The Determinant of a Special Vandermonde Matrix

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  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 $i_1<i_2<\cdots<i_{n-2}<i_{n-1}$.

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

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 $\mathbf {A}$.