Determinant of matrix? How can we calculate the determinant of this $\,pn\times pn\,$ matrix. I have tried at my best level, and still am not able to come up with a solution. The matrix $a_{ij}$ entry is defined as
$$
a_{ij}
 =
\begin{cases}
c & \text{if }~ i = j\\
0 & \text{if }~ pk \leq i,j < p(k+1) \text{ for some }k\\
1 & \text{otherwise}
\end{cases}
$$
Edit: choice of $k$ is arbitrary.
 A: Your matrix can be written as $A=cI+B$, where $I$ is the identity matrix and $B$ is a matrix consisting of $n\times n$ blocks of size $p\times p$, with the blocks along the diagonal filled with $0$s and the remaining blocks filled with $1$s. Thus the eigenvalues of $A$ can be determined by determining the eigenvalues of $B$.
Any vector with non-zero entries in only one of the blocks and entries summing to $0$ is annihilated by $B$. There are $n(p-1)$ linearly independent such vectors, so $B$ has $n(p-1)$ eigenvalues $0$. The remaining eigenspaces are spanned by the $n$ vectors $v_k$ filled with $1$s in block $k$ and $0$s elsewhere. We can write $B=E-D$, with $E$ the matrix filled with $1$s, and $D$ a matrix with blocks of $1$s along the diagonal and $0$s elsewhere. Then the vector $v$ filled with $1$s is an eigenvector of $E$ with eigenvalue $np$ and of $D$ with eigenvalue $p$, so it is an eigenvector of $B$ with eigenvalue $p(n-1)$. All linear combinations of the $v_k$ that are orthogonal to $v$ are eigenvectors of $E$ with eigenvalue $0$ and eigenvectors of $D$ with eigenvalue $p$, so they are eigenvectors of $B$ with eigenvalue $-p$. There are $n-1$ linearly independent vectors of this kind.
In total, $B$ has $n(p-1)$ eigenvalues $0$, one eigenvalue $p(n-1)$ and $n-1$ eigenvalues $-p$. Since the corresponding eigenvalues of $A$ are shifted by $c$, we have
$$
\det A=c^{n(p-1)}(c-p)^{n-1}(c+p(n-1))\;.
$$
