Elementary operations, echelon form and rank

Row-reducing a matrix to echelon form, reading the rank off the staircase, and the minor method as a cross-check.

Module contents

The rank of a matrix measures how much genuinely independent information it carries. It is the number that decides whether a system of equations has one solution, many, or none.

Elementary operations

Three moves are permitted on rows (and the same three on columns):

Operation	Notation
Interchange two rows	$R_{i} \leftrightarrow R_{j}$
Multiply a row by a non-zero scalar	$R_{i} \to k R_{i}, k \neq = 0$
Add a multiple of one row to another	$R_{i} \to R_{i} + k R_{j}$

Matrices $A$ and $B$ are equivalent, written $A \sim B$ , if one can be obtained from the other by such operations. Elementary operations never change the rank. That is the entire basis of the method.

Echelon form

A matrix is in row echelon form when:

All zero rows sit at the bottom.
In each non-zero row, the first non-zero entry (the pivot or leading entry) lies strictly to the right of the pivot in the row above.
All entries below a pivot are zero.

The pivots form a descending staircase:

$20003500 - 1 270 4130$

Reduced row echelon form (RREF) goes further: every pivot is $1$ , and every pivot is the only non-zero entry in its column. Gauss–Jordan elimination produces this.

Rank

$ρ (A) = the number of non-zero rows in the echelon form of A$

Equivalently, the number of pivots.

Written $ρ (A)$ or $r (A)$ or $rank (A)$ .

Facts about rank

$ρ (A) \leq min (m, n)$ for a matrix of order $m \times n$ .
$ρ (A) = 0$ if and only if $A = O$ .
$ρ (A^{T}) = ρ (A)$ — row rank equals column rank, always.
For a square matrix of order $n$ : $ρ (A) = n ⟺ ∣ A ∣ \neq = 0 ⟺ A$ is non-singular. Such a matrix is said to have full rank.
$ρ (A) < n$ for a square singular matrix.
$ρ (A B) \leq min (ρ (A), ρ (B))$ .

Method 1 — Echelon form (preferred)

Get a non-zero entry into the top-left position, swapping rows if necessary.
Use it to clear everything below it in that column.
Move down one row and right one column; repeat.
Count the non-zero rows.

Work with integers where possible. Adding multiples of rows is nearly always cheaper than scaling to make a leading $1$ .

Method 2 — Minors

$ρ (A) = r$ means:

there exists at least one non-zero minor of order $r$ , and
every minor of order $r + 1$ (and higher) is zero.

Start from the largest possible order and work down until you find a non-zero minor.

This is quick for small matrices and for confirming an answer, but the number of minors explodes with size, so echelon form is the practical method.

Normal (canonical) form

Using both row and column operations, any matrix of rank $r$ can be reduced to

$[I_{r} O O O]$

called the normal form of $A$ . The order of the identity block is the rank. This is sometimes asked directly.

Worked examples

3 solved

Every step is shown, in the order you would write it in an answer book.

Example 1

Rank by reduction to echelon form

Question

Find the rank of $A = 1232463710$ .

Use the pivot $a_{11} = 1$ to clear column 1
$R_{2} \to R_{2} - 2 R_{1}$ : $(2 - 2, 4 - 4, 7 - 6) = (0, 0, 1)$

$R_{3} \to R_{3} - 3 R_{1}$ : $(3 - 3, 6 - 6, 10 - 9) = (0, 0, 1)$

$A \sim 100200311$
Clear below the next pivot
The next pivot is the $1$ in row 2, column 3. Apply $R_{3} \to R_{3} - R_{2}$ :

$A \sim 100200310$

This is echelon form: pivots at $(1, 1)$ and $(2, 3)$ , staircase descending, zero row at the bottom.
Count non-zero rows
Two non-zero rows, so $ρ (A) = 2$ .
Cross-check with minors
Rank $2$ predicts $∣ A ∣ = 0$ and some non-zero $2 \times 2$ minor.

$∣ A ∣ = 1 (40 - 42) - 2 (20 - 21) + 3 (12 - 12) = - 2 + 2 + 0 = 0 ✓$

$1237 = 7 - 6 = 1 \neq = 0 ✓$

Both conditions hold, confirming $ρ (A) = 2$ .

Answer

$ρ (A) = 2$

NoteColumn 2 is exactly 2 × column 1 — that dependence is what costs the matrix a unit of rank.

Example 2

Rank of a rectangular matrix

Question

Find the rank of $A = 123246 - 1 13 3 - 2 - 7$ .

Note the ceiling
$A$ is $3 \times 4$ , so $ρ (A) \leq min (3, 4) = 3$ .
Clear column 1
$R_{2} \to R_{2} - 2 R_{1}$ : $(0, 0, 1 - (- 2), - 2 - 6) = (0, 0, 3, - 8)$

$R_{3} \to R_{3} - 3 R_{1}$ : $(0, 0, 3 - (- 3), - 7 - 9) = (0, 0, 6, - 16)$

$A \sim 100200 - 1 36 3 - 8 - 16$
Clear below the second pivot
Pivot is the $3$ at position $(2, 3)$ . Apply $R_{3} \to R_{3} - 2 R_{2}$ :

$(0, 0, 6 - 6, - 16 + 16) = (0, 0, 0, 0)$

$A \sim 100200 - 1 30 3 - 8 0$
Count
Two non-zero rows: $ρ (A) = 2$ .

Answer

$ρ (A) = 2$

Example 3

Rank depending on a parameter

Question

Find the values of $k$ for which $A = 123 2 k 6 369$ has rank $1$ , rank $2$ , and rank $3$ .

Test for rank 3 first
Rank $3$ requires $∣ A ∣ \neq = 0$ . But row 3 $= 3 \times$ row 1 regardless of $k$ , so by the proportional-rows property

$∣ A ∣ = 0 for every k$

Rank $3$ is impossible.
Reduce
$R_{2} \to R_{2} - 2 R_{1}$ and $R_{3} \to R_{3} - 3 R_{1}$ :

$A \sim 100 2 k - 4 0 300$
Split on the pivot in row 2
If $k - 4 \neq = 0$ , i.e. $k \neq = 4$ : two non-zero rows, so $ρ (A) = 2$ .

If $k = 4$ : row 2 vanishes, leaving one non-zero row, so $ρ (A) = 1$ .

Answer

$ρ (A) = 1$ when $k = 4$ ; $ρ (A) = 2$ for every $k \neq = 4$ ; rank $3$ never occurs.

NoteWith a parameter, always reduce first and split into cases only at the point where a pivot could vanish.

Practice problems

6 with solutions

Work each one on paper first. The full solution — not just the answer — is one click away.

Problem 1Basic

Find the rank of $A = [1236]$ by both methods.

Show solution

Echelon. $R_{2} \to R_{2} - 2 R_{1}$ gives $[1030]$ — one non-zero row.

$ρ (A) = 1$

Minors. The only order- $2$ minor is $∣ A ∣ = 1 (6) - 3 (2) = 0$ . There is a non-zero order- $1$ minor (for instance $∣1∣ = 1$ ).

Largest non-zero minor has order $1$ , so $ρ (A) = 1$ ✓

Answer

ρ (A) = 1

Problem 2Basic

Reduce $A = 021142310$ to echelon form and state its rank.

Show solution

The $(1, 1)$ entry is $0$ , so swap first: $R_{1} \leftrightarrow R_{3}$ .

$120241013$

$R_{2} \to R_{2} - 2 R_{1}$ :

$100201013$

The pivot pattern is out of order, so swap $R_{2} \leftrightarrow R_{3}$ :

$100210031$

This is echelon form with three pivots.

$ρ (A) = 3$

Being $3 \times 3$ of full rank, $A$ is non-singular. (Indeed the echelon form is upper triangular with determinant $1 \cdot 1 \cdot 1 = 1$ , and the two swaps make $∣ A ∣ = (- 1)^{2} \cdot 1 = 1$ .)

Answer

ρ (A) = 3

(full rank, non-singular)

Problem 3Exam level

Find the rank of $A = 12312461369148121$ .

Show solution

Rows $2$ and $3$ are $2 \times$ and $3 \times$ row $1$ , so they will vanish immediately.

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 3 R_{1}$ , $R_{4} \to R_{4} - R_{1}$ :

$A \sim 1000 200 - 1 300 - 2 400 - 3$

Move the non-zero row up: $R_{2} \leftrightarrow R_{4}$ .

$A \sim 1000 2 - 1 00 3 - 2 00 4 - 3 00$

Echelon form with two non-zero rows.

$ρ (A) = 2$

Answer

ρ (A) = 2

Problem 4Exam level

Find the rank of $A = 2022 - 1 335 347114156$ .

Show solution

Pivot $a_{11} = 2$ . Apply $R_{3} \to R_{3} - R_{1}$ and $R_{4} \to R_{4} - R_{1}$ :

$R_{3}$ : $(0, 3 - (- 1), 7 - 3, 5 - 4) = (0, 4, 4, 1)$
$R_{4}$ : $(0, 5 - (- 1), 11 - 3, 6 - 4) = (0, 6, 8, 2)$

$A \sim 2000 - 1 346 34484112$

Pivot is now the $3$ at $(2, 2)$ . Clear below: $R_{3} \to 3 R_{3} - 4 R_{2}$ , $R_{4} \to R_{4} - 2 R_{2}$ .

$R_{3}$ : $(0, 12 - 12, 12 - 16, 3 - 4) = (0, 0, - 4, - 1)$
$R_{4}$ : $(0, 6 - 6, 8 - 8, 2 - 2) = (0, 0, 0, 0)$

$A \sim 2000 - 1 300 34 - 4 0 41 - 1 0$

Echelon form with three non-zero rows.

$ρ (A) = 3$

(Scaling $R_{3}$ by $3$ is legitimate — multiplying a row by a non-zero scalar preserves rank. It keeps the arithmetic in integers.)

Answer

ρ (A) = 3

Problem 5Exam level

For what value of $λ$ does $A = 30 - 3 15 λ 48 - 4$ have rank less than $3$ ?

Show solution

Rank $< 3$ for a $3 \times 3$ matrix means $∣ A ∣ = 0$ .

Expand along column 1:

$∣ A ∣ = 3 5 λ 8 - 4 - 0 + (- 3) 1548$

$= 3 (- 20 - 8 λ) - 3 (8 - 20)$

$= - 60 - 24 λ + 36 = - 24 - 24 λ$

Setting $∣ A ∣ = 0$ :

$- 24 - 24 λ = 0 \Rightarrow λ = - 1$

Confirm the rank at $λ = - 1$ . Apply $R_{3} \to R_{3} + R_{1}$ :

$300150480$

Two non-zero rows, so $ρ (A) = 2$ when $λ = - 1$ , and $ρ (A) = 3$ otherwise.

Answer

λ = - 1

, giving

ρ (A) = 2

Problem 6Challenge

Reduce $A = 110 12 - 1 23 - 1$ to normal form $[I_{r} O O O]$ and state its rank.

Show solution

Row operations first. $R_{2} \to R_{2} - R_{1}$ :

$100 11 - 1 21 - 1$

$R_{3} \to R_{3} + R_{2}$ :

$100110210$

$R_{1} \to R_{1} - R_{2}$ :

$100010110$

Now column operations. $C_{3} \to C_{3} - C_{1} - C_{2}$ :

Column 3 becomes $(1 - 1 - 0, 1 - 0 - 1, 0) = (0, 0, 0)$ .

$100010000 = [I_{2} O O O]$

The identity block has order $2$ , so

$ρ (A) = 2$

AnswerNormal form

[I_{2} O O O]

, so

ρ (A) = 2