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 $$\lambda: \N^* \to \N^*$$ be a permutation on $\N^*$.

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

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

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

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

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

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

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 $$\sgn \left({\lambda_1, \lambda_2, \lambda_3}\right)$$ 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.