Question

lf $\left|\begin{array}{ccc}{x}^n& x^{n+2}& x^{n+3}\\ y^n& y^{n+2}& y^{n+3}\\ z^n& z^{n+2}& z^{n+3}\end{array}\right|$ $=(x-y)(y-z\left)\right(z-x)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$, then the value of $n$ is

• A. $-1$
• B. $-2$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$-1$

$\left|\begin{array}{ccc}{x}^n& x^{n+2}& x^{n+3}\\ y^n& y^{n+2}& y^{n+3}\\ z^n& z^{n+2}& z^{n+3}\end{array}\right|=x^n{y}^n{z}^n\left|\begin{array}{ccc}1& x^2& x^3\\ 1& y^2& y^3\\ 1& z^2& z^3\end{array}\right|$

By row trasformation,

$r_1\to r_1-r_2$ and $r_3\to r_3-r_2$

$=(xyz{)}^n\left|\begin{array}{ccc}0& (x^2-y^2)& (x^3-y^3)\\ 1& y^2& y^3\\ 0& (z^2-y^2)& (z^3-y^3)\end{array}\right|=(xyz{)}^n(x-y)(z-y)\left|\begin{array}{ccc}0& (x+y)& (x^2+xy+y^2)\\ 1& y^2& y^3\\ 0& (z+y)& (z^2+zy+y^2)\end{array}\right|$

$=\left(xyz{)}^n\right(x-y\left)\right(z-y\left)\right[-1(x+y)(z^2+zy+y^2)-(z+y)(x^2+xy+y^2)]$

$=\left(xyz{)}^n\right(x-y\left)\right(y-z\left)\right[x{z}^2+xyz+x{y}^2+y{z}^2+z{y}^2+y^3-[z{x}^2+xyz+z{y}^2+y{x}^2+x{y}^2+y^3]]$

$=\left(xyz{)}^n\right(x-y\left)\right(y-z\left)\right(x{z}^2-z{x}^2-y{x}^2+y{z}^2)$

$=\left(xyz{)}^n\right(x-y\left)\right(y-z\left)\right(zx(z-x)+y(z-x)(z+x))$

$=\left(xyz{)}^n\right(x-y\left)\right(y-z\left)\right(z-x\left)\right(xz+yz+xy)$

$=\left(xyz{)}^{n+1}\right(x-y\left)\right(y-z\left)\right(z-x)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

So, $(x+y+z{)}^{n+1}=1$

$n+1=0$

$n=-1$