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Prove the following.
Using properties of determinants, prove: |x x² 1 + px³; y y² 1 + py³; z z² 1 + pz³| = (1 + pxyz)(x…
Using properties of determinants, prove:
$\begin{vmatrix} x & x^2 & 1 + px^3 \\ y & y^2 & 1 + py^3 \\ z & z^2 & 1 + pz^3 \end{vmatrix} = (1 + pxyz)(x - y)(y - z)(z - x)$, where $p$ is any scalar.
Given: $\begin{vmatrix} x & x^2 & 1 + px^3 \\ y & y^2 & 1 + py^3 \\ z & z^2 & 1 + pz^3 \end{vmatrix}$, where $p$ is any scalar
To show: $(1 + pxyz)(x - y)(y - z)(z - x)$
Answer
Answer
Split $C_3$ as a sum of two columns:
$\Delta = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + p\begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix}$.
In the second, take $x, y, z$ common from $R_1, R_2, R_3$. In the first, move $C_3$ to the front by two column swaps (sign unchanged).
$\Delta = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} + pxyz\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = (1 + pxyz)\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$.
Apply $R_1 \to R_1 - R_2$, $R_2 \to R_2 - R_3$: the rows become $(0, x-y, x^2-y^2)$, $(0, y-z, y^2-z^2)$, $(1, z, z^2)$. Take $(x-y)$ and $(y-z)$ common:
$= (1 + pxyz)(x-y)(y-z)\begin{vmatrix} 0 & 1 & x+y \\ 0 & 1 & y+z \\ 1 & z & z^2 \end{vmatrix}$.
Expand along $C_1$: $= (1+pxyz)(x-y)(y-z)\,[(y+z) - (x+y)]$ and $(y+z)-(x+y) = z - x$.
Hence $\Delta = (1 + pxyz)(x-y)(y-z)(z-x)$.
- Write $C_3$ as the sum of $(1,1,1)^T$ and $p(x^3, y^3, z^3)^T$ and split the determinant into two, by the sum property of columns.
- Take $x, y, z$ common from $R_1, R_2, R_3$ of the second determinant, leaving $pxyz\,|1\ x\ x^2; 1\ y\ y^2; 1\ z\ z^2|$.
- Move $C_3$ of the first determinant to the front by two swaps (sign unchanged), giving the same Vandermonde determinant.
- $\Delta = (1 + pxyz)\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$.
- Apply $R_1 \to R_1 - R_2$, $R_2 \to R_2 - R_3$ and take $(x-y)$, $(y-z)$ common, using $x^2-y^2 = (x-y)(x+y)$.
- Expand along $C_1$ to get $(y+z)-(x+y) = z - x$.
- Hence $\Delta = (1 + pxyz)(x-y)(y-z)(z-x)$.