Determinant of Matrix Product/Proof 2

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:
 * $\map \det {\mathbf A} = 0$

Also if $\mathbf A$ is not invertible then neither is $\mathbf A \mathbf B$.

Indeed, if $\mathbf A \mathbf B$ has an inverse $\mathbf C$, then:
 * $\mathbf A \mathbf B \mathbf C = \mathbf I$

whereby $\mathbf B \mathbf C$ is a right inverse of $\mathbf A$.

It follows by Left or Right Inverse of Matrix is Inverse that in that case $\mathbf B \mathbf C$ is the inverse of $A$.

It follows that:


 * $\map \det {\mathbf A \mathbf B} = 0$

Thus:
 * $0 = 0 \times \map \det {\mathbf B}$

that is:
 * $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$

Proof of case $2$
Assume $\mathbf A$ is invertible.

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

Let $\mathbf A = \mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1$.

So:
 * $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1 \mathbf B}$

It remains to be shown that for any square matrix $\mathbf D$ of order $n$:
 * $\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \map \det {\mathbf D}$

Let $e_i \paren {\mathbf I} = \mathbf E_i$ for all $i \in \closedint 1 k$.

Then:

and:

From Determinant of Unit Matrix:


 * $\map \det {\mathbf E} = \alpha$

Hence:
 * $\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \map \det {\mathbf D}$

Therefore:
 * $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$

as required.