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MATH 1554 Course Notes - Fall 2026

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Section 1.2 Row Reduction and Echelon Forms

Worksheet Worksheet

 
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Definition 1.13. Echelon Form and Row Row Reduced Echelon Form.

A rectangular matrix is in echelon form (REF) if:
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  1. All zero rows (if any) are at the bottom.
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  2. The first nonzero entry (or leading entry) of a row is to the right of any leading entry in the row above it.
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  3. All entries below a leading entry are zero.
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A matrix in echelon form is in row reduced echelon form (RREF) if additionally:
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  1. All leading entries equal \(1\text{.}\)
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  2. Each leading entry is the only nonzero entry in its column.
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Example 1.14.

Let \(\blacksquare\) denote a nonzero number, and \(\ast\) denote any number.
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\(\begin{bmatrix} 0 \amp \blacksquare \amp \ast \amp \ast \amp \ast \amp \ast \amp \ast \amp \ast \\ 0 \amp 0 \amp 0 \amp \blacksquare \amp \ast \amp \ast \amp \ast \amp \ast \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp \blacksquare \amp \ast \amp \ast \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp \blacksquare \amp \ast \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}\) is in REF and \(\begin{bmatrix} 0 \amp 1 \amp \ast \amp 0 \amp \ast \amp 0 \amp 0 \amp \ast \\ 0 \amp 0 \amp 0 \amp 1 \amp \ast \amp 0 \amp 0 \amp \ast \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 1 \amp 0 \amp \ast \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 1 \amp \ast \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}\) is in RREF.
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Exercise Group.

Determine whether each matrix is in RREF, REF, or neither.
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1.
\(\begin{bmatrix} 1 \amp 0 \\ 0 \amp 2 \end{bmatrix}\)
Answer.
REF
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2.
\(\begin{bmatrix} 0 \amp 0 \\ 0 \amp 0 \end{bmatrix}\)
Answer.
RREF
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3.
\(\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}\)
Answer.
Neither
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4.
\(\begin{bmatrix} 0 \amp 6 \amp 3 \amp 0 \end{bmatrix}\)
Answer.
REF
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5.
\(\begin{bmatrix} 1 \amp 17 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\)
Answer.
RREF
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Definition 1.15. Pivot Position and Pivot Column.

A pivot position in a matrix \(A\) is a location in \(A\) that corresponds to a leading \(1\) in the RREF of \(A\text{.}\) A pivot column is a column of \(A\) that contains a pivot position.
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Example 1.16.

Express the matrix in row reduced echelon form and identify the pivot columns.
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\begin{equation*} \left[ \begin{array}{rrrr} 0 \amp -3 \amp -6 \amp 4 \\ -1 \amp -2 \amp -1 \amp 3 \\ -2 \amp -3 \amp 0 \amp 3 \end{array} \right] \end{equation*}
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Solution.
\(\begin{eqnarray*} \begin{bmatrix} 0 \amp -3 \amp -6 \amp 4 \\ -1 \amp -2 \amp -1 \amp 3 \\ -2 \amp -3 \amp 0 \amp 3 \end{bmatrix} \amp \overset{\text{Swap $R_1$ and $R_2$}}{\Huge \sim} \amp \begin{bmatrix} -1 \amp -2 \amp -1 \amp 3 \\ 0 \amp -3 \amp -6 \amp 4 \\ -2 \amp -3 \amp 0 \amp 3 \end{bmatrix} \\ \amp \overset{-R_1 \rightarrow R_1}{\Huge \sim} \amp \begin{bmatrix} 1 \amp 2 \amp 1 \amp -3 \\ 0 \amp -3 \amp -6 \amp 4 \\ -2 \amp -3 \amp 0 \amp 3 \end{bmatrix} \\ \amp \overset{R_3 + 2R_1 \rightarrow R_3}{\Huge \sim} \amp \begin{bmatrix} 1 \amp 2 \amp 1 \amp -3 \\ 0 \amp -3 \amp -6 \amp 4 \\ 0 \amp 1 \amp 2 \amp -3 \end{bmatrix} \\ \amp \overset{\text{Swap $R_2$ and $R_3$}}{\Huge \sim} \amp \begin{bmatrix} 1 \amp 2 \amp 1 \amp -3 \\ 0 \amp 1 \amp 2 \amp -3 \\ 0 \amp -3 \amp -6 \amp 4 \end{bmatrix} \\ \amp \underset{R_3 + 3R_2 \rightarrow R_3}{\overset{R_1 - 2R_2 \rightarrow R_1}{\Huge \sim}} \amp \begin{bmatrix} 1 \amp 0 \amp -3 \amp 3 \\ 0 \amp 1 \amp 2 \amp -3 \\ 0 \amp 0 \amp 0 \amp -5 \end{bmatrix} \\ \amp \overset{-\frac{1}{5}R_3 \rightarrow R_3}{\Huge \sim} \amp \begin{bmatrix} 1 \amp 0 \amp -3 \amp 3 \\ 0 \amp 1 \amp 2 \amp -3 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \\ \amp \underset{R_2 + 3R_3 \rightarrow R_2}{\overset{R_1 - 3R_3 \rightarrow R_1}{\Huge \sim}} \amp \begin{bmatrix} 1 \amp 0 \amp -3 \amp 0 \\ 0 \amp 1 \amp 2 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \end{eqnarray*}\)
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Therefore, we can see that the pivot columns are columns 1, 2, and 4.
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Row Reduction Algorithm (Steps to RREF).

The algorithm below produces a matrix in RREF.
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  • Step 1a: Swap the first row with a lower one so the leftmost nonzero entry is in the first row.
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  • Step 1b: Scale the first row so that its leading entry equals \(1\text{.}\)
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  • Step 1c: Use row replacement so all entries above and below this \(1\) are \(0\text{.}\)
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  • Step 2a: Swap the second row with a lower one so the leftmost nonzero entry below the first row is in the second row.
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  • Step 2b: Scale the second row so that its leading entry equals \(1\text{.}\)
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  • Step 2c: Use row replacement so all entries above and below this \(1\) are \(0\text{.}\)
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  • Continue: Repeat these steps for each subsequent row until the matrix is in RREF.
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Example 1.17.

Use the algorithm above to convert the following matrix into RREF.
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\begin{equation*} \left[ \begin{array}{rrrr} 0 \amp 2 \amp 7 \amp 6 \\ 3 \amp 6 \amp 0 \amp -3 \\ 2 \amp 8 \amp 12 \amp 6 \end{array} \right] \end{equation*}
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Solution.
\(\begin{eqnarray*} \begin{bmatrix} 0 \amp 2 \amp 7 \amp 6 \\ 3 \amp 6 \amp 0 \amp -3 \\ 2 \amp 8 \amp 12 \amp 6 \end{bmatrix} \amp \overset{\text{Swap $R_1$ and $R_2$}}{\Huge \sim} \amp \begin{bmatrix} 3 \amp 6 \amp 0 \amp -3 \\ 0 \amp 2 \amp 7 \amp 6 \\ 2 \amp 8 \amp 12 \amp 6 \end{bmatrix} \\ \amp \underset{\frac{1}{2}R_3 \rightarrow R_3}{\overset{\frac{1}{3}R_1 \rightarrow R_1}{\Huge \sim}} \amp \begin{bmatrix} 1 \amp 2 \amp 0 \amp -1 \\ 0 \amp 2 \amp 7 \amp 6 \\ 1 \amp 4 \amp 6 \amp 3 \end{bmatrix} \\ \amp \overset{R_3 - R_1 \rightarrow R_3}{\Huge \sim} \amp \begin{bmatrix} 1 \amp 2 \amp 0 \amp -1 \\ 0 \amp 2 \amp 7 \amp 6 \\ 0 \amp 2 \amp 6 \amp 4 \end{bmatrix} \\ \amp \overset{\text{Swap $R_2$ and $R_3$}}{\Huge \sim} \amp \begin{bmatrix} 1 \amp 2 \amp 0 \amp -1 \\ 0 \amp 2 \amp 6 \amp 4 \\ 0 \amp 2 \amp 7 \amp 6 \end{bmatrix} \\ \amp \overset{\frac{1}{2}R_2 \rightarrow R_2}{\Huge \sim} \amp \begin{bmatrix} 1 \amp 2 \amp 0 \amp -1 \\ 0 \amp 1 \amp 3 \amp 2 \\ 0 \amp 2 \amp 7 \amp 6 \end{bmatrix} \\ \amp \underset{R_3 - 2R_2 \rightarrow R_3}{\overset{R_1 - 2R_2 \rightarrow R_1}{\Huge \sim}} \amp \begin{bmatrix} 1 \amp 0 \amp -6 \amp -5 \\ 0 \amp 1 \amp 3 \amp 2 \\ 0 \amp 0 \amp 1 \amp 2 \end{bmatrix} \\ \amp \underset{R_2 - 3R_3 \rightarrow R_2}{\overset{R_1 + 6R_3 \rightarrow R_1}{\Huge \sim}} \amp \begin{bmatrix} 1 \amp 0 \amp 0 \amp 7 \\ 0 \amp 1 \amp 0 \amp -4 \\ 0 \amp 0 \amp 1 \amp 2 \end{bmatrix}\end{eqnarray*}\)
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Definition 1.18. Pivot Variables and Free Variables.

Variables in a system that correspond to pivot column in the augmented matrix are called pivot variables. The other variables in the system are called free variables. This is because, if the system is consistent, then any choice of the free variables will lead to a valid solution.
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Example 1.19.

Determine the pivot variables and the free variables for the system represented below.
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\begin{equation*} \left[ \begin{array}{rrrrr|r} 1 \amp 3 \amp 0 \amp 7 \amp 2 \amp 4 \\ 0 \amp 0 \amp 2 \amp 0 \amp 4 \amp -2 \\ 0 \amp 0 \amp 0 \amp -3 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}
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Solution.
Notice that the pivots are in columns 1, 3, and 4. So the pivot variables are \(x_1, x_3, \text{ and } x_4\text{.}\) Therefore, the free variables are \(x_2 \text{ and } x_5\text{.}\)
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