General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:

$$
A =
\begin{pmatrix}
a_{1,1} &amp; a_{1,2} &amp; a_{1,3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
a_{2,1} &amp; a_{2,2} &amp; a_{2,3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
a_{3,1} &amp; a_{3,2} &amp; a_{3,3} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; a_{4,4} &amp; a_{4,5} &amp; a_{4,6} &amp; 0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; a_{5,4} &amp; a_{5,5} &amp; a_{5,6} &amp; 0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; a_{6,4} &amp; a_{6,5} &amp; a_{6,6} &amp; 0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; a_{7,7} &amp; a_{7,8} &amp; a_{7,9}\\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; a_{8,7} &amp; a_{8,8} &amp; a_{8,9}\\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; a_{9,7} &amp; a_{9,8} &amp; a_{9,9}\\
\end{pmatrix}
$$

Question.

What about its determinant $|A|$?.

Another question

I was wondering that maybe matrix $A$ can be expressed as a product of particular matrices to have such a structure... maybe using these matrices:

$$
A_1 = 
\begin{pmatrix}
a_{1,1} &amp; a_{1,2} &amp; a_{1,3}\\
a_{2,1} &amp; a_{2,2} &amp; a_{2,3}\\
a_{3,1} &amp; a_{3,2} &amp; a_{3,3}\\
\end{pmatrix}
$$

$$
A_2 = 
\begin{pmatrix}
a_{4,4} &amp; a_{4,5} &amp; a_{4,6}\\
a_{5,4} &amp; a_{5,5} &amp; a_{5,6}\\
a_{6,4} &amp; a_{6,5} &amp; a_{6,6}\\
\end{pmatrix}
$$

$$
A_2 = 
\begin{pmatrix}
a_{7,7} &amp; a_{7,8} &amp; a_{7,9}\\
a_{8,7} &amp; a_{8,8} &amp; a_{8,9}\\
a_{9,7} &amp; a_{9,8} &amp; a_{9,9}\\
\end{pmatrix}
$$

I can arrange $A$ as a compination of those: $A = f(A_1,A_2,A_3)$

Kronecker product

One possibility can be the Kronecker product:

$$
A=
\begin{pmatrix}
1 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 0\\
\end{pmatrix} \otimes A_1 +
\begin{pmatrix}
0 &amp; 0 &amp; 0\\
0 &amp; 1 &amp; 0\\
0 &amp; 0 &amp; 0\\
\end{pmatrix} \otimes A_2 +
\begin{pmatrix}
0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 1\\
\end{pmatrix} \cdot A_3
$$

But what about the determinant??? There are sums in this case which is not good...
First write
$$\left[ \begin{array}{cccc} A_1 \hspace{-5pt} &amp;&amp;&amp; \\ &amp; A_2 \hspace{-5pt} &amp;&amp; \\[-3pt] &amp;&amp; \ddots \hspace{-5pt} &amp; \\ &amp;&amp;&amp; A_k \end{array} \right] = \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &amp;&amp;&amp; \\ &amp; \text{I}_{n_2} \hspace{-5pt} &amp;&amp; \\[-3pt] &amp;&amp; \ddots \hspace{-5pt} &amp; \\ &amp;&amp;&amp; \text{I}_{n_k} \end{array} \right] \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &amp;&amp;&amp; \\ &amp; A_2 \hspace{-5pt} &amp;&amp; \\[-3pt] &amp;&amp; \ddots \hspace{-5pt} &amp; \\ &amp;&amp;&amp; \text{I}_{n_k} \end{array} \right] \dots \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &amp;&amp;&amp; \\ &amp; \text{I}_{n_2} \hspace{-5pt} &amp;&amp; \\[-3pt] &amp;&amp; \ddots \hspace{-5pt} &amp; \\ &amp;&amp;&amp; A_k \end{array} \right] $$
Also,
$$\det \left( \left[ \begin{array}{ccccc} \text{I}_{n_1} \hspace{-5pt} &amp;&amp;&amp;&amp; \\[-3pt] &amp; \ddots \hspace{-5pt} &amp;&amp;&amp; \\ &amp;&amp; A_j \hspace{-5pt} &amp;&amp; \\[-3pt] &amp;&amp;&amp; \ddots \hspace{-5pt} &amp; \\ &amp;&amp;&amp;&amp; \text{I}_{n_k} \end{array} \right] \right) = \det (A_j)$$ 
which can be seen by using the cofactor formula and repeatedly expanding along a row or column with all 0's and one 1

$$ \implies \det \left( \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &amp;&amp;&amp; \\ &amp; A_2 \hspace{-5pt} &amp;&amp; \\[-3pt] &amp;&amp; \ddots \hspace{-5pt} &amp; \\ &amp;&amp;&amp; A_k \end{array} \right] \right) = \det (A_1) \det (A_2) \cdots \det (A_k)$$
A &quot;functorial approach&quot; using the exterior product: If $\phi: V \rightarrow V$ is an endomorphism of a vector space, you may calculate the determinant of the endomorphism $\phi$ as the induced map
$$\wedge^n (\phi): \wedge^n V \rightarrow \wedge^n V$$
where $n:=\dim(V)$. Since $\wedge^n$ is a functor you get a canonical map $\wedge^l (\phi)$ for any integer $l \geq 1$.
It follows $\wedge^n (\phi)$ is an endomorphism of a one dimensional vector space $\wedge^n V$ and hence it is given as multiplication with a number $a$. The number $a$ is the determinant: $a=\det(\phi)$ of the map $\phi$. If you choose a basis $B$ of $V$ and the matrix of $\phi$ in this basis is a matrix $A$, it follows $a=det(A)$ is the determinant of the matrix $A$.
There is a formula:
$$\wedge^{n_1+n_2}(V_1\oplus V_2) \cong \wedge^{n_1}V_1 \otimes \wedge^{n_2}V_2,$$
where $n_i:=\dim(V_i)$. Let
\begin{align*} \phi= \begin{pmatrix} \phi_1 &amp; 0 \\ 0 &amp; \phi_2 \end{pmatrix} \end{align*}
where $\phi_i$ is an endomorphism of $V_i$. It &quot;follows&quot;
$$\det(\phi)=\wedge^{n_1+n_2}(\phi) \cong \wedge^{n_1}(\phi_1) \otimes \wedge^{n_2}(\phi_2)$$
But the tensor product $\wedge^{n_1}V_1 \otimes \wedge^{n_2}V_2$ is a one dimensional vector space and any linear endomorphism of such a space is given (in a basis) as multiplication with a number. Choosing a basis it follows the endomorphism $\wedge^{n_1}(\phi_1) \otimes \wedge^{n_2}(\phi_2)$ is multiplication with the number
$$\det(A_1)\det(A_2),$$
where $A_i$ is a matrix of $\phi_i$ in a basis $B_i$ for $V_i$.
Question: &quot;But what about the determinant???&quot;
By induction it follows that if $M$ is a matrix with square matrices $A_i$ along the diagonal, you get the formula
$$\det(M)=\det(A_1) \cdots \det(A_n).$$