Row Reduction and Echelon Form

Quick answer

A linear system is recorded as an augmented matrix, one row per equation. Three row operations — swapping two rows, scaling a row, and adding a multiple of one row to another — change the matrix without changing the solution set. Applying them systematically produces echelon form, where each leading entry sits to the right of the one above, and then reduced echelon form, where each pivot is 1 and is alone in its column. The solution can then be read straight off the rows.

What you'll learn

Write a linear system as an augmented matrix
Apply the three row operations correctly
Reduce a matrix to echelon and reduced echelon form
Identify pivot positions and read off a solution

The system as a matrix

Solving a linear system means combining equations until each unknown stands alone. The variable names take no part in that work, so record only the coefficients. The system

\begin{aligned} x + 2y &= 5\\ 3x - y &= 1 \end{aligned}

becomes the augmented matrix, one row per equation, with a bar marking the right-hand side:

\left[\begin{array}{cc|c} 1 & 2 & 5\\ 3 & -1 & 1 \end{array}\right]

Three row operations rewrite it without changing what it says:

Swap two rows.
Scale a row by a nonzero number.
Replace a row by itself plus a multiple of another row.

Why row operations keep every solution

Each operation is a move on whole equations, and each is reversible. Swapping two equations changes their order, and a solution of a set of equations does not depend on the order. Multiplying an equation by $c \ne 0$ gives an equation with the same solutions, since dividing by $c$ returns the original. Adding a multiple of one equation to another produces an equation that every common solution already satisfies, and subtracting the same multiple undoes it. Every row operation can be undone by another row operation, so the new system has exactly the same solution set as the old one — no solutions lost, none gained.

Example 1 — Two equations

Solve $x + 2y = 5$ and $3x - y = 1$ .

Start from the augmented matrix and clear the $3$ below the first pivot with $R_2 - 3R_1$ :

\left[\begin{array}{cc|c} 1 & 2 & 5\\ 3 & -1 & 1 \end{array}\right] \longrightarrow \left[\begin{array}{cc|c} 1 & 2 & 5\\ 0 & -7 & -14 \end{array}\right]

Scale the second row by $-\tfrac{1}{7}$ , then clear the $2$ above it with $R_1 - 2R_2$ :

\left[\begin{array}{cc|c} 1 & 2 & 5\\ 0 & 1 & 2 \end{array}\right] \longrightarrow \left[\begin{array}{cc|c} 1 & 0 & 1\\ 0 & 1 & 2 \end{array}\right]

The rows now read $x = 1$ and $y = 2$ .

x + 2y = 5
3x − y = 1

The two equations meet at (1, 2)

Echelon form

Working down the matrix, the goal is a staircase shape.

A matrix is in echelon form when every all-zero row sits at the bottom. Each leading nonzero entry, called a pivot, must also stand strictly to the right of the pivot in the row above.

Going further gives the form that answers the question outright.

A matrix is in reduced row echelon form when it is in echelon form, every pivot is $1$ , and every pivot is the only nonzero entry in its column.

Each matrix has exactly one reduced row echelon form, whatever route the row operations take to get there.

The algorithm. Work left to right. In the current column, pick a nonzero entry at or below the current row and swap it up. Scale it to $1$ , then use it to clear every other entry in its column. Move down one row and right at least one column, and repeat.

Worked examples

Example 2 — Three equations, three unknowns

Solve $x + y + z = 4$ , $2x - y + 3z = 3$ and $3x + 2y - z = 6$ .

\left[\begin{array}{ccc|c} 1 & 1 & 1 & 4\\ 2 & -1 & 3 & 3\\ 3 & 2 & -1 & 6 \end{array}\right] \xrightarrow{\;R_2 - 2R_1,\; R_3 - 3R_1\;} \left[\begin{array}{ccc|c} 1 & 1 & 1 & 4\\ 0 & -3 & 1 & -5\\ 0 & -1 & -4 & -6 \end{array}\right]

Swap the last two rows to bring a $-1$ up, scale it to $1$ , and clear below:

\left[\begin{array}{ccc|c} 1 & 1 & 1 & 4\\ 0 & 1 & 4 & 6\\ 0 & 0 & 13 & 13 \end{array}\right]

The last row says $z = 1$ . Back-substituting gives $y = 6 - 4 = 2$ and $x = 4 - 2 - 1 = 1$ . Clearing upward instead produces the reduced form, with the solution in the last column:

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

Check in the original equations: $1 + 2 + 1 = 4$ ✓, $2 - 2 + 3 = 3$ ✓ and $3 + 4 - 1 = 6$ ✓.

Common mistakes

Practice problems

Write the system $2x + y = 7$ , $x - y = -1$ as an augmented matrix.

Answer

$\left[\begin{array}{cc|c} 2 & 1 & 7\\ 1 & -1 & -1 \end{array}\right]$

Full solution

One row per equation, with the coefficients of $x$ and $y$ and then the constant.
Solve that system by row reduction.

Answer

$(2, 3)$

Full solution

Swapping the rows puts a $1$ in the first pivot: then $R_2 - 2R_1$ gives $3y = 9$ , so $y = 3$ and $x = -1 + 3 = 2$ .
Solve $x - 2y + z = 0$ , $2x + y - z = 5$ , $3x - y + 2z = 5$ .

Answer

$(2, 1, 0)$

Full solution

Clearing the first column gives $5y - 3z = 5$ and $5y - z = 5$ . Subtracting, $2z = 0$ , so $z = 0$ , $y = 1$ and $x = 2y - z = 2$ .
What does a row $\left[\begin{array}{ccc|c} 0 & 0 & 0 & 4 \end{array}\right]$ tell you?

Answer

The system has no solution.

Full solution

The row says $0 = 4$ . No values of the variables make that true, so the system is inconsistent.
Is $\left[\begin{array}{ccc} 1 & 0 & 2\\ 0 & 3 & 1\\ 0 & 0 & 0 \end{array}\right]$ in echelon form? In reduced echelon form?

Answer

Echelon yes; reduced no

Full solution

The staircase and the zero row at the bottom are correct, but the second pivot is $3$ rather than $1$ .
Put $\left[\begin{array}{cc|c} 1 & 4 & 9\\ 2 & 7 & 16 \end{array}\right]$ in reduced row echelon form.

Answer

$\left[\begin{array}{cc|c} 1 & 0 & 1\\ 0 & 1 & 2 \end{array}\right]$

Full solution

$R_2 - 2R_1$ gives $\left[\begin{array}{cc|c} 1 & 4 & 9\\ 0 & -1 & -2 \end{array}\right]$ . Scaling by $-1$ and then $R_1 - 4R_2$ finishes it, giving $x = 1$ and $y = 2$ .
How many pivots does $\left[\begin{array}{cccc} 0 & 1 & 2 & 0\\ 0 & 0 & 0 & 5\\ 0 & 0 & 0 & 0 \end{array}\right]$ have, and in which columns?

Answer

Two pivots, in columns $2$ and $4$

Full solution

The leading entries are the $1$ in row 1 column 2 and the $5$ in row 2 column 4.
Which row operation turns $\left[\begin{array}{cc} 1 & 3\\ 4 & 5 \end{array}\right]$ into $\left[\begin{array}{cc} 1 & 3\\ 0 & -7 \end{array}\right]$ ?

Answer

$R_2 - 4R_1$

Full solution

Subtracting four times the first row from the second gives $4 - 4 = 0$ and $5 - 12 = -7$ .
A system of three equations in three unknowns row reduces to a matrix with pivots in columns 1, 2 and 3. How many solutions does it have?

Answer

Exactly one

Full solution

Every variable has a pivot and no pivot sits in the last column, so the reduced form reads off a single value for each variable.
A student solves a system, reaches $\left[\begin{array}{cc|c} 1 & 2 & 5\\ 0 & 0 & 0 \end{array}\right]$ , and reports “no solution”. What went wrong?

Hint

What does the second row say?

Answer

The row says $0 = 0$ , which is always true. The system has infinitely many solutions.

Full solution

A zero row on both sides carries no information: one equation was a multiple of the other. No solution requires a pivot in the last column, as in $\left[\begin{array}{cc|c} 0 & 0 & 1 \end{array}\right]$ . Here the surviving equation $x + 2y = 5$ is satisfied by a whole line of pairs.

Frequently asked questions

What are the three elementary row operations?

Swap two rows; multiply a row by a nonzero number; add a multiple of one row to another row. Each is reversible, so none changes the solution set.

What is echelon form?

Every all-zero row sits at the bottom, and each leading nonzero entry is strictly to the right of the leading entry in the row above.

What is reduced row echelon form?

Echelon form with every leading entry equal to 1 and every other entry in a pivot column equal to 0. A matrix has exactly one reduced echelon form.

What is a pivot?

The leading nonzero entry of a row in echelon form. Its position is a pivot position, and its column is a pivot column.

How do you know a system has no solution?

A row of the form [0 0 … 0 | c] with c nonzero says 0 = c, which is false, so the system is inconsistent.

What to learn next

Solution Sets of Linear SystemsRead the shape of a solution set from the pivots: none, exactly one, or infinitely many, written in parametric vector form as a particular solution plus directions.

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