UPSC NDA II 2018 Mathematics Solved Paper

Concept:Using elementary row transformations simplifies the determinant without changing its value.Explanation:We start with the given determinant equation:$\begin{vmatrix} a-x & c & b \\ c & b-x & a \\ b & a & c-x \end{vmatrix} = 0$Apply the row operation $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$.This gives:$\begin{vmatrix} a+b+c-x & a+b+c-x & a+b+c-x \\ c & b-x & a \\ b & a & c-x \end{vmatrix} = 0$Factor the common term $(a+b+c-x)$ from the first row:$(a+b+c-x) \begin{vmatrix} 1 & 1 & 1 \\ c & b-x & a \\ b & a & c-x \end{vmatrix} = 0$Thus, either $(a+b+c-x)=0$ or the remaining determinant equals zero.Solving $(a+b+c-x)=0$ gives $x = a+b+c$.Given that $a+b+c = 0$, we have $x = 0$.Therefore, one solution is $x=0$.Answer:$x = 0$ (Option D)