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 $\struct {R, +, \circ}$.

Let $\mathbf C = \sqbrk c_n = \mathbf A \mathbf B$.

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:


 * $\alpha \map \det {\mathbf A'} = \map \det {\mathbf A}$
 * $\alpha \map \det {\mathbf C'} = \map \det {\mathbf C}$

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

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

From Square Matrix is Row Equivalent to Triangular Matrix, it follows that $\mathbf B^\intercal$ can be converted into a lower triangular matrix $\paren {\mathbf B^\intercal}'$ 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 $\paren {\mathbf C'}^\intercal$.

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

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


 * $\beta \map \det {\paren {\mathbf B^\intercal}'} = \map \det {\mathbf B^\intercal}$
 * $\beta \map \det {\mathbf C''} = \map \det {\paren {\mathbf C'}^\intercal}$

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

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

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

Combinining these results shows that:
 * $\map \det {\paren {\mathbf B^\intercal}' \paren {\mathbf A'}^\intercal} = \map \det {\paren {\mathbf B^\intercal}'} \map \det {\paren {\mathbf A'}^\intercal}$

Then: