The Determinant: A Scalar Invariant of Square Matrices

The determinant of a square matrix is the unique scalar-valued function characterized by alternation and multilinearity. We construct it via the Leibniz formula, prove the product formula det(AB) = det(A)det(B), and derive Cramer's rule for solving linear systems.

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March 1, 2026

1. Permutations and Their Signs

Definition 1 (Permutation).

A bijection

σ : {1, \dots, n} \to {1, \dots, n}

is called a permutation of degree

n

. The set of all such permutations forms a group under composition, denoted

S_{n}

, with

∣ S_{n} ∣ = n!

Definition 2 (Transpositions and the sign of a permutation).

A permutation that interchanges exactly two elements and fixes all others is called a transposition. Every permutation can be written as a product of transpositions. If

σ

decomposes into an even number of transpositions, we set

sgn (σ) = + 1

; if an odd number,

sgn (σ) = - 1

. The value

sgn (σ)

is called the sign (or signature) of

σ

Theorem 3.

For any permutations

σ, τ \in S_{n}

sgn (σ τ) = sgn (σ) \cdot sgn (τ) .

Proof.

Write

σ

as a product of

p

transpositions and

τ

as a product of

q

transpositions. Then

σ τ

is a product of

p + q

transpositions, so

sgn (σ τ) = (- 1)^{p + q} = (- 1)^{p} (- 1)^{q} = sgn (σ) \cdot sgn (τ)

. □

2. The Definition of the Determinant

Definition 4 (Determinant).

Let

A = (a_{ij})

be an

n \times n

matrix. The determinant of

A

is the scalar

det A = σ \in S_{n} \sum sgn (σ) i = 1 \prod n a_{i, σ (i)} .

Example 5.

For

n = 2

det (a c b d) = a d - b c

For

n = 3

(Sarrus' rule):

det a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33} = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{13} a_{22} a_{31} - a_{12} a_{21} a_{33} - a_{11} a_{23} a_{32}

3. Alternating and Multilinear Properties of the Determinant

Theorem 6 (Fundamental properties of the determinant).

The determinant satisfies the following properties with respect to its rows (and, by symmetry, its columns):

Multilinearity. The determinant is linear in each row separately.
Alternation. Interchanging two rows reverses the sign of the determinant.
Normalization. $det I_{n} = 1$ .

Proof.

(1) Multilinearity: Suppose the

i

-th row of

A

has the form

a_{i} = c a_{i}^{'} + d a_{i}^{''}

. In the formula

det A = \sum_{σ} sgn (σ) \prod_{k} a_{k, σ (k)}

, only the factor coming from the

i

-th row is affected, so by linearity of products we obtain

det A = c det A^{'} + d det A^{''}

, where

A^{'}

and

A^{''}

are the matrices obtained by replacing the

i

-th row with

a_{i}^{'}

and

a_{i}^{''}

respectively.

(2) Alternation: Let

A^{'}

be the matrix obtained from

A

by swapping rows

i

and

j

. The substitution

σ \mapsto σ \circ (ij)

in the defining sum introduces a factor of

sgn ((ij)) = - 1

, giving

det A^{'} = - det A

(3) Normalization: The

(i, σ (i))

entry of

I_{n}

1

σ = id

and otherwise at least one factor in the product is

0

. Hence

det I_{n} = sgn (id) \cdot 1 = 1

. □

Theorem 7.

The following consequences hold:

If two rows of $A$ are equal, then $det A = 0$ .
If some row of $A$ is the zero vector, then $det A = 0$ .
Adding a scalar multiple of one row to another leaves the determinant unchanged.
Scaling a single row by $c$ multiplies the determinant by $c$ .
$det A^{T} = det A$ .

Proof.

(1) If rows

i

and

j

are equal, swapping them leaves the matrix unchanged, so

det A = - det A

by alternation, whence

det A = 0

(2) Set

c = 0

in the multilinearity property.

(3) The operation

R_{i} \to R_{i} + c R_{j}

yields, by multilinearity,

det A + c det A^{'}

, where

A^{'}

has rows

i

and

j

equal; by (1),

det A^{'} = 0

(4) This is immediate from multilinearity.

(5) In the defining formula, replacing

σ

σ^{- 1}

and observing that

sgn (σ^{- 1}) = sgn (σ)

yields

det A^{T} = det A

. □

4. Cofactor Expansion

Definition 8 (Cofactors).

Let

M_{ij}

denote the determinant of the

(n - 1) \times (n - 1)

submatrix obtained by deleting the

i

-th row and

j

-th column of

A

. The quantity

\tilde{a}_{ij} = (- 1)^{i + j} M_{ij}

is called the

(i, j)

cofactor of

A

Theorem 9 (Cofactor expansion).

For any fixed row index

i

(expansion along the

i

-th row):

det A = j = 1 \sum n a_{ij} \tilde{a}_{ij} .

For any fixed column index

j

(expansion along the

j

-th column):

det A = i = 1 \sum n a_{ij} \tilde{a}_{ij} .

Proof.

We prove the row expansion. Group the terms of

det A = \sum_{σ \in S_{n}} sgn (σ) \prod_{k = 1}^{n} a_{k, σ (k)}

according to the value

σ (i) = j

. For each such

j

, the remaining assignment

{1, \dots, n} ∖ {i} \to {1, \dots, n} ∖ {j}

is an

(n - 1)

-permutation

σ^{'}

, and one checks by counting transpositions that

sgn (σ) = (- 1)^{i + j} sgn (σ^{'})

. Therefore

det A = j = 1 \sum n a_{ij} (- 1)^{i + j} σ^{'} \sum sgn (σ^{'}) k \neq = i \prod a_{k, σ^{'} (k)} = j = 1 \sum n a_{ij} \tilde{a}_{ij} .

The column expansion follows by combining

det A = det A^{T}

with the row expansion. □

Example 10.

Let

A = 201 1 - 1 0 321

. Expanding along the first row gives

det A = 2 \cdot det (- 1 0 21) - 1 \cdot det (0121) + 3 \cdot det (01 - 1 0)

= 2 (- 1) - 1 (- 2) + 3 (1) = - 2 + 2 + 3 = 3.

5. The Product Formula

Theorem 11 (Multiplicativity of the determinant).

For any

n \times n

matrices

A

and

B

det (A B) = det A \cdot det B .

Proof.

det A = 0

, then

A

is singular, so

rank (A B) \leq rank A < n

, hence

A B

is also singular, and both sides equal zero.

det A \neq = 0

, then

A

is invertible and can be written as a product of elementary matrices:

A = E_{1} E_{2} \dots E_{k}

. One verifies from the basic properties that

det (E_{i} B) = det E_{i} \cdot det B

for each elementary matrix

E_{i}

. The general result follows by induction. □

Theorem 12.

A square matrix

A

is invertible if and only if

det A \neq = 0

. In that case,

det (A^{- 1}) = (det A)^{- 1}

Proof.

A

is invertible, then

A A^{- 1} = I

, so the product formula gives

det A \cdot det (A^{- 1}) = 1

. It follows that

det A \neq = 0

and

det (A^{- 1}) = (det A)^{- 1}

Conversely, if

det A \neq = 0

but

A

were singular, then

rank A < n

and row reduction would produce a zero row, forcing

det A = 0

— a contradiction. □

6. Cramer's Rule

Theorem 13 (Cramer's rule).

Let

A

be an invertible

n \times n

matrix. The unique solution of

A x = b

is given by

x_{i} = \frac{det A _{i}}{det A} (i = 1, \dots, n),

where

A_{i}

is the matrix formed by replacing the

i

-th column of

A

with

b

Proof.

Since

A

is invertible, the solution is

x = A^{- 1} b

. Applying the formula

A^{- 1} = \frac{1}{d e t A} A^{T}

(where

A^{T}

is the transpose of the cofactor matrix), we obtain

x_{i} = \frac{1}{det A} j = 1 \sum n \tilde{a}_{ji} b_{j} = \frac{det A _{i}}{det A},

the last equality holding because

\sum_{j = 1}^{n} b_{j} \tilde{a}_{ji}

is precisely the cofactor expansion of

det A_{i}

along its

i

-th column. □

Theorem 14 (The adjugate formula for the inverse).

For an invertible matrix

A

A^{- 1} = \frac{1}{det A} A^{T},

where

A^{T}

denotes the transpose of the cofactor matrix (the adjugate of

A

Proof.

Let

A^{T} = (\tilde{a}_{ji})

. The

(i, k)

entry of

A A^{T}

(A A^{T})_{ik} = j = 1 \sum n a_{ij} \tilde{a}_{kj} .

When

i = k

, this is the cofactor expansion of

det A

along the

i

-th row, so it equals

det A

. When

i \neq = k

, it is the determinant of the matrix obtained by replacing the

k

-th row of

A

with a copy of the

i

-th row; since two rows are equal, this determinant vanishes. Therefore

A A^{T} = (det A) I

. Since

det A \neq = 0

, dividing both sides by

det A

gives

A \cdot \frac{1}{d e t A} A^{T} = I

, and one checks

\frac{1}{d e t A} A^{T} \cdot A = I

in the same way. □

Linear Algebra Algebra Textbook Determinant Cofactor Expansion Cramer's Rule

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Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

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The Determinant: A Scalar Invariant of Square Matrices

Folio Official

March 1, 2026

1. Permutations and Their Signs

Definition 1 (Permutation).

A bijection

σ : {1, \dots, n} \to {1, \dots, n}

is called a permutation of degree

n

. The set of all such permutations forms a group under composition, denoted

S_{n}

, with

∣ S_{n} ∣ = n!

Definition 2 (Transpositions and the sign of a permutation).

A permutation that interchanges exactly two elements and fixes all others is called a transposition. Every permutation can be written as a product of transpositions. If

σ

decomposes into an even number of transpositions, we set

sgn (σ) = + 1

; if an odd number,

sgn (σ) = - 1

. The value

sgn (σ)

is called the sign (or signature) of

σ

Theorem 3.

For any permutations

σ, τ \in S_{n}

sgn (σ τ) = sgn (σ) \cdot sgn (τ) .

Proof.

Write

σ

as a product of

p

transpositions and

τ

as a product of

q

transpositions. Then

σ τ

is a product of

p + q

transpositions, so

sgn (σ τ) = (- 1)^{p + q} = (- 1)^{p} (- 1)^{q} = sgn (σ) \cdot sgn (τ)

. □

2. The Definition of the Determinant

Definition 4 (Determinant).

Let

A = (a_{ij})

be an

n \times n

matrix. The determinant of

A

is the scalar

det A = σ \in S_{n} \sum sgn (σ) i = 1 \prod n a_{i, σ (i)} .

Example 5.

For

n = 2

det (a c b d) = a d - b c

For

n = 3

(Sarrus' rule):

det a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33} = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{13} a_{22} a_{31} - a_{12} a_{21} a_{33} - a_{11} a_{23} a_{32}

3. Alternating and Multilinear Properties of the Determinant

Theorem 6 (Fundamental properties of the determinant).

The determinant satisfies the following properties with respect to its rows (and, by symmetry, its columns):

Multilinearity. The determinant is linear in each row separately.
Alternation. Interchanging two rows reverses the sign of the determinant.
Normalization. $det I_{n} = 1$ .

Proof.

(1) Multilinearity: Suppose the

i

-th row of

A

has the form

a_{i} = c a_{i}^{'} + d a_{i}^{''}

. In the formula

det A = \sum_{σ} sgn (σ) \prod_{k} a_{k, σ (k)}

, only the factor coming from the

i

-th row is affected, so by linearity of products we obtain

det A = c det A^{'} + d det A^{''}

, where

A^{'}

and

A^{''}

are the matrices obtained by replacing the

i

-th row with

a_{i}^{'}

and

a_{i}^{''}

respectively.

(2) Alternation: Let

A^{'}

be the matrix obtained from

A

by swapping rows

i

and

j

. The substitution

σ \mapsto σ \circ (ij)

in the defining sum introduces a factor of

sgn ((ij)) = - 1

, giving

det A^{'} = - det A

(3) Normalization: The

(i, σ (i))

entry of

I_{n}

1

σ = id

and otherwise at least one factor in the product is

0

. Hence

det I_{n} = sgn (id) \cdot 1 = 1

. □

Theorem 7.

The following consequences hold:

If two rows of $A$ are equal, then $det A = 0$ .
If some row of $A$ is the zero vector, then $det A = 0$ .
Adding a scalar multiple of one row to another leaves the determinant unchanged.
Scaling a single row by $c$ multiplies the determinant by $c$ .
$det A^{T} = det A$ .

Proof.

(1) If rows

i

and

j

are equal, swapping them leaves the matrix unchanged, so

det A = - det A

by alternation, whence

det A = 0

(2) Set

c = 0

in the multilinearity property.

(3) The operation

R_{i} \to R_{i} + c R_{j}

yields, by multilinearity,

det A + c det A^{'}

, where

A^{'}

has rows

i

and

j

equal; by (1),

det A^{'} = 0

(4) This is immediate from multilinearity.

(5) In the defining formula, replacing

σ

σ^{- 1}

and observing that

sgn (σ^{- 1}) = sgn (σ)

yields

det A^{T} = det A

. □

4. Cofactor Expansion

Definition 8 (Cofactors).

Let

M_{ij}

denote the determinant of the

(n - 1) \times (n - 1)

submatrix obtained by deleting the

i

-th row and

j

-th column of

A

. The quantity

\tilde{a}_{ij} = (- 1)^{i + j} M_{ij}

is called the

(i, j)

cofactor of

A

Theorem 9 (Cofactor expansion).

For any fixed row index

i

(expansion along the

i

-th row):

det A = j = 1 \sum n a_{ij} \tilde{a}_{ij} .

For any fixed column index

j

(expansion along the

j

-th column):

det A = i = 1 \sum n a_{ij} \tilde{a}_{ij} .

Proof.

We prove the row expansion. Group the terms of

det A = \sum_{σ \in S_{n}} sgn (σ) \prod_{k = 1}^{n} a_{k, σ (k)}

according to the value

σ (i) = j

. For each such

j

, the remaining assignment

{1, \dots, n} ∖ {i} \to {1, \dots, n} ∖ {j}

is an

(n - 1)

-permutation

σ^{'}

, and one checks by counting transpositions that

sgn (σ) = (- 1)^{i + j} sgn (σ^{'})

. Therefore

det A = j = 1 \sum n a_{ij} (- 1)^{i + j} σ^{'} \sum sgn (σ^{'}) k \neq = i \prod a_{k, σ^{'} (k)} = j = 1 \sum n a_{ij} \tilde{a}_{ij} .

The column expansion follows by combining

det A = det A^{T}

with the row expansion. □

Example 10.

Let

A = 201 1 - 1 0 321

. Expanding along the first row gives

det A = 2 \cdot det (- 1 0 21) - 1 \cdot det (0121) + 3 \cdot det (01 - 1 0)

= 2 (- 1) - 1 (- 2) + 3 (1) = - 2 + 2 + 3 = 3.

5. The Product Formula

Theorem 11 (Multiplicativity of the determinant).

For any

n \times n

matrices

A

and

B

det (A B) = det A \cdot det B .

Proof.

det A = 0

, then

A

is singular, so

rank (A B) \leq rank A < n

, hence

A B

is also singular, and both sides equal zero.

det A \neq = 0

, then

A

is invertible and can be written as a product of elementary matrices:

A = E_{1} E_{2} \dots E_{k}

. One verifies from the basic properties that

det (E_{i} B) = det E_{i} \cdot det B

for each elementary matrix

E_{i}

. The general result follows by induction. □

Theorem 12.

A square matrix

A

is invertible if and only if

det A \neq = 0

. In that case,

det (A^{- 1}) = (det A)^{- 1}

Proof.

A

is invertible, then

A A^{- 1} = I

, so the product formula gives

det A \cdot det (A^{- 1}) = 1

. It follows that

det A \neq = 0

and

det (A^{- 1}) = (det A)^{- 1}

Conversely, if

det A \neq = 0

but

A

were singular, then

rank A < n

and row reduction would produce a zero row, forcing

det A = 0

— a contradiction. □

6. Cramer's Rule

Theorem 13 (Cramer's rule).

Let

A

be an invertible

n \times n

matrix. The unique solution of

A x = b

is given by

x_{i} = \frac{det A _{i}}{det A} (i = 1, \dots, n),

where

A_{i}

is the matrix formed by replacing the

i

-th column of

A

with

b

Proof.

Since

A

is invertible, the solution is

x = A^{- 1} b

. Applying the formula

A^{- 1} = \frac{1}{d e t A} A^{T}

(where

A^{T}

is the transpose of the cofactor matrix), we obtain

x_{i} = \frac{1}{det A} j = 1 \sum n \tilde{a}_{ji} b_{j} = \frac{det A _{i}}{det A},

the last equality holding because

\sum_{j = 1}^{n} b_{j} \tilde{a}_{ji}

is precisely the cofactor expansion of

det A_{i}

along its

i

-th column. □

Theorem 14 (The adjugate formula for the inverse).

For an invertible matrix

A

A^{- 1} = \frac{1}{det A} A^{T},

where

A^{T}

denotes the transpose of the cofactor matrix (the adjugate of

A

Proof.

Let

A^{T} = (\tilde{a}_{ji})

. The

(i, k)

entry of

A A^{T}

(A A^{T})_{ik} = j = 1 \sum n a_{ij} \tilde{a}_{kj} .

When

i = k

, this is the cofactor expansion of

det A

along the

i

-th row, so it equals

det A

. When

i \neq = k

, it is the determinant of the matrix obtained by replacing the

k

-th row of

A

with a copy of the

i

-th row; since two rows are equal, this determinant vanishes. Therefore

A A^{T} = (det A) I

. Since

det A \neq = 0

, dividing both sides by

det A

gives

A \cdot \frac{1}{d e t A} A^{T} = I

, and one checks

\frac{1}{d e t A} A^{T} \cdot A = I

in the same way. □

Linear Algebra Algebra Textbook Determinant Cofactor Expansion Cramer's Rule

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

The Determinant: A Scalar Invariant of Square Matrices

1. Permutations and Their Signs

2. The Definition of the Determinant

3. Alternating and Multilinear Properties of the Determinant

4. Cofactor Expansion

5. The Product Formula

6. Cramer's Rule

Share your expertise with the world

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The Determinant: A Scalar Invariant of Square Matrices

1. Permutations and Their Signs

2. The Definition of the Determinant

3. Alternating and Multilinear Properties of the Determinant

4. Cofactor Expansion

5. The Product Formula

6. Cramer's Rule

Share your expertise with the world

More from Folio Official

Matrices and Representation of Linear Maps

Jordan Normal Form: Beyond Diagonalization

The Singular Value Decomposition: Structure of Arbitrary Matrices

Diagonalization: Simplifying Matrices by Choice of Basis