Determinant of Matrix Product/Proof 1

Proof
This proof assumes that $\mathbf A$ and $\mathbf B$ are $n \times n$-matrices over a commutative ring with unity $\left({R, +, \circ}\right)$.

Let $\mathbf C = \left[{c}\right]_n = \mathbf A \mathbf B$.

From Square Matrix is Row Equivalent to Triangular Matrix, it follows that $\mathbf A$ can be converted into a upper triangular matrix $\mathbf A'$ by a finite sequence of elementary row operations $\hat o_1, \ldots, \hat o_{m'}$.

Let $\mathbf C'$ denote the matrix that results from using $\hat o_1, \ldots, \hat o_{m'}$ on $\mathbf C$.

From Elementary Row Operations Commute with Matrix Multiplication, it follows that $\mathbf C' = \mathbf A' \mathbf B$.

Effect of Sequence of Elementary Row Operations on Determinant shows that there exists $\alpha \in R$ such that:


 * $\alpha \det \left({\mathbf A'}\right) = \det \left({\mathbf A}\right)$
 * $\alpha \det \left({\mathbf C'}\right) = \det \left({\mathbf C}\right)$

Let $\mathbf B^\intercal$ be the transpose of $B$.

From Transpose of Matrix Product, it follows that:
 * $\left({\mathbf C'}\right)^\intercal = \left({\mathbf A' \mathbf B}\right)^\intercal = \mathbf B^\intercal \left({\mathbf A'}\right)^\intercal$

From Square Matrix is Row Equivalent to Triangular Matrix, it follows that $\mathbf B^\intercal$ can be converted into a lower triangular matrix $\left({\mathbf B^\intercal}\right)'$ by a finite sequence of elementary row operations $\hat p_1, \ldots, \hat p_{m''}$.

Let $\mathbf C$ denote the matrix that results from using $\hat p_1, \ldots, \hat p_{m}$ on $\left({\mathbf C'}\right)^\intercal$.

From Elementary Row Operations Commute with Matrix Multiplication, it follows that:
 * $\mathbf C'' = \left({\mathbf B^\intercal}\right)' \left({\mathbf A'}\right)^\intercal$

Effect of Sequence of Elementary Row Operations on Determinant shows that there exists $\beta \in R$ such that:


 * $\beta \det \left({\left({\mathbf B^\intercal}\right)'}\right) = \det \left({\mathbf B^\intercal}\right)$
 * $\beta \det \left({\mathbf C''}\right) = \det \left({ \left({\mathbf C'}\right)^\intercal }\right)$

From Transpose of Upper Triangular Matrix is Lower Triangular, it follows that $\left({\mathbf A'}\right)^\intercal$ is a lower triangular matrix.

Then Product of Triangular Matrices shows that $\left({\mathbf B^\intercal}\right)' \left({\mathbf A'}\right)^\intercal$ is a lower triangular matrix whose diagonal elements are the products of the diagonal elements of $\left({\mathbf B^\intercal}\right)'$ and $\left({\mathbf A'}\right)^\intercal$.

From Determinant of Triangular Matrix, we have that $\det \left({\left({\mathbf A'}\right)^\intercal}\right)$, $\det \left({\left({\mathbf B^\intercal}\right)' }\right)$, and $\det \left({\left({\mathbf B^\intercal}\right)' \left({\mathbf A'}\right)^\intercal }\right)$ are equal to the product of their diagonal elements.

Combinining these results shows that:
 * $\det \left({\left({\mathbf B^\intercal}\right)' \left({\mathbf A'}\right)^\intercal}\right) = \det \left({\left({\mathbf B^\intercal}\right)'}\right) \det \left({\left({\mathbf A'}\right)^\intercal }\right)$

Then: