Q102 / 180 · COMEDK 2014 · Mathematics

is

  • Correct answer
    A
    0
  • B
    $4 a b c$
  • C
    $4(a b c)^{-1}$
  • D
    $4\left[a b c+(a b c)^{-1}\right]$

Solution

Given determinant is $$ \left|\begin{array}{ccc} 4 & 4 & 4 \\ \left(a+a^{-1}\right)^{2} & \left(b+b^{-1}\right)^{2} & \left(c+c^{-1}\right)^{2} \\ \left(a-a^{-1}\right)^{2} & \left(b-b^{-1}\right)^{2} & \left(c-c^{-1}\right)^{2} \end{array}\right| $$ Using operation $R_{2} \rightarrow R_{2}-R_{3}$, we get $\left|\begin{array}{ccc}4 & 4 & 4 \\ 4 a a^{-1} & 4 b b^{-1} & 4 c c^{-1} \\ \left(a-a^{-1}\right)^{2} & \left(b-b^{-1}\right)^{2} & \left(c-c^{-1}\right)^{2}\end{array}\right|$ $=\left|\begin{array}{ccc}4 & 4 & 4 \\ 4 & 4 & 4 \\ \left(a-a^{-1}\right)^{2} & \left(b-b^{-1}\right)^{2} & \left(c-c^{-1}\right)^{2}\end{array}\right|=0$ [ $R_{1}$ and $R_{2}$ are identical rows]

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

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