Question

If n is an integer and if $ \begin{vmatrix} x^n& x^{n+2}& x^{n+3} \\[0.3em] y^n &y^{n+2} & y^{n+3} \\[0.3em] z^n & z^{n+2}&z^{n+3} \end{vmatrix}=(x-y\,(y-z)\,(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ then n equals

• A. 1
• B. -1
• C. 2
• D. None of these.

Answer 1

Answer: (B)

$ \begin{vmatrix} x^n& x^{n+2}& x^{n+3} \\[0.3em] y^n &y^{n+2} & y^{n+3} \\[0.3em] z^n & z^{n+2}&z^{n+3} \end{vmatrix} = x^ny^nz^n \begin{vmatrix} 1& x^2& x^3 \\[0.3em] 1 &y^2 & y^3 \\[0.3em] 1 & z^{2}&z^{3} \end{vmatrix} $
$ x^ny^nz^n \begin{vmatrix} 1& x^2& x^3 \\[0.3em] 1 &y^2-x^3 & y^3-x^3 \\[0.3em] 0 & z^{2} - x^2&z^{3} - x^3 \end{vmatrix} $
(By operating $R_3 - R_1 \, R_2 - R_1$)
= $ x^ny^nz^n (y -x)(z-x)\begin{vmatrix} 1& x^2& x^3 \\[0.3em] 0 &x+y & x^2 +y^2+xy \\[0.3em] 0 & x+z &x^2 +z^2 +xz \end{vmatrix} $
= $x^n \, y^n \, z^n \, (x - y) (x - z) (x^3 + xz^2 + x^2z + yz^2 + yz^2 +xyz- x^3 - xy^2 - x^2y - zx^2 - zy^2 - xyz)$
= $x^n \, y^n \, z^n \, (x - y) (x - z) [xz^2 + yz^2 - xy^2 -2y^2]$
= $ x^ny^nz^n(x- y) (x - z) [x(z^2 -y^2) + yz(z-y)] $
= $x^ny^nz^n (x- y) (x- z)(z - y) (xy + yz + zx)$
= $(x^ny^nz^n)(x -y)(x- z) (z - y) (xy + yz + zx)$
= $x^{n + 1} \, y^{n +1} \, z^{n+1} \, (x - y) (y - Z) (z - X) $
$\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$
Comparing with given value of determinant $n$ + 1
= 0 $\Rightarrow $ n = - 1.