Why is $\det \begin{pmatrix} x_1-x_0 & (x_1-x_0)x_1 & \cdots & (x_1-x_0)x_1^{n-1} \\ \vdots & \vdots & & \vdots \\ x_n-x_0 & (x_n-x_0)x_n & \cdots & (x_n-x_0)x_n^{n-1} \end{pmatrix} $

equal to

$$(x_1-x_0) \cdot (x_2-x_0)\cdots(x_n-x_0) \cdot \det\begin{pmatrix} 1 & x_1 & \cdots & x_1^{n-1} \\ \vdots & \vdots & & \vdots \\ 1 & x_n & ... & x_n^{n-1} \end{pmatrix}$$

I don't see how we can factor this out.


You extract the factor $x_1-x_0$ from the first line of the matrix, you extract the factor $x_2-x_0$ from the second line and so on. Don't forget that $\det$ is a multilinear map.

  • $\begingroup$ I would like to see the rule for that $\endgroup$ – user528814 Apr 21 '18 at 15:59
  • 1
    $\begingroup$ The rule is: if $A$ and $B$ are square matrices of the same size, if $\lambda\in\mathbb R$ and if each line of $A$ is equal to the corresponding line of $B$, except that the $i^\text{th}$ line of $A$ is the $i^\text{th}$ line of $B$ times $\lambda$, then $\det(A)=\lambda\det(B)$. $\endgroup$ – José Carlos Santos Apr 21 '18 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy