# Determinant of Matrix Product/General Case

## Theorem

Let $\mathbf A_1, \mathbf A_2, \cdots, \mathbf A_n$ be square matrices of order $n$, where $n > 1$.

Then:

$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_n} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_n}$

where $\det \mathbf A$ denotes the determinant of $\mathbf A$.

## Proof

Proof by induction:

### Basis for the Induction

$n = 2$ holds by Determinant of Matrix Product.

So shown for base case.

### Induction Hypothesis

This is our induction hypothesis:

$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k}$

Now we need to show true for $n = k + 1$:

$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k \mathbf A_{k + 1} } = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k} \map \det {\mathbf A_{k + 1} }$

### Induction Step

This is our induction step:

 $\ds \map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k \mathbf A_{k + 1} }$ $=$ $\ds \map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k} \map \det {\mathbf A_{k + 1} }$ Determinant of Matrix Product $\ds$ $=$ $\ds \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k} \map \det {\mathbf A_{k + 1} }$ Induction hypothesis

The result follows by induction:

$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_n} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_n}$

$\blacksquare$