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Row Reduced Echelon Form of a Matrix
😊 Hello, students! Today, we unlock one of the most powerful tools in linear algebra: the row reduced echelon form (RREF). It’s the “simplest” version of a matrix that preserves all the information about solutions to linear systems. By the end of this lesson, you’ll be able to reduce any matrix to RREF — and understand why each step matters.
1. What Is a Leading Entry?
In a non-zero row of a matrix, the leading entry is the leftmost non-zero entry.
In the row \( [0\ 0\ 3\ -1\ 2] \), the leading entry is \( 3 \) (in column 3).
2. Row Echelon Form (REF)
A matrix is in row echelon form (REF) if it satisfies all of the following:
- All zero rows (if any) are at the bottom.
- Each leading entry of a row is to the right of the leading entry of the row above it.
- All entries below a leading entry are zero.
Example of REF:
Notice: leading entries at (1,1) and (2,3); zero row at bottom; zeros below each pivot.
3. Reduced Row Echelon Form (RREF)
A matrix is in reduced row echelon form (RREF) if it is in REF and satisfies two more conditions:
- Every leading entry is 1 (called a pivot).
- Each leading 1 is the only non-zero entry in its column.
Example of RREF:
Every pivot is 1, and all entries above and below each pivot are 0.
4. Why Do We Care About RREF?
RREF reveals everything about a linear system \( A\mathbf{x} = \mathbf{b} \):
- Consistency: Is there a solution?
- Uniqueness: Exactly one, or infinitely many?
- Rank: Number of pivots = rank of the matrix.
- Basis for row/column space: Pivot rows/columns form bases.
5. How to Reduce a Matrix to RREF
Use elementary row operations only:
- Row swap: \( R_i \leftrightarrow R_j \)
- Scaling: \( R_i \to kR_i \), \( k \ne 0 \)
- Replacement: \( R_i \to R_i + kR_j \), \( i \ne j \)
Example: Reduce \[ A = \begin{bmatrix} 2 & 4 & 6 \\ 1 & 1 & 1 \\ 3 & 5 & 7 \end{bmatrix} \] to RREF.
Step 1: Make leading entry in row 1 equal to 1:
\( R_1 \to \frac{1}{2}R_1 \)
Step 2: Eliminate below pivot (col 1):
\( R_2 \to R_2 – R_1 \), \( R_3 \to R_3 – 3R_1 \)
Step 3: Make pivot in row 2:
\( R_2 \to -R_2 \)
Step 4: Eliminate below (and above!) pivot in col 2:
\( R_3 \to R_3 + R_2 \), then \( R_1 \to R_1 – 2R_2 \)
✅ This is RREF! Note the zero row at bottom and pivots only in cols 1 and 2.
6. Rank from RREF
The rank of a matrix \( A \), denoted \( \text{rank}(A) \), is the number of non-zero rows in its RREF (i.e., number of pivots).
In the example above, RREF has 2 non-zero rows ⇒ \( \text{rank}(A) = 2 \).
7. Applications of RREF
Solving \( A\mathbf{x} = \mathbf{0} \)
From RREF, identify:
- Pivot variables (corresponding to pivot columns)
- Free variables (non-pivot columns)
8. Summary: REF vs RREF
| Property | REF | RREF |
|---|---|---|
| Zero rows at bottom | ✅ | ✅ |
| Staircase pattern | ✅ | ✅ |
| Zeros below pivots | ✅ | ✅ |
| Pivots = 1 | ❌ (not required) | ✅ |
| Zeros above pivots | ❌ | ✅ |
| Uniqueness | ❌ (many possible) | ✅ (only one) |
9. Self-Check: Test Your Understanding
- Is the following matrix in RREF? \[ \begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \]
- What is the rank of the matrix in (1)?
- Reduce \( \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix} \) to RREF.
Answers: (1) Yes! (2) 2 (3) \( \begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix} \)
Row Reduced Echelon Form – Questions & Full Solutions
😊 Hello, students! Below are all questions from your documents on row reduced echelon form (RREF), each with a full solution. Use this to master Gaussian elimination, pivots, and consistency!
Q1. Determine which of the following matrices are in **row reduced echelon form (RREF)**:
Solution:
Recall RREF requires:
- Leading entry in each non-zero row is 1 (pivot).
- Each pivot is the only non-zero entry in its column.
- Pivots move strictly right as we go down.
- Zero rows at bottom.
- (a) ❌ Not RREF — column 3 has non-zero entries above and below pivot (no pivot in col 3).
- (b) ❌ Not RREF — leading 1 in row 1, but column 2 and 3 have non-zero entries in same row — and no pivot in col 2.
- (c) ✅ RREF — all pivots = 1, zeros above and below, staircase pattern, no zero rows.
Q2. Reduce the matrix to RREF:
Solution:
Step 1: Eliminate below pivot in col 1:
- \( R_2 \to R_2 – 2R_1 \)
- \( R_3 \to R_3 – R_1 \)
Step 2: Eliminate below and above pivot in col 2:
- \( R_3 \to R_3 + R_2 \)
- \( R_1 \to R_1 – 2R_2 \)
✅ This is RREF. Pivot columns: 1 and 2. Rank = 2.
Q3. The following matrix is in RREF. What is its rank?
Solution:
Count non-zero rows: there are **3**.
✅ Rank = 3.
Q4. Solve the system using RREF:
Solution:
Augmented matrix:
Row reduce:
- \( R_2 \to R_2 – 2R_1 \), \( R_3 \to R_3 + R_1 \)
- \( R_3 \to R_3 – R_2 \)
- Back-substitute to RREF
Let \( z = t \) (free variable). Then:
- \( y = 2 – t \)
- \( x = -1 + t \)
✅ General solution: \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} + t \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}, \quad t \in \mathbb{R} \]
Q5. Reduce to RREF and determine consistency:
Solution:
Row ops:
- \( R_2 \to R_2 – 2R_1 \) → row becomes \([0\ 0\ 0\ |\ 1]\)
- \( R_3 \to R_3 – 3R_1 \) → row becomes \([0\ 0\ 0\ |\ 0]\)
Last non-zero row: \( 0 = 1 \) → **inconsistent**.
✅ No solution.
Q6. Reduce to RREF and solve:
Solution:
Swap \( R_1 \leftrightarrow R_2 \), then eliminate:
✅ Unique solution: \( x = 2,\ y = 1,\ z = 0 \).
Q7. Find the RREF and rank of:
Solution:
All rows are multiples of row 1:
- \( R_2 \to R_2 – 2R_1 \)
- \( R_3 \to R_3 + R_1 \)
✅ Rank = 1.
📌 Key Takeaways:
- RREF is unique for any matrix.
- Rank = number of pivots = non-zero rows in RREF.
- System is consistent ⇔ no row of form \([0\ \cdots\ 0\ |\ c]\) with \( c \ne 0 \).
- Free variables → infinitely many solutions.