Q99 / 155 · WBJEE 2021 · Mathematics

The determinant $\left|\begin{array}{ccc}a^{2}+10 & a b & a c \\ a b & b^{2}+10 & b c \\ a c & b c & c^{2}+10\end{array}\right|$ is

  • A
    divisible by 10 but not by 100
  • Correct answer
    B
    divisible by 100
  • C
    not divisible by 100
  • D
    not divisible by 10

Solution

$=\frac{1}{a b c}\left|\begin{array}{ccc}a\left(a^{2}+10\right) & a^{2} b & a^{2} c \\ a b^{2} & b\left(b^{2}+10\right) & b^{2} c \\ a c^{2} & b c^{2} & c\left(c^{2}+10\right)\end{array}\right|=\frac{a b c}{a b c}\left|\begin{array}{ccc}a^{2}+10 & a^{2} & a^{2} \\ b^{2} & b^{2}+10 & b^{2} \\ c^{2} & c^{2} & c^{2}+10\end{array}\right|$ $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$ $=\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+10\right)\left|\begin{array}{ccc}1 & 1 & 1 \\ \mathrm{~b}^{2} & \mathrm{~b}^{2}+10 & \mathrm{~b}^{2} \\ \mathrm{c}^{2} & \mathrm{c}^{2} & \mathrm{c}^{2}+10\end{array}\right|$ $=\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+10\right)\left|\begin{array}{ccc}1 & 0 & 0 \\ \mathrm{~b}^{2} & 10 & 0 \\ \mathrm{c}^{2} & 0 & 10\end{array}\right|\left[\begin{array}{c}\mathrm{C}_{2}-\mathrm{C}_{1} \\ \mathrm{C}_{3}-\mathrm{C} 1\end{array}\right]$ $=\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+10\right) 100$

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

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