Q70 / 170 · Manipal (MET) 2017 · Mathematics

If $\Delta_1=\left|\begin{array}{lll}x & a & b \\ b & x & a \\ a & b & x\end{array}\right|$ and $\Delta_2=\left|\begin{array}{ll}x & b \\ a & x\end{array}\right|$ are the given determinants, then

  • A
    $\Delta_1=3\left(\Delta_2\right)^2$
  • Correct answer
    B
    $\left(\frac{d}{d x}\right)\left(\Delta_1\right)=3 \Delta_2$
  • C
    $\left(\frac{d}{d x}\right)\left(\Delta_1\right)=3\left(\Delta_2\right)^2$
  • D
    $\Delta_1=3\left(\Delta_2\right)^{3 / 2}$

Solution

$\begin{aligned} & \text { Given, } \Delta_1=\left|\begin{array}{lll}x & a & b \\ b & x & a \\ a & b & x\end{array}\right|, \Delta_2=\left|\begin{array}{ll}x & b \\ a & x\end{array}\right| \\ & \begin{aligned} & \therefore \frac{d}{d x}\left(\Delta_1\right) \\ &=\left|\begin{array}{lll}1 & 0 & 0 \\ b & x & a \\ a & b & x\end{array}\right|+\left|\begin{array}{lll}x & a & b \\ 0 & 1 & 0 \\ a & b & x\end{array}\right|+\left|\begin{array}{lll}x & a & b \\ b & x & a \\ 0 & 0 & 1\end{array}\right| \\ & \quad=\left|\begin{array}{ll}x & a \\ b & x\end{array}\right|+\left|\begin{array}{ll}x & b \\ a & x\end{array}\right|+\left|\begin{array}{ll}x & a \\ b & x\end{array}\right|=3 \Delta_2 \\ & \therefore \frac{d}{d x}\left(\Delta_1\right)=3 \Delta_2\end{aligned}\end{aligned}$

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

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