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Let $$p (x) = \begin{vmatrix} 1 & x & x & \dots & x & x \\ x & 1 & x & \dots & x & x \\ x & x & 1 & \dots & x & x \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ x & x & x & \dots & 1 & x \\ x & x & x & \dots & x & 1 \\ \end{vmatrix} .$$

How to find the (multiple) zeroes, the degree of the polynomial and the initial coefficient, all depending on the natural number $b$?

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    $\begingroup$ What $b$?There is no $b$ in your question. $\endgroup$ – Alex M. Mar 28 '16 at 11:33
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Add all columns (from second one) to the first one and then substract first row from all the rest:

$$\begin{vmatrix}1+(n-1)x&x&x&\ldots&x\\0&1-x&0&\ldots&0\\0&0&1-x&\ldots&0\\..&..&..&..&..\\0&0&0&\ldots&1-x\end{vmatrix}=\left(1+(n-1)x\right)(1-x)^{n-1}$$

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Hint: You can determine the characteristic polynomial of the corresponding matrix. Note that the matrix is symmetric and therefore diagonalizable. Now $-x + 1$ is an eigenvalue with a geometric multiplicity of at least $n - 1$ and the sum of all eigenvalues [according to their algebraic multiplicity] is equal to the trace of the matrix.

Once you've calculated the characteristic polynomial you can determine $p(x)$ by plugging in $\lambda = 0$.

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  • $\begingroup$ n-1 zeros are only 1 ? $\endgroup$ – Algebra 2015 Mar 28 '16 at 11:06
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    $\begingroup$ I don't understand your question. $\endgroup$ – Dominik Mar 28 '16 at 11:08
  • $\begingroup$ I do not understand, which numbers are zeros ? $\endgroup$ – Algebra 2015 Mar 28 '16 at 11:11
  • $\begingroup$ You first need to calculate the characteristic polynomial, then you will be able to calculate the zeroes of $p$. $\endgroup$ – Dominik Mar 28 '16 at 11:14
  • $\begingroup$ Actually, you don't really need to exhibit the characteristic polynomial, as the determinant is simply the product of eigenvalues (but both ways to view this are equivalent, of course). @Algebra2015 see also this: math.stackexchange.com/questions/507641/… $\endgroup$ – xxx Mar 28 '16 at 11:51

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