JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Determinants

The set $S=\{\sqrt{n}:1\le n\le 50,n\text{odd}\}$ contains the twenty-five values $\sqrt{1},\sqrt{3},\dots ,\sqrt{49}$. Expanding the determinant along the first row, $|A|=1(1\cdot 1-0\cdot 0)-0+a((-1)\cdot 0-1\cdot (-a))=1+a^2$. For a third-order matrix $\det (\mathrm{adj}A)=|A{|}^2$, so each term of the sum is $(1+a^2{)}^2$, and since $a=\sqrt{n}$ we have $a^2=n$. Hence ${\sum }_{a\in S}\det (\mathrm{adj}A)={\sum }_{n=1,3,\dots ,49}(1+n{)}^2$. As $n$ runs through the odd numbers, $1+n$ runs through the even numbers $2,4,\dots ,50$, that is $2k$ for $k=1$ to $25$, so the sum becomes ${\sum }_{k=1}^{25}(2k{)}^2=4{\sum }_{k=1}^{25}{k}^2=4\cdot \frac{25\cdot 26\cdot 51}{6}=4\cdot 5525=22100$. Writing this as $100\lambda$ gives $\lambda =221$.