# Determinant/Field/Recursively/Multiplication theorem/Section
## Theorem

Let $K$ denote a field, and $n\in \mathbb {N} _{+}$. Then for matrices $A,B\in \operatorname {Mat} _{n}(K)$, the relation
    $\det {\left(A\circ B\right)}=\det A\cdot \det B\,$

### Proof

We fix the matrix $B$.
Suppose first that $\det B=0$. Then, due to fact the matrix $B$ is not invertibler and therefore, also $A\circ B$ is not invertible. Hence, $\det A\circ B=0$.
Suppose now that $B$ is invertible. In this case, we consider the well-defined mapping
    $\delta \colon \operatorname {Mat} _{n}(K)\longrightarrow K,A\longmapsto (\det A\circ B)(\det B)^{-1}.$
We want to show that this mapping equals the mapping $A\mapsto \det A$, by showing that it fulfills all the properties which, according to fact, characterize the determinant. If $z_{1},\ldots ,z_{n}$ denote the rows of $A$, then $\delta (A)$ is computed by applying the determinant to the rows $z_{1}B,\ldots ,z_{n}B$, and then by multiplying with $(\det B)^{-1}$. Hence the multilinearity and the alternating property follows from exercise. If we start with $A=E_{n}$, then $A\circ B=B$ and thus
    $\delta (E_{n})=(\det B)\cdot (\det B)^{-1}=1\,.$
$\Box$

## Theorem

Let $K$ denote a field, and let $M$ denote an $n\times n$-matrix over $K$. Then
    $\det M=\det {M^{\text{tr}}}\,.$

### Proof

If $M$ is not invertible, then, due to fact, the determinant is $0$ and the rank is smaller than $n$. This does also hold for the transposed matrix, so that its determinant is again $0$. So suppose that $M$ is invertible. We reduce the statement in this case to the corresponding statement for the elementary matrices, which can be verified directly, see exercise. Because of fact, there exist elementary matrices $E_{1},\ldots ,E_{s}$ such that
    $D=E_{s}\cdots E_{1}M\,$
is a diagonal matrix. Due to exercise, we have
    ${D^{\text{tr}}}={M^{\text{tr}}}{E_{1}^{\text{tr}}}\cdots {E_{s}^{\text{tr}}}\,$
and
    ${M^{\text{tr}}}={D^{\text{tr}}}({E_{s}^{\text{tr}}})^{-1}\cdots ({E_{1}^{\text{tr}}})^{-1}\,.$
The diagonal matrix $D$ is not changed under transposing it. Since the determinants of the elementary matrices are also not changed under transposition, we get, using fact,
    ${\begin{aligned}\det {M^{\text{tr}}}&=\det {\left({D^{\text{tr}}}({E_{s}^{\text{tr}}})^{-1}\cdots ({E_{1}^{\text{tr}}})^{-1}\right)}\\&=\det {D^{\text{tr}}}\cdot \det({E_{s}^{\text{tr}}})^{-1}\cdots \det({E_{1}^{\text{tr}}})^{-1}\\&=\det D\cdot \det {\left(E_{s}^{-1}\right)}\cdots \det {\left(E_{1}^{-1}\right)}\\&=\det {\left(E_{1}^{-1}\right)}\cdots \det {\left(E_{s}^{-1}\right)}\cdot \det D\cdot \\&=\det {\left(E_{1}^{-1}\cdots E_{s}^{-1}\cdot D\right)}\\&=\det M.\end{aligned}}$
$\Box$

## Corollary

Let $K$ be a field, and let $M={\left(a_{ij}\right)}_{ij}$ be an $m\times n$-matrix over $K$. For $i,j\in \{1,\ldots ,n\}$, let $M_{ij}$ be the matrix which arises from $M$, by leaving out the $i$-th row and the $j$-th column. Then (for $n\geq 2$ and for every fixed $i$ and $j$)
    $\det M=\sum _{i=1}^{n}(-1)^{i+j}a_{ij}\det M_{ij}=\sum _{j=1}^{n}(-1)^{i+j}a_{ij}\det M_{ij}\,.$

### Proof

For $j=1$, the first equation is the recursive definition of the determinant. From that statement, the case $i=1$ follows, due to fact. By exchanging columns and rows, the statement follows in full generality, see exercise.
$\Box$