Determinant of order n of a matrix

Question

Do you have some ideas, please, to calculate this determinant? I've tried applying some recurrence method by expanding along the last column, but I haven't been successful. I appreciate any help you can provide.

Matrix:

Answer 1

$ |M_n| ~ = ~ ? $

Define

$v_{ij}:=$ $ \begin{cases} a_i b_j & \mbox{i < j} \\ a_j b_i & \mbox{i $\geq$ j} \end{cases} $ , i=1,...,n-1, j=1,...,n

$v_j:=(v_{ij})_{i=1,...,n-1}\in \mathbb{R}^{n-1}$ , j=1,...,n $~~$ : $j$th column of matrix $ M_n $ without last row

$w:=(a_i)_{i=1,...,n-1}\in \mathbb{R}^{n-1}$

$V:=(v_j)_{j=1,...,n} \in \mathbb{R}^{(n-1)\times n} ~~$ : matrix $M_n$ without last row for fixed n

$V_k:=(v_j)_{j\in\{1,...,n\}\setminus\{k\}} \in \mathbb{R}^{(n-1)\times (n-1)} ~~$ : matrix $V$ without $k$th column

expand using last row:

$(1) ~~ |M_n| = \begin{vmatrix} v_1 & v_2 & v_3 & ... & v_{n-1} & v_n \\ a_1 b_n & a_2 b_n & a_3 b_n & ... & a_{n-1} b_n & a_n b_n \end{vmatrix}= \sum_{j=1}^n (-1)^{n+j}a_jb_n |V_j| \ $

$(2) ~~ \sum_{j=1}^n (-1)^{n+j}a_jb_n |V_j| = \sum_{j=1}^{n-2} (-1)^{n+j}a_jb_n |V_j| + \sum_{j=n-1}^{n} (-1)^{n+j}a_jb_n |V_j| $

$(3) ~~ |V_j| = \begin{vmatrix} ... & v_{n-1} & v_n \\ \end{vmatrix} \ ~~~ \forall j=1,...,n-2$

$v_{n-1} = b_{n-1} w ~ $ and $ ~ v_n = b_n w ~ \Rightarrow ~ |V_j| = b_{n-1} b_n \begin{vmatrix} ... & w & w \\ \end{vmatrix} = 0 \ ~~~ \forall j=1,...,n-2$

$\Rightarrow ~ (4) ~~ \sum_{j=1}^{n-2} (-1)^{n+j}a_jb_n |V_j| = 0 $

$(1), ~ (2), ~ (4) \Rightarrow ~ |M_n| = (-1)^{2n-1}a_{n-1}b_n |V_{n-1}| + (-1)^{2n}a_nb_n |V_n|$

$(5) ~~ |M_n| = -a_{n-1}b_n |V_{n-1}| + a_nb_n |V_n|$

$|V_{n-1}| = \begin{vmatrix} v_1 & v_2 & ... & v_{n-2} & v_n \\ \end{vmatrix} = \begin{vmatrix} v_1 & v_2 & ... & v_{n-2} & b_n w \\ \end{vmatrix} $

$|V_{n-1}| = {b_n \over b_{n-1}} \begin{vmatrix} v_1 & v_2 & ... & v_{n-2} & b_{n-1} w \\ \end{vmatrix} $

$|V_{n}| = \begin{vmatrix} v_1 & v_2 & ... & v_{n-2} & v_{n-1} \\ \end{vmatrix} = \begin{vmatrix} v_1 & v_2 & ... & v_{n-2} & b_{n-1} w \\ \end{vmatrix} $

$ ~~~~~~~~~~~~~~~~~~~~~ \Downarrow$

$(6) ~~ |M_n| = -a_{n-1} {b_n^2 \over b_{n-1}} |V_n| + a_nb_n |V_n| = (a_n - {b_n \over b_{n-1}} a_{n-1} ) b_n ~ |V_n| $

note $ V_n = M_{n-1} $

$(7) ~~ |M_n| = (a_n - {b_n \over b_{n-1}} a_{n-1} ) b_n ~ |M_{n-1}| $

setting $~M_0 \equiv (1), ~ a_0 \equiv 0 ~~ and ~~ b_0 \equiv 1 $

$(8) ~~ |M_n| = \prod_{k=1}^n (a_k - {b_k \over b_{k-1}} a_{k-1} ) b_k ~~~~~ \forall n \in \mathbb{N} ~~~~~~~~ \square $