<s>[INST] General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:
$$
A =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & a_{4,4} & a_{4,5} & a_{4,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{5,4} & a_{5,5} & a_{5,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{6,4} & a_{6,5} & a_{6,6} & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & a_{7,7} & a_{7,8} & a_{7,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{8,7} & a_{8,8} & a_{8,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
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:
$$
A_1 = 
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}\\
\end{pmatrix}
$$
$$
A_2 = 
\begin{pmatrix}
a_{4,4} & a_{4,5} & a_{4,6}\\
a_{5,4} & a_{5,5} & a_{5,6}\\
a_{6,4} & a_{6,5} & a_{6,6}\\
\end{pmatrix}
$$
$$
A_2 = 
\begin{pmatrix}
a_{7,7} & a_{7,8} & a_{7,9}\\
a_{8,7} & a_{8,8} & a_{8,9}\\
a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
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:
$$
A=
\begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_1 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_2 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1\\
\end{pmatrix} \cdot A_3
$$
But what about the determinant??? There are sums in this case which is not good...
 [/INST] First write
$$\left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] = \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \dots \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] $$
Also,
$$\det \left( \left[ \begin{array}{ccccc} \text{I}_{n_1} \hspace{-5pt} &&&& \\[-3pt] & \ddots \hspace{-5pt} &&& \\ && A_j \hspace{-5pt} && \\[-3pt] &&& \ddots \hspace{-5pt} & \\ &&&& \text{I}_{n_k} \end{array} \right] \right) = \det (A_j)$$ 
which can be seen by using the cofactor formula and repeatedly expanding along a row or column with all 0's and one 1
$$ \implies \det \left( \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] \right) = \det (A_1) \det (A_2) \cdots \det (A_k)$$
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