# 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:

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

Also if $\mathbf A$ is not invertible then neither is $\mathbf A \mathbf B$. Indeed, if $\mathbf{AB}$ has an inverse $\mathbf C$, then $\mathbf{ABC} = \mathbf{I}$, whereby $\mathbf{BC}$ is a right inverse of $\mathbf A$. It follows by Left or Right Inverse of Matrix is Inverse that in that case $\mathbf{BC}$ is the inverse of $A$.

It follows that:

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

Thus:

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

$\Box$

### 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:

$\det \left({\mathbf A\mathbf B}\right) = \det \left({\mathbf E_{k}\mathbf E_{k-1} \cdots \mathbf E_{1} \mathbf B}\right)$

It remains to be shown that for any square matrix $\mathbf D$ of order $n$:

$\det \left({\mathbf E \mathbf D}\right) = \det \left({\mathbf E}\right) \cdot \det\left({\mathbf D}\right)$

Let $e_i \left({\mathbf I}\right) = \mathbf E_i$ for all $i \in [1,2,\cdots,k]$, then using Elementary Row Operations by Matrix Multiplication and Effect of Sequence of Elementary Row Operations on Determinant yields

$\det \left({\mathbf {ED}}\right) = \det \left({\mathbf E_k \mathbf E_{k-1} \cdots \mathbf {E_1D}}\right) = \det \left({e_ke_{k-1} \cdots e_1 \left({\mathbf D}\right)}\right) = \alpha \det \left({\mathbf D}\right)$
$\det \left({\mathbf E}\right) = \det \left({\mathbf E_k \mathbf E_{k-1} \cdots \mathbf {E_1I}}\right) = \det \left({e_ke_{k-1} \cdots e_1 \left({\mathbf I}\right)}\right) = \alpha \det \left({\mathbf I}\right)$
$\det \left({\mathbf E}\right) = \alpha$

And so $\det \left({\mathbf E \mathbf D}\right) = \det \left({\mathbf E}\right) \cdot \det\left({\mathbf D}\right)$

$\Box$

Therefore:

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

as required.

$\blacksquare$