# Krattenthaler's Identity

## Theorem

$\begin{vmatrix} \left({x + q_2}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + p_2}\right) \\ \left({y + q_2}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + p_2}\right) \\ \left({z + q_2}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + p_2}\right) \end{vmatrix} = \left({x - y}\right) \left({x - z}\right) \left({y - z}\right) \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right)$

where $\left\vert{\, \cdot \,}\right\vert$ denotes determinant.

## Proof

 $\displaystyle$  $\displaystyle \begin{vmatrix} \left({x + q_2}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + p_2}\right) \\ \left({y + q_2}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + p_2}\right) \\ \left({z + q_2}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + p_2}\right) \end{vmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin{vmatrix} \left({x + q_2}\right) \left({x + q_3}\right) & \left({p_1 - q_2}\right) \left({x + q_3}\right) & \left({p_1 - q_3}\right) \left({x + p_2}\right) \\ \left({y + q_2}\right) \left({y + q_3}\right) & \left({p_1 - q_2}\right) \left({y + q_3}\right) & \left({p_1 - q_3}\right) \left({y + p_2}\right) \\ \left({z + q_2}\right) \left({z + q_3}\right) & \left({p_1 - q_2}\right) \left({z + q_3}\right) & \left({p_1 - q_3}\right) \left({z + p_2}\right) \end{vmatrix}$ Multiple of Row Added to Row of Determinant $\displaystyle$ $=$ $\displaystyle \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \begin{vmatrix} \left({x + q_2}\right) \left({x + q_3}\right) & x + q_3 & x + p_2 \\ \left({y + q_2}\right) \left({y + q_3}\right) & y + q_3 & y + p_2 \\ \left({z + q_2}\right) \left({z + q_3}\right) & z + q_3 & z + p_2 \end{vmatrix}$ Determinant with Row Multiplied by Constant $\displaystyle$ $=$ $\displaystyle \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \begin{vmatrix} x \left({x + q_3}\right) & x + q_3 & p_2 - q_3 \\ y \left({y + q_3}\right) & y + q_3 & p_2 - q_3 \\ z \left({z + q_3}\right) & z + q_3 & p_2 - q_3 \end{vmatrix}$ Multiple of Row Added to Row of Determinant $\displaystyle$ $=$ $\displaystyle \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right) \begin{vmatrix} x \left({x + q_3}\right) & x + q_3 & 1\\ y \left({y + q_3}\right) & y + q_3 & 1\\ z \left({z + q_3}\right) & z + q_3 & 1 \end{vmatrix}$ Determinant with Row Multiplied by Constant $\displaystyle$ $=$ $\displaystyle \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right) \begin{vmatrix} x^2 & x & 1\\ y^2 & y & 1\\ z^2 & z & 1 \end{vmatrix}$ Multiple of Row Added to Row of Determinant $\displaystyle$ $=$ $\displaystyle \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right) \left({x - y}\right) \left({y - z}\right) \left({x - z}\right)$ Vandermonde Determinant

$\blacksquare$

## Source of Name

This entry was named for Christian Friedrich Krattenthaler.