Question

If $\begin{vmatrix}x^{n}&x^{n+2}&x^{n+3}\\ y^{n} &y^{n+2}&y^{n+3}\\ z^{n} &z^{n+2}&z^{n+3}\end{vmatrix} = (y - z) (z - x) (x - y) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$ then n is equal to :

• A. 2
• B. -2
• C. -1
• D. 1

Answer 1

Answer: (C)

Given: $\begin{vmatrix}x^{n}&x^{n+2}&x^{n+3}\\ y^{n} &y^{n+2}&y^{n+3}\\ z^{n} &z^{n+2}&z^{n+3}\end{vmatrix} $
$ = \left(x - y\right) \left(y - z\right) \left(z - x\right) \left[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right] $
Let $ A = \begin{vmatrix}x^{n}&x^{n+2}&x^{n+3}\\ y^{n} &y^{n+2}&y^{n+3}\\ z^{n} &z^{n+2}&z^{n+3}\end{vmatrix} $
Take $x^n, y^n$ and $z^n$ common from the rows $R_1, R_2$ and $R_3$ respectively. $A = x^{n}y^{ n}z^{n} \begin{vmatrix}1&x^{2}&x^{3}\\ 1 &y^{2}&y^{3}\\ 1 &z^{2}&z^{3}\end{vmatrix} $
Operate $R_{2}\to R_{2} - R_{1}$ and $ R_{3} \to R_{3} - R_{1} $
$A = x^{n}y^{n}z^{n} \begin{vmatrix}1&x^{2}&x^{3}\\ 0&y^{2}-x^{2}&y^{3}-x^{3}\\ 0 &z^{2}-x^{2}&z^{3}-x^{3}\end{vmatrix} $
Take (y - x) (z - x) common from the rows $R_2$ and $R_3$ respectively:
Then we get $A = x^ny^nz^n (y - x) (z - x)$
$\begin{vmatrix}1&x^{2}&x^{3}\\ 0 &y+x&x^{2}+y^{2}+xy\\ 0 &z+x&z^{2}+x^{2}+zx\end{vmatrix}$
$ =x^{n}y^{n}z^{n}\left(y-x\right)\left(z-x\right)\left[yz^{2}+x^{2}y+xyz+xz^{2}+x^{3}+x^{2}z-x^{2}z-y^{2}z-xyz-x^{3}-xy^{2}-x^{2}y\right]$
$ =x^{n}y^{n}z^{n}\left(x-y\right)\left(y-z\right)\left(z-x\right)\left[xy+yz+zx\right] $
$= \left[x^{n+1} y^{n+1}z^{n} +x^{n}y^{n+1}z^{n+1}+x^{n+1}y^{n}z^{n+1}\right] $
$ \left(x-y\right)\left(y-z\right)\left(z-x\right) $
$= x^{n+1}. y^{n+1}. z^{n+1}\left(x-y\right)\left(y-z\right)\left(z-x\right) $
$ \left[\frac{1}{x} + \frac{1}{y}+\frac{1}{z}\right] $
But given
$A = \left(y-z\right)\left(z - x\right) \left(x - y\right) \left[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right] $
So, $x^{n+1}y^{n+1}z^{n+1} = 1 = x^{0}y^{0}z^{0}$
$\Rightarrow \ n+1 = 0 \Rightarrow n = - 1 $.