Let S = {sqrt{n} : 1 leq n leq 50, n text{ is | vedclass

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(B) The set $S$ consists of square roots of odd integers from $1$ to $50$. Thus,$S = \{\sqrt{1}, \sqrt{3}, \dots, \sqrt{49}\}$. The number of terms in

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$221$

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(B) The set $S$ consists of square roots of odd integers from $1$ to $50$. Thus,$S = \{\sqrt{1}, \sqrt{3}, \dots, \sqrt{49}\}$. The number of terms in $S$ is $25$.For the matrix $A = \begin{bmatrix} 1 & 0 & a \ -1 & 1 & 0 \ -a & 0 & 1 \end{bmatrix}$,the determinant is $|A| = 1(1 - 0) - 0 + a(0 - (-a)) = 1 + a^2$.We know that $\operatorname{det}(\operatorname{adj} A) = |A|^{n-1}$,where $n$ is the order of the matrix. Here $n=3$,so $\operatorname{det}(\operatorname{adj} A) = |A|^2 = (1 + a^2)^2$.We need to calculate $\sum_{a \in S} (1 + a^2)^2$. Since $a = \sqrt{n}$,$a^2 = n$. The sum becomes $\sum_{n \in \{1, 3, \dots, 49\}} (1 + n)^2$.Let $n = 2k - 1$ for $k = 1, 2, \dots, 25$. Then $1 + n = 1 + 2k - 1 = 2k$.The sum is $\sum_{k=1}^{25} (2k)^2 = 4 \sum_{k=1}^{25} k^2$.Using the formula $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$,we get $4 	imes \frac{25(26)(51)}{6} = 4 	imes 25 	imes 13 	imes 17 = 22100$.Given $\sum \operatorname{det}(\operatorname{adj} A) = 100 \lambda$,we have $22100 = 100 \lambda$,which implies $\lambda = 221$.

Let $S = \{\sqrt{n} : 1 \leq n \leq 50, n \text{ is odd}\}$. Let $a \in S$ and $A = \begin{bmatrix} 1 & 0 & a \\ -1 & 1 & 0 \\ -a & 0 & 1 \end{bmatrix}$. If $\sum_{a \in S} \operatorname{det}(\operatorname{adj} A) = 100 \lambda$,then $\lambda$ is equal to:

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