# Determinant of Matrix Product/Proof 2

## Theorem

Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be a square matrices of order $n$.

Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.

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

Then:

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

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:

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

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:

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

$\Box$

### 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 $n$:

$\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \cdot \map \det {\mathbf D}$

Let $e_i \paren {\mathbf I} = \mathbf E_i$ for all $i \in \closedint 1 k$, then using Elementary Row Operations as Matrix Multiplications and Effect of Sequence of Elementary Row Operations on Determinant yields:

$\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E_k \mathbf E_{k - 1} \dotsm \mathbf {E_1} \mathbf D} = \map \det {e_k e_{k - 1} \cdots e_1 \paren {\mathbf D} } = \alpha \map \det {\mathbf D}$
$\map \det {\mathbf E} = \map \det {\mathbf E_k \mathbf E_{k - 1} \cdots \mathbf {E_1} \mathbf I} = \map \det {e_k e_{k - 1} \cdots e_1 \paren {\mathbf I} } = \alpha \map \det {\mathbf I}$
$\map \det {\mathbf E} = \alpha$

And so $\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \cdot \map \det {\mathbf D}$

$\Box$

Therefore:

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

as required.

$\blacksquare$