The Determinant: A Scalar Invariant of Square Matrices

Beginning with permutations and their signs, we define the determinant via the Leibniz formula and establish its fundamental properties: multilinearity, alternation, and normalization. We then develop cofactor expansion, the product formula det(AB) = det(A)det(B), the adjugate matrix, and Cramer's rule.

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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 Fundamental Properties

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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1 フォロワー·105 記事

The Determinant: A Scalar Invariant of Square Matrices

1 Permutations and Their Signs

2 The Definition of the Determinant

3 Fundamental Properties

4 Cofactor Expansion

5 The Product Formula

6 Cramer's Rule

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