Q151 / 270 · NDA 2011 (Phase 1) · Mathematics

What is the value of the determinant $\left|\begin{array}{ccc}x+1 & x+2 & x+4 \\ x+3 & x+5 & x+8 \\ x+7 & x+10 & x+14\end{array}\right|$ ?

  • A
    $x+2$
  • B
    $x^2+2$
  • C
    2
  • Correct answer
    D
    -2

Solution

$\begin{aligned} & \left|\begin{array}{ccc}x+1 & x+2 & x+4 \\ x+3 & x+5 & x+5 \\ x+7 & x+10 & x+14\end{array}\right| \\ & c_1 \rightarrow c_2-c_1, c_3-c_2\end{aligned}$ $\left|\begin{array}{ccc}-1 & -2 & x+4 \\ -2 & -3 & x+8 \\ -3 & -4 & x+14\end{array}\right|$ $\begin{aligned} & =\left|\begin{array}{ccc}1 & 2 & x+4 \\ 2 & 3 & x+8 \\ 3 & 4 & x+14\end{array}\right| \\ & =\mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3-\mathrm{R}_2\end{aligned}$ $\begin{aligned} & \left|\begin{array}{ccc}1 & 2 & x+4 \\ 1 & 1 & 4 \\ 1 & 1 & 6\end{array}\right| \\ = & \mathrm{R}_1-\mathrm{R}_2, \mathrm{R}_2-\mathrm{R}_3\end{aligned}$ $\begin{aligned} & \left|\begin{array}{ccc}0 & 1 & x \\ 0 & 0 & -2 \\ 1 & 1 & 6\end{array}\right| \\ = & 1\{(-2)\}=-2 .\end{aligned}$

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

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