Question

Let $x=bc-a^2$, $y=ca-b^2$, $z=ab-c^2$, $r=a^2+b^2+c^2$, $s=bc+ca+ab$ and
${\Delta }_1=\left|\begin{array}{ccc}x& y& z\\ y& z& x\\ z& x& y\end{array}\right|$, ${\Delta }_2=\left|\begin{array}{ccc}r& s& s\\ s& r& s\\ s& s& r\end{array}\right|$
then

• A. ${\Delta }_1={\Delta }_2$
• B. ${\Delta }_1^2={\Delta }_2$
• C. ${\Delta }_1={\Delta }_2^2$
• D. ${\Delta }_1+{\Delta }_2=0$

Answer 1

The correct option is A ${\Delta }_1={\Delta }_2$

${\Delta }_1=\left|\begin{array}{ccc}x& y& z\\ y& z& x\\ z& x& y\end{array}\right|$

Applying $C_1\to C_1+C_2+C_3$

${\Delta }_1=\left|\begin{array}{ccc}x+y+z& y& z\\ y+z+z& z& x\\ z+x+y& x& y\end{array}\right|=(x+y+z)\left|\begin{array}{ccc}1& y& z\\ 1& z& x\\ 1& x& y\end{array}\right|$

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

${\Delta }_1=(x+y+z)\left|\begin{array}{ccc}1& y& z\\ 0& z-y& x-z\\ 0& x-y& y-z\end{array}\right|=(x+y+z)(-{(y-z)}^2-(x-z)(x-y))$

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

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

$=(a^2+b^2+c^2){(a^2+b^2+c^2-bc-ac-ab)}^2$

${\Delta }_2=\left|\begin{array}{ccc}r& s& s\\ s& r& s\\ s& s& r\end{array}\right|$

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

${\Delta }_2=\left|\begin{array}{ccc}r& s& s\\ 0& r-s& 0\\ 0& 0& r-s\end{array}\right|=r{(r-s)}^2=(a^2+b^2+c^2){(a^2+b^2+c^2-bc-ac-ab)}^2$

Therefore, ${\Delta }_1={\Delta }_2$