# Definition:Determinant

## Definition

### Determinant of Matrix

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

That is, let:

$\quad \mathbf A = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$

### Definition 1

Let $\lambda: \N_{> 0} \to \N_{> 0}$ be a permutation on $\N_{>0}$.

The determinant of $\mathbf A$ is defined as:

$\ds \map \det {\mathbf A} := \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} } = \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{n \map \lambda n}$

where:

the summation $\ds \sum_\lambda$ goes over all the $n!$ permutations of $\set {1, 2, \ldots, n}$
$\map \sgn \lambda$ is the sign of the permutation $\lambda$.

### Definition 2

The determinant of $\mathbf A$ is defined as follows:

For $n = 1$, the order $1$ determinant is defined as:

$\begin {vmatrix} a_{1 1} \end {vmatrix} = a_{1 1}$

Thus the determinant of an order $1$ matrix is that element itself.

For $n > 1$, the determinant of order $n$ is defined recursively as:

$\quad \ds \map \det {\mathbf A} := \begin {vmatrix} a_{1 1} & a_{1 2} & a_{1 3} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & a_{2 3} & \cdots & a_{2 n} \\ a_{3 1} & a_{3 2} & a_{3 3} & \cdots & a_{3 n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & a_{n 3} & \cdots & a_{n n} \\ \end {vmatrix} = a_{1 1} \begin {vmatrix} a_{2 2} & a_{2 3} & \cdots & a_{2 n} \\ a_{3 2} & a_{3 3} & \cdots & a_{3 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 2} & a_{n 3} & \cdots & a_{n n} \\ \end {vmatrix} - a_{1 2} \begin {vmatrix} a_{2 1} & a_{2 3} & \cdots & a_{2 n} \\ a_{3 1} & a_{3 3} & \cdots & a_{3 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 3} & \cdots & a_{n n} \\ \end {vmatrix} + \cdots + \paren {-1}^{n + 1} a_{1 n} \begin {vmatrix} a_{2 1} & a_{2 2} & \cdots & a_{2, n - 1} \\ a_{3 1} & a_{3 3} & \cdots & a_{3, n - 1} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 3} & \cdots & a_{n, n - 1} \\ \end {vmatrix}$

### Determinant of Linear Operator

Let $V$ be a finite-dimensional vector space over a field $K$.

Let $A: V \to V$ be a linear operator of $V$.

The determinant $\map \det A$ of $A$ is defined to be the determinant of any matrix of $A$ relative to some basis.

## Rows and Columns

### Row of Determinant

Let $\mathbf D$ be a determinant.

The rows of $\mathbf D$ are the lines of elements reading across the page.

### Column of Determinant

Let $\mathbf D$ be a determinant.

The columns of $\mathbf D$ are the lines of elements reading across the page.

## Examples

### Determinant of Order 1

This is the trivial case:

$\begin {vmatrix} a_{1 1} \end {vmatrix} = a_{1 1}$

Thus the determinant of an order $1$ matrix is that element itself.

### Determinant of Order 2

 $\ds \begin {vmatrix} a_{1 1} & a_{1 2} \\ a_{2 1} & a_{2 2} \end{vmatrix}$ $=$ $\ds \map \sgn {1, 2} a_{1 1} a_{2 2} + \map \sgn {2, 1} a_{1 2} a_{2 1}$ $\ds$ $=$ $\ds a_{1 1} a_{2 2} - a_{1 2} a_{2 1}$

### Determinant of Order 3

Let:

$\quad \map \det {\mathbf A} = \begin {vmatrix} a_{1 1} & a_{1 2} & a_{1 3} \\ a_{2 1} & a_{2 2} & a_{2 3} \\ a_{3 1} & a_{3 2} & a_{3 3} \end {vmatrix}$

Then:

 $\ds \map \det {\mathbf A}$ $=$ $\ds a_{1 1} \begin {vmatrix} a_{2 2} & a_{2 3} \\ a_{3 2} & a_{3 3} \end {vmatrix} - a_{1 2} \begin {vmatrix} a_{2 1} & a_{2 3} \\ a_{3 1} & a_{3 3} \end {vmatrix} + a_{1 3} \begin {vmatrix} a_{2 1} & a_{2 2} \\ a_{3 1} & a_{3 2} \end{vmatrix}$ $\ds$ $=$ $\ds \map \sgn {1, 2, 3} a_{1 1} a_{2 2} a_{3 3}$ $\ds$  $\, \ds + \,$ $\ds \map \sgn {1, 3, 2} a_{1 1} a_{2 3} a_{3 2}$ $\ds$  $\, \ds + \,$ $\ds \map \sgn {2, 1, 3} a_{1 2} a_{2 1} a_{3 3}$ $\ds$  $\, \ds + \,$ $\ds \map \sgn {2, 3, 1} a_{1 2} a_{2 3} a_{3 1}$ $\ds$  $\, \ds + \,$ $\ds \map \sgn {3, 1, 2} a_{1 3} a_{2 1} a_{3 2}$ $\ds$  $\, \ds + \,$ $\ds \map \sgn {3, 2, 1} a_{1 3} a_{2 2} a_{3 1}$ $\ds$ $=$ $\ds a_{1 1} a_{2 2} a_{3 3}$ $\ds$  $\, \ds - \,$ $\ds a_{1 1} a_{2 3} a_{3 2}$ $\ds$  $\, \ds - \,$ $\ds a_{1 2} a_{2 1} a_{3 3}$ $\ds$  $\, \ds + \,$ $\ds a_{1 2} a_{2 3} a_{3 1}$ $\ds$  $\, \ds + \,$ $\ds a_{1 3} a_{2 1} a_{3 2}$ $\ds$  $\, \ds - \,$ $\ds a_{1 3} a_{2 2} a_{3 1}$

and thence in a single expression as:

$\ds \map \det {\mathbf A} = \frac 1 6 \sum_{i \mathop = 1}^3 \sum_{j \mathop = 1}^3 \sum_{k \mathop = 1}^3 \sum_{r \mathop = 1}^3 \sum_{s \mathop = 1}^3 \sum_{t \mathop = 1}^3 \map \sgn {i, j, k} \map \sgn {r, s, t} a_{i r} a_{j s} a_{k t}$

where $\map \sgn {i, j, k}$ is the sign of the permutation $\tuple {i, j, k}$ of the set $\set {1, 2, 3}$.

The values of the various instances of $\map \sgn {\lambda_1, \lambda_2, \lambda_3}$ are obtained by applications of Parity of K-Cycle.

## Also see

• Results about determinants can be found here.