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There is a similar question like this. And this is extension of this question

How can we calculate the determinant of this $\,pn-1\times pn-1\,$ 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 $$ a_{ij} = \begin{cases} a & \text{if }~ i = j\\ 0 & \text{if }~ pk \leq i,j < p(k+1) \text{ for some }k\\ b & \text{otherwise} \end{cases} $$

It would be very nice if I can get a general formula and a clear elaboration is much appreciated.

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Alen, I'm not going to write it all out for a third time if you don't make more of an effort at pointing out where you're having trouble applying my method from the other answer and the long comment thread under it to this case. What did you do in order to apply it? How far did you get? What precisely kept you from proceeding after that? Which eigenvectors and eigenvalues did you find, which space was left where you weren't able to find the eigenvalues? – joriki May 8 '13 at 7:34
@joriki:"we need to deal with the last component separately. All eigenvectors in the linked question except for the one filled with 1s sum to 0. Thus we can append a 0 to them to obtain eigenvectors of the present matrix." This part is giving me trouble. – Alen May 8 '13 at 8:44
That's a lot of text. Which part is the problem? Do you understand which eigenvectors I'm referring to? Do you understand why they sum to $0$? Do you understand what I mean by appending a $0$ to them? Do you understand why that yields eigenvectors of the present matrix? Communication is very inefficient if I have to keep guessing what you understand and what you don't understand; that leaves all the work of explaining everything all over again on my side; but it's your question, not mine; you should try to do more of the work. – joriki May 8 '13 at 8:48
"Do you understand what I mean by appending a 0 to them?" ans: no – Alen May 8 '13 at 8:49
@joriki:You are helping me very much and I am grateful for that. I don't understand what do you mean by appending a 0 to them. And how that provides eigenvector of the present matrix. – Alen May 8 '13 at 8:52

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