Row echelon form and reduced row echelon form: Worked examples

Before anything else, one promise about vocabulary. A matrix is just a grid of numbers. When we write a system of equations as a grid, we drop the letters $x, y, z$ and keep only their coefficients, plus the constants after the $=$ sign in a final column separated by a bar. That grid is the augmented matrix. A pivot is the first non-zero number in a row, reading left to right. A pivot column is a column that contains a pivot. Everything below flows from those three words.

The scenario matrix

Any linear system, once you push it to RREF, lands in exactly one of these outcome classes. The number of pivots versus the number of unknowns decides everything.

Cell	Situation	What the RREF looks like	Solution set
C1	pivots = unknowns, no pivot in last column	full identity block on the left	exactly one point
C2	pivots < unknowns, no pivot in last column	some column has no pivot	infinitely many (free variables)
C3	pivot in the last column	a row reads $0 = 1$	no solution (inconsistent)
C4	a zero row appears	$[00 \dots 0]$ at the bottom	rank drops — links to C1/C2
C5	need a row swap (a zero blocks the pivot slot)	must reorder before eliminating	any of the above
C6	more equations than unknowns (tall)	often forces C3 or C1	over-determined
C7	more unknowns than equations (wide)	always at least one free variable	forced C2
C8	word problem → build the matrix yourself	your choice	real answer
C9	exam twist: a parameter $k$ splits cases	pivot depends on $k$	branch by $k$
C10	degenerate: the whole matrix is zero	all zero rows	every vector solves it

The examples below hit every cell. Watch the label.

Example 1 — Cell C1: one clean point (the square, easy case)

Forecast: three equations, three unknowns, nothing looks degenerate — guess one unique point. Jot a guess for $x$ before reading on.

Augmented matrix: $121 1 - 1 2 11 - 1 632$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - R_{1}$ . Why this step? The top-left $1$ is our first pivot. To isolate $x$ in equation 1 only, we kill every $x$ below it. $100 1 - 3 1 1 - 1 - 2 6 - 9 - 4$
Swap $R_{2} \leftrightarrow R_{3}$ . Why this step? The second pivot column is column 2. A pivot of $+ 1$ (from old $R_{3}$ ) is friendlier than $- 3$ ; swapping avoids fractions early. (This is a mini-preview of Cell C5.) $100 11 - 3 1 - 2 - 1 6 - 4 - 9$
$R_{3} \to R_{3} + 3 R_{2}$ . Why this step? Kill the $- 3$ below the second pivot so $y$ leaves equation 3. $100110 1 - 2 - 7 6 - 4 - 21 (REF)$
$R_{3} \to - \frac{1}{7} R_{3}$ , giving the leading $1$ in row 3. Then clean upward: $R_{2} \to R_{2} + 2 R_{3}$ , $R_{1} \to R_{1} - R_{3}$ , then $R_{1} \to R_{1} - R_{2}$ . Why this step? This is Gauss–Jordan: make each pivot $1$ , then zero the entries above each pivot so the answer is read off. $100010001123 (RREF)$

Read off: $x = 1, y = 2, z = 3$ .

Verify: $1 + 2 + 3 = 6$ ✓; $2 (1) - 2 + 3 = 3$ ✓; $1 + 4 - 3 = 2$ ✓. Three pivots, three unknowns, no pivot in the last column → C1 confirmed.

Example 2 — Cell C2 & C7: a free variable, a whole line of answers

Forecast: 3 unknowns but only 2 equations — wide (Cell C7). There aren't enough equations to pin all three. Guess: infinitely many solutions.

$[12241349]$

$R_{2} \to R_{2} - 2 R_{1}$ . Why this step? Kill $x$ below the first pivot. $[10201141] (REF)$
$R_{1} \to R_{1} - R_{2}$ . Why this step? The second pivot sits in column 3 (column 2 got skipped — its entry is $0$ ). Clear above it for RREF. $[10200131] (RREF)$

Pivots are in columns $1$ and $3$ . Column $2$ has no pivot ⇒ $y$ is a free variable. Set $y = t$ : $x = 3 - 2 t, y = t, z = 1.$

Verify: pick $t = 1$ : $x = 1, y = 1, z = 1$ . Check eq 1: $1 + 2 + 1 = 4$ ✓; eq 2: $2 + 4 + 3 = 9$ ✓. Pick $t = 0$ : $x = 3, y = 0, z = 1$ : $3 + 0 + 1 = 4$ ✓; $6 + 0 + 3 = 9$ ✓. Two different points both work → a line → C2/C7 confirmed. See the picture below.

Figure — Row echelon form and reduced row echelon form

Example 3 — Cell C3 & C6: inconsistent, no solution

Forecast: three equations, two unknowns — tall (Cell C6). Two equations already fix a point; the third might disagree. Guess: watch for a contradiction.

$111 1 - 1 2 317$

$R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - R_{1}$ . Why this step? Kill $x$ below the first pivot. $100 1 - 2 1 3 - 2 4$
$R_{2} \to - \frac{1}{2} R_{2}$ , then $R_{3} \to R_{3} - R_{2}$ . Why this step? Make the second pivot a clean $1$ , then kill below it. $100110313$
Look at row 3: it reads $0 x + 0 y = 3$ , i.e. $0 = 3$ . Why this matters? A pivot in the last (constants) column is the algebra shouting "impossible!". No numbers $x, y$ make $0$ equal $3$ .

Verify: From rows 1–2 alone: $y = 1$ , $x = 2$ . Test in equation 3: $2 + 2 (1) = 4 \neq = 7$ . The third line misses the point where the first two meet → no solution, C3/C6 confirmed.

Example 4 — Cell C5: the row swap you cannot skip

Forecast: the very first coefficient is $0$ . You cannot use a zero as a pivot (you'd divide by it later). Guess: we must swap rows first, then it's a normal one-point solve.

$012211112569$

Swap $R_{1} \leftrightarrow R_{2}$ . Why this step? The pivot slot (row 1, column 1) holds a $0$ . Swap up a row with a non-zero entry there so we have a real pivot. $102121112659$
$R_{3} \to R_{3} - 2 R_{1}$ . Why this step? Kill the $x$ below the new pivot. $100 12 - 1 110 65 - 3$
$R_{2} \to \frac{1}{2} R_{2}$ , then $R_{3} \to R_{3} + R_{2}$ . Why this step? Turn the second pivot into $1$ , then clear below it. $100110 1 \frac{1}{2} \frac{1}{2} 6 \frac{5}{2} - \frac{1}{2}$
$R_{3} \to 2 R_{3}$ ( $z = - 1$ ), then clean upward: $R_{2} \to R_{2} - \frac{1}{2} R_{3}$ , $R_{1} \to R_{1} - R_{3}$ , then $R_{1} \to R_{1} - R_{2}$ . Why this step? Gauss–Jordan finish to read off the answer. $100010001 43 - 1 (RREF)$

Read off: $x = 4, y = 3, z = - 1$ .

Verify: eq 1: $2 (3) + (- 1) = 5$ ✓; eq 2: $4 + 3 - 1 = 6$ ✓; eq 3: $2 (4) + 3 + 2 (- 1) = 9$ ✓. C5 confirmed — without the swap we'd have been stuck.

Example 5 — Cell C4 & C2: a genuine zero row appears

Forecast: equation 2 looks suspiciously like $2 \times$ equation 1. If a row is a copy of another, elimination will annihilate it into all zeros. Guess: a zero row and a free variable.

$121 12 - 1 121240$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - R_{1}$ . Why this step? Kill $x$ below the first pivot. $100 10 - 2 100 20 - 2$

Row 2 became $[0000]$ — a zero row (Cell C4). It carries no information (it's $0 = 0$ , always true). Move it to the bottom.

Swap $R_{2} \leftrightarrow R_{3}$ , then $R_{2} \to - \frac{1}{2} R_{2}$ . Why this step? Zero rows belong at the bottom; scale the surviving pivot to $1$ . $100110100210$
$R_{1} \to R_{1} - R_{2}$ . Why this step? Clear above the second pivot (column 2) for RREF. $100010100110 (RREF)$

Pivots in columns $1, 2$ ; column $3$ has none ⇒ $z$ is free. Set $z = t$ : $x = 1 - t, y = 1, z = t .$

Verify: $t = 0$ : $(1, 1, 0)$ : eq 1 $1 + 1 + 0 = 2$ ✓, eq 3 $1 - 1 + 0 = 0$ ✓. $t = 2$ : $(- 1, 1, 2)$ : eq 1 $- 1 + 1 + 2 = 2$ ✓, eq 3 $- 1 - 1 + 2 = 0$ ✓. Rank $= 2$ (two pivots, one zero row) $< 3$ unknowns → free variable → C4/C2 confirmed.

Example 6 — Cell C10: the degenerate all-zero right side (homogeneous)

Forecast: everything on the right is $0$ . Then $x = y = z = 0$ always works (the trivial solution). The only question is whether there are others. Guess: yes, because equation 2 is $2 \times$ equation 1 — not enough real equations.

$[1224 - 1 - 2 00]$

$R_{2} \to R_{2} - 2 R_{1}$ . Why this step? Kill $x$ below the pivot. $[1020 - 1 0 00] (RREF)$

The second row vanished entirely. One pivot (column 1), so $y$ and $z$ are both free. With $y = s, z = t$ : $x = - 2 s + t, y = s, z = t .$

Verify: $(s, t) = (0, 0)$ gives $(0, 0, 0)$ — the trivial solution, guaranteed for any homogeneous system. $(s, t) = (1, 0)$ : $(- 2, 1, 0)$ : $- 2 + 2 - 0 = 0$ ✓. $(s, t) = (0, 1)$ : $(1, 0, 1)$ : $1 + 0 - 1 = 0$ ✓. Infinitely many → a whole plane through the origin → C10 confirmed. Note: a homogeneous system can never be inconsistent, because the all-zero vector is always a solution.

Example 7 — Cell C8: a word problem you must translate

Forecast: three baskets, three unknown prices — smells like Cell C1, one clean answer. Guess a price for the apple first.

Translate each basket into an equation, then a matrix row: $⎩ ⎨ ⎧ a + b + c = 10 2 a + b + 0 c = 12 a + 0 b + 3 c = 14 \Rightarrow 121110103101214$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - R_{1}$ . Why this step? Standard: clear below the first pivot ( $a$ ). $100 1 - 1 - 1 1 - 2 2 10 - 8 4$
$R_{2} \to - R_{2}$ , then $R_{3} \to R_{3} + R_{2}$ . Why this step? Make pivot $2$ positive $1$ , then clear below. $10011012410812$
$R_{3} \to \frac{1}{4} R_{3}$ ( $c = 3$ ), then clean upward: $R_{2} \to R_{2} - 2 R_{3}$ , $R_{1} \to R_{1} - R_{3}$ , then $R_{1} \to R_{1} - R_{2}$ . Why this step? Gauss–Jordan to read prices directly. $100010001523 (RREF)$

Read off: $a = \$ 5,\ b = $2,\ c = $3$.

Verify (with units): Basket A $= 5+2+3 = \$ 10 $✓ . B a s k e tB$ = 2(5)+2 = $12 $✓ . B a s k e tC$ = 5 + 3(3) = $14$ ✓. All prices positive and sensible → C8 confirmed.

Example 8 — Cell C9: the exam twist with a parameter $k$

Forecast: the coefficient $k$ hides in the matrix. Somewhere a pivot will depend on $k$ ; when that pivot collapses to $0$ , the behaviour flips. Guess: one special "bad" value of $k$ .

$[11 1 k 12]$

$R_{2} \to R_{2} - R_{1}$ . Why this step? Clear $x$ below the pivot. $[10 1 k - 1 11]$
Inspect the entry $k - 1$ in the pivot-2 slot. Why this step? Whether row 2 gives a real second pivot depends on $k - 1$ being non-zero. This is the branch point.

Case (a): $k \neq = 1$ . Then $k - 1 \neq = 0$ is a genuine pivot. Two pivots, two unknowns, no pivot in the last column ⇒ unique solution. Solving: $y = \frac{1}{k - 1}$ , and $x = 1 - y = \frac{k - 2}{k - 1}$ .
Case (b/c): $k = 1$ . Row 2 becomes $[00 ∣ 1]$ , i.e. $0 = 1$ ⇒ pivot in the last column ⇒ no solution (Cell C3). (There is no " $k$ " here that gives infinitely many, because when $k = 1$ the constant $1$ makes the two lines parallel but distinct.)

Verify: take $k = 3$ : formula gives $y = \frac{1}{2}$ , $x = \frac{1}{2}$ . Check: $x + y = 1$ ✓; $x + 3 y = \frac{1}{2} + \frac{3}{2} = 2$ ✓. Take $k = 0$ : $y = \frac{1}{- 1} = - 1$ , $x = \frac{- 2}{- 1} = 2$ . Check: $2 + (- 1) = 1$ ✓; $2 + 0 (- 1) = 2$ ✓. And $k = 1$ : equations $x + y = 1$ , $x + y = 2$ can't both hold → no solution ✓. C9 confirmed.

Recall Which cell is which — self-test

"3 equations, 3 unknowns, RREF is the identity block" is which cell? ::: C1 — unique solution. A column with no pivot means what? ::: A free variable → infinitely many solutions (C2/C7/C10). A pivot in the augmented last column means what? ::: Inconsistent, no solution (C3). Why must you sometimes swap rows before eliminating? ::: A $0$ in the pivot slot can't be a pivot (you'd divide by zero) — swap a non-zero row up (C5). Can a homogeneous system (all zeros on the right) ever be inconsistent? ::: No — the all-zero vector always solves it (C10). At a parameter value where a pivot becomes $0$ , what do you do? ::: Analyse that value as a separate case; don't trust the general formula (C9).

Connections

Every example here ran Gaussian Elimination to REF, then Gauss-Jordan Elimination to RREF.
Counting pivots = Rank of a Matrix (Examples 5, 6 have rank $<$ unknowns).
The outcome classes are the heart of Solving Systems of Linear Equations.
Free variables (Examples 2, 5, 6) are formalised in Free and Basic Variables.
A homogeneous system with only the trivial solution ties to Linear Independence of the columns.
Row swaps and scalings (Examples 1, 4) are each an elementary matrix acting on the left.
Augmenting with the identity instead of constants gives Matrix Inverse via RREF.
Back to the parent topic.