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 ${\Delta }_1+{\Delta }_2=0$

${\Delta }_1=\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$

Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$

${\Delta }_1=\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-a\\ a^2& b^2-a^2& c^2-a^2\end{array}\right|=(b-a)(c-a)\left|\begin{array}{ccc}1& 0& 0\\ a& 1& 1\\ a^2& b+a& c+a\end{array}\right|=(b-a)(c-a)(c-b)$

And

${\Delta }_2=\left|\begin{array}{ccc}1& bc& a\\ 1& ca& b\\ 1& ab& c\end{array}\right|$

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

${\Delta }_2=\left|\begin{array}{ccc}1& bc& a\\ 0& c(a-b)& b-a\\ 0& b(a-c)& c-a\end{array}\right|=(b-a)(c-a)\left|\begin{array}{ccc}1& bc& a\\ 0& -c& 1\\ 0& -b& 1\end{array}\right|=-(b-a)(c-a)(c-b)$

Therefore,

${\Delta }_1+{\Delta }_2=0$

Hence, option 'A' is correct.