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I really struggle with this problem, how do you calculate the determinant of matrix $A \in \mathbb{R}^{n \times n}$, whose expression is $$ \begin{pmatrix} 2 & 1& ...& 1\\ 1& 2& ...& 1\\ 1& 1& 2& ...\\ 1& 1& ...& 2 \end{pmatrix} ? $$

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Did you try some induction? – DonAntonio Feb 2 '14 at 13:35
A more general problem which can be solved by very similar methods:… and… – Martin Sleziak Dec 16 '14 at 15:46
up vote 5 down vote accepted


Add all the columns to the first one then subtract the first row from the other rows. Develop now according to the first column.

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Hint: Find all the nonzero eigenvalues of $A - I$ with their multiplicities and recall that commuting matrices are simultaneously triangularizable.

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I don't get the simultaneously triangularizable part. Certainly one does not need that rather deep fact to see that the eigenvalues (with algebraic multiplicities) of $A-I$ are obtained from those of $A$ by subtracting $1$ from each eigenvalue? – Marc van Leeuwen Feb 2 '14 at 13:58
@MarcvanLeeuwen Oh you're right, I don't know why I used this reasoning. Thank you. – user33321 Feb 2 '14 at 14:43

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