Determinant of a Certain 3 by 3 Block Matrix with Scaled Identity Blocks What is the determinant or/and eigenvalues of the following 3 by 3 block matrix:
$$\left[\begin{array}{ccc} \frac{3}{4}I & \frac{1}{4}I & \frac{1}{4}I \\ \frac{1}{4}I & \frac{3}{4}I & -\frac{1}{4}I \\ \frac{1}{4}I & -\frac{1}{4}I & \frac{3}{4}I \end{array}\right]$$
Where $I$ is the identity matrix of fixed size. 
 A: The block matrix (let us denote by $M$) can be expressed as the Kronecker product of matrices $A$ and $I$ (the fixed size identity matrix, of dimension $n$) as follows:-
$$M=A\otimes  I$$
where $A$ is the $3\times3$ matrix:-
$$A=\left[\begin{array}{ccc} \frac{3}{4} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ \frac{1}{4} & -\frac{1}{4} & \frac{3}{4} \end{array}\right]$$
From the linked Wikipedia entry, for square matrices $A$ of size $m$ and $B$ of size $n$ the determinant of their Kronecker product is given by
$$|A\otimes B|=|A|^n|B|^m$$
Applying this to the matrix in question, results in
$$|M|=|A|^n|I|^3=\left(\frac{1}{4}\right)^n\times1=\frac{1}{4^n}$$
The eigenvalues of $A$ are
$\lambda_1=\frac{1}{4}, \lambda_2=1, \lambda_3=1$, while the eigenvalues of $I$ are $\mu_j=1$ for $j\in\{1,2,..,n\}$, leading to the eigenvalues of $M$ being $\lambda_i\mu_j$ for $i\in\{1,2,3\}$ and $j\in\{1,2,..,n\}$. 
Thus, $M$ will have $n$ eigenvalues of value $\frac{1}{4}$ and $2n$ eigenvalues of value $1$.
