Question

Using the property of determinants and with out expanding prove that
$\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|=0$

Answer 1

$\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|=0Applying{c}_1⟶c_1+c_2$

$=\left|\begin{array}{ccc}c-b& b-c& c-a\\ a-c& c-a& a-b\\ b-a& a-b& b-c\end{array}\right|$

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

$Astwocolumnareidenticalthen$

$\therefore (a-b)(a-c)(c-b)=0$