Question

Without expanding the determinant, prove that $\left|\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right|$ $=\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|$

Answer 1

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

Multiplying and dividing $R_1$ by $a$ , $R_2$ by $b$ and $R_3$ by $c$

$=\frac{1}{abc}\left|\begin{array}{ccc}{a}^2& a^3& abc\\ b^2& b^3& abc\\ c^2& c^3& abc\end{array}\right|$

Taking $abc$ common from $C_3$

$=\frac{abc}{abc}\left|\begin{array}{ccc}{a}^2& a^3& 1\\ b^2& b^3& 1\\ c^2& c^3& 1\end{array}\right|$

$=\left|\begin{array}{ccc}{a}^2& a^3& 1\\ b^2& b^3& 1\\ c^2& c^3& 1\end{array}\right|$

$c_1\leftrightarrow c_3$

$=-\left|\begin{array}{ccc}1& a^3& a^2\\ 1& b^3& b^2\\ 1& c^3& c^2\end{array}\right|$ (When two rows/columns of a determinant are interchanged, then the value of determinant differs by a negative sign. )

$c_2\leftrightarrow c_3$

$=\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|$