Question

If $\left|\begin{array}{ccc}{x}^n& x^{x+2}& x^{x+4}\\ y^n& y^{n+2}& y^{n+4}\\ z^n& z^{n+2}& z^{n+4}\end{array}\right|=(\frac{1}{y^2}-\frac{1}{x^2})(\frac{1}{z^2}-\frac{1}{y^2})(\frac{1}{x^2}-\frac{1}{z^2})$ then n=

• A. -4
• B. 4
• C. 2
• D. -2

Answer 1

The correct option is A

-4

$\Delta =\left|\begin{array}{ccc}{x}^n& x^{x+2}& x^{x+4}\\ y^n& y^{n+2}& y^{n+4}\\ z^n& z^{n+2}& z^{n+4}\end{array}\right|=x^n{y}^n{z}^n\left|\begin{array}{ccc}1& x^2& x^4\\ 1& y^2& y^4\\ 1& z^2& z^4\end{array}\right|$
$\Delta =x^n{y}^n{z}^n\left|\begin{array}{ccc}1& x^2& x^4\\ 0& y^2-z^2& y^4-z^4\\ 0& z^2-x^2& z^4-x^4\end{array}\right|(R_2\to R_2-R_3,R_3\to R_3-R_1)$
$\Delta =x^n{y}^n{z}^n\left[\right(y^2-z^2\left)\right(z^2+x^2\left)\right(z^2-x^2)-(z^2-x^2\left)\right(y^2+z^2\left)\right(y^2-z^2\left)\right]\Delta =x^n{y}^n{z}^n(y^2-z^2)(z^2-x^2)(x^2+z^2-y^2-z^2)\Delta =x^n{y}^n{z}^n(x^2-y^2)(y^2-z^2)(z^2-x^2)Giventhat\Delta =(\frac{1}{y^2}-\frac{1}{x^2})(\frac{1}{z^2}-\frac{1}{y^2})(\frac{1}{x^2}-\frac{1}{z^2})$
$\Delta =\left(\frac{x^2-y^2}{x^2{y}^2}\right)\left(\frac{y^2-z^2}{y^2{z}^2}\right)\left(\frac{z^2-x^2}{x^2{z}^2}\right)\Delta =x^{-4}{y}^{-4}{z}^{-4}(x^2-y^2)(y^2-z^2)(z^2-x^2)$

Hence n =-4