Question

Find

$2\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2-bc& b^2-ca& c^2-ab\end{array}\right|=$

• A. $0$
• B. $2$
• C. $4$
• D. $-4$

Answer 1

The correct option is A $0$

Let $△=2\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2-bc& b^2-ca& c^2-ab\end{array}\right|$

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

$△=2\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-a\\ a^2-bc& b^2-a^2-ca+bc& c^2-ab-a^2+bc\end{array}\right|$

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

taking $(b-a)$ common from $C_2$ and $(c-a)$ from $C_3$, we get

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

Expanding it along $R_1,$ we get

$△=2(b-a)(c-a)(a+b+c-(a+b+c\left)\right)=0$