Row echelon form and reduced row echelon form

WHAT are these forms?

WHY does the staircase work? (first-principles derivation)

The only tools allowed are the three elementary row operations, because each preserves the solution set:

HOW to reach REF — Gaussian elimination

1. Find the leftmost column with a non-zero entry → pivot column.
2. Swap a non-zero entry up to the pivot position (avoid dividing by $0$).
3. Use Replace to zero out everything below the pivot.
4. Ignore that row+column, recurse on the submatrix below-right.

HOW to reach RREF — Gauss–Jordan elimination

Do everything above, then going from the bottom pivot upward: 5. Scale each pivot row so the pivot becomes $1$. 6. Use Replace to zero out entries above each pivot too.

Worked Example 1 — to REF then RREF

Solve $\begin{cases} x + 2y + z = 6\\ 2x + 5y + 3z = 15\\ x + 3y + 3z = 11 \end{cases}$

Augmented matrix: $\left[\begin{array}{ccc|c} 1&2&1&6\\ 2&5&3&15\\ 1&3&3&11\end{array}\right]$

Step A: $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - R_1$. Why this step? The pivot is the top-left $1$; we kill the entries below it so $x$ disappears from equations 2 and 3. $\left[\begin{array}{ccc|c} 1&2&1&6\\ 0&1&1&3\\ 0&1&2&5\end{array}\right]$

Step B: $R_3 \to R_3 - R_2$. Why this step? The next pivot is the $1$ in row 2, column 2. Kill the entry below it so $y$ leaves equation 3. $\left[\begin{array}{ccc|c} 1&2&1&6\\ 0&1&1&3\\ 0&0&1&2\end{array}\right] \quad \textbf{(REF — staircase done!)}$

From REF, back-substitute: $z=2$, then $y+z=3\Rightarrow y=1$, then $x+2y+z=6\Rightarrow x=2$.

Now to RREF (clean above pivots, bottom-up):

Step C: $R_2 \to R_2 - R_3$, $R_1 \to R_1 - R_3$. Why? Clear the column of pivot in row 3. $\left[\begin{array}{ccc|c} 1&2&0&4\\ 0&1&0&1\\ 0&0&1&2\end{array}\right]$

Step D: $R_1 \to R_1 - 2R_2$. Why? Clear above the row-2 pivot. $\left[\begin{array}{ccc|c} 1&0&0&2\\ 0&1&0&1\\ 0&0&1&2\end{array}\right] \quad \textbf{(RREF)}$ Read off directly: $x=2,\ y=1,\ z=2$. ✅

Worked Example 2 — free variables & infinite solutions

$\left[\begin{array}{ccc|c} 1&2&3&4\\ 2&4&7&9\end{array}\right]$

Step A: $R_2 \to R_2 - 2R_1$. Why? Kill below the leading $1$. $\left[\begin{array}{ccc|c} 1&2&3&4\\ 0&0&1&1\end{array}\right] \ (\text{REF})$

Step B: $R_1 \to R_1 - 3R_2$ (clean above pivot in column 3). $\left[\begin{array}{ccc|c} 1&2&0&1\\ 0&0&1&1\end{array}\right]\ (\text{RREF})$

Pivot columns: $1,3$. Column $2$ has no pivot ⇒ $y$ is a free variable. $x = 1 - 2y$, $z = 1$, $y$ free. Infinitely many solutions, a whole line. Why? Fewer pivots than variables ⇒ leftover freedom.

Worked Example 3 — inconsistent system

$\left[\begin{array}{cc|c} 1&1&2\\ 2&2&5\end{array}\right] \xrightarrow{R_2 \to R_2-2R_1} \left[\begin{array}{cc|c} 1&1&2\\ 0&0&1\end{array}\right]$ Row 2 reads $0=1$. A pivot in the augmented (last) column ⇒ no solution. Why this matters: a pivot in the constants column is the algebraic shout "contradiction!".

Flashcards

Connections

• Gaussian Elimination and Gauss-Jordan Elimination
• Rank of a Matrix (= number of pivots)
• Solving Systems of Linear Equations
• Free and Basic Variables
• Matrix Inverse via RREF (augment with $I$)
• Linear Independence (pivot columns are independent)
• Elementary Matrices (each row op = left-multiplication)

Concept Map

Hinglish (regional understanding)