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

Given $$S = \{\sqrt{n} : 1 \leqslant n \leqslant 50, n \text{ is odd}\}$$ and the matrix:

$$ A = \begin{bmatrix} 1 & 0 & a \\ -1 & 1 & 0 \\ -a & 0 & 1 \end{bmatrix} $$

Find $$\det(A)$$.

Expanding along the first row:

$$ \det(A) = 1(1 \cdot 1 - 0 \cdot 0) - 0(-1 \cdot 1 - 0 \cdot (-a)) + a(-1 \cdot 0 - 1 \cdot (-a)) $$

$$ = 1(1) - 0 + a(0 + a) = 1 + a^2 $$

For an $$n \times n$$ matrix, $$\det(\text{adj } A) = (\det A)^{n-1}$$.

Here $$n = 3$$, so:

$$ \det(\text{adj } A) = (\det A)^2 = (1 + a^2)^2 $$

Since $$a = \sqrt{n}$$ where $$n$$ is odd, $$a^2 = n$$.

$$ \det(\text{adj } A) = (1 + n)^2 $$

The odd values of $$n$$ from 1 to 50 are: $$1, 3, 5, 7, \ldots, 49$$.

There are 25 such values.

$$ \sum_{a \in S} \det(\text{adj } A) = \sum_{\substack{n=1 \\ n \text{ odd}}}^{49} (1+n)^2 = \sum_{\substack{n=1 \\ n \text{ odd}}}^{49} (1+n)^2 $$

When $$n = 1, 3, 5, \ldots, 49$$, we get $$(1+n) = 2, 4, 6, \ldots, 50$$.

$$ = \sum_{k=1}^{25} (2k)^2 = 4\sum_{k=1}^{25} k^2 = 4 \cdot \frac{25 \cdot 26 \cdot 51}{6} $$

$$ = 4 \cdot \frac{33150}{6} = 4 \cdot 5525 = 22100 $$

We are given $$\sum = 100\lambda$$:

$$ 22100 = 100\lambda $$

$$ \lambda = 221 $$

The answer is Option B: 221.