JEE MAIN - Mathematics (2003 - No. 32)
If $$\left| {\matrix{
a & {{a^2}} & {1 + {a^3}} \cr
b & {{b^2}} & {1 + {b^3}} \cr
c & {{c^2}} & {1 + {c^3}} \cr
} } \right| = 0$$ and vectors $$\left( {1,a,{a^2}} \right),\,\,$$
$$\left( {1,b,{b^2}} \right)$$ and $$\left( {1,c,{c^2}} \right)\,$$ are non-coplanar, then the product $$abc$$ equals :
$$\left( {1,b,{b^2}} \right)$$ and $$\left( {1,c,{c^2}} \right)\,$$ are non-coplanar, then the product $$abc$$ equals :
$$0$$
$$2$$
$$-1$$
$$1$$
Explanation
$$\left| {\matrix{
a & {{a^2}} & {1 + {a^3}} \cr
b & {{b^2}} & {1 + {b^3}} \cr
c & {{c^2}} & {1 + {c^3}} \cr
} } \right| = 0$$
$$ \Rightarrow \left| {\matrix{ a & {{a^2}} & 1 \cr b & {{b^2}} & 1 \cr c & {{c^2}} & 1 \cr } } \right| + \left| {\matrix{ a & {{a^2}} & {{a^3}} \cr b & {{b^2}} & {{b^3}} \cr c & {{c^2}} & {{c^3}} \cr } } \right| = 0$$
$$ \Rightarrow \left( {1 + abc} \right)\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| = 0$$
As $$\,\,\,\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| \ne 0$$ (given condition)
$$\therefore$$ $$abc=-1$$
$$ \Rightarrow \left| {\matrix{ a & {{a^2}} & 1 \cr b & {{b^2}} & 1 \cr c & {{c^2}} & 1 \cr } } \right| + \left| {\matrix{ a & {{a^2}} & {{a^3}} \cr b & {{b^2}} & {{b^3}} \cr c & {{c^2}} & {{c^3}} \cr } } \right| = 0$$
$$ \Rightarrow \left( {1 + abc} \right)\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| = 0$$
As $$\,\,\,\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| \ne 0$$ (given condition)
$$\therefore$$ $$abc=-1$$
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