UPSC NDA 2 2021 Mathematics Solved Paper

Concept:The determinant of a matrix can be simplified using the given relationships between $x$, $y$, and $z$.Explanation:First, substitute the given expressions for $x$, $y$, and $z$ into the determinant.We have:$x = \frac{a}{b-c}$, $y = \frac{b}{c-a}$, $z = \frac{c}{a-b}$.The determinant is:$\begin{vmatrix} 1 & -x & x \\ 1 & 1 & -y \\ 1 & z & 1 \end{vmatrix}$Replace $x$, $y$, $z$ with their fractions.The resulting determinant becomes:$\begin{vmatrix} 1 & -\frac{a}{b-c} & \frac{a}{b-c} \\ 1 & 1 & -\frac{b}{c-a} \\ 1 & \frac{c}{a-b} & 1 \end{vmatrix}$Now expand the determinant using the first row.$\Delta = 1 \cdot \left(1 \cdot 1 - \left(-\frac{b}{c-a}\right) \cdot \frac{c}{a-b}\right) - \left(-\frac{a}{b-c}\right) \cdot \left(1 \cdot 1 - \left(-\frac{b}{c-a}\right) \cdot 1\right) + \frac{a}{b-c} \cdot \left(1 \cdot \frac{c}{a-b} - 1 \cdot 1\right)$Simplify each term carefully.First term: $1 \cdot \left(1 + \frac{bc}{(c-a)(a-b)}\right)$Second term: $+\frac{a}{b-c} \cdot \left(1 + \frac{b}{c-a}\right)$Third term: $+\frac{a}{b-c} \cdot \left(\frac{c}{a-b} - 1\right)$Combine the second and third terms:$\frac{a}{b-c} \left[1 + \frac{b}{c-a} + \frac{c}{a-b} - 1\right] = \frac{a}{b-c} \left[\frac{b}{c-a} + \frac{c}{a-b}\right]$Convert to a common denominator: $\frac{a}{b-c} \cdot \frac{b(a-b) + c(c-a)}{(c-a)(a-b)}$Simplify numerator: $b(a-b) + c(c-a) = ab - b^2 + c^2 - ac = a(b-c) - (b^2 - c^2) = a(b-c) - (b-c)(b+c) = (b-c)(a - b - c)$So the sum of second and third terms becomes $\frac{a}{b-c} \cdot \frac{(b-c)(a-b-c)}{(c-a)(a-b)} = \frac{a(a-b-c)}{(c-a)(a-b)}$Now add the first term: $1 + \frac{bc}{(c-a)(a-b)}$The total expression is:$\Delta = 1 + \frac{bc}{(c-a)(a-b)} + \frac{a(a-b-c)}{(c-a)(a-b)}$Combine the fractions over common denominator $(c-a)(a-b)$:$1 + \frac{bc + a(a-b-c)}{(c-a)(a-b)} = 1 + \frac{bc + a^2 - ab - ac}{(c-a)(a-b)}$Rewrite $1$ as $\frac{(c-a)(a-b)}{(c-a)(a-b)}$:$\Delta = \frac{(c-a)(a-b) + bc + a^2 - ab - ac}{(c-a)(a-b)}$Expand $(c-a)(a-b) = a c - bc - a^2 + ab$So numerator becomes: $(ac - bc - a^2 + ab) + bc + a^2 - ab - ac = 0$Thus $\Delta = 0$.Answer:The value of the determinant is $0$, which corresponds to option A.