Session 06 - Systems of Linear Equations

Mathematics for Master Students

Dr. Tobias Vlcek

Recap & Your Questions

Session 05 in a Nutshell

First-order condition: optima hide where $f'(x) = 0$, solve it for the candidates
Second-order condition: $f''(x^*) > 0$ a minimum, $f''(x^*) < 0$ a maximum, a flat tangent alone is never enough
On a closed interval, compare stationary points and endpoints, a boundary can win
Profit peaks where marginal revenue = marginal cost
The EOQ $q^* = \sqrt{2DK/h}$ balances ordering against holding costs

Common Issues in the Homework

Endpoints forgotten (P4): on $[0, 3]$ the interior stationary point gave the minimum $-11$; the true maximum $f(0) = 5$ sits at the boundary, the endpoints are always candidates
The cap ignored (P5): the FOC returned $q = 60$, but the plant makes at most $50$. Since $\pi'(50) = 10 > 0$, profit is still rising, the best feasible output is $q = 50$, not $60$
A sign slip in the SOC: $f''(x^*) < 0$ is a maximum (concave), not a minimum, check the sign before naming the type

Which tasks gave you trouble? Bring them up now. We take the time to work through them before moving on.

Warm-Up: Recipe Check

Question

Minimize the cost $C(q) = q^2 - 8q + 20$. Where is the minimum, and how do you know it is a minimum?

FOC: $C'(q) = 2q - 8 = 0 \;\Rightarrow\; q = 4$
SOC: $C''(q) = 2 > 0$ → convex → a minimum, confirmed
Value: $C(4) = 16 - 32 + 20 = 4$, lowest cost at $q = 4$

Today’s Plan

From one equation to a system: several conditions on several unknowns at once
Gaussian elimination: a clean recipe for three unknowns
How many solutions? one, none, or infinitely many
Linear systems in business: equilibrium and production planning
A look back: the toolkit you built across all six sessions

From One Equation to a System

What Makes an Equation Linear?

An equation is linear when every unknown appears on its own, to the first power, nothing more exotic.

Linear: $2x + 3y = 13$, or $x_1 + 2x_2 + 2x_3 = 16$, a weighted sum of the variables
Not linear: $x^2 + y = 4$ (a square), $\;xy = 6$ (a product), $\;\sqrt{x} + y = 1$ (a root)
The rule: no powers, no products of variables, no variables under roots or in exponents
Linear systems are the ones we can solve with pure, mechanical bookkeeping

One Equation, Two Unknowns = a Line

A single equation like $x + y = 5$ has infinitely many solutions: $(1,4)$, $(2,3)$, $(3,2)$, …, every point on one line.

A System Pins It Down

One condition leaves a whole line of choices. Add a second condition and the freedom shrinks.

A system of linear equations asks several conditions to hold at the same time
Business rarely has just one constraint: labour and material and budget
Each equation is a line; a solution must sit on every line at once
Two lines usually cross at a single point, one shared answer

A Two-Product Workshop

A furniture workshop makes chairs ($x$) and tables ($y$) and wants to use its resources fully this week:

Wood: a chair takes $2$ units, a table takes $3$; $13$ units are in stock $\;\Rightarrow\; 2x + 3y = 13$
Labour: each item takes one day of work; $5$ days are available $\;\Rightarrow\; x + y = 5$
Two conditions, two unknowns, a $2 \times 2$ system:

\[2x + 3y = 13 \qquad\qquad x + y = 5\]

Method 1: Substitution

Solve one equation for one variable, then substitute into the other.

From the labour equation: $\;x = 5 - y$
Put that into the wood equation: $\;2(5 - y) + 3y = 13$
Expand: $\;10 - 2y + 3y = 13 \;\Rightarrow\; 10 + y = 13$
So $\;y = 3$, and back-substituting $\;x = 5 - 3 = 2$
Solution: $3$ tables and $2$ chairs, i.e. $(x, y) = (2, 3)$

Method 2: Elimination

Scale the equations so one variable cancels when you subtract.

Multiply the labour equation by $2$: $\;2x + 2y = 10$
Subtract it from the wood equation: \[(2x + 3y) - (2x + 2y) = 13 - 10 \;\Rightarrow\; y = 3\]
Back-substitute: $\;x + 3 = 5 \;\Rightarrow\; x = 2$
Same solution $(2, 3)$, two roads, one destination

Where the Lines Meet

Always check: $2(2) + 3(3) = 4 + 9 = 13$ ✓ and $2 + 3 = 5$ ✓, the point sits on both lines.

Your Turn: Linear or Not?

Question

Is $\;3xy = 12\;$ a linear equation in $x$ and $y$?

No, it multiplies the two unknowns together, and $xy$ is a product of variables
Linear equations only allow each variable on its own, to the first power
A quick test: could you write it as (number)$\cdot x$ + (number)$y = $ number? Here you cannot

Three Unknowns: Gaussian Elimination

A Bigger System

For three unknowns, substitution gets messy. Our example for this section:

\[x + y + z = 9, \qquad x + 2y + 2z = 16, \qquad 2x + y + 3z = 17\]

Three conditions, three unknowns, and a lot of repetitive writing ahead. What we need first is a compact bookkeeping shorthand.

A Tidy Way to Write a System

The system is just its numbers, the augmented matrix (coefficients, then the right-hand side):

\[\left[\begin{array}{ccc|c} 1 & 1 & 1 & 9 \\ 1 & 2 & 2 & 16 \\ 2 & 1 & 3 & 17 \end{array}\right]\]

Each row is an equation; the bar just separates the left from the right side. No matrix algebra, pure bookkeeping.

Three Allowed Moves

We reshape the system with three row operations, each keeps the solution unchanged:

Swap two rows (reorder the equations)
Scale a row by a non-zero number (multiply or divide an equation)
Add a multiple of one row to another (combine two equations)

The goal: create zeros below the diagonal, a triangular staircase we can read off from the bottom up.

Forward Elimination: Clear the First Column

Start from the augmented matrix and knock out the $x$-terms below the top row:

\[\left[\begin{array}{ccc|c} 1 & 1 & 1 & 9 \\ 1 & 2 & 2 & 16 \\ 2 & 1 & 3 & 17 \end{array}\right] \;\xrightarrow[\;R_3 \,\to\, R_3 - 2R_1\;]{\;R_2 \,\to\, R_2 - R_1\;}\; \left[\begin{array}{ccc|c} 1 & 1 & 1 & 9 \\ 0 & 1 & 1 & 7 \\ 0 & -1 & 1 & -1 \end{array}\right]\]

$R_2 - R_1$: $\;(1,2,2\,|\,16) - (1,1,1\,|\,9) = (0,1,1\,|\,7)$
$R_3 - 2R_1$: $\;(2,1,3\,|\,17) - 2(1,1,1\,|\,9) = (0,-1,1\,|\,-1)$

Forward Elimination: Clear the Second Column

Now use the new second row to clear the entry beneath it, then scale to a leading $1$:

\[\left[\begin{array}{ccc|c} 1 & 1 & 1 & 9 \\ 0 & 1 & 1 & 7 \\ 0 & -1 & 1 & -1 \end{array}\right] \;\xrightarrow[\;R_3 \,\to\, R_3 / 2\;]{\;R_3 \,\to\, R_3 + R_2\;}\; \left[\begin{array}{ccc|c} 1 & 1 & 1 & 9 \\ 0 & 1 & 1 & 7 \\ 0 & 0 & 1 & 3 \end{array}\right]\]

$R_3 + R_2$: $\;(0,-1,1\,|\,-1) + (0,1,1\,|\,7) = (0,0,2\,|\,6)$
$R_3 / 2$: $\;(0,0,2\,|\,6) \div 2 = (0,0,1\,|\,3)$

Back-Substitution

The matrix is now triangular, zeros fill the lower-left corner. Read it from the bottom up:

\[x + y + z = 9, \qquad y + z = 7, \qquad z = 3\]

Bottom row: $\;z = 3$
Middle row: $\;y + 3 = 7 \;\Rightarrow\; y = 4$
Top row: $\;x + 4 + 3 = 9 \;\Rightarrow\; x = 2$
Solution: $(x, y, z) = (2, 4, 3)$

Always Verify

Plug the triple $(2, 4, 3)$ back into the original three equations:

$x + y + z = 2 + 4 + 3 = 9$ ✓
$x + 2y + 2z = 2 + 8 + 6 = 16$ ✓
$2x + y + 3z = 4 + 4 + 9 = 17$ ✓

Common Mistake

When you apply a row operation, act on the whole row, the right-hand side included. Subtracting $2R_1$ from $R_3$ but forgetting to also do $17 - 2 \cdot 9 = -1$ silently corrupts the whole solve.

Your Turn: Does Swapping Hurt?

Question

Halfway through, you swap two rows of the augmented matrix. Does that change the system’s solution?

No, swapping rows just reorders the equations; the set of conditions is identical
All three row operations are reversible and solution-preserving
Swapping is often useful: it can bring a convenient $1$ up to the pivot spot

How Many Solutions?

One, None, or Infinitely Many

A linear system has exactly one of three fates, there is no other possibility:

Unique solution, the conditions pin down one answer (our examples so far)
No solution, the conditions contradict each other
Infinitely many solutions, a condition is redundant, leaving freedom

Geometry tells the story cleanly for two lines in the plane.

Two Lines: No Solution

$x + y = 4$ and $x + y = 6$ are parallel, same slope, different heights. They never meet, so no pair $(x, y)$ satisfies both.

Two Lines: Infinitely Many

$2x + 2y = 8$ is just $x + y = 4$ doubled, the same line. Every point on it solves both; there are infinitely many solutions.

Parameterizing the Free Case

When one equation adds nothing new, we describe all solutions with a free parameter.

For $x + y = 4$ with a redundant partner:

Let $y = t$ be free, any real number
Then $x = 4 - t$
Solution set: $(x, y) = (4 - t,\; t)$, one point for each $t$
E.g. $t = 0 \to (4, 0)$, $\;t = 1 \to (3, 1)$, $\;t = 4 \to (0, 4)$

Three Unknowns: Planes Instead of Lines

With three unknowns, each equation is a plane in space, but the same three fates return:

Unique: three planes meet at a single point (our Gaussian example)
None: the planes have no common point, like parallel walls
Infinitely many: they share a whole line (or plane) of points
The geometry grows, but elimination handles all of it the same way

How Elimination Announces the Case

You do not need the picture, the triangular form tells you which fate you have:

A clean staircase with a leading $1$ in every column → unique solution
A row like $\;0 \; = \; 5\;$ (all coefficients zero, non-zero right side) → no solution
A row like $\;0 \; = \; 0\;$ (an entire row vanishes) → infinitely many, a free parameter

Read the Zero Row Carefully

Common Mistake

A vanished row $\;0 = 0\;$ means the equation was redundant, the system has infinitely many solutions, not “no solution”. “No solution” only appears as $\;0 = (\text{non-zero})$, a genuine contradiction. Read the right-hand side before you decide.

Your Turn: Reading the Outcome

Question

During elimination a row collapses to $\;0 = 0$. Does the system have a unique solution, no solution, or infinitely many?

$\;0 = 0\;$ is always true, that equation carried no new information
Fewer real conditions than unknowns leaves a variable free
So the system has infinitely many solutions, parameterize the free one

Linear Systems in Business

Market Equilibrium

A market clears where supply meets demand, quantity offered equals quantity wanted.

For a product at price $p$, suppose

Demand: $q = 120 - 2p$, buyers want less as the price rises
Supply: $q = 3p - 30$, sellers offer more as the price rises
Equilibrium is the price and quantity solving both at once, a $2 \times 2$ system

Solving for Equilibrium

Both express $q$, so set them equal, or eliminate $q$ from the system \[q + 2p = 120, \qquad q - 3p = -30.\]

Subtract the second from the first: $\;5p = 150 \;\Rightarrow\; p = 30$
Back-substitute: $\;q = 120 - 2(30) = 60$
Equilibrium: price €30, quantity $60$ units
Check the supply side: $\;3(30) - 30 = 60$ ✓

Supply Meets Demand

Below €30 buyers outnumber sellers (shortage); above it sellers outnumber buyers (surplus). The lines cross where the market balances.

Your Turn: Above the Equilibrium

Question

The price is set at €40, above the equilibrium of €30. Is there excess supply or excess demand?

Demand at €40: $\;120 - 80 = 40$; supply at €40: $\;3 \cdot 40 - 30 = 90$
Sellers offer $90$ but buyers want only $40$, a surplus of $50$ units
Above equilibrium there is excess supply; price pressure pushes back down toward €30

Where the 3×3 Came From

Remember the system we solved by elimination? It was a production plan all along.

A workshop makes three products $x, y, z$ and wants to use every resource fully:

Packaging: one unit each, $9$ available $\;\Rightarrow\; x + y + z = 9$
Labour: $1, 2, 2$ hours per product, $16$ hours $\;\Rightarrow\; x + 2y + 2z = 16$
Machine: $2, 1, 3$ hours per product, $17$ hours $\;\Rightarrow\; 2x + y + 3z = 17$

Reading the Production Plan

We already found the unique solution by Gaussian elimination: \[(x, y, z) = (2, 4, 3)\]

Make 2 of product $x$, 4 of product $y$, 3 of product $z$
Every resource is used to the last unit: $9$ packaging, $16$ labour, $17$ machine hours
One tidy elimination turned three resource limits into a concrete production schedule
The same recipe scales to the large planning systems behind real logistics software

Voluntary Mock Exam

To round off the tutorial, there is a voluntary mock exam covering Sessions 1–6.

Adapted from the original KLU course’s exam, real exam-style questions, no grade attached
Date: to be announced, watch the tutorial page
How to prepare: rework the tasks, retry the self-test quizzes, and keep the cheatsheets beside you
Find it here: Mock Exam

Closing

Key Takeaways

A linear equation is a plain weighted sum of unknowns, one such equation in two unknowns is a line
Solve a $2 \times 2$ system by substitution or elimination, both give the same crossing point
Gaussian elimination reduces a $3 \times 3$ system to triangular form, then back-substitutes
Every system has one, none, or infinitely many solutions, the triangular form reveals which
Equilibrium and production planning are linear systems in business dress

That’s it for today.

Any Questions?

Until the Mock Exam

Work through the Tasks: systems drills with worked solutions
Check yourself with the Self-Test quiz
Keep the Cheatsheet handy, elimination in a nutshell
Then put it all together on the Mock Exam

The Toolkit You Built

Six sessions, one steadily growing toolkit for the math behind your M.Sc. courses:

Notation & logic, then sets, the language to state ideas precisely
Functions: how one quantity depends on another, and how to read a graph
Differentiation: rates of change and the shape of a curve
Optimization: turning a flat tangent into the best decision
Linear systems: solving for several unknowns at once

You are ready for the mock exam, and for the math your degree will lean on.

Literature & Further Reading

These sessions cover the essentials, textbooks offer more depth and practice
Sydsaeter & Hammond: Essential Mathematics for Economic Analysis
Jacques: Mathematics for Economics and Business
Full recommendations on the tutorial’s literature page