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.
Best Answer
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: $$ \det (\alpha)=(1+vA^{-1}u^T)\det (A)=\left(1+\sum_i\frac {b_i }{a_i}\right)\prod_i a_i. $$