Question

State true or false.

The given equation $\left|\begin{array}{ccc}yz-x^2& zx-y^2& xy-z^2\\ zx-y^2& xy-z^2& yz-x^2\\ xy-z^2& yz-x^2& zx-y^2\end{array}\right|$ is divisible by $(x+y+z)$.

• A. True
• B. False

Answer 1

The correct option is A True

Consider $\Delta =\left|\begin{array}{ccc}yz-x^2& zx-y^2& xy-z^2\\ zx-y^2& xy-z^2& yz-x^2\\ xy-z^2& yz-x^2& zx-y^2\end{array}\right|$

Applying $C_1\to C_1+C_2+C_3$

$\Delta =\left|\begin{array}{ccc}xy+yz+zx-x^2-y^2-z^2& zx-y^2& xy-z^2\\ xy+yz+zx-x^2-y^2-z^2& xy-z^2& yz-x^2\\ xy+yz+zx-x^2-y^2-z^2& yz-x^2& zx-y^2\end{array}\right|$

$=(xy+yz+zx-x^2-y^2-z^2)\left|\begin{array}{ccc}1& zx-y^2& xy-z^2\\ 1& xy-z^2& yz-x^2\\ 1& yz-x^2& zx-y^2\end{array}\right|$

Applying $R_1\to R_1-R_3$ and $R_2\to R_2-R_3$

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

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

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

$=(xy+yz+zx-x^2-y^2-z^2){(x+y+z)}^2[(x-y)(y-x)-(x-z)(y-z)]$

$=(xy+yz+zx-x^2-y^2-z^2){(x+y+z)}^2[xy+yz+zx-x^2-y^2-z^2]$

$={(xy+yz+zx-x^2-y^2-z^2)}^2{(x+y+z)}^2$

$\therefore \Delta ={(xy+yz+zx-x^2-y^2-z^2)}^2{(x+y+z)}^2$

$\therefore \Delta$ is divisible by $(x+y+z)$ and quotient is ${(xy+yz+zx-x^2-y^2-z^2)}^2(x+y+z)$

The given statement is true.