2.3 Gaussian elimination and RREF

Why elimination matters

When students first meet Gaussian elimination, it can feel like a pile of rules:

• swap rows,
• multiply a row,
• subtract a row,
• keep going until the matrix looks “nice.”

That is not the right mental model.

Gaussian elimination is really a reading strategy. You start with a system that is hard to read, then you use row operations to make the logical structure of the system more visible. Each row operation should answer a question such as:

• Which variable am I trying to isolate next?
• Which entry is blocking me?
• Which move will create more zeros without changing the solution set?

If you keep asking those questions, elimination becomes much less mechanical.

Intuition first: what are we trying to build?

The dream shape is not “a random simpler matrix.” The dream shape is a matrix whose important columns are easy to spot and whose equations are easy to read from bottom to top.

In practice, that means:

1. pick a pivot entry,
2. clear the entries below it,
3. move to the smaller submatrix that remains,
4. then clean up above the pivots if you want full reduced form.

So elimination is a controlled way of building a staircase of pivots.

The first nonzero entry in a nonzero row is called a pivot or leading entry. A column containing a pivot is a pivot column. Columns without a pivot correspond to free variables.

That vocabulary matters because it tells you what kind of solution set to expect later.

This is why elimination is legitimate. We are reorganizing information, not inventing a new problem.

One full elimination path

Elimination starts from an augmented matrix and repeatedly clears the entries beneath a pivot. We will follow that idea on a small system:

$$\begin{aligned} x + 2y + 2z &= 4, \\ x + 3y + 3z &= 5, \\ 2x + 6y + 5z &= 6. \end{aligned}$$

Its augmented matrix is

$$\begin{bmatrix} 1 & 2 & 2 & 4 \\ 1 & 3 & 3 & 5 \\ 2 & 6 & 5 & 6 \end{bmatrix}.$$

Notice the teaching pattern:

• below-pivot clearing creates REF,
• above-pivot clearing creates RREF,
• RREF is the easiest form to read directly.

Try the same path interactively

The stepper below keeps the same logic, but slows it down. At each stage, look at three things:

1. which pivot is active,
2. which row operation is being applied, and
3. what becomes easier to read afterward.

A longer assignment-style reduction

Short examples are useful for learning the mechanics, but assignment problems usually test whether you can keep the whole reduction organized. The main danger is not the size of the numbers. The main danger is losing the purpose of each row operation.

Consider the augmented matrix

$$C= \begin{bmatrix} 1 & 2 & 0 & 1 & 0 & 2 & 1 \\ 1 & 2 & 1 & 0 & 1 & 2 & 3 \\ 2 & 4 & 1 & 1 & 2 & 5 & 1 \\ 5 & 10 & 3 & 2 & 4 & 11 & 8 \end{bmatrix}.$$

The first column already has a usable pivot in row 1. Clearing below that pivot gives

$$\begin{bmatrix} 1 & 2 & 0 & 1 & 0 & 2 & 1 \\ 0 & 0 & 1 & -1 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 & 2 & 1 & -1 \\ 0 & 0 & 3 & -3 & 4 & 1 & 3 \end{bmatrix}.$$

Now row 2 supplies the next pivot. Use it to remove the matching entries below:

$$R_3\leftarrow R_3-R_2, \qquad R_4\leftarrow R_4-3R_2.$$

After also clearing the duplicate row created by the next pivot, one clean echelon stage is

$$\begin{bmatrix} 1 & 2 & 0 & 1 & 0 & 2 & 1 \\ 0 & 0 & 1 & -1 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & 1 & -3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}.$$

This is the point where you should pause and read the structure: pivot columns are columns 1, 3, and 5; free columns are columns 2, 4, and 6. To reach RREF, clear the entry above the pivot in column 5:

$$R_2\leftarrow R_2-R_3.$$

The final reduced form is

$$C'= \begin{bmatrix} 1 & 2 & 0 & 1 & 0 & 2 & 1 \\ 0 & 0 & 1 & -1 & 0 & -1 & 5 \\ 0 & 0 & 0 & 0 & 1 & 1 & -3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}.$$

Let

$$x_2=u,\qquad x_4=v,\qquad x_6=w.$$

Then the pivot equations read

$$\begin{aligned} x_1+2x_2+x_4+2x_6 &= 1,\\ x_3-x_4-x_6 &= 5,\\ x_5+x_6 &= -3. \end{aligned}$$

So every solution has the form

$$x= \begin{bmatrix} 1\\0\\5\\0\\-3\\0 \end{bmatrix} +u \begin{bmatrix} -2\\1\\0\\0\\0\\0 \end{bmatrix} +v \begin{bmatrix} -1\\0\\1\\1\\0\\0 \end{bmatrix} +w \begin{bmatrix} -2\\0\\1\\0\\-1\\1 \end{bmatrix}.$$

This example illustrates the real purpose of a long reduction: not merely to produce zeros, but to expose the pivot/free-variable pattern from which the whole solution set follows.

How to read an RREF matrix

Once a matrix is in RREF, the main reading questions are:

• Does every variable column have a pivot?
• Is there any free variable?
• Is there a contradiction row?

Common mistakes

Quick checks

Exercises

Exercise 1

Start from

$$\begin{bmatrix} 1 & 1 & 2 & 3 \\ 2 & 3 & 5 & 8 \\ 0 & 1 & 1 & 2 \end{bmatrix}.$$

What is the most natural first row operation if your goal is to clear the entry below the first pivot?

Exercise 2

Consider the matrix

$$\begin{bmatrix} 1 & 0 & 2 & 4 \\ 0 & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}.$$

Does this represent a system with a unique solution, infinitely many solutions, or no solution?

Exercise 3

Start from

$$\begin{bmatrix} 1 & 3 & 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}.$$

Which single row operation turns this matrix into RREF?

Exercise 4

For the RREF matrix

$$\begin{bmatrix} 1 & 2 & 0 & 1 & 0 & 2 & 1 \\ 0 & 0 & 1 & -1 & 0 & -1 & 5 \\ 0 & 0 & 0 & 0 & 1 & 1 & -3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix},$$

set $x_2=u$, $x_4=v$, and $x_6=w$. Write $x_1$, $x_3$, and $x_5$ in terms of u, v, and w.

Read this first

This page builds directly on 2.2 Augmented matrices and row operations. If you are still unsure why row operations preserve the solution set, go back to that note first before practicing longer elimination paths.