Mathematics · Complete guide

Rank of a matrix, all three ways

Rank is one number that tells you how much genuinely new information a matrix carries. This guide builds the idea from a picture, then works the three methods exams expect (minors, echelon form and normal form) on real-looking questions, and ends with the part most students skip: what rank says about a system of equations. The lab lets you type any matrix and replay every step.

Start here

What rank actually measures

Think of each row of a matrix as a piece of information. Some rows add something new; others are just copies or combinations of rows you already have. The rank is the number of rows that genuinely add something new: the number of linearly independent rows. Surprisingly, counting columns instead always gives the same number.

In $[1224]$ the second row is just twice the first. It tells you nothing new, so the rank is 1, not 2. Picture each row as an arrow from the origin and move the sliders below.

A = [3112]

Rank 2

Two directions: rank 2

The rows point in genuinely different directions, so together they span a real area. That area is |det A| = |3×2 − 1×1| = 5. A non-zero determinant means neither row can be built from the other.

Key idea

Two arrows that point in different directions span an area, and that area is the determinant. When one row is a multiple of the other, the area collapses to a line, the determinant is 0 and the rank drops. Every method in this guide is a systematic way of detecting that collapse in bigger matrices.

The definition exams use

ρ (A) = r ⟺ some r \times r minor of A \neq = 0 and every (r + 1) \times (r + 1) minor = 0

ρ(A) and r(A) are both common notation for the rank of A.

Toolkit

Facts and moves to know first

Facts

For an $m \times n$ matrix, $0 \leq ρ (A) \leq min (m, n)$ .
$ρ (A) = 0$ only for the zero matrix. Any non-zero matrix has rank at least 1.
$ρ (A^{T}) = ρ (A)$ : row rank equals column rank.
A square $n \times n$ matrix has rank $n$ exactly when $∣ A ∣ \neq = 0$ (it is non-singular).
$ρ (I_{n}) = n$ , and $ρ (k A) = ρ (A)$ for any $k \neq = 0$ .
$ρ (A B) \leq min {ρ (A), ρ (B)}$ and $ρ (A + B) \leq ρ (A) + ρ (B)$ .
$ρ (A^{T} A) = ρ (A A^{T}) = ρ (A)$ for real matrices.

Elementary operations: they never change the rank

Swap two rows: $R_{i} \leftrightarrow R_{j}$ .
Scale a row by a non-zero number: $R_{i} \to k R_{i}, k \neq = 0$ .
Add a multiple of another row: $R_{i} \to R_{i} + k R_{j}$ .
The same three moves on columns ( $C_{i}$ ) are also allowed.

Matrices linked by these moves are called equivalent, written $A \sim B$ . They have the same rank, which is why we can simplify first and count afterwards. Note that $\sim$ , not $=$ : the matrices themselves are different.

Method 1

Method 1: minors

A minor of order k is the determinant of a $k \times k$ square block you get by keeping any k rows and any k columns. This method follows the definition directly.

Start with the largest possible order, $min (m, n)$ . For a square matrix that is just $∣ A ∣$ .
If any minor of that order is non-zero, stop: that order is the rank.
If every one is zero, drop to the next order down and repeat.

You only need one non-zero minor to fix the rank, but you need all minors of the next order up to be zero, which is why starting at the top saves work.

Worked question · minors

Find the rank of

A = 121243365

Largest order: 3. There is only one 3×3 minor, |A|.
$∣ A ∣ = 1 (4 \cdot 5 - 6 \cdot 3) - 2 (2 \cdot 5 - 6 \cdot 1) + 3 (2 \cdot 3 - 4 \cdot 1) = 2 - 8 + 6 = 0$
So the rank is less than 3. Look for a non-zero 2×2 minor.
Rows 1 and 2 are proportional (row 2 = 2 × row 1), so every minor built from them alone is 0. Take rows 1 and 3, columns 1 and 2:
$1123 = 1 \cdot 3 - 2 \cdot 1 = 1 \neq = 0$
Conclude.
There is a non-zero minor of order 2, and the only minor of order 3 is zero.

Answer

ρ (A) = 2

When to use it

Minors are fastest for 2×2 and 3×3 matrices, and they are the natural method when the matrix contains an unknown like k: set the determinant to zero and solve. For anything bigger, the number of minors explodes (a 4×4 matrix has 16 minors of order 3), so switch to echelon form.

Method 2

Method 2: echelon form

Use row operations to push the matrix into row echelon form, a staircase shape, then simply count the rows that are not all zero.

What echelon form looks like

All-zero rows sit at the bottom.
The first non-zero entry of each row (its pivot) is further right than the pivot in the row above.
Every entry below a pivot is zero.

The routine

Get a non-zero pivot at the top of the first column (swap rows if you need to; a 1 is ideal).
Use it to turn every entry below it into 0 with $R_{i} \to R_{i} - k R_{1}$ .
Move one row down and one column right, and repeat.
Rank = number of non-zero rows (= number of pivots).

* 000 * * 00 * * 00 * * * 0

An echelon form with three pivots (boxed) and one zero row: rank 3. The pivots step to the right, but not always by exactly one column.

Worked question · echelon form

Find the rank of the 3×4 matrix

A = 123246 - 1 13 3 - 2 - 7

Clear column 1 below the pivot 1.
$R_{2} \to R_{2} - 2 R_{1}, R_{3} \to R_{3} - 3 R_{1}$
$A \sim 100200 - 1 36 3 - 8 - 16$
Column 2 has no pivot below row 1, so move on to column 3.
The next pivot is the 3 in row 2. $R_{3} \to R_{3} - 2 R_{2}$
$A \sim 100200 - 1 30 3 - 8 0$
Count the non-zero rows.
Two non-zero rows, one zero row. (Row 3 was secretly R₁ + R₂ all along.)

Answer

ρ (A) = 2

Why it works

Row operations never change the rank, and in echelon form the non-zero rows are obviously independent: each has a pivot in a column where every row below it has a 0. So counting them counts the rank of the original matrix.

Method 3

Method 3: normal form

Here you keep going past echelon form, using row and column operations, until the matrix is as simple as it can possibly be: an identity block in the top-left corner and zeros everywhere else. This is the normal form (or canonical form), and the size of the identity block is the rank.

Every matrix can be reduced to one of these

I_{r}

[I_{r} O]

[I_{r} O]

[I_{r} O O O]

⟹ ρ (A) = r

O stands for a block of zeros of whatever size fills the matrix.

Worked question · normal form

Reduce

A = 110 12 - 1 23 - 1

to normal form and find its rank.

Row operations first.
$R_{2} \to R_{2} - R_{1}$ gives row 2 = (0, 1, 1). Then $R_{3} \to R_{3} + R_{2}$ gives row 3 = (0, 0, 0).
$A \sim 100110210$
Column operations clear row 1.
$C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - 2 C_{1}$
$A \sim 100010010$
Clear row 2.
$C_{3} \to C_{3} - C_{2}$
$A \sim 100010000 = [I_{2} O O O]$

Answer

Normal form

[I_{2} O O O]

, so

ρ (A) = 2

Echelon vs normal form

Echelon form already tells you the rank; normal form just finishes the tidying. Exams ask for it by name (“reduce to normal form”), so know the target shapes. A handy fact: there are always non-singular matrices P and Q with $P A Q$ equal to the normal form. P records the row operations and Q the column operations.

Interactive

The rank lab: any matrix, every step

Type a matrix up to 4×4, or start from one of the worked questions, then step through the working for each method. Boxed entries are pivots; blue entries are the ones the last operation changed. All the arithmetic is exact, with fractions and no rounding.

Start from

Rows

Columns

Whole numbers, decimals or fractions like 3/4.

Rank

Rank-deficient by 1 · max possible 3

Show every step

Start · 1 of 4

121243365

The matrix as given.

Strategy

Which method, when, and exam shortcuts

Method	Operations	Best for	Watch out
Minors	Determinants only	2×2 and 3×3 matrices, and anything with an unknown k	Needs every minor of the next order up to be zero
Echelon form	Row operations	Rectangular or larger matrices, and systems of equations	Count non-zero rows, not non-zero entries
Normal form	Row and column operations	When the question asks for it by name	Column operations are fine for rank, but not when solving equations

Spot it in 5 seconds

Rows or columns that are multiples of each other, a row that is the sum of two others, or a whole row of zeros each drop the rank. A matrix whose rows are all multiples of one row has rank 1.

Outer product

A non-zero column times a non-zero row, $u v^{T}$ , always has rank 1: every row is a multiple of $v^{T}$ . So $[12] [34]$ has rank 1.

Products

$ρ (A B) \leq min {ρ (A), ρ (B)}$ . A 3×2 times a 2×3 is a 3×3 matrix, but its rank is at most 2, so its determinant must be 0. No multiplying needed.

Exam favourite · a parameter

For which values of k does

A = 111 1 k 1 11 k

have rank 3, 2 or 1?

Find the determinant as a function of k.
$∣ A ∣ = 1 (k^{2} - 1) - 1 (k - 1) + 1 (1 - k) = k^{2} - 2 k + 1 = (k - 1)^{2}$
Rank 3 whenever the determinant is non-zero.
That is every k except k = 1.
At k = 1, look again.
Every row becomes (1, 1, 1): all three rows are identical, so only one counts.

Answer

Rank 3 for

k \neq = 1

, rank 1 at

k = 1

, and rank 2 is never possible. A “determinant = 0, so rank 2” reflex gets this wrong.

Why it matters

Rank and systems of equations

Write a system of m equations in n unknowns as $A X = B$ , and form the augmented matrix $[A ∣ B]$ by tacking the right-hand side on as an extra column. Comparing the two ranks tells you everything about the solutions without solving anything (the Rouché–Capelli theorem).

ρ (A) \neq = ρ ([A ∣ B])

No solution

Inconsistent. Row reduction produces a row like 0 = 5.

ρ (A) = ρ ([A ∣ B]) = n

Exactly one solution

Every unknown is pinned down.

ρ (A) = ρ ([A ∣ B]) = r < n

Infinitely many solutions

$n - r$ unknowns are free to take any value.

Homogeneous systems, AX = 0

These are always consistent, since X = 0 works. They have a non-trivial solution exactly when $ρ (A) < n$ ; for a square matrix that means $∣ A ∣ = 0$ . The number of free unknowns, $n - ρ (A)$ , is called the nullity, and $rank + nullity = n$ .

Worked question · consistency

For which k does the system

x + y = 2, x + (k^{2} - 3) y = k

have no solution, infinitely many, or exactly one?

When is A rank-deficient?
$∣ A ∣ = 11 1 k^{2} - 3 = k^{2} - 4$ , which is zero at $k = \pm 2$ . For every other k, $ρ (A) = ρ ([A ∣ B]) = 2 = n$ : exactly one solution.
k = 2.
The second equation becomes $x + y = 2$ , the same as the first. $ρ (A) = ρ ([A ∣ B]) = 1 < 2$ : infinitely many solutions.
k = −2.
Now it is $x + y = - 2$ , contradicting $x + y = 2$ . $[A ∣ B] \sim [1010 2 - 4]$ , so $ρ (A) = 1 \neq = 2 = ρ ([A ∣ B])$ : no solution.

Answer

No solution at

k = - 2

; infinitely many at

k = 2

; a unique solution for every other k.

Practice

Exercises, with answers

0 of 15 correct so far

A. Find the rank

Type a whole number. Use any method; the solution shows the quickest one.

Problem 1 · spot the pattern

A = 132264396

ρ(A)

Problem 2 · the classic 1 to 9

A = 147258369

|A|ρ(A)

Problem 3 · rectangular

A = 112123134145

ρ(A)

Problem 4 · a 4×4

A = 1012011121351234

ρ(A)

B. Unknowns and systems

Problem 5 · make it rank 1

For what value of k does

[23 4 k]

have rank 1?

Problem 6 · consistency

The system

x + y + z = 3, x + 2 y + 3 z = 6, 2 x + 3 y + 4 z = b

has solutions for exactly one value of b. Find it, and the number of free unknowns when it does.

bNumber of free unknowns

C. Concept checks

One attempt each. Read the explanation even when you get it right.

1The largest possible rank of a 3 × 5 matrix is:

2A is a 4 × 4 matrix with |A| = 0. Then:

3Which of these can change the rank of a matrix?

4A is 3 × 2 and B is 2 × 3. The 3 × 3 matrix AB:

5The homogeneous system AX = 0 in n unknowns has a non-trivial solution if and only if:

6For a system in 3 unknowns, ρ(A) = ρ([A | B]) = 2. The system has:

7If u is a non-zero 3 × 1 column and v a non-zero 3 × 1 column, the rank of uvᵀ is:

Revise

One-page revision sheet

Rank in one line

The number of linearly independent rows (equivalently, columns). At most min(m, n). Zero only for the zero matrix.

Minors in one breath

Largest order first. One non-zero minor fixes the rank at that order, provided all larger minors are zero.

Echelon in one breath

Row operations to a staircase. Rank = number of non-zero rows = number of pivots.

Normal form in one breath

Row and column operations to $[I_{r} O O O]$ . The size of the identity block is the rank.

Square matrices

$∣ A ∣ \neq = 0 ⟺ ρ (A) = n ⟺ A^{- 1}$ exists. $∣ A ∣ = 0$ only tells you the rank is below n, not what it is.

Systems

Ranks differ: no solution. Equal to n: one. Equal but below n: infinitely many, with n − r free unknowns. AX = 0 has a non-trivial solution iff ρ(A) < n.

Watch out for

The mistakes examiners are fishing for

Reading |A| = 0 as “rank = n − 1”. It only means the rank is below n; the k-matrix above drops straight from 3 to 1.
Counting non-zero entries instead of non-zero rows in echelon form.
Multiplying a row by 0. That is not an elementary operation, and it can destroy rank.
Looking for a determinant of a rectangular matrix. It has none; use its largest square minors, of order min(m, n).
Using column operations while solving equations. They keep the rank but scramble the unknowns. For systems, stick to row operations.
Demanding every 2×2 minor be non-zero for rank 2. One is enough.

Try it on real questions

Ten past-paper questions on rank

From CUET-PG, IIT JAM, JNU and the University of Hyderabad: rank directly, a rank condition with an unknown, consistency of linear systems, and the rank facts about AᵀA.

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

Practice these →All questions on this topic →More resources →

Rank of a matrix, all three ways

What rank actually measures

Two directions: rank 2

Facts and moves to know first

Method 1: minors

Method 2: echelon form

What echelon form looks like

The routine

Method 3: normal form

The rank lab: any matrix, every step

Which method, when, and exam shortcuts

Rank and systems of equations

Exercises, with answers

A. Find the rank

B. Unknowns and systems

C. Concept checks

One-page revision sheet

Rank in one line

Minors in one breath

Echelon in one breath

Normal form in one breath

Square matrices

Systems

The mistakes examiners are fishing for

Ten past-paper questions on rank

Read next

Rank of a matrix, all three ways

What rank actually measures

Two directions: rank 2

Facts and moves to know first

Method 1: minors

Method 2: echelon form

What echelon form looks like

The routine

Method 3: normal form

The rank lab: any matrix, every step

Which method, when, and exam shortcuts

Rank and systems of equations

Exercises, with answers

A. Find the rank

B. Unknowns and systems

C. Concept checks

One-page revision sheet

Rank in one line

Minors in one breath

Echelon in one breath

Normal form in one breath

Square matrices

Systems

The mistakes examiners are fishing for

Ten past-paper questions on rank

Read next