Question

Using properties of determinants prove that :
$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ bc& ca& ab\end{array}\right|=(a-b)(b-c)(c-a)$.

Answer 1

Consider $\Delta =\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ bc& ca& ab\end{array}\right|$ apply $C_1\to C_1-C_3$ ,$C_2\to C_2-C_3$

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

take $(a-c)$ common from $C_1$ and $b-c$ from $C_2$

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

expand by $R_1$ we get

$\Delta =(a-c)(b-c)(b-a)$

$=(a-b)(b-c)(c-a)$

Hence proved

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