linear algebra - 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...

Answer

As stated as a comment, the result is here. It really makes block diagonal matrices wonderful, hence finding canonical forms important.