Q47 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

$\left|\begin{array}{ccc}\frac{-b c}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & \frac{-a c}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & \frac{-a b}{c^2}\end{array}\right|=$

  • A
    0
  • Correct answer
    B
    4
  • C
    -1
  • D
    $\frac{a^2+b^2+c^2}{a^2 b^2 c^2}$

Solution

$\Delta=\left|\begin{array}{ccc}\frac{-b c}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & \frac{-a c}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & \frac{-a b}{c^2}\end{array}\right|$ $\Delta=\frac{1}{a^2}\left|\begin{array}{ccc}-b c & a c & a b \\ \frac{c}{b} & \frac{-a c}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & \frac{-a b}{c^2}\end{array}\right|$ $\mathrm{R}_2 \rightarrow b^2 \mathrm{R}_2, \mathrm{R}_3 \rightarrow c^2 \mathrm{R}_3$ $\Delta=\frac{1}{a^2 b^2 c^2}\left|\begin{array}{ccc}-b c & a c & a b \\ b c & -a c & a b \\ b c & a c & -a b\end{array}\right|$ $\mathrm{R}_2 \rightarrow \mathrm{R}_2+\mathrm{R}_1 ; \mathrm{R}_3 \rightarrow \mathbb{R}_3+\mathrm{R}_1$ $\Delta=\frac{1}{a^2 b^2 c^2}\left|\begin{array}{ccc}-b c & a c & a b \\ 0 & 0 & 2 a b \\ 0 & 2 a c & 0\end{array}\right|$ $\Delta=\frac{1}{a^2 b^2 c^2}\left[-b c\left(-4 a^2 b c\right)\right] ; \Delta=4$

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

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