Question

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

• A. $0$
• B. $\sum a^2(b-c)$
• C. $2\sum a^2(b-c)$
• D. $-2\sum ab(a-b)$

Answer 1

The correct option is A $0$

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

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

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

$=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$ $-\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|$ [Taking common $abc$ from $C_3$ ]

$=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$ $+\left|\begin{array}{ccc}a& 1& a^2\\ b& 1& b^2\\ c& 1& c^2\end{array}\right|$ [Interchanging $C_3$ and $C_2$ of second matrix]

$=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$ $-\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$ [Interchanging $C_1$ and $C_2$ of second matrix]

$=0$