[Question 2014a01] Answer 2 (the post n-1 column splitting method is provided by Guo Xiaojun)

[Question 2014a01] Answer 2 (the post n-1 column splitting method is provided by Guo Xiaojun)

\ [| A | =\begin {vmatrix} 1 & X_1 ^ 2-ax_1 & X_1 ^ 3-ax_1 ^ 2 & \ cdots & X_1 ^ n-ax_1 ^ {n-1} \ 1 & X_2 ^ 2-ax_2 & X_2 ^ 3-ax_2 ^ 2 & \ cdots & X_2 ^ n-ax_2 ^ {n-1 }\\ vdots & \ vdots \ 1 & x_n ^ 2-ax_n & x_n ^ 3-ax_n ^ 2 & \ cdots & x_n ^ n-ax_n ^ {n-1} \ end {vmatrix }. \]

Starting from the second column, each column can be recorded as "1" and "2", splitting the rear \ (n-1 \) column, respectively, it can be split into the sum of \ (2 ^ {n-1} \) determinant. let's consider these factors. Suppose the column \ (I \) is the first column to select "1" from left to right, which is determined by the nature of the determinant, to make the split determine a non-zero value, select "1" for the column \ (I + 1, \ cdots, n \). In this case, the column \ (2, \ cdots, i-1 \) columns are selected "2", so each column can be extracted from the public factor \ (-A \), the determinant can be extracted \ (-) {I-1 }\). according to the above analysis

\ [| A | =\begin {vmatrix} 1 & X_1 ^ 2 & X_1 ^ 3 & \ cdots & X_1 ^ n \ 1 & X_2 ^ 2 & X_2 ^ 3 &\ cdots & X_2 ^ n \ vdots & \ vdots \ 1 & x_n ^ 2 & x_n ^ 3 & \ cdots & x_n ^ n \ end {vmatrix} + (-) \ begin {vmatrix} 1 & X_1 & X_1 ^ 3 & \ cdots & X_1 ^ n \ 1 & X_2 & X_2 ^ 3 & \ cdots & X_2 ^ n \ vdots & \ vdots & \ vdots \ 1 & x_n ^ 3 & \ cdots & x_n ^ n \ end {vmatrix} \]

\ [+ \ Cdots + (-) ^ {n-1} \ begin {vmatrix} 1 & X_1 & X_1 ^ 2 & \ cdots & X_1 ^ {n-1} \ 1 & X_2 & X_2 ^ 2 & \ cdots & X_2 ^ {n-1 }\\\ vdots & \ vdots \ 1 & x_n ^ 2 & \ cdots & x_n ^ {n-1} \ end {vmatrix }. \]

Consider \ (| A | \) as the \ (n + 1 \) Order Determinant expanded by the first line,

\ [| A | =\begin {vmatrix} 0 &-1 &-A & \ cdots &-a ^ {n-1} \ 1 & X_1 & X_1 ^ 2 & \ cdots & X_1 ^ n \ 1 & X_2 & X_2 ^ 2 & \ cdots & X_2 ^ n \ vdots & \ vdots \ 1 & x_n & x_n ^ 2 & \ cdots & x_n ^ n \ end {vmatrix }. \]

When \ (A \ NEQ 0 \), we have (the next step is to split by the first column ):

\ [| A | =-\ frac {1} {A} \ begin {vmatrix} 1 + (-1) & A ^ 2 & \ cdots & A ^ n \ 1 + 0 & X_1 & X_1 ^ 2 & \ cdots & X_1 ^ n \ 1 + 0 & X_2 & X_2 ^ 2 & \ cdots & X_2 ^ n \ vdots & \ vdots \ 1 + 0 & x_n ^ 2 & \ cdots & x_n ^ n \ end {vmatrix} \]

\ [=-\ Frac {1} {A} \ big (\ prod _ {1 \ Leq I <J \ Leq n} (x_j-x_ I) \ prod _ {I = 1} ^ N (x_i-a)-\ prod _ {1 \ Leq I <J \ Leq n} (x_j-x_ I) \ prod _ {I = 1} ^ nx_ I \ big) \]

\ [= \ Frac {1} {A} \ prod _ {1 \ Leq I <J \ Leq n} (x_j-x_ I) \ big (\ prod _ {I = 1} ^ nx_ I-\ prod _ {I = 1} ^ N (x_i-a) \ big ). \]

When \ (a = 0 \), the order escalation method and the Vander monde determining factor can be used to find

\ [| A | = \ prod _ {1 \ Leq I <J \ Leq n} (x_j-x_ I) \ big (\ sum _ {I = 1} ^ nx_1 \ cdots \ hat {x} _ I \ cdots x_n \ big ). \ quad \ Box \]

[Question 2014a01] Answer 2 (the post n-1 column splitting method is provided by Guo Xiaojun)