Question

If ${\Delta }_1=\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$, ${\Delta }_2=\left|\begin{array}{ccc}1& bc& a\\ 1& ca& b\\ 1& ab& c\end{array}\right|$, then

• A. ${\Delta }_1+{\Delta }_2=0$
• B. ${\Delta }_1+2{\Delta }_2=0$
• C. ${\Delta }_1={\Delta }_2$
• D. none of these

Answer 1

The correct option is A ${\Delta }_1+{\Delta }_2=0$

${\Delta }_1=\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
${\Delta }_1=\left|\begin{array}{ccc}0& 0& 1\\ a-b& b-c& c\\ a^2-b^2& b^2-c^2& c^2\end{array}\right|$
$=(a-b)(b-c)\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& c\\ a+b& b+c& c^2\end{array}\right|$
$\Rightarrow {\Delta }_1=(a-b)(b-c)(c-a)$
${\Delta }_2=\left|\begin{array}{ccc}1& bc& a\\ 1& ca& b\\ 1& ab& c\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$=\left|\begin{array}{ccc}0& (b-a)c& a-b\\ 0& (c-b)a& b-c\\ 1& ab& c\end{array}\right|$
$=(a-b)(b-c)\left|\begin{array}{ccc}0& -c& 1\\ 0& -a& 1\\ 1& ab& c\end{array}\right|$
$=(a-b)(b-c)(a-c)$
$\Rightarrow {\Delta }_2=-(a-b)(b-c)(c-a)$
Hence,${\Delta }_2=-{\Delta }_1$
${\Delta }_1+{\Delta }_2=0$