Q46 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
$\operatorname{det}\left[\begin{array}{ccc}
\frac{a^2+b^2}{c} & c & c \\
a & \frac{b^2+c^2}{a} & a \\
b & b & \frac{c^2+a^2}{b}
\end{array}\right]=$
- A$(a-b)(b-c)(c-a)$
- B$(a+b)(b+c)(c+a)$
- C$2 a b c$
- Correct answerD$4 a b c$
Solution
$\left|\begin{array}{ccc}\frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b}\end{array}\right|$
$\begin{aligned} & =\frac{1}{a b c}\left|\begin{array}{ccc}a^2+b^2 & c^2 & c^2 \\ a^2 & b^2+c^2 & a^2 \\ b^2 & b^2 & a^2+c^2\end{array}\right| \\ & =\frac{1}{a b c}\left|\begin{array}{ccc}a^2+b^2 & c^2 & c^2 \\ a^2-b^2 & c^2 & -c^2 \\ b^2 & b^2 & a^2+c^2\end{array}\right| \quad\left(R_2 \rightarrow R_2-R_1\right) \\ & =\frac{1}{a b c}\left|\begin{array}{ccc}2 a^2 & 2 c^2 & 0 \\ a^2-b^2 & c^2 & -c^2 \\ b^2 & b^2 & a^2+c^2\end{array}\right| \quad\left(R_{;}, \gt R_1+R_2\right)\end{aligned}$
$\begin{aligned} & =\frac{1}{a b c} \times\left[2 a^2\left(a^2 c^2+c^4+b^2 c^2\right)\right. \\ & \left.\quad-2 c^2\left(a^4+c^2 a^2-a^2 b^2-b^2 c^2+b^2 c^2\right)\right] \\ & =\frac{1}{a b c}\left[2 a^2 c^2\left(a^2+b^2+c^2\right)-2 a^2 c^2\left(a^2+c^2-b^2\right)\right] \\ & =\frac{2 a^2 c^2}{a b c} \times 2 b^2=4 a b c\end{aligned}$
Hence, the correct answer is option: 'D'.