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Calculate the determinant of the matrix, which is obtained by crossing out crossing out columns i and j.

$$ \left(\begin{array}{ccccc} 1 & x_{1} & \cdots & x_{1}^{n} & x_{1}^{n+1} \\ 1 & x_{2} & \cdots & x_{2}^{n} & x_{2}^{n+1} \\ \vdots & \vdots & \vdots & \vdots \\ 1 & x_{n} & \cdots & x_{n}^{n} & x_{n}^{n+1} \end{array}\right) $$

Alex_Lesley
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  • What have you tried? What do you know? Where is this question from? I am voting to close this question for a lack of context. – ViktorStein Mar 25 '20 at 18:11
  • I added two lines and tried to expand the determinant into these two lines – Alex_Lesley Mar 25 '20 at 18:12
  • This is called the Vandermonde matrix. – ViktorStein Mar 25 '20 at 18:13
  • See here. – Ragib Zaman Mar 25 '20 at 18:15
  • this is not the determinant of Vandermond – Alex_Lesley Mar 25 '20 at 18:16
  • this matrix has size n × (n + 2) – Alex_Lesley Mar 25 '20 at 18:17
  • Yes but after you delete two columns your matrix has the form addressed in the answer by David E Speyer that I linked to above. – Ragib Zaman Mar 25 '20 at 18:19
  • Add two rows, $1,x,x^2,...,x^{n+1}$ and $1,y,y^2,...,y^{n+1}$ on top. The determinant of the resulting matrix is a Vandermonde of the numbers $x,y,x_1,x_2,...,x_n$. So, it is equal to $\prod_i(x_i-x)\prod_i(x_i-y)\prod_{i<j}(x_j-x_i)$. On the other hand, expanding the determinant according to the Laplace formula corresponding to the first two rows you get that it is equal to a polynomial $\sum_{i,j}D_{i,j}(x^iy^j-x^jy^i)$, where $D_{i,j}$ is the corresponding minor obtained deleting the first two rows and columns $i$ and $j$. –  Mar 25 '20 at 18:28
  • Now you can find those coefficients from the first expression using Vieta's formulas. $D_{i,j}$ is your determinant up to the sign in Laplace's expansion. –  Mar 25 '20 at 18:28
  • So, it is going to be something like $s_is_j\prod_{i<j}(x_j-x_i)$, where $s_i$ is the sum of products of $i$ of the $x_1,x_2,...,x_n$. –  Mar 25 '20 at 18:34
  • @ViktorGlombik this matrix has size n × (n + 2) – Alex_Lesley Mar 26 '20 at 10:20
  • This question is quite clearly distinct to and not answered by the "duplicate". Questions get closed for being duplicates when they are not far too often on Math.SE these days... – Ragib Zaman Mar 26 '20 at 17:23

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