User:Anghel/Sandbox

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 Row Equivalent to Triangular Matrix, it follows that $\mathbf A$ can be converted into a 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 Effect of Sequence of Elementary Row Operations on Determinant, it follows that there exists $k \in R$ such that:


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

Similarly, $\mathbf B$ can be converted into a triangular matrix $\mathbf B'$ 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 $\mathbf C'$.

Then, there exists $l \in R$ such that:
 * $l\det \left({\mathbf B'}\right) = \det \left({\mathbf B}\right)$
 * $kl\det \left({\mathbf C''}\right) = k\det \left({\mathbf C'}\right) = \det \left({\mathbf C}\right)$

From Product of Triangular Matrices, it follows that $\mathbf A' \mathbf B'$ is a triangular matrix whose diagonal elements are the products of the diagonal elements of $\mathbf A'$ and $\mathbf B'$.

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

Combinining these two results shows that


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

Then: