Determinant of Matrix Product/General Case

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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:

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

$\blacksquare$