The Timpson Lyceum · Linear Algebra (2026)

Systems of Linear Equations and Echelon Forms – Lecture 1 (Linear Algebra)

Overview

This lecture begins the course proper, taking the familiar problem of finding where two lines cross and rebuilding it as a general procedure that works in any number of variables. It starts by pinning down exactly what makes an equation linear, then shows that a system of linear equations always falls into one of three cases — no solution, exactly one, or infinitely many — with no other possibility. The central move is to strip a system down to its coefficients and record it as a matrix, so that solving becomes a matter of manipulating rows rather than rewriting variables. Three elementary row operations drive the whole process, and running them to completion produces echelon form and then reduced echelon form. From those forms the lecture reads off the pivot positions, separates basic variables from free variables, and states the criterion that decides existence and uniqueness for any system at all.

Key Concepts

What makes an equation linear

A linear equation in the variables $x_1, x_2, \dots, x_n$ is any equation that can be written in the form

a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b

where the $a_i$ and $b$ are constants. The defining test is whether the equation can be rearranged into that shape — not whether it is already presented that way.

Every variable appears to the first power, multiplied only by a constant. Variables are never multiplied by each other and never appear inside a power, root, or other function.
With many variables in play, subscripted names $x_1, x_2, x_3, \dots$ replace $x, y, z$ , which would otherwise run out.

Equation	Linear?	Reason
$2x + 9y = 11$	Yes	Constant times variable, summed, equal to a constant
$19x_1 + 3 = -7x_2 + 1$	Yes	Rearranges to $19x_1 + 7x_2 = -2$
$y = x^2 + 1$	No	$x$ carries an exponent of $2$ — a parabola, not a line
$x_1 x_2 = 4$	No	The variables multiply each other; it cannot be split into $a_1x_1 + a_2x_2$

"Linear" does not mean "line." The solution set of a single linear equation is a flat object one dimension below the space it lives in: a line in $\mathbb{R}^2$ , a plane in $\mathbb{R}^3$ , and a flat three-dimensional surface in $\mathbb{R}^4$ .

Systems and their three outcomes

A system of linear equations is a collection of linear equations in the same variables. To solve it is to find every assignment of values satisfying all the equations simultaneously.
A system has exactly one of three outcomes, and there is no fourth:
- No solution — the system is called inconsistent. In two dimensions, parallel lines such as $y = 2x$ and $y = 2x + 1$ .
- Exactly one solution — lines crossing at a single point, as with $y = 2x$ and $y = -x + 3$ meeting at $(1,2)$ .
- Infinitely many solutions — the equations describe the same object, as with $y = 2x$ and $2y = 4x$ .
A system can never have exactly two solutions, or exactly seventeen. The count is always $0$ , $1$ , or infinite.

Matrix notation

Discarding the variable names and keeping only the numbers, a system becomes a rectangular array. For

\begin{aligned} x + y &= 3 \\ 2x - y &= 0 \end{aligned}

the coefficient matrix records only the left-hand sides, while the augmented matrix appends the constants from the right:

\begin{aligned} \text{coefficient:}\quad \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix} \qquad \text{augmented:}\quad \left[\begin{array}{cc|c} 1 & 1 & 3 \\ 2 & -1 & 0 \end{array}\right] \end{aligned}

The vertical bar marking the equals sign is a convenience, not part of the notation; context usually makes clear which kind of matrix is meant. Most work in this course uses augmented matrices, since they carry the full equations.
Two matrices are row equivalent when one can be turned into the other by row operations. Row equivalent augmented matrices represent systems with the same solution set.

The three elementary row operations

Replacement — add a multiple of one row to another row. This is ordinary elimination, and it is the operation that does the real work.
Interchange — swap two rows. Writing the equations in a different order cannot change which values satisfy them.
Scaling — multiply a row by a nonzero constant. The restriction matters: multiplying through by zero destroys the equation and changes the solution set.

Echelon form and reduced echelon form

A matrix is in echelon form when every nonzero row begins with a leading nonzero entry that has nothing but zeros beneath it, and each successive leading entry sits strictly to the right of the one above, so the nonzero pattern steps downward. Any all-zero rows sit at the bottom. The name is apt — echelon means a stepped or stair-like arrangement.
A matrix is in reduced echelon form when, in addition, every leading entry is $1$ and each leading $1$ is the only nonzero entry in its column.
Reaching echelon form is the forward phase of the algorithm — work down the diagonal, clearing zeros below each leading entry. Reaching reduced echelon form adds a backward phase, clearing zeros above each leading $1$ . Always use the current pivot row to clear its own column; reusing an already-finished row undoes prior work.

Theorem 1 (uniqueness of the reduced echelon form). Every matrix is row equivalent to one and only one matrix in reduced echelon form. Existence follows from the algorithm itself, which must terminate because each pass permanently retires one row and one column of a finite array. Uniqueness is proved by induction on the number of columns: deleting the last column of two reduced echelon forms of the same matrix makes them identical by the induction hypothesis, and examining whether the final column is a pivot column forces the last columns to agree as well.

Pivots, basic variables, and free variables

A pivot position is a location that holds a leading $1$ in the reduced echelon form — equivalently, a leading entry in any echelon form. A pivot column is a column containing a pivot position.
Pivot positions are a property of the matrix itself. Different sequences of row operations still produce the same pivot positions, because the reduced echelon form is unique.
Variables in pivot columns are basic variables; variables in non-pivot columns are free variables. A free variable may be assigned any value whatever, and each such choice determines the basic variables uniquely.
Writing the solution with the basic variables expressed in terms of the free ones gives a parametric description of the solution set — the free variables play the role that $t$ plays in a parametric equation.

Theorem 2 (existence and uniqueness). A linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot column — equivalently, if and only if an echelon form has no row of the form

\begin{bmatrix} 0 & 0 & \cdots & 0 & b \end{bmatrix} \quad \text{with } b \neq 0

Such a row asserts $0 = b$ for some nonzero $b$ , which is never true, so the system has no solution. When a system is consistent, it has a unique solution if there are no free variables, and infinitely many solutions if there is at least one free variable.

A quick count often settles matters before any arithmetic: with more unknowns than equations, at least one variable must be free, so the system is either inconsistent or has infinitely many solutions — a unique solution is impossible. Two planes in $\mathbb{R}^3$ can be parallel or meet in a line, but they can never meet in a single point.

Worked Example

Solve the system

\begin{aligned} x_1 - 2x_2 + x_3 &= 0 \\ 2x_2 - 8x_3 &= 8 \\ 5x_1 - 5x_3 &= 10 \end{aligned}

Write the augmented matrix and drive it to reduced echelon form. The forward phase clears each pivot column below the diagonal; the backward phase clears above.

\begin{aligned} \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 2 & -8 & 8 \\ 5 & 0 & -5 & 10 \end{array}\right] &\sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 2 & -8 & 8 \\ 0 & 10 & -10 & 10 \end{array}\right] && R_3 \to R_3 - 5R_1 \\ &\sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & -4 & 4 \\ 0 & 10 & -10 & 10 \end{array}\right] && R_2 \to \tfrac{1}{2} R_2 \\ &\sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & -4 & 4 \\ 0 & 0 & 30 & -30 \end{array}\right] && R_3 \to R_3 - 10R_2 \\ &\sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & -4 & 4 \\ 0 & 0 & 1 & -1 \end{array}\right] && R_3 \to \tfrac{1}{30} R_3 \\ &\sim \left[\begin{array}{ccc|c} 1 & -2 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1 \end{array}\right] && R_1 \to R_1 - R_3,\; R_2 \to R_2 + 4R_3 \\ &\sim \left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1 \end{array}\right] && R_1 \to R_1 + 2R_2 \end{aligned}

Every column of the coefficient part is a pivot column, so there are no free variables and the solution is unique. Reading the final matrix row by row gives $x_1 = 1$ , $x_2 = 0$ , $x_3 = -1$ , that is $(1, 0, -1)$ . Substituting back confirms all three equations: $1 - 0 + (-1) = 0$ , $0 + 8 = 8$ , and $5 + 5 = 10$ .

For contrast, a system whose echelon form contains the row

\left[\begin{array}{ccc|c} 0 & 0 & 0 & 15 \end{array}\right]

asserts $0 = 15$ and is therefore inconsistent — no amount of further work will produce a solution.

Summary

A linear equation is one expressible as $a_1x_1 + \cdots + a_nx_n = b$ ; variables appear to the first power only, never multiplied together. "Linear" describes the algebraic form, not a line.
Every system of linear equations has no solution, exactly one solution, or infinitely many — never any other count.
Recording a system as an augmented matrix reduces solving to three row operations: replacement, interchange, and scaling by a nonzero constant. These never change the solution set.
Echelon form places a leading nonzero entry with zeros below it in each successive row; reduced echelon form additionally makes every leading entry a $1$ that is alone in its column. Every matrix has exactly one reduced echelon form.
Pivot columns identify the basic variables; the remaining variables are free, and expressing the basic variables in terms of the free ones gives a parametric description of the solution set.
A system is consistent exactly when the rightmost column of the augmented matrix is not a pivot column. A consistent system is unique when it has no free variables and infinite when it has at least one.