Question

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

Answer 1

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

Applying:- $R_1\to R_1\to R_2$

$R_2\to R_2\to R_3$

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

$\Rightarrow$ Expanding the determinant

$\Rightarrow$ $1\left[\right(a-b\left)\right(-a\left)\right(b-c)-(b-c\left)\right(-c\left)\right(a-b\left)\right]$

$\Rightarrow (a-b)(b-c)(c-a)$

Hence proved.