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Consider real numbers $a_i,b_i\in\mathbb{R}$ and the following determinant

$$ \alpha_n= \begin{vmatrix} a_1 + b_1 & b_1 & b_1 & b_1&\cdots & b_1 \\ b_2 & a_2 + b_2 & b_2 & b_2 & \cdots & b_2 \\ b_3 & b_3 & a_3 + b_3 & b_3 & \cdots & b_3 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ b_n & b_n & b_n & b_n & \cdots & a_n + b_n\end{vmatrix} $$ My attempt was to substitute column 1 with column 1 - column 2 $(C_1 \to C_1 - C_2)$. Then do the same with others columns: $C_2 \to C_2 - C_3$,..., $C_n\to C_n - C_1$. After all these transformations, arise this determinant $$ \alpha_n =\begin{vmatrix} a_1 &0& 0 &0&\cdots & -a_1 \\ -a_2 &a_2 & 0 & 0 & \cdots & 0 \\ 0 & -a_3 & a_3 & 0 & \cdots &0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & a_n \end{vmatrix} $$ So it seems that $\alpha_n$ only depends on $a_i$. However, computing the case $n=2$, gives $$ \alpha_2 = \begin{vmatrix} a_1 +b_1 &b_1 \\ b_2 & a_2 + b_2 \end{vmatrix} = a_1a_2 + a_1b_2 + a_2b_1. $$ What am I doing wrong? How can I compute a closed form for $\alpha_n$? Thank you in advance.

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    $\begingroup$ The last operation on $C_n$ doesn't seem correct, as $C_1$ has been modified before. $\endgroup$
    – Damien
    Feb 3 '21 at 9:30
  • $\begingroup$ Thank you for your comment! $\endgroup$
    – Senna
    Feb 3 '21 at 10:12
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As your error was already clarified in other answers here only a hint to solution will be given.

I will assume $a_i\ne0$. Observe that your matrix is in form: $$ \alpha=A+u^Tv, $$ where $$A=\operatorname {diag}(a_1,a_2,\dots,a_n),\quad u=(b_1,b_2,\dots,b_n),\quad v=(1,1,\dots,1). $$ Then by matrix determinant lemma: $$ \det (\alpha)=(1+vA^{-1}u^T)\det (A)=\left(1+\sum_i\frac {b_i }{a_i}\right)\prod_i a_i. $$

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  • $\begingroup$ This answers my 2nd question in a very nice way. Thank you! +1 $\endgroup$
    – Senna
    Feb 3 '21 at 15:01
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Note that while performing row or column operations on determinants, it is necessary to preserve atleast one column/row, we cannot change all of them at once.

In your case, you cannot do $C_n \to C_n-C_1$.

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  • $\begingroup$ Thank you for your answer! $\endgroup$
    – Senna
    Feb 3 '21 at 10:12

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