Vandermonde Matrix Identity for Cauchy Matrix/Examples/3x3

Example of Vandermonde Matrix Identity for Cauchy Matrix
Illustrate $3\times 3$ case for Vandermonde Matrix Identity for Cauchy Matrix and Value of Cauchy Determinant:

Then:

Details:

Define Vandermonde matrices


 * $V_x = \begin {pmatrix}

1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 & x_2^2 & x_3^2 \\ \end {pmatrix}, \quad V_y = \begin {pmatrix} 1 & 1 & 1 \\ y_1 & y_2 & y_3 \\ y_1^2 & y_2^2 & y_3^2 \\ \end {pmatrix}$

Define polynomials:


 * $\map p x = \paren {x - x_1} \paren {x - x_2} \paren {x - x_3}$


 * $\map {p_1} x = \paren {x - x_2} \paren {x - x_3},

\quad \map {p_2} x = \paren {x - x_1} \paren {x - x_3}, \quad \map {p_3} x = \paren {x - x_1} \paren {x - x_2}$ Define invertible diagonal matrices:


 * $P = \begin {pmatrix}

\map {p_1} {x_1} & 0       & 0 \\ 0       & \map {p_2} {x_2} & 0 \\ 0       & 0        & \map {p_3} {x_3} \\ \end {pmatrix}, \quad Q = \begin {pmatrix} \map p {y_1} & 0       & 0 \\ 0       & \map p {y_2} & 0 \\ 0       & 0        & \map p {y_3} \\ \end {pmatrix}$

Then:
 * $P = \begin {pmatrix}

\paren {x_1 - x_2} \paren {x_1 - x_3} & 0                                  & 0 \\ 0                                    & \paren {x_2 - x_1}\paren {x_2 - x_3} & 0 \\ 0                                    & 0                                   & \paren {x_3 - x_1}\paren {x_3 - x_2} \\ \end {pmatrix}, \quad Q = \begin {pmatrix} \paren {y_1 - x_1}\paren {y_1 - x_2}\paren {y_1 - x_3} & 0       & 0 \\ 0       &  \paren {y_2 - x_1}\paren {y_2 - x_2}\paren {y_2 - x_3} & 0 \\ 0       & 0        &  \paren {y_3 - x_1}\paren {y_3 - x_2}\paren {y_3 - x_3} \\ \end {pmatrix}$

Determinant of Diagonal Matrix implies


 * $\map \det P = \paren {x_1 - x_2} \paren {x_1 - x_3} \paren {x_2 - x_1} \paren {x_2 - x_3} \paren {x_3 - x_1} \paren {x_3 - x_2}$


 * $\map \det Q = \paren {y_1 - x_1} \paren {y_1 - x_2} \paren {y_1 - x_3} \paren {y_2 - x_1} \paren {y_2 - x_2} \paren {y_2 - x_3} \paren {y_3 - x_1} \paren {y_3 - x_2} \paren {y_3 - x_3}$

Vandermonde Determinant implies


 * $\map \det {V_x} = \paren {x_3 - x_2} \paren {x_3 - x_1} \paren {x_2 - x_1}$


 * $\map \det {V_y } = \paren {y_3 - y_2} \paren {y_3 - y_1} \paren {y_2 - y_1}$

Determinant of Matrix Product and Definition:Inverse Matrix imply


 * $\map \det {V_x^{-1} } = \dfrac 1 {\map \det {V_x} }, \quad \map \det {Q^{-1} } = \dfrac 1 {\map \det Q}$

Then:

Insert the four determinant equations and simplify to obtain the equation for $3 \times 3$ $\map \det C$.