Determinant of Matrix Product/Proof 2

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
Consider two cases:


 * $(1): \quad \mathbf A$ is not invertible.


 * $(2): \quad \mathbf A$ is invertible.

Proof of case $(1)$:

Assume $\mathbf A$ is not invertible.

Then $\det \left({\mathbf A}\right) = 0$.

Also if $\mathbf A$ is not invertible then neither is $\mathbf A \mathbf B$, and so


 * $\det\left({\mathbf A \mathbf B}\right) = 0$

Thus:
 * $0 = 0 \cdot \det\left({\mathbf B}\right)$


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

Proof of case $(2)$:

Assume $\mathbf A$ is invertible.

Then $\mathbf A$ is a product of elementary matrices, $\mathbf E$.

Let $\mathbf A = \mathbf E^{k} \mathbf E^{k-1} \cdots \mathbf E^{1}$.

So:
 * $\det \left({\mathbf A\mathbf B}\right) = \det \left({\mathbf E^{k}\mathbf E^{k-1} \cdots \mathbf E^{1} \mathbf B}\right)$

But for any matrix $\mathbf D$:
 * $\det \left({\mathbf E \mathbf D}\right) = \det \left({\mathbf E}\right) \cdot \det\left({\mathbf D}\right)$

by Effect of Elementary Row Operations on Determinant.

Therefore:
 * $\det \left({\mathbf A \mathbf B}\right) = \det \left({\mathbf E^{k}}\right) \det\left({\mathbf E^{k-1}}\right) \cdots \det\left({\mathbf E^{1}}\right) \det\left({\mathbf B}\right)$


 * $\det\left({\mathbf A \mathbf B}\right) = \det\left({\mathbf E^{k} \mathbf E^{k-1} \ldots \mathbf E^{1}}\right) \det \left({\mathbf B}\right)$


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

as required.