UPSC NDA I 2021 Mathematics Solved Paper

Concept:For a $2 \times 2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, its value is $ad - bc$.To maximize the value, choose the largest possible numbers for $a$ and $d$ and the smallest possible for $b$ and $c$.Explanation:All prime numbers less than $20$ are: $2, 3, 5, 7, 11, 13, 17, 19$.We need four different primes for $p, q, r, s$ in $\begin{vmatrix} p & q \\ r & s \end{vmatrix}$.The determinant is $ps - qr$.To maximize $ps - qr$, assign the two largest primes ($19$ and $17$) to $p$ and $s$, and the two smallest primes ($2$ and $3$) to $q$ and $r$.One such assignment is $p = 17$, $q = 2$, $r = 3$, $s = 19$.Thus, the determinant becomes $17 \times 19 - 2 \times 3 = 323 - 6 = 317$.Any other arrangement using these same four numbers gives the same value because $ps - qr$ is unchanged when swapping $p$ with $s$ or $q$ with $r$.Hence, the maximum possible determinant is $317$.Answer:$317$