Here is a more elaborate, clear, and educational solution to the given problem.

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1. Key Concepts and Formulas

Before diving into the solution, let's recall some fundamental properties of matrices and determinants that will be crucial for solving this problem:

• Determinant of Adjoint Matrix: For any square matrix $A$ of order $n \times n$, the determinant of its adjoint matrix, denoted as $\text{adj } A$, is given by the formula: $\det(\text{adj } A) = (\det A)^{n-1}$ In this problem, the matrix $A$ is a $3 \times 3$ matrix, so $n=3$. Therefore, for this specific case, we will use $\det(\text{adj } A) = (\det A)^{3-1} = (\det A)^2$.

• Sum of Squares of First N Natural Numbers: The sum of the squares of the first $N$ natural numbers is given by the formula: $\sum_{k=1}^{N} k^2 = 1^2 + 2^2 + 3^2 + \dots + N^2 = \frac{N(N+1)(2N+1)}{6}$

These two formulas will be the backbone of our solution.

2. Understanding the Set S

The problem defines the set $S$ as: $S = \{\sqrt{n} : 1 \le n \le 50 \text{ and n is odd}\}$ This means we need to find all odd integers $n$ between 1 and 50 (inclusive), and then take their square roots to form the elements of $S$.

Let's list the odd integers $n$ in the given range: $n \in \{1, 3, 5, 7, \dots, 49\}$

Now, forming the elements of $S$ by taking the square root of each $n$: $S = \{\sqrt{1}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \dots, \sqrt{49}\}$ This set contains 25 elements, as there are 25 odd numbers between 1 and 49. (Number of terms = $\frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1 = \frac{49-1}{2} + 1 = 24/2 + 1 = 12+1 = 25$).

3. Calculating the Determinant of Matrix A

The given matrix is: A = \left[ {\matrix{ 1 & 0 & a \cr { - 1} & 1 & 0 \cr { - a} & 0 & 1 \cr } } \right] To find $\det A$, we will expand along the first row. The formula for the determinant of a $3 \times 3$ matrix \left[ {\matrix{ p & q & r \cr s & t & u \cr v & w & x \cr } } \right] is $p(tx-uw) - q(sx-uv) + r(sw-tv)$.

Following this, for matrix $A$: \det A = 1 \cdot \det \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right] - 0 \cdot \det \left[ {\matrix{ { - 1} & 0 \cr { - a} & 1 \cr } } \right] + a \cdot \det \left[ {\matrix{ { - 1} & 1 \cr { - a} & 0 \cr } } \right] Let's calculate each $2 \times 2$ determinant:

• $1 \cdot (1 \cdot 1 - 0 \cdot 0) = 1 \cdot (1 - 0) = 1$
• $0 \cdot (\dots)$ will be $0$.
• $a \cdot ((-1) \cdot 0 - 1 \cdot (-a)) = a \cdot (0 - (-a)) = a \cdot a = a^2$

Combining these terms: $\det A = 1 - 0 + a^2 = a^2 + 1$ This result is crucial for the next step.

4. Calculating the Determinant of Adjoint A

Now we use the property of the determinant of an adjoint matrix. Since $A$ is a $3 \times 3$ matrix ($n=3$), we have: $\det(\text{adj } A) = (\det A)^{3-1} = (\det A)^2$ Substituting the value of $\det A = a^2+1$ that we just calculated: $\det(\text{adj } A) = (a^2+1)^2$ This expression gives us the determinant of the adjoint matrix for any $a \in S$.

5. Evaluating the Summation

The problem asks us to find the sum $\sum_{a\, \in \,S}^{} {\det (adj\,A)}$. Substituting the expression we just found for $\det(\text{adj } A)$: $\sum_{a\, \in \,S}^{} {\det (adj\,A)} = \sum_{a\, \in \,S}^{} {({a^2} + 1)^2}$ Now, we iterate through each element $a$ in the set $S = \{\sqrt{1}, \sqrt{3}, \sqrt{5}, \dots, \sqrt{49}\}$ and substitute it into the expression $(a^2+1)^2$:

• For $a = \sqrt{1}$: The term is $((\sqrt{1})^2 + 1)^2 = (1 + 1)^2 = 2^2$.
• For $a = \sqrt{3}$: The term is $((\sqrt{3})^2 + 1)^2 = (3 + 1)^2 = 4^2$.
• For $a = \sqrt{5}$: The term is $((\sqrt{5})^2 + 1)^2 = (5 + 1)^2 = 6^2$.
• ...
• For $a = \sqrt{49}$: The term is $((\sqrt{49})^2 + 1)^2 = (49 + 1)^2 = 50^2$.

So, the summation becomes: $\sum_{a\, \in \,S}^{} {({a^2} + 1)^2} = 2^2 + 4^2 + 6^2 + \dots + 50^2$ This is a sum of squares of even numbers. We can factor out $2^2$ from each term: $2^2(1^2) + 2^2(2^2) + 2^2(3^2) + \dots + 2^2(25^2)$ $= 2^2 (1^2 + 2^2 + 3^2 + \dots + 25^2)$ $= 4 (1^2 + 2^2 + 3^2 + \dots + 25^2)$ Now we can use the formula for the sum of the squares of the first $N$ natural numbers, where $N=25$: $1^2 + 2^2 + \dots + 25^2 = \frac{25(25+1)(2 \cdot 25+1)}{6}$ $= \frac{25 \cdot 26 \cdot 51}{6}$ Let's simplify this expression: $= \frac{25 \cdot (2 \cdot 13) \cdot (3 \cdot 17)}{2 \cdot 3}$ $= 25 \cdot 13 \cdot 17$ $= 25 \cdot 221$ $= 5525$ Now, substitute this back into our summation: $\sum_{a\, \in \,S}^{} {\det (adj\,A)} = 4 \cdot (25 \cdot 221) = 100 \cdot 221$ So, the total sum is $22100$.

6. Determining the Value of $\lambda$

The problem states that $\sum_{a\, \in \,S}^{} {\det (adj\,A) = 100\lambda}$. We have calculated the sum to be $100 \cdot 221$. Therefore, we can set up the equation: $100\lambda = 100 \cdot 221$ Dividing both sides by 100, we get: $\lambda = 221$

7. Tips for Success and Common Mistakes to Avoid

• Memorize Key Formulas: Ensure you know the determinant properties (especially $\det(\text{adj } A)$ and $\det(AB)$) and standard series summation formulas (sum of first $N$ natural numbers, sum of squares, sum of cubes).
• Careful with Set Definition: Always carefully interpret the definition of sets like $S$. List out a few terms to confirm your understanding before proceeding.
• Determinant Calculation: Be meticulous when calculating determinants. A single sign error can propagate and lead to an incorrect final answer. Expanding along a row or column with zeros can simplify calculations.
• Algebraic Manipulation: Pay attention to algebraic steps, especially when factoring out common terms from a series.
• Recognize Patterns: In summations, look for patterns. Here, recognizing $2^2 + 4^2 + \dots + 50^2$ as $4(1^2 + 2^2 + \dots + 25^2)$ is crucial.

8. Summary and Key Takeaway

This problem is a good example of how JEE problems often combine multiple mathematical concepts. Here, we used:

1. Matrix and Determinant Properties: Specifically, the relationship between the determinant of a matrix and the determinant of its adjoint.
2. Set Theory Interpretation: Correctly identifying the elements of set $S$.
3. Series and Summation: Recognizing and applying the formula for the sum of squares of natural numbers.

The ability to seamlessly transition between these topics and perform accurate calculations is key to solving such problems efficiently. The final answer is $\lambda = 221$, which corresponds to option (B).