UPSC NDA I 2022 Mathematics Solved Paper

Concept:The sum of two determinants simplifies to a negative sum of squares, always negative.Explanation:Start with $\Delta_1 = \begin{vmatrix} 1 & p & q \\ 1 & q & r \\ 1 & r & p \end{vmatrix}$.Apply row operations: $R_2 \to R_2 - R_1$, $R_3 \to R_3 - R_1$.Then $\Delta_1 = \begin{vmatrix} 1 & p & q \\ 0 & q-p & r-q \\ 0 & r-p & p-q \end{vmatrix}$.Expand: $\Delta_1 = (q-p)(p-q) - (r-p)(r-q)$.Simplify: $\Delta_1 = -(p-q)^2 - (r-p)(r-q)$.Expand: $\Delta_1 = -p^2 + 2pq - q^2 - r^2 + qr + pr - pq$Combine: $\Delta_1 = -p^2 - q^2 - r^2 + pq + pr + qr$. ---(1)Now $\Delta_2 = \begin{vmatrix} 1 & 1 & 1 \\ q & r & p \\ r & p & q \end{vmatrix}$.Apply column operations: $C_2 \to C_2 - C_1$, $C_3 \to C_3 - C_1$.Then $\Delta_2 = \begin{vmatrix} 1 & 0 & 0 \\ q & r-q & p-q \\ r & p-r & q-r \end{vmatrix}$.Expand: $\Delta_2 = (r-q)(q-r) - (p-q)(p-r)$.Simplify: $\Delta_2 = -(q-r)^2 - (p-q)(p-r)$.Expand: $\Delta_2 = -q^2 - r^2 + 2qr - p^2 + pq + pr - qr$Combine: $\Delta_2 = -p^2 - q^2 - r^2 + pq + pr + qr$. ---(2)Add (1) and (2):$\Delta_1 + \Delta_2 = -2(p^2+q^2+r^2) + 2(pq+pr+qr)$.Rewrite as $-\left[(p-q)^2 + (q-r)^2 + (r-p)^2\right]$.Since squares are non-negative, the sum is always negative (p ≠ q ≠ r ensures non-zero).Answer:$\Delta_1 + \Delta_2$ is always negative.