If $\left| \begin{matrix} a & a^{2} & 1 + a^{3} \ b & b^{2} & 1 + b^{3} \ c & c^{2} & 1 + c^{3} \end{matrix} ight|$0 and vectors $\left( 1,a,a^{2} ight),(1,b,b^{2})$and $(1,c,c^{2})$are non-coplanar, then the product$\ abc$ equals

Question

If $\left| \begin{matrix} a & a^{2} & 1 + a^{3} \ b & b^{2} & 1 + b^{3} \ c & c^{2} & 1 + c^{3} \end{matrix} ight|$0 and vectors $\left( 1,a,a^{2} ight),(1,b,b^{2})$and $(1,c,c^{2})$are non-coplanar, then the product$\ abc$ equals

Dr. Abubacker · Accepted Answer

(b)
$\left| \begin{matrix}
a & a^{2} & 1 + a^{3} \
b & b^{2} & 1 + b^{3} \
c & c^{2} & 1 + c^{3}
\end{matrix} ight| = \left| \begin{matrix}
a & a^{2} & 1 \
b & b^{2} & 1 \
c & c^{2} & 1
\end{matrix} ight| + \left| \begin{matrix}
a & a^{2} & a^{3} \
b & b^{2} & b^{3} \
c & c^{2} & c^{3}
\end{matrix} ight| = 0$
$\Rightarrow \left| \begin{matrix}
a & a^{2} & 1 \
b & b^{2} & 1 \
c & c^{2} & 1
\end{matrix} ight| + abc\left| \begin{matrix}
a & a^{2} & 1 \
b & b^{2} & 1 \
c & c^{2} & 1
\end{matrix} ight| = 0$
$\Rightarrow (1 + abc)\left| \begin{matrix}
a & a^{2} & 1 \
b & b^{2} & 1 \
c & c^{2} & 1
\end{matrix} ight| = 0$
$\left\lbrack 	herefore\left| \begin{matrix}
a & a^{2} & 1 \
b & b^{2} & 1 \
c & c^{2} & 1
\end{matrix} ight| 
eq 0 ightbrack$
$\Rightarrow \ 1 + abc = 0$
$\Rightarrow abc = - 1$

Q. If $\left| \begin{matrix} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{matrix} \right|$0 and vectors $\left( 1,a,a^{2} \right),(1,b,b^{2})$and $(1,c,c^{2})$are non-coplanar, then the product$\ abc$ equals

Explanation

Reviewed by Dr. Abubacker

Want more practice?