To solve the problem step by step, we will follow the instructions given in the video transcript and derive the required value of \( \lambda \).
### Step 1: Define the Set \( S \)
The set \( S \) is defined as:
\[
S = \{ \sqrt{n} : 1 \leq n \leq 50 \text{ and } n \text{ is odd} \}
\]
The odd integers from 1 to 50 are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49.
Thus, the elements of \( S \) are:
\[
S = \{ \sqrt{1}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \sqrt{9}, \sqrt{11}, \sqrt{13}, \sqrt{15}, \sqrt{17}, \sqrt{19}, \sqrt{21}, \sqrt{23}, \sqrt{25}, \sqrt{27}, \sqrt{29}, \sqrt{31}, \sqrt{33}, \sqrt{35}, \sqrt{37}, \sqrt{39}, \sqrt{41}, \sqrt{43}, \sqrt{45}, \sqrt{47}, \sqrt{49} \}
\]
### Step 2: Calculate the Determinant of Matrix \( A \)
The matrix \( A \) is given as:
\[
A = \begin{pmatrix}
1 & 0 & a \\
-1 & 1 & 0 \\
-a & 0 & 1
\end{pmatrix}
\]
To find the determinant \( \text{det}(A) \), we use the formula for the determinant of a 3x3 matrix:
\[
\text{det}(A) = 1 \cdot (1 \cdot 1 - 0 \cdot 0) - 0 \cdot (-1 \cdot 1 - 0 \cdot (-a)) + a \cdot (-1 \cdot 0 - 1 \cdot (-a))
\]
Calculating this gives:
\[
\text{det}(A) = 1 - 0 + a^2 = 1 + a^2
\]
### Step 3: Calculate the Adjoint of Matrix \( A \)
The adjoint of a matrix \( A \) can be expressed in terms of its determinant:
\[
\text{adj}(A) = \text{det}(A) \cdot A^{-1}
\]
However, we are interested in \( \text{det}(\text{adj}(A)) \). The determinant of the adjoint of a matrix is given by:
\[
\text{det}(\text{adj}(A)) = (\text{det}(A))^{n-1}
\]
where \( n \) is the order of the matrix. Here, \( n = 3 \), so:
\[
\text{det}(\text{adj}(A)) = (1 + a^2)^{3-1} = (1 + a^2)^2
\]
### Step 4: Sum Over All Elements in \( S \)
Now we need to sum \( \text{det}(\text{adj}(A)) \) over all \( a \in S \):
\[
\sum_{a \in S} \text{det}(\text{adj}(A)) = \sum_{a \in S} (1 + a^2)^2
\]
Calculating \( (1 + a^2)^2 \):
\[
(1 + a^2)^2 = 1 + 2a^2 + a^4
\]
Thus, we need to calculate:
\[
\sum_{a \in S} (1 + 2a^2 + a^4) = \sum_{a \in S} 1 + 2\sum_{a \in S} a^2 + \sum_{a \in S} a^4
\]
The number of elements in \( S \) is 25 (since there are 25 odd numbers from 1 to 49), so:
\[
\sum_{a \in S} 1 = 25
\]
### Step 5: Calculate \( \sum_{a \in S} a^2 \) and \( \sum_{a \in S} a^4 \)
1. **Calculate \( \sum_{a \in S} a^2 \)**:
\[
\sum_{k=1}^{25} (2k-1)^2 = \sum_{k=1}^{25} (4k^2 - 4k + 1) = 4\sum_{k=1}^{25} k^2 - 4\sum_{k=1}^{25} k + 25
\]
Using the formulas:
\[
\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
\]
For \( n = 25 \):
\[
\sum_{k=1}^{25} k = \frac{25 \cdot 26}{2} = 325
\]
\[
\sum_{k=1}^{25} k^2 = \frac{25 \cdot 26 \cdot 51}{6} = 5525
\]
Thus:
\[
\sum_{a \in S} a^2 = 4 \cdot 5525 - 4 \cdot 325 + 25 = 22100 - 1300 + 25 = 20825
\]
2. **Calculate \( \sum_{a \in S} a^4 \)**:
\[
\sum_{k=1}^{25} (2k-1)^4 = \sum_{k=1}^{25} (16k^4 - 32k^3 + 24k^2 - 8k + 1)
\]
Using the formulas for \( \sum k^3 \) and \( \sum k^4 \):
\[
\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2
\]
For \( n = 25 \):
\[
\sum_{k=1}^{25} k^3 = \left( \frac{25 \cdot 26}{2} \right)^2 = 105625
\]
Similarly, \( \sum_{k=1}^{25} k^4 \) can be calculated using:
\[
\sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}
\]
For \( n = 25 \):
\[
\sum_{k=1}^{25} k^4 = \frac{25 \cdot 26 \cdot 51 \cdot 197}{30} = 5525 \cdot 197 = 1087975
\]
### Step 6: Final Calculation
Putting it all together:
\[
\sum_{a \in S} (1 + 2a^2 + a^4) = 25 + 2 \cdot 20825 + 1087975
\]
Calculating this gives:
\[
= 25 + 41650 + 1087975 = 1124650
\]
### Step 7: Set Equal to \( 100\lambda \)
Given that:
\[
\sum_{a \in S} \text{det}(\text{adj}(A)) = 100\lambda
\]
We have:
\[
1124650 = 100\lambda
\]
Thus:
\[
\lambda = \frac{1124650}{100} = 11246.5
\]
### Final Answer
\[
\lambda = 11246.5
\]
