# Determinant/Field/Recursively/Introduction/Section

## Definition

Let ${\displaystyle {}K}$ be a field, and let ${\displaystyle {}M}$ denote an ${\displaystyle {}n\times n}$-matrix over ${\displaystyle {}K}$ with entries ${\displaystyle {}a_{ij}}$. For ${\displaystyle {}i\in \{1,\ldots ,n\}}$, let ${\displaystyle {}M_{i}}$ denote the ${\displaystyle {}(n-1)\times (n-1)}$-matrix, which arises from ${\displaystyle {}M}$, when we remove the first column and the ${\displaystyle {}i}$-th row. Then one defines recursively the determinant of ${\displaystyle {}M}$ by

${\displaystyle {}\det M={\begin{cases}a_{11}\,,&{\text{ for }}n=1\,,\\\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det M_{i}&{\text{ for }}n\geq 2\,.\end{cases}}\,}$

The determinant is only defined for square matrices. For small ${\displaystyle {}n}$, the determinant can be computed easily.

## Example

For a ${\displaystyle {}2\times 2}$-matrix

${\displaystyle {}M={\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,,}$

we have

${\displaystyle \det {\begin{pmatrix}a&b\\c&d\end{pmatrix}}=ad-cb.}$

## Example

For a ${\displaystyle {}3\times 3}$-matrix ${\displaystyle {}M={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}}$, we have

${\displaystyle \det {\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}.}$

This is called the rule of Sarrus.

## Lemma

For an upper triangular matrix

${\displaystyle {}M={\begin{pmatrix}b_{1}&\ast &\cdots &\cdots &\ast \\0&b_{2}&\ast &\cdots &\ast \\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&b_{n-1}&\ast \\0&\cdots &\cdots &0&b_{n}\end{pmatrix}}\,,}$
we have
${\displaystyle {}\det M=b_{1}b_{2}\cdots b_{n-1}b_{n}\,.}$
In particular, for the

identity matrix we get ${\displaystyle {}\det E_{n}=1}$.

### Proof

This follows with a simple induction directly from the recursive definition of the determinant.

${\displaystyle \Box }$