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Please hint me. ‎How ‎can I ‎calculate ‎determinant ‎of ‎matrix ‎‎$‎X‎$‎?‎ \begin{equation*}‎ ‎\mathbf{X}=\left(‎ \begin{array}{ccc}‎ A&B&‎\cdots&B\\‎ B&A&‎\cdots& B\\‎ \vdots & \vdots & \ddots &\vdots\\‎ B&B&‎\cdots&A‎ \end{array}\right) \end{equation*}‎‎ ‎where ‎‎$ B‎‎‎‎$‎denote ‎the ‎matrix ‎each ‎of ‎whose ‎entries ‎is ‎‎$‎+1$ and ‎$‎A$ is the diagonal matrix whose its entry is ‎$‎a$‎, ‎$ A ‎‎$ ‎and ‎$‎B‎$ ‎are squre matrices with the same order‎‎‎‎‎.‎

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Let $A$ and $B$ be $n\times n$ matrices, and so let $X$ be a $kn\times kn$ matrix.

Let $X_0$ be the matrix obtained from $X$ by setting $A$ to be the zero matrix. The eigenvalues of $X_0$ are $kn-n$ (with multiplicity $1$), $-n$ (with multiplicity $k-1$), and $0$ (with multiplicity $kn-k$).

Now, $X=X_0+aI_{kn}$, where $I_{kn}$ is the $kn\times kn$ identity matrix. Hence $X$ has eigenvalues $kn-n+a$ (with multiplicity $1$), $a-n$ (with multiplicity $k-1$) and $a$ (with multiplicity $kn-k$). Since $\det(X)$ is the product of the eigenvalues of $X$, we get $$\det(X)=(kn-n+a)(a-n)^{k-1}a^{kn-k}.$$

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  • $\begingroup$ thanks very very much. Do you allow me that I use your answer in my paper? $\endgroup$ – M.Mehranian Jan 23 '15 at 14:23
  • $\begingroup$ In a research paper? $\endgroup$ – Casteels Jan 23 '15 at 14:24
  • $\begingroup$ yes, I want it for the proof of a part of theorem $\endgroup$ – M.Mehranian Jan 23 '15 at 14:28
  • $\begingroup$ Sure no problem. $\endgroup$ – Casteels Jan 23 '15 at 14:32
  • $\begingroup$ Ok, I didn't know, I do it now. $\endgroup$ – M.Mehranian Jan 23 '15 at 14:49

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