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Mathematics Question on Determinants

If n is an integer and if ∣xnxn+2xn+3\[0.3em]ynyn+2yn+3\[0.3em]znzn+2zn+3∣=(x−y (y−z) (z−x)(1x+1y+1z) \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

Explanation

Solution

∣xnxn+2xn+3\[0.3em]ynyn+2yn+3\[0.3em]znzn+2zn+3∣=xnynzn∣1x2x3\[0.3em]1y2y3\[0.3em]1z2z3∣\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} xnynzn∣1x2x3\[0.3em]1y2−x3y3−x3\[0.3em]0z2−x2z3−x3∣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 R3−R1 R2−R1R_3 - R_1 \, R_2 - R_1) = xnynzn(y−x)(z−x)∣1x2x3\[0.3em]0x+yx2+y2+xy\[0.3em]0x+zx2+z2+xz∣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} = xn yn zn (x−y)(x−z)(x3+xz2+x2z+yz2+yz2+xyz−x3−xy2−x2y−zx2−zy2−xyz)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) = xn yn zn (x−y)(x−z)[xz2+yz2−xy2−2y2]x^n \, y^n \, z^n \, (x - y) (x - z) [xz^2 + yz^2 - xy^2 -2y^2] = xnynzn(x−y)(x−z)[x(z2−y2)+yz(z−y)]x^ny^nz^n(x- y) (x - z) [x(z^2 -y^2) + yz(z-y)] = xnynzn(x−y)(x−z)(z−y)(xy+yz+zx)x^ny^nz^n (x- y) (x- z)(z - y) (xy + yz + zx) = (xnynzn)(x−y)(x−z)(z−y)(xy+yz+zx)(x^ny^nz^n)(x -y)(x- z) (z - y) (xy + yz + zx) = xn+1 yn+1 zn+1 (x−y)(y−Z)(z−X)x^{n + 1} \, y^{n +1} \, z^{n+1} \, (x - y) (y - Z) (z - X) (1x+1y+1z)\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) Comparing with given value of determinant nn + 1 = 0 ⇒\Rightarrow n = - 1.