Question

The value of the determinant $\left|\begin{array}{ccc}1& a& a^2-bc\\ 1& b& b^2-ca\\ 1& c& c^2-ab\end{array}\right|$ is .....................

Answer 1

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

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

Consider,

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

taking $abc$ common from $C_3$ gives

$=abc\left|\begin{array}{ccc}1& a& \frac{1}{a}\\ 1& b& \frac{1}{b}\\ 1& c& \frac{1}{c}\end{array}\right|$

multiplying $C_1$ with $a$,$C_2$ with $b$ and $C_3$ with $c$ gives

$=\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|$
interchanging columns gives

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

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