<s>[INST] Determinant of Matrix with each entry being a diagonal matrix Let $\mathbf{I}_n$ be the identity matrix with size $n$ by $n.$
Consider the $n$ by $n$ matrix
$ \mathbf{A} = \begin{pmatrix}
a_{11} & a_{12} & \cdots &a_{1n} \\
a_{21} & a_{22}  &\cdots & a_{2n} \\
 \cdots\\
a_{n1} & a_{n2}  &\cdots & a_{nn} 
\end{pmatrix}  $
Define a $n^2$ by $n^2$ matrix as follows:
$ \mathbf{B} = \begin{pmatrix}
a_{11}\mathbf{I}_n & a_{12}\mathbf{I}_n & \cdots &a_{1n}\mathbf{I}_n \\
a_{21}\mathbf{I}_n & a_{22}\mathbf{I}_n  &\cdots & a_{2n}\mathbf{I}_n \\
 \cdots\\
a_{n1}\mathbf{I}_n & a_{n2}\mathbf{I}_n  &\cdots & a_{nn}\mathbf{I}_n 
\end{pmatrix}  $
For example when $n = 2,$
$ \mathbf{A} = \begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} 
\end{pmatrix}  $
and
$ \mathbf{B} = \begin{pmatrix}
a_{11}&0 & a_{12}&0 \\
0&a_{11} & 0 & a_{12}\\
a_{21}&0 & a_{22} &0\\
0&a_{21} & 0& a_{22}
\end{pmatrix}  $
Then, we have $$det(\mathbf{B}) =(a_{11}a_{22}-a_{12}a_{21} )^2=(det(\mathbf{A}))^2$$
The question is that: Do we have that $$det(\mathbf{B}) = (det(\mathbf{A}))^n$$ for any $n = 1, 2, 3, ...$?
I have verified this for $n=2$ and $3$ (of course it trivially holds when $n=1$), but do not have much idea about larger $n.$
 [/INST] The answer is yes. One way to see this is as a consequence of the properties of the Kronecker product. The matrix $B$ can be written in the form $B = A \otimes I_n$. From the properties of the Kronecker product, we have
$$
\det(A \otimes I_n) = \det(A)^n \cdot \det(I_n)^{n} = \det(A)^n \cdot 1.
$$
Alternatively, a direct approach is to note that there exists a Permutation matrix $P$ such that $P^TBP$ is the block-diagonal matrix
$$
P^TBP = \pmatrix{A\\ & \ddots \\&& A}.
$$
From there, note that the determinant of a block-diagonal matrix is the product of the determinants of its blocks, which is to say that
$$
\det(B) = \det(P^TBP) = \det(A) \cdots \det(A) = \det(A)^n.
$$
</s>