Reduced Row Echelon Form (RREF)

The Ultimate Answer Sheet

In our previous lessons, we learned to use Gaussian Elimination to transform a matrix into Row Echelon Form (REF). This gave us an upper-triangular system that was easy to solve using back-substitution.

REF is powerful, but it's not the final destination. It gives us a system that is easy for a human to solve. But what if we wanted a form so simple that the answers are just... there, with no substitution required?

This is the purpose of Reduced Row Echelon Form (RREF). It's the cleanest, most unique form of a matrix, and it acts as the ultimate answer sheet for a linear system.

From REF to RREF: The Extra Steps

To be in Reduced Row Echelon Form, a matrix must first be in Row Echelon Form. Then, it must satisfy two additional, stricter conditions:

  1. Every pivot entry must be 1.

    We achieve this by scaling each pivot row. For example, if a row is `[0, 2, 4 | 6]`, we would divide the entire row by 2 to make the pivot a 1, resulting in `[0, 1, 2 | 3]`.

  2. Every entry above each pivot must be 0.

    This is the "upwards" elimination phase. After clearing everything below the pivots (Gaussian Elimination), we go back and clear everything above them, starting from the rightmost pivot and working our way left.

Example: The Full Journey
Let's take our simple, unique-solution system from a previous lesson and take it all the way to RREF.

Step 1: Start and Achieve REF

This is standard Gaussian Elimination, clearing entries below the pivots.

[21154−60−2−2729]→Elim[21150−8−2−120012]\left[\begin{array}{ccc|c} 2 & 1 & 1 & 5 \\ 4 & -6 & 0 & -2 \\ -2 & 7 & 2 & 9 \end{array}\right] \xrightarrow{\text{Elim}} \left[\begin{array}{ccc|c} 2 & 1 & 1 & 5 \\ 0 & -8 & -2 & -12 \\ 0 & 0 & 1 & 2 \end{array}\right]

Step 2: Scale All Pivots to 1

We divide each row by its pivot value (2, -8, and 1).

[10.50.52.5010.251.50012]\left[\begin{array}{ccc|c} 1 & 0.5 & 0.5 & 2.5 \\ 0 & 1 & 0.25 & 1.5 \\ 0 & 0 & 1 & 2 \end{array}\right]

Step 3: Eliminate Entries ABOVE the Pivots

This is "upwards" elimination, starting from the rightmost pivot.

Use the pivot in R3 to clear entries above it:

  • R2 → R2 - 0.25 * R3
  • R1 → R1 - 0.5 * R3

The matrix becomes:

[10.501.501010012]\left[\begin{array}{ccc|c} 1 & 0.5 & 0 & 1.5 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}\right]

Now use the pivot in R2 to clear the entry above it:

  • R1 → R1 - 0.5 * R2
Why is RREF So Important?
  • Uniqueness: Every matrix has one and only one unique Reduced Row Echelon Form. This makes RREF a definitive "fingerprint" for a matrix.
  • Readability: For systems with a unique solution, the RREF gives you the answer directly. For systems with infinite solutions, it makes the relationship between pivot and free variables exceptionally clear.
  • Theoretical Power: RREF is the ultimate tool for understanding a matrix's fundamental properties. It tells you the rank, the basis for the column space, and the basis for the null space, all in one neat package.

Up Next: LU Decomposition