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Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc} -x & a_2&a_3&a_4&\cdots &a_n\\ a_{1} & -x & a_3&a_4&\cdots &a_n\\ a_1&a_{2} & -x &a_4&\cdots &a_n\\ \vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\ a_1&a_{2} & a_3&a_4&\cdots & -x\\ \end{array}\end{bmatrix}$, then how to find the $\det (M)$?
Proof: First I started by taking the $a_i$ from each $i$-th columns, then$ |M|=\prod_{i=1}^{n}{a_i} \begin{vmatrix}\begin{array}{ccccccc} \frac{-x}{a_1} & 1&1&1&\cdots &1\\ {1} & \frac{-x}{a_2} & 1&1&\cdots &1\\ 1&1 & \frac{-x}{a_3} &1&\cdots &1\\ \vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\ 1&1 & 1&1&\cdots & \frac{-x}{a_n}\\ \end{array}\end{vmatrix}$. After this is there any easiest way to find the determinant.

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closed as off-topic by user21820, Paul Frost, Mars Plastic, Daniele Tampieri, Javi Aug 18 at 13:35

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    $\begingroup$ What is your try? $\endgroup$ – Rajat Sep 24 '15 at 10:51
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    $\begingroup$ Up to $(-1)^n$, it's the characteristic polynomial of $M(0)$, if that helps. $\endgroup$ – lhf Sep 24 '15 at 11:07
  • $\begingroup$ Trying low values of $n$ will give you good idea of the general form. $\endgroup$ – lhf Sep 24 '15 at 11:08
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Subtract the first row from each of the other rows. Most of the terms are now zero, and you can expand across the first row. Each product misses one of the $(x+a_i)$ factors, replaced by $a_i$. So the determinant is $$(-1)^n\prod_i(x+a_i)\left[\frac x{x+a_1}-\frac {a_2}{x+a_2}-\frac {a_3}{x+a_3}...\right]\\ =(-1)^n\prod_i(x+a_i)\left[1-\sum_i\frac{a_i}{x+a_i}\right]$$

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  • $\begingroup$ Concise and simple! +1 $ $ $\endgroup$ – user1551 Sep 24 '15 at 11:39
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Let $D=-\operatorname{diag}(x+a_1,\,\ldots,\,x+a_n)$. Then $M=D+ea^T$. Using the rank-1 update formula for determinant, we have $\det M=(1+a^TD^{-1}e)\det(D)$. After some work, you should be able to prove that the determinant is $$(-1)^n\left[\prod_i (x+a_i)-\sum_ia_i\prod_{j\ne i}(x+a_j)\right].$$

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The empirical formula I got from considering $n=2,3,4$ in Wolfram Alpha is $$ (-1)^n(x^n-\sum_{k=2}^n (k-1)\sigma_k x^{n-k}) $$ where $\sigma_k$ is the $k$-th elementary symmetric polynomial in $a_1,\dots,a_n$.

I don't see how this follows at once from the other answers.

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  • $\begingroup$ even I also got the similar formula by using wolfram Mathematica, but I don't know how to write a proof. $\endgroup$ – L S B. user255259 Sep 24 '15 at 13:33
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Hint: $\det(M)$ is a polynomial in $x$ and $M$ is clearly singular if $x=-a_k$ for some $k$

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    $\begingroup$ I thought the exact same thing, but it's not. For $n=2$, the polynomial is $x^2-a_1a_2$. And for $n=1$, it's $-x$, not $x+a_1$. $\endgroup$ – lhf Sep 24 '15 at 11:15
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    $\begingroup$ I also tripped over this. For some reason, when I set the first $x$ to $a_1$, I subconsciously substituted the other $x$s by $a_2$ up to $a_n$, and mistakenly thought that the columns became linearly dependent. I could realise what was wrong only after I stared at the matrix for five minutes. $\endgroup$ – user1551 Sep 24 '15 at 11:47

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