Definition:Determinant/Matrix

Definition
Let $\mathbf A = \left[{a}\right]_{n}$ be a square matrix of order $n$.

That is, let:
 * $\mathbf A = \begin{bmatrix}

a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}$

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

Then the determinant of $\mathbf A$ is defined as:


 * $\displaystyle \det \left({\mathbf A}\right) := \sum_{\lambda} \left({\operatorname{sgn} \left({\lambda}\right) \prod_{k=1}^n a_{k \lambda \left({k}\right)}}\right) = \sum_{\lambda} \operatorname{sgn} \left({\lambda}\right) a_{1 \lambda \left({1}\right)} a_{2 \lambda \left({2}\right)} \cdots a_{n \lambda \left({n}\right)}$

where:
 * the summation $\displaystyle \sum_\lambda$ goes over all the $n!$ permutations of $\left\{{1, 2, \ldots, n}\right\}$.
 * $\operatorname{sgn} \left({\lambda}\right)$ is the sign of the permutation $\lambda$.

When written out in full, it is denoted by:
 * $\det \left({\mathbf A}\right) = \begin{vmatrix}

a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix}$

Alternatively the notation $\left|{\mathbf A}\right|$ can be used for $\det \left({\mathbf A}\right)$ but this may be prone to ambiguity.

Determinant of Order 1
This is the trivial case:

$\begin{vmatrix} a_{11} \end{vmatrix} = \operatorname{sgn} \left({1}\right) a_{1 1} = a_{1 1}$

Thus the determinant of a single number is that number itself.

Determinant of Order 2
$\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = \operatorname{sgn} \left({1, 2}\right) a_{1 1} a_{2 2} + \operatorname{sgn} \left({2, 1}\right) a_{1 2} a_{2 1} = a_{1 1} a_{2 2} - a_{1 2} a_{2 1}$

Determinant of Order 3
Let $\det \left({\mathbf A}\right) = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$.

Then:

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

Note
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 matrices, we can discuss a determinant of order $n$.

Comment
It can be seen that the actual calculation of the value of a determinant is a long process, especially for large matrices. You seem to have to calculate as many terms as there are elements in the set of permutations of $n$ elements, which is $n!$.

However, the Expansion Theorem for Determinants puts paid to the necessity of this.