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Given a $n\times n$ matrix $$A= \begin{pmatrix} a_{1}+p&p&p&p&\cdots & p&p&p\\ p&a_{2}+p&p&p&\cdots&p&p&p\\ \vdots & \vdots & \vdots & \vdots &\ddots & \vdots & \vdots & \vdots &\\ p&p&p&p&\cdots&p&a_{n-1}+p&p\\ p&p&p&p&\cdots&p&p&a_{n}+p \end{pmatrix} $$

we have
$$\det A = \left(\prod_{i=1}^{n} a_{i}\right)\left(1+p\sum_{i=1}^{n} \dfrac{1}{a_i}\right)$$

My attempt was to prove that by induction which turned out to be tedious.

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    $\begingroup$ To reduce it to the second factor, multiply the $j$th row by $a_ja_j^{-1}$ and then "pull out" each $a_j$ in front of the determinant. $\endgroup$ – Friedrich Philipp Mar 20 '16 at 4:58
  • $\begingroup$ You could use that $\det$ is linear in each column. Then, it is a simple induction. $\endgroup$ – user251257 Mar 20 '16 at 5:19
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Subtract the bottom row from the others to get $$\mathrm{det} \begin{pmatrix} a_1 & 0 & ... & -a_n \\ 0 & a_2 & ... & -a_n \\ ... & ... & ... & ... \\ p & p & ... & a_n + p \end{pmatrix}.$$ Then column expansion along the first column: $$a_1 \cdot \mathrm{det} \begin{pmatrix} a_2 & 0 & ... & -a_n \\ 0 & a_3 & ... & -a_n \\ ... & ... & ... & ... \\ p & p & ... & a_n + p \end{pmatrix} + (-1)^{n-1} p \cdot \mathrm{det} \begin{pmatrix} 0 & 0 & ... & -a_n \\ a_2 & 0 & ... & -a_n \\ ... & ... & ... & ... \\ 0 & 0 & ... & -a_n \end{pmatrix}.$$ The first determinant is known by induction from the case $n-1$. The second is upper-triangular after permuting the rows.

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