Q169 / 270 · NDA 2018 (Phase 2) · Mathematics
If $a+b+c=0$, then one of the solution of
$\left|\begin{array}{ccc}
a-x & c & b \\
c & b-x & a \\
b & a & c-x
\end{array}\right|=0 \text { is }$
- A$\mathrm{x}=\mathrm{a}$
- B$\mathrm{x}=\sqrt{\frac{3\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)}{2}}$
- C$x=\sqrt{\frac{2\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)}{3}}$
- Correct answerD$\mathrm{x}=0$
Solution
$\begin{aligned}
\left|\begin{array}{ccc}
a-x & c & b \\
c & b-x & a \\
b & a & c-x
\end{array}\right|=0 \\
-\left|\begin{array}{ccc}
x-(a+b+c) & c & b \\
x-(a+b+c) & b-x & a \\
x-(a+b+c) & a & c-x
\end{array}\right|=0 \\
\Rightarrow\left|\begin{array}{ccc}
x & c & b \\
x & b-x & a \\
x & a & c-x
\end{array}\right|=0 \\
\left\{\text { Applying } \mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3\right. \text { \} } \\
\left|\begin{array}{ccc}
x & c & b \\
0 & c+x-b & b-a \\
0 & c-a & b+x-c
\end{array}\right|=0 \\
\text { \{Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \text { and } \mathrm{R}_3 \rightarrow \mathrm{R}_1-\mathrm{R}_3 \text { \} } \\
x\{(c+x-b)(b+x-c)-(b-a)(c-a)\}=0 \\
x\left\{\left(x^2-(b-c)^2-b c+a c+a b-a^2\right\}=0\right. \\
x\left(x^2-a^2-b^2-c^2+a b+b c+c a\right)=0 \\
x\left\{x^2-(a-b)^2-(b-c)^2-(c-a)^2\right\}=0 \\
\begin{array}{ll}
\therefore & x=0 \\
\end{array}
\end{aligned}$
Hence, the correct answer is option: 'D'.