Definition:Determinant/Matrix

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Definition

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

That is, let:

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


Definition 1

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


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

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

where:

the summation $\displaystyle \sum_\lambda$ 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:


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


In Full

When written out in full, the determinant of $\mathbf A$ is denoted:

$\map \det {\mathbf A} = \begin {vmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {vmatrix}$


Order

The order of a determinant is defined as the order of the square matrix on which it is defined.


Also denoted as

The notation $\size {\mathbf A}$ can be used for $\map \det {\mathbf A}$ but this may be prone to ambiguity.

Some sources omit the brackets: $\det \mathbf A$.

Where ambiguity does not result, either style is acceptable on $\mathsf{Pr} \infty \mathsf{fWiki}$.


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

\(\displaystyle \begin {vmatrix} a_{1 1} & a_{1 2} \\ a_{2 1} & a_{2 2} \end{vmatrix}\) \(=\) \(\displaystyle \map \sgn {1, 2} a_{1 1} a_{2 2} + \map \sgn {2, 1} a_{1 2} a_{2 1}\)
\(\displaystyle \) \(=\) \(\displaystyle a_{1 1} a_{2 2} - a_{1 2} a_{2 1}\)


Determinant of Order 3

Let:

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


Then:

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

and thence in a single expression as:

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

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 defined as

While a determinant is a number which is associated with a square matrix, the use of the term for the actual array itself is frequently seen.

Thus we can discuss, for example, the elements, columns and rows of a determinant.

So, similarly to square matrix, we can discuss a determinant of order $n$.


Also see

  • Results about determinants can be found here.


Historical Note

The theory of determinants was advanced significantly by Augustin Louis Cauchy.


Sources