[Question 2014a02] Answer 2 (sum + split method, provided by Zhang chengchun)

[Question 2014a02] Answer 2 (Sum Method +Splitting Method provided by Zhang chengchun)

Add the second column, \ (\ cdots \), and \ (n \) of the determinant \ (D_n \) to the first column.

\ [D_n = \ begin {vmatrix} \ sum _ {I = 1} ^ na_ I + (n-2) a_1 & a_1 + A_2 & \ cdots & a_1 + A _ {n-1} & a_1 + a_n \ sum _ {I = 1} ^ na_ I + (n-2) a_2 & 0 & \ cdots & A_2 + A _ {n-1} & A_2 + a_n \ vdots & \ vdots \ sum _{ I = 1} ^ na_ I + (n-2) A _ {n-1} & A _ {n-1} + A_2 & \ cdots & 0 & A _ {n-1} + a_n \ sum _ {I = 1} ^ na_ I + (n-2) a_n & a_n + A_2 & \ cdots & a_n + A _ {n-1} & 0 \ end {vmatrix }. \]

Split the first column of the above determining factor

\ [D_n = \ sum _ {I = 1} ^ na_ I \ begin {vmatrix} 1 & a_1 + A_2 & \ cdots & a_1 + A _ {n-1} & a_1 + a_n \ \ 1 & 0 & \ cdots & A_2 + A _ {n-1} & A_2 + a_n \ vdots & \ vdots \ 1 & _ {n-1} + A_2 & \ cdots & 0 & A _ {n-1} + a_n \ 1 & a_n + A_2 & \ cdots & a_n + A _ {n-1} & 0 \ end {vmatrix} \]

\ [+ (N-2) \ begin {vmatrix} A_1 & a_1 + A_2 & \ cdots & a_1 + A _ {n-1} & a_1 + a_n \ A_2 & 0 & \ cdots & A_2 + A _ {n-1} & A_2 + a_n \ vdots & \ vdots \ A _ {n-1} & A _ {n-1} + A_2 & \ cdots & 0 & A _ {n-1} + a_n \ a_n & a_n + A_2 & \ cdots & a_n + A _ {n-1} & 0 \ end {vmatrix }. \]

The first determinant on the right of the preceding formula: multiply the first column by \ (-a_ I \) to the column \ (I \), \ (I = 2, \ cdots, n \), then, a public factor is proposed from each row. The second determinant on the right of the above formula: multiply the first column by \ (-1 \) to the column \ (I \), respectively, then, the public factor \ (I = 2, \ cdots, n \) is proposed from each column.

\ [D_n = \ sum _ {I = 1} ^ na_ I \ prod _ {I = 1} ^ na_ I \ begin {vmatrix} \ frac {1} {A_1} & 1 & \ cdots & 1 & 1 \ frac {1} {A_2} &-1 & \ cdots & 1 & 1 \ vdots & \ vdots \ frac {1} {A _ {n-1} & 1 & \ cdots &-1 & 1 \ frac {1} {a_n} & 1 &\ cdots & 1 &-1 \ end {vmatrix} \]

\ [+ (N-2) \ prod _ {I = 2} ^ na_ I \ begin {vmatrix} A_1 & 1 & \ cdots & 1 & 1 \ A_2 &-1 & \ cdots & 1 & 1 \\ \ vdots & \ vdots \ A _ {n-1} & 1 & \ cdots &-1 & 1 \ a_n & 1 & \ cdots & 1 &-1 \ end {vmatrix }. \]

On the right side of the preceding formula, the two determine factors are the first line multiplied by \ (-1 \) and added to the \ (I \) line, \ (I = 2, \ cdots, n \), which can be changed to the Claw Type deciding factor:

\ [D_n = \ sum _ {I = 1} ^ na_ I \ prod _ {I = 1} ^ na_ I \ begin {vmatrix} \ frac {1} {A_1} & 1 & \ cdots & 1 & 1 \ frac {1} {A_2}-\ frac {1} {A_1} &-2 & \ cdots & 0 & 0 \ vdots & \ vdots & \ vdots \ frac {1} {A _ {n-1}-\ frac {1} {A_1} & 0 & \ cdots & -2 & 0 \ frac {1} {a_n}-\ frac {1} {A_1} & 0 & \ cdots & 0 &-2 \ end {vmatrix} \]

\ [+ (N-2) \ prod _ {I = 2} ^ na_ I \ begin {vmatrix} A_1 & 1 & \ cdots & 1 & 1 \ a_2-a_1 &-2 & \ cdots & 0 & 0 \\ \ vdots & \ vdots \ A _ {n-1}-A_1 & 0 & \ cdots &-2 & 0 \ a_n-a_1 & 0 &\ cdots & 0 &-2 \ end {vmatrix }. \]

On the right side of the above formula, the two determine factors are: the \ (I \) line is multiplied by \ (\ frac {1} {2} \) and added to the first line, \ (I = 2, \ cdots, n \), available

\ [D_n = \ sum _ {I = 1} ^ na_ I \ prod _ {I = 1} ^ na_ I \ begin {vmatrix} \ frac {1} {2} (\ sum _ {I = 1} ^ n \ frac {1} {a_ I }) -\ frac {N-2} {2} \ frac {1} {A_1} & 0 & \ cdots & 0 \ frac {1} {A_2}-\ frac {1} {A_1} &-2 & \ cdots & 0 & 0 \ vdots & \ vdots \ frac {1} {_ {n-1 }}- \ frac {1} {A_1} & 0 & \ cdots &-2 & 0 \ frac {1} {a_n}-\ frac {1} {A_1} & 0 & \ cdots & 0 &-2 \ end {vmatrix} \]

\ [+ (N-2) \ prod _ {I = 2} ^ na_ I \ begin {vmatrix} \ frac {1} {2} (\ sum _ {I = 1} ^ Na _ I) -\ frac {N-2} {2} A_1 & 0 & \ cdots & 0 & 0 \ a_2-a_1 &-2 & \ cdots & 0 & 0 \ vdots & \ vdots & \ vdots \ A _ {n-1}-A_1 & 0 & \ cdots &-2 & 0 \ a_n-a_1 & 0 & \ cdots & 0 &-2 \ end {vmatrix} \]

\ [= (-2) ^ {N-2} \ prod _ {I = 1} ^ na_ I \ Bigg (n-2) ^ 2-\ big (\ sum _ {I = 1} ^ na_ I \ big) \ big (\ sum _ {I = 1} ^ n \ frac {1} {a_ I} \ big) \ Bigg ). \ quad \ Box \]

[Question 2014a02] Answer 2 (sum + split method, provided by Zhang chengchun)