If x, y, z are different and <img alt="\bigtriangleup = \begin{vmatrix} x & x^{2} &x^{3}-1 \\ y& y^{2} &y^{3}-1 \\ z&z^{2} & z^{3}-1 \end{vmatrix}= 0" src="https://entrancecorner.codecogs.com/gif.latex?%5Cbigtriangleup%20%3

If x, y, z are different and $\bigtriangleup = \begin{vmatrix} x & x^{2} &x^{3}-1 \\ y& y^{2} &y^{3}-1 \\ z&z^{2} & z^{3}-1 \end{vmatrix}= 0$ , then using properties of determinants, show that $xyz= 1\cdot$

Answers (1)

$\bigtriangleup = \begin{vmatrix} x & x^{2} &x^{3}-1 \\ y& y^{2} &y^{3}-1 \\ z& z^{2} &z^{3}-1 \end{vmatrix}= 0$
$\Rightarrow \bigtriangleup = \begin{vmatrix} x & x^{2} &x^{3} \\ y& y^{2} &y^{3} \\ z& z^{2} &z^{3} \end{vmatrix}+\begin{vmatrix} x & x^{2} &-1 \\ y& y^{2} &-1 \\ z& z^{2} &-1 \end{vmatrix}= 0$
Taking x, y, z common from R_1,R₂ and R₃respectively in the first determinant . Also taking -1 common from C₃ be the second determinant
$\Rightarrow xyz\begin{vmatrix} 1 & x &x^{2} \\ 1& y&y^{2} \\ 1& z &z^{2} \end{vmatrix}-\begin{vmatrix} x & x^{2}&1 \\ y& y^{2}&1 \\ z& z^{2} &1 \end{vmatrix}= 0$
In second determinant C₁ $\leftrightarrow$ C₃
$\Rightarrow xyz\begin{vmatrix} 1 & x &x^{2} \\ 1& y&y^{2} \\ 1& z &z^{2} \end{vmatrix}-\begin{vmatrix} 1 & x^{2}&x \\ 1& y^{2}&y \\ 1& z^{2} &z \end{vmatrix}= 0$
In second determinant C₂ $\leftrightarrow$ C3
$\Rightarrow xyz\begin{vmatrix} 1 & x &x^{2} \\ 1& y&y^{2} \\ 1& z &z^{2} \end{vmatrix}-\begin{vmatrix} 1 & x&x^{2} \\ 1& y&y ^{2}\\ 1& z &z^{2} \end{vmatrix}= 0$
$\Rightarrow \left ( xyz-1 \right )\begin{vmatrix} 1 & x &x^{2} \\ 1& y&y^{2} \\ 1& z &z^{2} \end{vmatrix}= 0$
By $R_{1}\rightarrow R_{1}-R_{3}\; \; ,R_{2}\rightarrow R_{2}-R_{3}$
$\Rightarrow \left ( xyz-1 \right )\begin{vmatrix} 0 & x-z &x^{2}-z^{2} \\ 0 &y-z &y^{2}-z^{2} \\ 1 &z &z^{2} \end{vmatrix}= 0$
Taking $\left ( x-z \right )\; and\; \left ( y-z \right )$ common from R₁ and R_{2 respectively.}
Expanding along C₁
$\Rightarrow \left ( xyz-1 \right )\left ( y-z \right )\left ( x-z \right )\left ( y+z-z-x \right )= 0$
$\Rightarrow \left ( xyz-1 \right )\left ( y-z \right )\left ( x-z \right )\left ( y-x \right )= 0$
AS $x\neq y\neq z\; \; \therefore \left ( xyz-1 \right )= 0\Rightarrow xyz= 1$
$\therefore xyz= 1$