LINEAR-ALGEBRA · Unit 8 · Video 3 · Interactive Practice

Zero, One, or Infinity — The Rank Classification

IKey Formulas

Formula	Name	Description
$r = \text{number of pivots}$	Rank	Count of independent equations after row reduction
$n - r$	Free variable gap	Controls uniqueness: if 0, solution is unique
$m - r$	Zero rows gap	Controls existence: if 0, every $b$ is solvable
$r \leq \min(m, n)$	Rank bound	Rank cannot exceed rows or columns

IIVisualization 1 — The Four Cases

Two yes/no questions — is $r = n$ ? is $r = m$ ? — sort every $m \times n$ matrix into one of four cases.

Rows m 3 Cols n 3 Rank r 3

classification

the two gaps

💡 Challenge: find the two settings where every $b$ is solvable — one giving a unique solution, one giving infinitely many.

IIIVisualization 2 — The Shape of R

The shape of $R$ — pivots, free columns, zero rows — is exactly the two gaps made visible.

Case

reduced form R

pivot free column zero row

the two gaps

IVVisualization 3 — Why Never Exactly Two

If $p$ and $q$ both solve $Ax = b$ , so does $p + t(q - p)$ for every real $t$ — how many solutions is that?

Parameter t 0

💡 Because $t$ ranges over all real numbers, a second solution instantly forces infinitely many — so a linear system has $0$ , $1$ , or $\infty$ solutions, never exactly two.

VQuiz Questions

Question 1

A $5 \times 3$ matrix $A$ has rank $r = 3$ . What are the two gaps and which case applies?

✅ Correct! With $r = n = 3$ , no free variables (unique if solvable). With $m - r = 2$ , there are conditions on $b$ .

❌ Not quite. Compute both gaps: $n - r$ and $m - r$ . Here $m = 5$ , $n = 3$ , $r = 3$ .

Show solution

Solution:

Given: $m = 5$ rows, $n = 3$ columns, $r = 3$ .

Uniqueness gap: $n - r = 3 - 3 = 0$ → no free variables → unique if solvable ✓
Existence gap: $m - r = 5 - 3 = 2$ → two zero rows → conditions on $b$ ⚠️

Since $r = n$ and $r < m$ , this is Case 2 (zero or one solution).

The matrix is tall-and-thin with full column rank: more equations than unknowns, so not every $b$ is reachable, but when a solution exists it's unique.

Question 2

True or False: If a $4 \times 4$ matrix $A$ has rank 3, then $Ax = b$ has infinitely many solutions for every $b$ .

✅ Correct! Since $m - r = 1$ , there's a condition on $b$ . Some $b$ 's give $\infty$ solutions, others give 0.

❌ Not quite. The statement says 'for every $b$ '. Check both gaps — does $m - r$ guarantee every $b$ works?

Show solution

Solution: False.

With $m = 4$ , $n = 4$ , $r = 3$ :

$n - r = 4 - 3 = 1$ → one free variable → not unique ✓
$m - r = 4 - 3 = 1$ → one zero row → condition on $b$ ⚠️

This is Case 4 ( $r < m$ and $r < n$ ): zero or infinitely many.

When solutions exist, the free variable gives infinitely many. But the zero row means some $b$ vectors cause a contradiction ( $0 = \text{nonzero}$ ), giving no solution. It is not true for every $b$ .

Question 3

After row reducing the augmented matrix $[A \mid b]$ , you obtain:

$\left[\begin{array}{cccc|c} 1 & 0 & -2 & 1 & 3 \\ 0 & 1 & 4 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$

How many solutions does this system have?

✅ Correct! The zero row gives $0 = 0$ (no contradiction), and 2 free variables produce a 2-parameter family of solutions.

❌ Not quite. Check: does the zero row create a contradiction? And count the free (non-pivot) columns.

Show solution

Solution: Infinitely many solutions.

From the augmented matrix:

$A$ is $3 \times 4$ ( $m = 3$ , $n = 4$ ), with $r = 2$ pivots (columns 1 and 2)
The last row reads $0 = 0$ → no contradiction → system is solvable ✓
$n - r = 4 - 2 = 2$ → 2 free variables ( $x_3$ and $x_4$ )

Since the system is solvable with 2 free variables, there are infinitely many solutions:

$x = \begin{bmatrix} 3 \\ -1 \\ 0 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} 2 \\ -4 \\ 1 \\ 0 \end{bmatrix} + x_4\begin{bmatrix} -1 \\ -1 \\ 0 \\ 1 \end{bmatrix}$

Question 4

Which statement correctly describes the roles of the two gaps $n - r$ and $m - r$ ?

✅ Correct! Free variables ( $n - r$ ) determine uniqueness; zero rows ( $m - r$ ) determine existence. Two independent levers!

❌ Not quite. Think: free variables let you choose different solutions (uniqueness), while zero rows restrict which $b$ 's work (existence).

Show solution

Solution: $n - r$ controls uniqueness, $m - r$ controls existence.

The two gaps act as independent levers:

$n - r$ = number of free variables (non-pivot columns)

$n - r = 0$ : no free variables → unique (when solvable)
$n - r > 0$ : free variables → infinitely many (when solvable)

$m - r$ = number of zero rows in $R$

$m - r = 0$ : no conditions → always solvable
$m - r > 0$ : conditions on $b$ → may be unsolvable

Remember: columns ( $n - r$ ) → uniqueness, rows ( $m - r$ ) → existence.

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