Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $-3$

Answer 1

The correct option is B $-1$

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

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

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

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

$=-\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|$ [ Taking transpose. Since determinants remains unaltered under transpose]

Comparing this with the given problem we have,

$\lambda =-1$.