Q68 / 170 · Manipal (MET) 2017 · Mathematics

If $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|=k(a-b)(b-c)(c-a)$, then $k$ is equal to

  • A
    -2
  • Correct answer
    B
    1
  • C
    2
  • D
    abc

Solution

$\begin{aligned} & \Delta=\left|\begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|=\left|\begin{array}{ccc} 1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2 \end{array}\right| \\ & \left(\text { using } R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3-R_1\right) \\ & =(b-a)(c-a)\left|\begin{array}{ccc} 1 & a & a^2 \\ 0 & 1 & b+a \\ 0 & 1 & c+a \end{array}\right| \\ & =(b-a)(c-a)(c+a-b-a) \\ & \Rightarrow(b-a)(c-a)(c-b) \\ & =k(a-b)(b-c)(c-a)(given) \\ & \Rightarrow k=1 \end{aligned}$

Hence, the correct answer is option: 'B'.

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