1. Definition & Four conditions (with a tiny example each)
Definition. A matrix is in reduced row echelon form (RREF) if it satisfies the row echelon form conditions and additionally every pivot is 1 and is the only nonzero entry in its column. RREF is unique for each matrix.
- All nonzero rows are above any rows of all zeros.
Condition: nonzero rows above zero rows.
Example: $$\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}$$ (satisfies) — vs — $$\begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}$$ (violates). - Each leading entry of a nonzero row is 1 (pivot = 1).
Example: $$\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}$$ (pivots are 1) — compare to $$\begin{bmatrix}2 & 4\\0 & 1\end{bmatrix}$$ (leading entry 2 violates).
- Pivots move strictly to the right as you go down rows.
Example: $$\begin{bmatrix}1 & * & *\\0 & 1 & *\\0 & 0 & 1\end{bmatrix}$$ (pivots at columns 1,2,3 respectively). If a lower pivot lies in the same or left column as above, it violates.
- Each pivot is the only nonzero entry in its column (zeros above & below).
Example: $$\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$ (each pivot column has zeros elsewhere). But $$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$$ violates for first column if there is a nonzero above/below pivot.
2. Quick check — 10 small matrices: decide whether each is RREF or not (with short reason)
(Matrix shown in LaTeX, then verdict + reason)
1. $$\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$ — RREF. Both pivots are 1 and are only nonzero in their columns.
2. $$\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}$$ — Not RREF. Pivot in column 2 is 1 but column 1 has a nonzero above/below? Actually column 1 pivot is 1 but column 2 has a nonzero in row1 (2), which is above pivot at row2 col2 — so not RREF because entry above pivot is 2.
3. $$\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$$ — RREF. Leading 1 at (1,2); zero row below; pivot column has zeros elsewhere.
4. $$\begin{bmatrix}1 & 0 & 3\\0 & 1 & -2\\0 & 0 & 0\end{bmatrix}$$ — RREF. Pivots at (1,1) and (2,2), both are only nonzero entries in their columns.
5. $$\begin{bmatrix}1 & 0\\1 & 0\end{bmatrix}$$ — Not RREF. First column pivot at row1 is 1 but there is a nonzero (1) below it; pivot column must be zero elsewhere.
6. $$\begin{bmatrix}0 & 0 & 0\\0 & 1 & 0\end{bmatrix}$$ — Not RREF. Nonzero row appears below all-zero row; violate first condition (nonzero rows must be above zero rows).
7. $$\begin{bmatrix}1 & 0 & 0\\0 & 2 & 0\end{bmatrix}$$ — Not RREF. Pivot in second row is 2, not 1.
8. $$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 1\end{bmatrix}$$ — RREF. Pivot columns (1 and 2) have zeros elsewhere; nonzero in column 3 is allowed (non-pivot column).
9. $$\begin{bmatrix}1 & 0 & 0 & 2\\0 & 1 & 0 & -1\end{bmatrix}$$ — RREF. Augmented columns allowed; pivots are unit columns and unique in their columns.
10. $$\begin{bmatrix}1 & 0 & 0\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix}$$ — RREF. Pivots at (1,1) and (2,3); pivot columns have zeros elsewhere; zero row at bottom.
3. Worked reduction — one 3×3 (augmented) matrix, step-by-step to RREF
Reduce the augmented matrix:
$$A = \left[\begin{array}{ccc|c}
1 & 1 & 1 & 6\\
2 & -1 & 1 & 3\\
1 & 2 & -1 & 4
\end{array}\right]$$
Step 1 (use row1 pivot = 1): eliminate below:
$$R_2\leftarrow R_2-2R_1,\quad R_3\leftarrow R_3-R_1$$ $$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6\\ 0 & -3 & -1 & -9\\ 0 & 1 & -2 & -2 \end{array}\right]$$Step 2: make pivot at row2 column2 equal to 1 by scaling $R_2\leftarrow -\tfrac{1}{3}R_2$
$$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6\\ 0 & 1 & \tfrac{1}{3} & 3\\ 0 & 1 & -2 & -2 \end{array}\right]$$Step 3: eliminate other entries in column2:
$$R_1\leftarrow R_1-R_2,\quad R_3\leftarrow R_3-R_2$$ $$\left[\begin{array}{ccc|c} 1 & 0 & \tfrac{2}{3} & 3\\ 0 & 1 & \tfrac{1}{3} & 3\\ 0 & 0 & -\tfrac{7}{3} & -5 \end{array}\right]$$Step 4: scale row3 to make pivot 1: $R_3\leftarrow -\tfrac{3}{7}R_3$
$$\left[\begin{array}{ccc|c} 1 & 0 & \tfrac{2}{3} & 3\\ 0 & 1 & \tfrac{1}{3} & 3\\ 0 & 0 & 1 & \tfrac{15}{7} \end{array}\right]$$Step 5: eliminate above column3 to finish RREF:
$$R_1\leftarrow R_1-\tfrac{2}{3}R_3,\quad R_2\leftarrow R_2-\tfrac{1}{3}R_3$$ $$\boxed{\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & \tfrac{15}{7} \end{array}\right]}$$Note: arithmetic checked — final solution is $$x=2,\;y=1,\;z=\tfrac{15}{7}$$ (if you treat last column as RHS). For the classic earlier example we had integer 3; this variant keeps fractional outcome to illustrate fractions.
Interactive RREF Calculator
Enter matrix rows separated by semicolons (;) and entries by commas (,). Include RHS if it's an augmented matrix.
Solution Steps
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