Worked examples — Pivot positions, free variables

Before starting, some reminders in plain words:

• $A$ is a grid of numbers ($m$ rows, $n$ columns). Each column stands for one unknown $x_1,\dots ,x_n$.
• $\mathbf{b}$ is the right-hand side: the column of numbers the equations must equal. So $A\mathbf{x}=\mathbf{b}$ means "each row of $A$, dotted with the unknowns, gives the matching entry of $\mathbf{b}$."
• The augmented matrix $[A\mid \mathbf{b}]$ glues $\mathbf{b}$ on as one extra column, separated by a vertical bar. Its last column (right of the bar) is what we call the "$\mathbf{b}$-column." It's just bookkeeping: the bar reminds us that column carries the answers, not an unknown.

• Row-reduce = apply those moves until each row's first nonzero number sits to the right of the row above (staircase shape). That staircase form is called echelon form.
• A pivot = the leading (first nonzero) entry of a nonzero row after reduction. The number of pivots is the rank $r$ (see Rank and the Rank-Nullity Theorem).

This whole page is a companion to Pivot positions, free variables.

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The scenario matrix

Every linear system falls into exactly one of these cells. Two independent switches decide the outcome: (A) Is the system consistent (does any solution exist)? (B) How many free variables $=n-r$?

#	Consistency	Free vars $n-r$	Shape of solution set	Special feature	Example
C1	Consistent	$0$	single point (unique)	square, full rank	Ex 1
C2	Consistent	$1$	a line	one free column	Ex 2
C3	Consistent	$2$	a plane	two free columns	Ex 3
C4	Inconsistent	(irrelevant)	empty (no solution)	$\mathbf{b}$-column is a pivot	Ex 4
C5	Consistent	$n$ (all free)	all of space	$A=0$, degenerate	Ex 5
C6	Consistent	$1$ (here $n=2,r=1$)	a line	more rows than cols, zero rows appear	Ex 6
C7	Consistent	$0$	unique point	more rows than cols, still full column rank	Ex 7
C8	Consistent	$1$	a line	word problem (real units)	Ex 8
C9	either	$0$ or contradiction	point or empty	exam twist: parameter $k$ flips the case	Ex 9

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Ex 1 — C1: unique solution (zero free variables)

Forecast: $2$ columns; the rows $(2,1)$ and $(1,3)$ are not proportional, so both columns will earn a pivot → $r=2=n$ → $n-r=0$, a single point.

1. Swap to get a leading $1$ up top: $R_1\leftrightarrow R_2$ → $\left[\begin{array}{ccc}1& 3& 5\\ 2& 1& 5\end{array}\right].$ Why this step? A leading $1$ makes the elimination arithmetic clean.
2. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{ccc}1& 3& 5\\ 0& -5& -5\end{array}\right].$ Why? Kill the entry below the first pivot to expose the staircase.
3. $R_2\leftarrow -\frac{1}{5}{R}_2$: gives $x_2=1$. Why? Scale the pivot to $1$ (one step toward RREF) so $x_2$ can be read off directly.
4. Back-substitute $x_1=5-3x_2=5-3=2$. Why? Row 1 solves its pivot variable using the now-known value.

Result: pivots in columns 1 and 2 ⇒ no free columns, $n-r=2-2=0$. Unique solution $(x_1,x_2)=(2,1)$.

Verify: $2(2)+1(1)=5$ ✓ and $1(2)+3(1)=5$ ✓. Both equations satisfied.

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Ex 2 — C2: one free variable (a line)

Forecast: $n=3$ unknowns but only $m=2$ rows, so $r\le 2$ and $n-r\ge 1$. Why? Each pivot lives in its own row, so pivots can't exceed the number of rows; with at least one non-pivot column, expect a line.

1. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 0& 2& 2\end{array}\right].$ Why? Zero out below the first pivot.
2. $R_2\leftarrow \frac{1}{2}{R}_2$: $\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 0& 1& 1\end{array}\right].$ Why? Make the second pivot a leading $1$.
3. Pivots sit in columns 1 and 3; column 2 has no pivot → $x_2$ is free. Set $x_2=t$. Why? Every non-pivot column is, by definition, a free parameter.
4. Row 2: $x_3=1$. Row 1: $x_1=4-2x_2-3x_3=4-2t-3=1-2t$. Why? Solve each pivot variable in terms of the free one.

Result: $n-r=3-2=1$ free variable ⇒ a line: $\mathbf{x}=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+t\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right].$

The figure below draws this line in the $x_1$–$x_2$ plane (the pivot value $x_3=1$ is fixed all along it). The coral dot is the particular solution $(1,0,1)$; the mint arrow is the direction $(-2,1,0)$ that the free variable $t$ sweeps — this direction is exactly the null-space vector. The butter dots are sample solutions at $t=-1$ and $t=1$; every point on the lavender line solves the system.

Figure Ex 2: one free variable traces a line — coral = particular solution, mint arrow = free-variable direction, butter dots = sample solutions.

Verify at $t=0$: $x_1+2x_2+3x_3=1+0+3=4$ ✓; $2+0+8=10$ ✓. At $t=1$: $\mathbf{x}=(-1,1,1)$: $-1+2+3=4$ ✓, $-2+4+8=10$ ✓.

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Ex 3 — C3: two free variables (a plane)

Forecast: $n=4$ unknowns and clearly $2$ staircase pivots, so $n-r=4-2=2$. Why? Two rows give at most two pivots and both are visibly present, leaving two non-pivot columns → a plane.

1. Read pivots: leading entries in column 1 (row 1) and column 3 (row 2). Why? They are the first nonzero entries of each row.
2. Non-pivot columns are 2 and 4 → $x_2=s$, $x_4=t$ free. Why? No leading entry lives there.
3. Row 2: $x_3=6-4x_4=6-4t$. Row 1: $x_1=5-2x_2-3x_4=5-2s-3t$. Why? Solve each pivot variable against the two free ones.

Result: $n-r=4-2=2$ ⇒ a plane: $\mathbf{x}=\left[\begin{array}{c}5\\ 0\\ 6\\ 0\end{array}\right]+s\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right]+t\left[\begin{array}{c}-3\\ 0\\ -4\\ 1\end{array}\right].$

Verify at $(s,t)=(0,0)$: eq1 $=5$ ✓, eq2 $=6$ ✓. At $(s,t)=(1,1)$: $\mathbf{x}=(0,1,2,1)$ → eq1 $0+2+0+3=5$ ✓, eq2 $2+4=6$ ✓.

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Ex 4 — C4: inconsistent (no solution, free count irrelevant)

Forecast: row 2 of $A$ is exactly $2\times$ row 1, but $\mathbf{b}$'s entries $1,5$ are not in that same ratio. Why suspect trouble? Proportional rows in $A$ demand proportional right-hand sides; a mismatch forces a contradiction ($0=$ nonzero).

1. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 3\end{array}\right].$ Why? Eliminate below the first pivot.
2. Row 2 reads $0x_1+0x_2=3$, i.e. $0=3$. Why does this end everything? The augmented column now holds a leading entry — the $\mathbf{b}$-column is itself a pivot column, the flag for inconsistency.

Result: no solution. Note column 2 has no pivot in $A$, so naively "there's a free variable" — but with zero solutions, that freedom describes nothing. Consistency is checked first.

Verify: $x_1+x_2=1$ forces $2x_1+2x_2=2\ne 5$. Contradiction confirmed — no $(x_1,x_2)$ works.

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Ex 5 — C5: the all-zero matrix (every variable free)

Forecast: the matrix imposes no constraint at all, so no column can hold a pivot → $r=0$ → $n-r=2$. Why? With rank $0$, every column is non-pivot, so every variable is free — the whole plane.

1. RREF of the zero matrix is the zero matrix: no nonzero rows ⇒ no pivots, so $r=0$. Why? A pivot is the first nonzero entry of a row; there are none.
2. Both columns are non-pivot ⇒ $x_1=s$, $x_2=t$ both free. Why? Every column lacks a leading entry.

Result: $n-r=2-0=2$ ⇒ solution set is all of ${\mathbb{R}}^2$: $\mathbf{x}=s\left[\begin{array}{c}1\\ 0\end{array}\right]+t\left[\begin{array}{c}0\\ 1\end{array}\right],s,t\in \mathbb{R}.$ This is the extreme of $n-r$: rank zero gives maximum freedom.

Verify: any $(s,t)$ gives $0=0$ in both rows ✓. Degenerate limiting case handled.

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Ex 6 — C6: more rows than columns, with a redundant (zero) row

Forecast: all three rows of $A$ are multiples of $(1,2)$, so only $r=1$ pivot survives → $n-r=2-1=1$, a line. Why? Proportional rows collapse to a single independent equation; the extras become zero rows, leaving one free column.

1. $R_2\leftarrow R_2-2R_1$, $R_3\leftarrow R_3-3R_1$: $\left[\begin{array}{ccc}1& 2& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right].$ Why? Eliminate below the first pivot; the two lower rows collapse.
2. One pivot (column 1); column 2 is free, $x_2=t$. There are two zero rows but only one free variable. Why the myth fails: free variables count non-pivot columns $=n-r=2-1=1$, not zero rows.
3. Row 1: $x_1=1-2x_2=1-2t$. Why? Solve the single pivot variable against the free one.

Result: a line $\mathbf{x}=\left[\begin{array}{c}1\\ 0\end{array}\right]+t\left[\begin{array}{c}-2\\ 1\end{array}\right]$. Two zero rows ≠ two free variables.

Verify at $t=0$: all three original equations give $1,2,3$ matching $\mathbf{b}$ ✓. At $t=1$: $\mathbf{x}=(-1,1)$ → $-1+2=1$, $-2+4=2$, $-3+6=3$ ✓.

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Ex 7 — C7: more rows than columns, full column rank (unique)

Forecast: the first two rows are independent, so $r=2=n$ → $n-r=0$, a unique point — even though there are more rows than columns. Why? Full column rank pins every variable; the third row must then be a redundant combination (and it had better be consistent).

1. $R_3\leftarrow R_3-R_1-R_2$: $\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& 0& 0\end{array}\right].$ Why? Cancel row 3 against the two pivots above.
2. Pivots in columns 1 and 2 ⇒ full column rank, no free variables ($n-r=2-2=0$). The extra row became zero and, crucially, its $\mathbf{b}$-side is also $0$ → consistent. Why check the $\mathbf{b}$-side? A nonzero there (like Ex 4) would kill it.

Result: unique solution $x_1=2$, $x_2=3$.

Verify: $1(2)+0=2$ ✓; $0+1(3)=3$ ✓; $1(2)+1(3)=5$ ✓. All three rows agree.

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Ex 8 — C8: word problem (real units, one free variable)

Forecast: $n=3$ unknowns but only $m=2$ constraints, so $r\le 2$ and $n-r\ge 1$ → a line of recipes. Why? Two equations can pin at most two variables; the third stays free (then real-world $\ge 0$ limits clip the line to a segment).

1. Model it: $A+O+G=10,2A+3O+5G=34.$ Matrix $\left[\begin{array}{cccc}1& 1& 1& 10\\ 2& 3& 5& 34\end{array}\right].$ Why? Each unknown is a column; each rule is a row.
2. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{cccc}1& 1& 1& 10\\ 0& 1& 3& 14\end{array}\right].$ Why? Eliminate below the first pivot.
3. Pivots in columns 1, 2; column 3 (grape) is free, $G=t$. Why? No leading entry in column 3.
4. Row 2: $O=14-3G=14-3t$. Row 1: $A=10-O-G=10-(14-3t)-t=-4+2t$. Why? Solve pivots against the free variable.

Result: $\mathbf{x}=(A,O,G)=(-4+2t,14-3t,t)$, $n-r=3-2=1$ ⇒ a line of recipes. Physically we also need $A,O,G\ge 0$: $A\ge 0\Rightarrow t\ge 2$; $O\ge 0\Rightarrow t\le \frac{14}{3}\approx 4.67$; so $2\le t\le \frac{14}{3}$ L of grape. Maths gives a line; units clip it to a segment.

Verify at $t=3$: $(A,O,G)=(2,5,3)$ → volume $2+5+3=10$ ✓, sugar $2(2)+3(5)+5(3)=4+15+15=34$ ✓, all $\ge 0$ ✓.

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Ex 9 — C9: exam twist (a parameter flips the case)

Forecast: eliminating puts $k-6$ in the second pivot slot, so $k=6$ is the one dangerous value. Why? When the pivot entry hits $0$ the rank drops; whether that gives "no solution" or "infinitely many" then hinges on the $\mathbf{b}$-column.

1. $R_2\leftarrow R_2-3R_1$: $\left[\begin{array}{ccc}1& 2& 1\\ 0& k-6& 2\end{array}\right].$ Why? Eliminate below the pivot; the $(2,2)$ entry becomes $k-6$.
2. Case $k\ne 6$: entry $k-6\ne 0$ → pivot in column 2 too. Both columns pivoted ⇒ $n-r=0$ ⇒ (i) unique solution. Why? Full column rank, consistent.
3. Case $k=6$: row 2 becomes $\left[00\mid 2\right]$, i.e. $0=2$ → the $\mathbf{b}$-column is a pivot ⇒ (ii) no solution. Why? Contradiction, like Ex 4.
4. Is (iii) ever reachable? Infinitely many needs a free column and consistency. Here $k=6$ is the only rank-drop value, but it makes $\mathbf{b}$ inconsistent, so (iii) never happens for this $\mathbf{b}$. Why note this? The exam bait is assuming "rank drop ⇒ infinitely many" — false unless the $\mathbf{b}$-side cooperates.

Result: unique for all $k\ne 6$; no solution at $k=6$; never infinitely many.

Verify solve the unique case: eliminate gives $x_2=\frac{2}{k-6}$, $x_1=1-2x_2$. At $k=8$: $x_2=1$, $x_1=-1$ → check $1(-1)+2(1)=1$ ✓, $3(-1)+8(1)=5$ ✓. At $k=6$: $0=2$ contradiction ✓.

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Recall Which cell am I in? (two questions)

First: is the $\mathbf{b}$-column a pivot column? ::: If yes → inconsistent (C4/Ex 9 case ii), stop. If no → consistent, continue. Then: how many non-pivot columns? ::: That count is $n-r$: $0$→point, $1$→line, $2$→plane, $n$→all of space.

Recall The trap all three mistakes share

Why do "zero rows = free vars", "free var ⇒ infinite solutions", and "rank drop ⇒ infinite" all fail? ::: Each ignores a separate switch — free variables count non-pivot columns (not zero rows), and any solution count needs consistency checked first via the $\mathbf{b}$-column.

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Connections

• Pivot positions, free variables
• Row Reduction and Echelon Forms
• Rank and the Rank-Nullity Theorem
• Null Space and Solution Sets of Ax=0
• Existence and Uniqueness of Solutions
• Basis and Dimension
• Linear Independence

Case Map (read this first, diagram second)

In plain words: to classify any system, run two checks in order.

1. Reduce the augmented matrix to echelon form.
2. Consistency check — if the $\mathbf{b}$-column holds a pivot (a row $[0\cdots 0\mid c]$ with $c\ne 0$), stop: no solution (cell C4).
3. If consistent, count non-pivot columns $=n-r$: $0$ free → a single point (C1/C7); $1$ free → a line (C2/C6/C8); $2$ free → a plane (C3); all $n$ free → the whole space (C5).

The diagram below is the same logic drawn as arrows.