1.6 Applications of Linear Systems

Theory: Applications of Linear Systems

Applications of Linear Systems (Lay §1.6)

All three standard applications turn a word problem into a linear system, then solve it with the recipe above. Typically the answer has a free variable, and you then pick a convenient value (or use a “nonnegative” condition).

Economic equilibrium (Leontief exchange model)

An economy is split into sectors; each sector’s output is shared among the sectors (an exchange table, whose columns sum to 1). We look for equilibrium prices where each sector’s income equals its expenditure.

Mini-example (Lay §1.6, Exercise 1 setup): Goods sells 80% of its output to Services and keeps 20%; Services sells 70% to Goods and keeps 30%. With prices $p_G, p_S$ , Goods’ expenditure is what it buys: $0.2p_G + 0.7p_S$ . Setting this equal to its income $p_G$ :

p_G = 0.2p_G + 0.7p_S \;\Longrightarrow\; 0.8p_G - 0.7p_S = 0 \;\Longrightarrow\; p_G = 0.875\,p_S

(The Services equation $p_S = 0.8p_G + 0.3p_S$ gives the same relation.) This is a homogeneous system with a free variable: e.g. $p_S = 80$ , $p_G = 70$ . Only the ratio of prices is determined.

Balancing chemical equations

Write one vector per compound listing the number of atoms of each element. The equation $x_1(\text{reactant}_1) + \cdots = x_k(\text{product}_1) + \cdots$ becomes a homogeneous system (move everything to the left side, flipping the signs of the product vectors). Solve, then choose the free variable to make all coefficients the smallest positive integers.

Network flow

A network consists of nodes (junctions) joined by branches carrying flows. The basic rule is conservation of flow:

\text{at each node:}\quad \text{flow in} = \text{flow out}

and also total flow into the network = total flow out of the network. Each node gives one linear equation. The general solution usually has free variables; a flow against the drawn arrow would be negative, so if the streets are one-way, the conditions $x_i \ge 0$ bound the free variables.

For network problems, the best way to present the answer is the ” $x_i = \ldots$ , $x_k$ is free” list (Lay §1.2 style), not the parametric vector form — it makes the nonnegativity constraints easy to read.

Solved Examples (Textbook Questions)

Exercise 1

(Adapted from Lay §1.6, Exercise 5)

Boron sulfide reacts with water to form boric acid and hydrogen sulfide gas. Balance the equation

\text{B}_2\text{S}_3 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{BO}_3 + \text{H}_2\text{S}

using the vector equation approach. [For each compound, construct a vector listing the numbers of atoms of boron, sulfur, hydrogen, and oxygen.]

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Solution

Atom vectors (order: B, S, H, O):

\text{B}_2\text{S}_3: \begin{bmatrix} 2 \\ 3 \\ 0 \\ 0 \end{bmatrix}, \quad \text{H}_2\text{O}: \begin{bmatrix} 0 \\ 0 \\ 2 \\ 1 \end{bmatrix}, \quad \text{H}_3\text{BO}_3: \begin{bmatrix} 1 \\ 0 \\ 3 \\ 3 \end{bmatrix}, \quad \text{H}_2\text{S}: \begin{bmatrix} 0 \\ 1 \\ 2 \\ 0 \end{bmatrix}

We need $x_1\,\text{B}_2\text{S}_3 + x_2\,\text{H}_2\text{O} = x_3\,\text{H}_3\text{BO}_3 + x_4\,\text{H}_2\text{S}$ . Move the product vectors to the left (flip their signs) to get a homogeneous system:

\left[\begin{array}{cccc|c} 2 & 0 & -1 & 0 & 0 \\ 3 & 0 & 0 & -1 & 0 \\ 0 & 2 & -3 & -2 & 0 \\ 0 & 1 & -3 & 0 & 0 \end{array}\right]

Row reduce. $R_2 \leftarrow R_2 - \tfrac32 R_1$ gives $[0 \ \ 0 \ \ \tfrac32 \ \ -1 \mid 0]$ . Swap $R_2 \leftrightarrow R_4$ , then $R_3 \leftarrow R_3 - 2R_2$ , then $R_4 \leftarrow R_4 - \tfrac12 R_3$ :

\sim \left[\begin{array}{cccc|c} 2 & 0 & -1 & 0 & 0 \\ 0 & 1 & -3 & 0 & 0 \\ 0 & 0 & 3 & -2 & 0 \\ 0 & 0 & \frac32 & -1 & 0 \end{array}\right] \sim \left[\begin{array}{cccc|c} 2 & 0 & -1 & 0 & 0 \\ 0 & 1 & -3 & 0 & 0 \\ 0 & 0 & 3 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]

$x_4$ is free. Continue to RREF: $R_3 \leftarrow \tfrac13 R_3$ ; $R_2 \leftarrow R_2 + 3R_3$ ; $R_1 \leftarrow R_1 + R_3$ , $R_1 \leftarrow \tfrac12 R_1$ :

\sim \left[\begin{array}{cccc|c} 1 & 0 & 0 & -\frac13 & 0 \\ 0 & 1 & 0 & -2 & 0 \\ 0 & 0 & 1 & -\frac23 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]

General solution: $x_1 = \tfrac13 x_4$ , $x_2 = 2x_4$ , $x_3 = \tfrac23 x_4$ , $x_4$ free. Choose $x_4 = 3$ (clears the fractions): $x_1 = 1$ , $x_2 = 6$ , $x_3 = 2$ , $x_4 = 3$ .

\boxed{\text{B}_2\text{S}_3 + 6\,\text{H}_2\text{O} \rightarrow 2\,\text{H}_3\text{BO}_3 + 3\,\text{H}_2\text{S}}

Check: B: $2 = 2$ ; S: $3 = 3$ ; H: $12 = 6 + 6$ ; O: $6 = 6$ ✓.

Exercise 2

(Adapted from Lay §1.6, Exercise 14)

Intersections in England are often built as one-way “roundabouts”. Traffic must travel in the directions shown. Around the circle, the branches are $A \xrightarrow{x_2} B \xrightarrow{x_3} C \xrightarrow{x_4} D \xrightarrow{x_5} E \xrightarrow{x_6} F \xrightarrow{x_1} A$ . The external roads are:

Intersection	External flow
A	100 out
B	50 in
C	120 out
D	150 in
E	80 out
F	100 in

Find the general solution of the network flow. Find the smallest possible value for $x_6$ .

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Solution

Check totals first: in $= 50 + 150 + 100 = 300$ , out $= 100 + 120 + 80 = 300$ ✓.

Flow in = flow out at each intersection:

Intersection	Flow in		Flow out	Equation
A	$x_1$	=	$x_2 + 100$	$x_1 - x_2 = 100$
B	$x_2 + 50$	=	$x_3$	$x_2 - x_3 = -50$
C	$x_3$	=	$x_4 + 120$	$x_3 - x_4 = 120$
D	$x_4 + 150$	=	$x_5$	$x_4 - x_5 = -150$
E	$x_5$	=	$x_6 + 80$	$x_5 - x_6 = 80$
F	$x_6 + 100$	=	$x_1$	$-x_1 + x_6 = -100$

Row reduce. Adding rows 1–5 to row 6 makes it $[0 \ 0 \ 0 \ 0 \ 0 \ 0 \mid 0]$ (the sum of all node equations is the “total flow” identity). Then back-substitute upward ( $R_4 \leftarrow R_4 + R_5$ , $R_3 \leftarrow R_3 + R_4$ , …):

\left[\begin{array}{cccccc|c} 1 & -1 & 0 & 0 & 0 & 0 & 100 \\ 0 & 1 & -1 & 0 & 0 & 0 & -50 \\ 0 & 0 & 1 & -1 & 0 & 0 & 120 \\ 0 & 0 & 0 & 1 & -1 & 0 & -150 \\ 0 & 0 & 0 & 0 & 1 & -1 & 80 \\ -1 & 0 & 0 & 0 & 0 & 1 & -100 \end{array}\right] \sim \left[\begin{array}{cccccc|c} 1 & 0 & 0 & 0 & 0 & -1 & 100 \\ 0 & 1 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & -1 & 50 \\ 0 & 0 & 0 & 1 & 0 & -1 & -70 \\ 0 & 0 & 0 & 0 & 1 & -1 & 80 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]

General flow pattern:

\begin{cases} x_1 = 100 + x_6 \\ x_2 = x_6 \\ x_3 = 50 + x_6 \\ x_4 = -70 + x_6 \\ x_5 = 80 + x_6 \\ x_6 \text{ is free} \end{cases}

Smallest $x_6$ : traffic must follow the arrows, so every $x_i \ge 0$ . The binding constraint is $x_4 = x_6 - 70 \ge 0$ , so $x_6 \ge 70$ . The minimum value of $x_6$ is 70 (then $x_4 = 0$ , $x_1 = 170$ , $x_2 = 70$ , $x_3 = 120$ , $x_5 = 150$ ).