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$\left|\begin{matrix} -x&a_2&\cdots&a_{n}\\ a_1&-x&\cdots&a_{n}\\ a_1&a_2&\cdots&a_{n}\\ \vdots&\vdots&\ddots&\vdots\\ a_1&a_2&\cdots&-x \end{matrix}\right|$

so i tried to find and expression $D_{n}=kD_{n-1}+tD_{n-2}$ by doing this $\left|\begin{matrix} -x&a_2&\cdots&a_{n}\\ a_1&-x&\cdots&a_{n}\\ a_1&a_2&\cdots&a_{n}\\ \vdots&\vdots&\ddots&\vdots\\ a_1&a_2&\cdots&a_n-x-a_n \end{matrix}\right|$ so we split it and get $\left|\begin{matrix} -x&a_2&\cdots&a_{n}\\ a_1&-x&\cdots&a_{n}\\ a_1&a_2&\cdots&a_{n}\\ \vdots&\vdots&\ddots&\vdots\\ a_1&a_2&\cdots&a_n \end{matrix}\right|$ which is an upper trianglular matrix hence the determinant is $\prod_{1\le i\le n}(-x-a_i)$

$\left|\begin{matrix} -x&a_2&\cdots&0\\ a_1&-x&\cdots&0\\ a_1&a_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ a_1&a_2&\cdots&-x-a_n \end{matrix}\right|$ now i expand it along the last column and get $(-x-a_n)D_{n-1}$ but the expression gets very complicated , can i please get a hint on a good way to solve it , i didnt see a better way by manipulating rows/columns(e.g. adding/subtracting all columns to the first but since i miss the $a_i$ in the $i$-th row i cant manipulate the determinant better).

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Subtract the last row from each preceding row to obtain

$$D_n(a_1, \ldots, a_n;x) = \left|\begin{matrix} -x&a_2&\cdots&a_{n}\\ a_1&-x&\cdots&a_{n}\\ \vdots&\vdots&\ddots&\vdots\\ a_1&a_2&\cdots&-x \end{matrix}\right| = \left|\begin{matrix} -x-a_1&0&\cdots&0&a_{n}+x\\ 0&-x-a_2&\cdots&0&a_{n}+x\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&-x-a_{n-1}&a_n+x\\ a_1&a_2&\cdots&a_{n-1}&-x \end{matrix}\right|$$

Now expand along the first column and iterate the procedure: $$= (-x-a_1)\left|\begin{matrix} -x-a_2&\cdots&0&a_{n}+x\\ \vdots&\ddots&\vdots&\vdots\\ 0&\cdots&-x-a_{n-1}&a_n+x\\ a_2&\cdots&a_{n-1}&-x \end{matrix}\right|+(-1)^{n+1}a_1\left|\begin{matrix} 0&\cdots&0&a_{n}+x\\ -x-a_2&\cdots&0&a_{n}+x\\ \vdots&\ddots&\vdots&\vdots\\ 0&\cdots&-x-a_{n-1}&a_n+x\\ a_2&\cdots&a_{n-1}&-x \end{matrix}\right|$$ \begin{align} &=(-x-a_1)D_{n-1}(a_2, \ldots, a_n;x)+a_1(-x-a_2)\cdots(-x-a_{n-1})\\ &=(-x-a_1)(-x-a_2)D_{n-2}(a_3, \ldots, a_n;x)+a_1(-x-a_2)\cdots(-x-a_{n-1})+a_2(-x-a_1)(-x-a_3)\cdots (-x-a_n)\\ &=\cdots\\ &=(-x-a_1)\cdots(-x-a_n)+\sum_{i=1}^n a_i(-x-a_1)\cdots(-x-a_{i-1})(-x-a_{i+1})\cdots (-x-a_n) \end{align}

which is the same as @user1551's result.

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  • $\begingroup$ @ancientmathematician Thanks, that was completely wrong. I got user1551's result now. $\endgroup$ – mechanodroid Jun 26 at 18:40
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Your matrix is $M=(-D-xI)+ue^T$ where $D=\operatorname{diag}(a_1,\ldots,a_n)$, $u=(a_1,\ldots,a_n)^T$ and $e=(1,\ldots,1)^T$. Using the identity $\det(A+uv^T)=\det(A)+v^T\operatorname{adj}(A)u$, we get \begin{aligned} \det(M) &=\det\left((-D-xI)+ue^T\right)\\ &=\det(-D-xI) + e^T\operatorname{adj}(-D-xI)\,u\\ &=\prod_{i=1}^n(-a_i-x) + \sum_{i=1}^na_i\prod_{j\ne i}(-a_j-x). \end{aligned}

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