Question

If $\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$ and vectors $(1,a,a^2),(1,b,b^2)$ and $(1,c,c^2)$ are non coplanar then $abc$ equals

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

The correct option is A $-1$

As vectors $(1,a,a^2),(1,b,b^2),(1,c,c^2)$ are non coplanar.

$\therefore \left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\ne 0$ .....(1)

Now, $\left|\begin{array}{ccc}a& a^2& a^3+1\\ b& b^2& b^3+1\\ c& c^2& c^3+1\end{array}\right|=0$

$\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|+\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|=0$

$abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|+\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|=0$

$C_1\leftrightarrow C_3,C_2\leftrightarrow C_3$ in second determinant

$\Rightarrow (1+abc)\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0$

$\Rightarrow (1+abc)=0$ by using (1)