# Determinants

## Definition

With each $n \times n$ matrix $A$ with $n \in \mathbb{N}$ it is possible to associate a scalar, the determinant of $A$ denoted by $\det (A)$ or $|A|$.

> *Definition*: let $A = (a_{ij})$ be an $n \times n$ matrix and let $M_{ij}$ denote the $(n-1) \times (n-1)$ matrix obtained from $A$ by deleting the row and column containing $a_{ij}$ with $n \in \mathbb{N}$ and $(i,j) \in \{1, \dots, n\} \times \{1, \dots, n\}$. The determinant of $M_{ij}$ is called the **minor** of $a_{ij}$. We define the **cofactor** of $A_{ij}$ of $a_{ij}$ by
>
> $$
>   A_{ij} = (-1)^{i+j} \det(M_{ij}).
> $$

This definition is necessary to formulate a definition for the determinant, as may be observed below.

> *Definition*: the **determinant** of an $n \times n$ matrix $A$ with $n \in \mathbb{N}$, denoted by $\det (A)$ or $|A|$ is a scalar associated with the matrix $A$ that is defined inductively as
>
> $$
>   \det (A) = \begin{cases}a_{11} &\text{ if } n = 1 \\ a_{11} A_{11} + a_{12} A_{12} + \dots + a_{1n} A_{1n} &\text{ if } n > 1\end{cases}
> $$
>
> where 
>
> $$
>   A_{1j} = (-1)^{1+j} \det (M_{1j})
> $$
>
> with $j \in \{1, \dots, n\}$ are the cofactors associated with the entries in the first row of $A$.

<br>

> *Theorem*: if $A$ is an $n \times n$ matrix with $n \in \mathbb{N} \backslash \{1\}$ then $\det(A)$ cam be expressed as a cofactor expansion using any row or column of $A$.

??? note "*Proof*:"

    Will be added later.

We then have for a $n \times n$ matrix $A$ with $n \in \mathbb{N} \backslash \{1\}$

$$
\begin{align*}
    \det(A) &= a_{i1} A_{i1} + a_{i2} A_{i2} + \dots + a_{in} A_{in}, \\
            &= a_{1j} A_{1j} + a_{2j} A_{2j} + \dots + a_{nj} A_{nj},
\end{align*}
$$

with $i,j \in \mathbb{N}$. 

For example, the determinant of a $4 \times 4$ matrix $A$ given by 

$$
    A = \begin{pmatrix} 0 & 2 & 3 & 0\\ 0 & 4 & 5 & 0\\ 0 & 1 & 0 & 3\\ 2 & 0 & 1 & 3\end{pmatrix}
$$

may be determined using the definition and the theorem above

$$
    \det(A) = 2 \cdot (-1)^5 \det\begin{pmatrix} 2 & 3 & 0\\ 4 & 5 & 0\\ 1 & 0 & 3\end{pmatrix} = -2 \cdot 3 \cdot (-1)^6 \det\begin{pmatrix} 2 & 3 \\ 4 & 5\end{pmatrix} = 12.
$$

## Properties of determinants

> *Theorem*: if $A$ is an $n \times n$ matrix then $\det (A^T) = \det (A)$. 

??? note "*Proof*:"

    It may be observed that the result holds for $n=1$. Assume that the results holds for all $k \times k$ matrices and that $A$ is a $(k+1) \times (k+1)$ matrix for some $k \in \mathbb{N}$. Expanding $\det (A)$ along the first row of $A$ obtains

    $$
        \det(A) = a_{11} \det(M_{11}) - a_{12} \det(M_{12}) + \dots + (-1)^{k+2} a_{1(k+1)} \det(M_{1(k+1)}),
    $$

    since the minors are all $k \times k$ matrices it follows from the principle of natural induction that

    $$
        \det(A) = a_{11} \det(M_{11}^T) - a_{12} \det(M_{12}^T) + \dots + (-1)^{k+2} a_{1(k+1)} \det(M_{1(k+1)}^T).
    $$

    The right hand side of the above equation is the expansion by minors of $\det(A^T)$ using the first column of $A^T$, therefore $\det(A^T) = \det(A)$.

> *Theorem*: if $A$ is an $n \times n$ triangular matrix with $n \in \mathbb{N}$, then the determinant of $A$ equals the product of the diagonal elements of $A$.

??? note "*Proof*:"

    Will be added later.

> *Theorem*: let $A$ be an $n \times n$ matrix
>
> 1. if $A$ has a row or column consisting entirely of zeros, then $\det(A) = 0$.
> 2. if $A$ has two identical rows or two identical columns, then $\det(A) = 0$.

??? note "*Proof*:"

    Will be added later.