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Consider the following Vandermonde matrix $$ V_n = \begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1} \end{pmatrix}. $$

It is well-known [1] that the determinant of $V_n$ is defined by $$ \displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right)\tag{1} $$

Let $V_{n-1}^{(i,j)}$ be a square matrix such that it is obtained by the removing $i$th row and $j$th column of $V_n$.

My question: Is it possible to get a closed-form for the determinant of $V_{n-1}^{(i,j)}$ similar to $(1)$.

My try: If $j=n$ then $V_{n-1}^{(i,n)}$ is a Vandermonde matrix and there is a closed-form for its determinant as $(1)$.

Thanks for any suggestions.

Edit: I think the general case of the proposed question is as follows; what is the closed-form of the determinant of the next matrix

$$ w_n = \begin{pmatrix} 1 & x_1^{i_1} & x_1^{i_2} & \cdots & x_1^{i_{n-2}} & x_1^{i_{n-1}} \\ 1 & x_2^{i_1} & x_2^{i_2} & \cdots & x_2^{i_{n-2}} & x_2^{i_{n-1}} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n^{i_1} & x_n^{i_2} & \cdots & x_n^{i_{n-2}} & x_n^{i_{n-1}} \end{pmatrix}. $$ where $i_t$, for $1\leq t \leq n-1$, are positive integer numbers such that $i_1<i_2<\cdots<i_{n-2}<i_{n-1}$.

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  • $\begingroup$ Pretty sure this is a duplicate of something posted a month ago or something like it..? $\endgroup$ – mathreadler Dec 26 '17 at 23:14
  • $\begingroup$ @mathreadler I appreciate to help me to find that question. Thsnks $\endgroup$ – Amin235 Dec 26 '17 at 23:20
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    $\begingroup$ The determinant of the general form of Vandermonde matrix is the product of the ordinary Vandermonde determinant and a Schur polynomial. $\endgroup$ – achille hui Jan 1 '18 at 22:45
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The question is to give the first minor of Vandermonde matrix.

Note that for all square matrix $\mathbf {A}$, ${\mathbf {A} \operatorname {adj} (\mathbf {A} )=\det(\mathbf {A} )\,\mathbf {I}}$, and $\operatorname {adj} (\mathbf {A} )=[{\mathbf {C} _{ij}]^{\mathsf {T}}=[(-1)^{i+j}\mathbf {M} _{ji}}]$ where $\mathbf {M} _{ij}$ is the $(i,j)$ minor of $\mathbf {A}$.

And here is the inverse of Vandermonde matrix.

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