • $x_4 \text{ is free}$

    Since the coefficients in a chemical equation must be whole numbers, we take $x_4 = 4$. This results in: * $x_1 = 1$ * $x_2 = 5$ * $x_3 = 3$ * $x_4 = 4$

    The balanced equation is: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$.

Network Flow

Network flow analysis involves studying the movement of quantities through a system of nodes and branches. This is applicable to traffic engineering, hydraulics, and electrical circuits.

• Definitions:

  • Network: A set of points called nodes connected by lines called branches.
  • Nodes: Connection points where flow is either entering or leaving.
  • Branches: The pathways between nodes, with flow direction and amount indicated.

• Fundamental Node Principle: For every node, the flow in must equal the flow out. Additionally, for the whole network, the total flow in must equal the total flow out.

• Traffic Flow Example (Baltimore): Consider the traffic flow over several one-way streets in downtown Baltimore represented in Figure 2. To determine the general flow pattern, we set up equations for each intersection (node).

  • Intersection A: $300 + 500 = x_1 + x_2$
  • Intersection B: $x_2 + x_4 = 300 + x_3$
  • Intersection C: $100 + 400 = x_4 + x_5$
  • Intersection D: $x_1 + x_5 = 600$

• Solving the Flow System: Row reduction of the associated augmented matrix derived from these equilibrium equations allows for the identification of the general flow pattern and any dependencies between the flow rates in different branches (free variables).