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

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


Determinant of Order 3

Let:

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


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 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 the first topic in linear algebra to be studied in any depth.

It was initiated by Gottfried Wilhelm von Leibniz in $1696$, then further developed by Étienne Bézout, Alexandre-Théophile Vandermonde, Gabriel Cramer, Joseph Louis Lagrange and Pierre-Simon de Laplace.

It was advanced significantly in the first half of the $19$th century by Augustin Louis Cauchy, Carl Gustav Jacob Jacobi and James Joseph Sylvester, who between them put them into the form with which we are now familiar.

The word determinant itself first appeared in Disquisitiones Arithmeticae by Carl Friedrich Gauss in $1801$.


Sources