Determinant of Matrix Product/Proof 1

Theorem
Let $\mathbf A = \left[{a}\right]_n$ and $\mathbf B = \left[{b}\right]_n$ be a square matrices of order $n$.

Let $\det \left({\mathbf A}\right)$ be the determinant of $\mathbf A$.

Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.

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

That is, the determinant of the product is equal to the product of the determinants.

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

Thus:
 * $\displaystyle \forall i, j \in \left[{1 \,.\,.\, n}\right]: c_{i j} = \sum_{l \mathop = 1}^n a_{i l} b_{l j}$

Then:

This gets messy very quickly, so we can try another approach.

From Square Matrix Row Equivalent to Triangular Matrix, it is possible to convert a square matrix into a triangular matrix by using elementary row operations.

From Effect of Elementary Row Operations on Determinant, this transformation does not change either the value or the sign of its determinant.

Thus:
 * $(1):\quad \exists \mathbf T_A$, an upper triangular matrix, such that $\det \left({\mathbf T_A}\right) = \det \left({\mathbf A}\right)$
 * $(2):\quad \exists \mathbf T_B$, an upper triangular matrix, such that $\det \left({\mathbf T_B}\right) = \det \left({\mathbf B}\right)$

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

From Product of Triangular Matrices, it also follows that $\mathbf T_A \mathbf T_B$ is an upper triangular matrix whose diagonal elements are the products of the diagonal elements of $\mathbf T_A$ and $\mathbf T_B$.

Thus it follows that:
 * $\det \left({\mathbf T_A \mathbf T_B}\right) = \det \left({\mathbf T_A}\right) \det \left({\mathbf T_B}\right)$

Thus it follows that:
 * $\det \left({\mathbf A \mathbf B}\right) = \det \left({\mathbf A}\right) \det \left({\mathbf B}\right)$