1.6 Applications of linear systems

Big idea

A linear system is a bookkeeping device. Whenever something is conserved, you get one equation for every place the books have to balance. Cars entering an intersection have to leave it. Atoms on the left of a reaction have to appear on the right. Money a sector spends has to come from what it earns. The unknowns are the quantities you cannot measure directly.

The algebra never changes. You name the unknowns, write one equation per balance point, and row reduce. What changes between applications is what the numbers mean and which solutions count as real answers.

That last part does most of the work. A system can have infinitely many solutions on paper while the situation allows only a few of them. Traffic cannot run backwards down a one-way street, so negative flows are out. A chemical equation needs whole-number coefficients, so fractions get scaled away. Solve the system first, then filter the solution set against the application.

Decoder

In a closed exchange economy, the equilibrium prices are the ones at which every sector’s income exactly covers its expenditures.

Each sector sells everything it produces and spends everything it earns. You are hunting for a price list under which no sector runs a surplus and no sector runs a deficit. Prices are the unknowns, not the quantities of goods.

Definitions and results

A flow network is a set of junctions joined by directed branches. Each branch carries an unknown amount. The conservation rule says that at every junction, total flow in equals total flow out. One junction gives one equation.

Adding all the junction equations gives a whole-network statement: everything that enters the network from outside also leaves it. That sum is a consequence of the other equations, which is why a network of $n$ junctions usually gives only $n - 1$ independent equations, and why flow networks are usually underdetermined. The extra constraints are physical, not algebraic: flows on one-way branches cannot be negative, and a branch has a capacity.

Balancing a chemical equation means choosing coefficients so that each element has the same number of atoms on both sides. Put one unknown on each compound and one equation on each element. Every equation has zero on the right, so the system is homogeneous and always has the useless all-zero solution. The useful answer is the solution with the smallest positive whole numbers.

A closed exchange model splits an economy into sectors. Column $j$ of the exchange matrix lists the fraction of sector $j$’s output that each sector buys, so every column adds to $1$. Writing $p_i$ for the price charged for sector $i$‘s total output, the income of sector $i$ is the sum of its shares of the other sectors’ output, each priced. Setting income equal to price for every sector gives the homogeneous system $(A - I)\mathbf{p} = \mathbf{0}$.

Because the columns of $A$ add to $1$, the rows of $A - I$ add to the zero row, so the system always has nonzero solutions. Its solution set is a line through the origin. Prices are therefore determined only up to a common scale: fix one sector’s price and every other price follows.

Worked examples

Traffic around one block

Four one-way streets form a loop: A to B, B to C, C to D, D to A. Cars enter the loop at A at 60 per hour and at C at 20 per hour. Cars leave the loop at B at 40 per hour and at D at 40 per hour. Let $x_1, x_2, x_3, x_4$ be the flows along A-to-B, B-to-C, C-to-D and D-to-A.

Balance each corner. In equals out:

$$ x_1 - x_4 = 60 \\ x_1 - x_2 = 40 \\ -x_2 + x_3 = 20 \\ x_3 - x_4 = 40 $$

Row reduce and the fourth equation turns into a row of zeros, leaving $x_4$ free:

$$ x_1 = 60 + x_4, \quad x_2 = 20 + x_4, \quad x_3 = 40 + x_4 $$

Traffic cannot be negative, so $x_4 \ge 0$ and every flow is at least its constant term. One measurement pins the rest down. If a counter on the D-to-A stretch reads 30 cars per hour, then $x_4 = 30$ and the solution is unique:

$$ x_1 = 90, \quad x_2 = 50, \quad x_3 = 70, \quad x_4 = 30 $$

Check all four corners. A: $60 + 30 = 90$. B: $90 = 50 + 40$. C: $50 + 20 = 70$. D: $70 = 30 + 40$. Every corner balances.

Burning aluminum

Aluminum and oxygen gas combine into aluminum oxide. Put an unknown on each compound:

$$ x_1 \, \mathrm{Al} + x_2 \, \mathrm{O}_2 \; \rightarrow \; x_3 \, \mathrm{Al}_2\mathrm{O}_3 $$

One equation per element. Aluminum: $x_1 = 2x_3$. Oxygen: $2x_2 = 3x_3$. As a homogeneous system,

$$ x_1 - 2x_3 = 0 \\ 2x_2 - 3x_3 = 0 $$

Take $x_3$ as the free variable. Then $x_1 = 2x_3$ and $x_2 = \tfrac{3}{2}x_3$. The smallest choice that clears the fraction is $x_3 = 2$, giving $x_1 = 4$ and $x_2 = 3$:

$$ 4 \, \mathrm{Al} + 3 \, \mathrm{O}_2 \; \rightarrow \; 2 \, \mathrm{Al}_2\mathrm{O}_3 $$

Check the atom counts. Aluminum: 4 on the left, $2 \times 2 = 4$ on the right. Oxygen: $3 \times 2 = 6$ on the left, $2 \times 3 = 6$ on the right. Balanced.

Prices in a three-sector economy

An economy has a farm, a mill and a power plant. Each column below is one producer’s output, split among the three buyers:

$$ A = \begin{bmatrix} 0.5 & 0.2 & 0.1 \\ 0.3 & 0.4 & 0.3 \\ 0.2 & 0.4 & 0.6 \end{bmatrix} $$

Read the first column as: the farm keeps half its own output, the mill buys $0.3$ of it, the power plant buys $0.2$ of it. Every column adds to $1$, so nothing is produced that nobody buys.

Let $p_1, p_2, p_3$ be the prices of the farm’s, the mill’s and the power plant’s annual output. The farm’s income is $0.5p_1 + 0.2p_2 + 0.1p_3$, and at equilibrium that equals $p_1$. Doing the same for the other two rows and moving everything to one side gives $(A - I)\mathbf{p} = \mathbf{0}$. Clearing decimals by multiplying each row by 10:

$$ -5p_1 + 2p_2 + p_3 = 0 \\ p_1 - 2p_2 + p_3 = 0 \\ p_1 + 2p_2 - 2p_3 = 0 $$

The second equation gives $p_1 = 2p_2 - p_3$. Substituting into the first gives $-8p_2 + 6p_3 = 0$, so $p_3 = \tfrac{4}{3}p_2$. Writing $p_2 = 3t$ makes $p_3 = 4t$ and $p_1 = 2t$, and the third equation checks out: $2t + 6t - 8t = 0$.

So the prices are forced into the ratio $2 : 3 : 4$. Fix the power plant’s output at 100 units of currency and the rest follows: the farm charges 50 and the mill charges 75. Check the mill’s books. Its income is $0.3(50) + 0.4(75) + 0.3(100) = 15 + 30 + 30 = 75$, which is exactly the price of its own output.

Practice

Model-building comes down to solving the system you wrote. Drill the solving step until it is automatic, so your attention stays on the modeling.

Videos

Watch how a network of junctions turns into a matrix with one row per branch and one column per node. The lecture goes further than this section does, but the first half is the traffic argument with currents in place of cars.

Quiz

Five systems with unique solutions, in the style the applications above produce.