Let $G$ be non-empty set and $\circ: G \times G \to G$ binary operation. If the following axioms are satisfied

1. associativity: $$(\forall a, b, c \in G) ((a \circ b) \circ c = a \circ (b \circ c)),$$
2. neutral element:$$(\exists e \in G) (\forall a \in G) (a \circ e = e \circ a),$$
3. inverse element:$$(\forall a \in G) (\exists a^{-1} \in G) (a \circ a^{-1} = a^{-1} \circ a),$$the pair $(G, \circ)$ is called a group.

Moreover, if $\circ$ is

• commutative: $$(\forall a, b \in G) (a \circ b = b \circ a),$$we say that a group $(G, \circ)$ is commutative (or Abelian) group.

Let $F$ be a non-empty set and $+:F\times F \to F$, $\cdot:F\times F \to F$ are binary operations. If the following axioms are satisfied

1. $(F, +)$ forms an Abelian group with identity $0$,
2. $(F \setminus {0}, \cdot)$ forms an Abelian group with identity $1$,
3. multiplication is distributive over addition:$$(\forall a, b, c \in F) \big(a \cdot (b + c) = a \cdot b + a \cdot c \land (a + b) \cdot c = a\cdot c + b \cdot a \big),$$then the triple $(F, +, \cdot)$ is a field.

Let $m, n \in \N$. Table of $m \cdot n$ numbers from field $F$ with $m$ rows and $n$ columns is called matrix of dimensions $m\times n$ and is denoted as

where $a_{11},\dots, a_{mn} \in F$.

The set of all matrices with size $m \times n$ is $F^{m, n}$.