Q34 / 50 · Manipal (MET) 2022 · Mathematics
If $\left|\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3\end{array}\right|=0$ 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 $a b c$ is _____
Correct answer: -1
Solution
We can rewrite,
$\begin{aligned}
& \left|\begin{array}{lll}
a & a^2 & 1+a^3 \\
b & b^2 & 1+b^3 \\
c & c^2 & 1+c^3
\end{array}\right|=\left|\begin{array}{lll}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right|+\left|\begin{array}{lll}
a & a^2 & a^3 \\
b & b^2 & b^3 \\
c & c^2 & c^3
\end{array}\right|=0 \\
& =\left|\begin{array}{lll}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right|+a b c\left|\begin{array}{lll}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right|=0 \\
& =(1+a b c)\left|\begin{array}{lll}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right|=0 \\
& =(1+a b c)=0 \quad\left[\because\left|\begin{array}{lll}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right| \neq 0\right] \\
& \Rightarrow \quad a b c=-1
\end{aligned}$
Hence, the correct answer is -1.