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Basic Concepts of Determinants

The Definition of a Determinant

The determinant is an important concept in linear algebra. It is not only a numerical characteristic of a matrix, but also has important applications in geometry, physics and other fields.

Second-Order Determinants

For a second-order matrix $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ , its determinant is defined as:

$|A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}$

The absolute value of a second-order determinant is the area of the parallelogram whose two adjacent sides are the given vectors
The sign of the determinant records the relative orientation of the two vectors (the right-hand rule)

Example: Compute $\begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix}$

Solution: $|A| = 2 \times 4 - 3 \times 1 = 8 - 3 = 5$

Third-Order Determinants

For a third-order matrix $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$ , its determinant is defined as:

$|A| = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$

The absolute value of a third-order determinant is the volume of the parallelepiped whose three adjacent sides are the given vectors, and the sign of the determinant records the relative orientation of the three vectors.

Expansion method: expand along the first row $|A| = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$

Example: Compute $\begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{vmatrix}$

Solution: Expand along the first row: $|A| = 1 \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2 \begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3 \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}$ $= 1(1 \times 0 - 4 \times 6) - 2(0 \times 0 - 4 \times 5) + 3(0 \times 6 - 1 \times 5)$ $= 1(0 - 24) - 2(0 - 20) + 3(0 - 5)$ $= -24 + 40 - 15 = 1$

n-th Order Determinants

For an n-th order matrix $A = (a_{ij})_{n \times n}$ , its determinant is defined as:

$|A| = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}$

Symbol notation

Symbol	Type	Pronunciation / Note	Meaning in This Article
$S_n$	Mathematical symbol	S n (symmetric group of degree n)	The set of all permutations of $n$ elements; the symmetric group is an important concept in combinatorics
$\sigma$	Greek letter	sigma	A permutation, that is, a way of rearranging the elements
$\text{sgn}(\sigma)$	Mathematical symbol	sign of sigma (sign function)	The sign function of a permutation, whose value is +1 or -1, used to decide the parity of a permutation
$M_{ij}$	Mathematical symbol	M i j (minor)	The minor of the entry $a_{ij}$

Expansion methods:

Expand along a row: $|A| = \sum_{j=1}^n (-1)^{i+j} a_{ij} M_{ij}$
Expand along a column: $|A| = \sum_{i=1}^n (-1)^{i+j} a_{ij} M_{ij}$

where $M_{ij}$ is the minor of the entry $a_{ij}$ .

Basic Properties of Determinants

Property 1: Invariance under Transposition

Property: $|A| = |A^T|$

Proof: in the definition of the determinant, rows and columns play symmetric roles.

Property 2: Swapping Two Rows (Columns) Changes the Sign

Property: interchanging two rows (columns) of a determinant changes its sign.

Example: $\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 1 \times 4 - 2 \times 3 = -2$

After interchanging the two rows: $\begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix} = 3 \times 2 - 4 \times 1 = 2$

Property 3: Linearity in Rows (Columns)

Property:

Multiplying a row (column) by a constant $k$ multiplies the determinant by $k$
A row (column) may be expressed as a linear combination of the other rows (columns)

Example: $\begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix} = 2 \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 2 \times 0 = 0$

Property 4: Adding a Multiple of a Row (Column)

Property: adding $k$ times one row (column) to another row (column) leaves the determinant unchanged.

Example: $\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = -2$

Add the first row to the second row: $\begin{vmatrix} 1 & 2 \\ 4 & 6 \end{vmatrix} = 1 \times 6 - 2 \times 4 = -2$

Property 5: A Zero Row (Column)

Property: if a determinant has a row (column) consisting entirely of zeros, then the determinant is zero.

Proof: expanding along that row (column), every term is zero.

Property 6: Proportional Rows (Columns)

Property: if two rows (columns) of a determinant are proportional, then the determinant is zero.

Example: $\begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} = 1 \times 4 - 2 \times 2 = 0$

because the second row is twice the first.

Special Determinants

Diagonal Determinants

A determinant in which every element off the main diagonal is zero.

Property: a diagonal determinant equals the product of the entries on the main diagonal.

Example: $\begin{vmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{vmatrix} = 2 \times 3 \times 4 = 24$

Triangular Determinants

The determinant of an upper-triangular or lower-triangular matrix.

Property: a triangular determinant equals the product of the entries on the main diagonal.

Example: $\begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{vmatrix} = 1 \times 4 \times 6 = 24$

Block Determinants

Property: for a block matrix $\begin{pmatrix} A & B \\ C & D \end{pmatrix}$ , if $A$ is invertible then:

$\begin{vmatrix} A & B \\ C & D \end{vmatrix} = |A| \cdot |D - CA^{-1}B|$

Exercises

Exercise 1

Compute the determinant $\begin{vmatrix} 3 & 1 \\ 2 & 4 \end{vmatrix}$ .

Reference Answer

determinantsecond-order determinantdeterminant calculation

Approach: Use the defining formula for a second-order determinant.

Steps:

The formula for a second-order determinant is $|A| = ad - bc$ , where $a, b$ are the two entries of the first row and $c, d$ are the two entries of the second row.
Substituting the matrix of this problem, $a = 3$ , $b = 1$ , $c = 2$ , $d = 4$ .
$|A| = 3 \times 4 - 1 \times 2 = 12 - 2 = 10$ .

Answer: the value of the determinant is 10.

Exercise 2

Compute the determinant $\begin{vmatrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 0 & 2 & 1 \end{vmatrix}$ .

Reference Answer

determinantthird-order determinantdeterminant expansion

Approach: Compute it by expanding along the first row.

Steps:

$|A| = 1 \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} - 0 \begin{vmatrix} 3 & 0 \\ 0 & 1 \end{vmatrix} + 2 \begin{vmatrix} 3 & 1 \\ 0 & 2 \end{vmatrix}$
$= 1(1 \times 1 - 0 \times 2) - 0 + 2(3 \times 2 - 1 \times 0)$
$= 1(1) + 2(6) = 1 + 12 = 13$

Answer: the value of the determinant is 13.

Exercise 3

Decide whether the determinant $\begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix}$ is zero, and explain why.

Reference Answer

determinantproperties of determinantsproportional rows

Approach: Compute the value of the determinant and analyse the reason.

Steps:

$|A| = 2 \times 2 - 4 \times 1 = 4 - 4 = 0$
The reason: the second row is $\frac{1}{2}$ times the first row, so the two rows are proportional.

Answer: the determinant is zero, because the two rows are proportional.

Exercise 4

Use the properties of determinants to simplify the computation of $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{vmatrix}$ .

Reference Answer

determinantproperties of determinantsproportional rows

Approach: Observe the structure of the determinant and use the properties to simplify.

Steps:

Observation: the second row is twice the first row, and the third row is three times the first row
By the relevant property: if two rows (columns) of a determinant are proportional, then the determinant is zero
Therefore this determinant is zero

Answer: the value of the determinant is 0.

Exercise 5

Let $A$ be a third-order square matrix with $|A| = 3$ . If the first row of $A$ is interchanged with the third row, what is the value of the new determinant?

Reference Answer

determinantproperties of determinantssign change when rows are swapped

Approach: Use the property that swapping two rows of a determinant changes its sign.

Steps:

The property of determinants: interchanging any two rows changes the sign of the determinant, that is, the new determinant equals the original determinant multiplied by $-1$ .
Here only the first row and the third row are interchanged, which is a single swap, so the new determinant $= -|A| = -3$ .
The result is negative, which shows that this swap changed the sign of the determinant.

Answer: the value of the new determinant is $-3$ .

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation / Note	Meaning in This Article
$\lvert A\rvert$	Mathematical symbol	determinant of A	The value of the determinant of the matrix A
$\begin{vmatrix} \end{vmatrix}$	Mathematical symbol	determinant brackets	Determinant notation
$\begin{pmatrix} \end{pmatrix}$	Mathematical symbol	matrix parentheses	Matrix notation
$S_n$	Mathematical symbol	symmetric group of degree n	The symmetric group of $n$ elements
$\sigma$	Greek letter	sigma	A permutation
$\text{sgn}(\sigma)$	Mathematical symbol	sign of sigma	The sign function of a permutation
$M_{ij}$	Mathematical symbol	minor	The minor of an entry
$a_{ij}$	Mathematical symbol	a i j	A matrix entry

Determinants

The Geometric Meaning of the Determinant

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Basic Concepts of Determinants

The Definition of a Determinant

Second-Order Determinants

Third-Order Determinants

n-th Order Determinants

Basic Properties of Determinants

Property 1: Invariance under Transposition

Property 2: Swapping Two Rows (Columns) Changes the Sign

Property 3: Linearity in Rows (Columns)

Property 4: Adding a Multiple of a Row (Column)

Property 5: A Zero Row (Column)

Property 6: Proportional Rows (Columns)

Special Determinants

Diagonal Determinants

Triangular Determinants

Block Determinants

Exercises

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Summary

Symbols Used in This Article

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