Let S = {√n : 1 ≤ n ≤ 50 and n is odd}. Let a ∈ S and A = [[1, 0, a], [-1, 1, 0], [-a, 0, 1]]. If Σ

Q: Wait, why does the determinant of the adjoint become $(\det A)^2$? Is it always squared?

That is a fantastic question! The general property is $\det(\text{adj } A) = (\det A)^{n-1}$, where $n$ is the order of the square matrix. In this problem, we are dealing with a $3 \times 3$ matrix, so $n=3$. Therefore, the exponent is $3-1=2$. If the matrix were $4 \times 4$, the exponent would be $3$. Always check the dimensions of your matrix before applying the formula!

Q: Why did the square root in $a = \sqrt{n}$ not cause issues in the final sum? I was worried about irrational numbers.

I love that you are thinking about the domain! Notice that the determinant calculation naturally leads to $a^2$. Since $a = \sqrt{n}$, then $a^2 = n$. The square root is essentially 'undone' by the squaring process inherent in the determinant formula. In JEE problems, often the 'scary' parts (like square roots) are designed to simplify perfectly. Trust the algebra!

Q: Is there a faster way to sum $2^2 + 4^2 + \dots + 50^2$ than adding them one by one?

Absolutely! Never add them manually—that is a trap for calculation errors. Factor out $2^2$ (which is 4) from every term. This leaves you with $4(1^2 + 2^2 + \dots + 25^2)$. Now you can use the standard formula $\frac{N(N+1)(2N+1)}{6}$ with $N=25$. It is elegant, fast, and much less prone to mistakes.

Question

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

Options:

(A) 218

(B) 221

(C) 663

(D) 1717

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Every great journey begins with understanding the terrain. We are given a set $S = \{\sqrt{n} : 1 \le n \le 50 	ext{ and } n 	ext{ is odd}\}$.

Before we touch the matrix, we must understand what we are summing over. The variable $n$ takes on odd values from $1$ to $50$, specifically $1, 3, 5, \dots, 49$.

If you count these, you will find there are exactly $25$ such values. Knowing that we have $25$ terms is our first victory, as it tells us that whatever expression we derive for $\det(	ext{adj } A)$, we will be summing it $25$ times.

### The Matrix and the Determinant

Now, let us look at our matrix $A$:

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

When you see a matrix with zeros, your heart should leap with joy. Zeros are gifts; they simplify our determinant expansion. Let us expand along the first row:

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

Simplifying this, we get:

$$\det(A) = 1 + a^2$$

It is remarkably clean. The complexity of the matrix collapses into a simple quadratic expression, which is the first sign that we are on the right track.

### The Adjoint Property—The JEE Secret Weapon

Here is where the problem tests your conceptual depth. We need $\det(	ext{adj } A)$. Many students panic here, thinking they need to calculate the inverse or the cofactor matrix, but that is the long, painful road.

Recall the property:

$$\det(	ext{adj } A) = (\det A)^{n-1}$$

Since our matrix is $3 	imes 3$, $n=3$. Therefore, $\det(	ext{adj } A) = (\det A)^2$. Substituting our previous result, we get:

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

Now, remember that $a = \sqrt{n}$. When we substitute this into our expression, the square root vanishes, leaving us with $(1 + n)^2$. The expression is now purely in terms of $n$, and the irrationality is gone.

### The Summation

We are now tasked with calculating:

$$\sum_{a \in S} \det(	ext{adj } A) = \sum_{n \in \{1, 3, \dots, 49\}} (1 + n)^2$$

Let us write out the first few terms to visualize the series:

For $n=1$: $(1+1)^2 = 2^2 = 4$

For $n=3$: $(1+3)^2 = 4^2 = 16$

For $n=5$: $(1+5)^2 = 6^2 = 36$

This is the sum of squares of even numbers: $2^2 + 4^2 + 6^2 + \dots + 50^2$.

### The Final Calculation

To sum this efficiently, we factor out $2^2$ from every term:

$$2^2(1^2 + 2^2 + 3^2 + \dots + 25^2)$$

We are left with $4 	imes \sum_{k=1}^{25} k^2$. Using the standard formula for the sum of squares, $\frac{N(N+1)(2N+1)}{6}$, where $N=25$:

$$	ext{Sum} = 4 	imes \frac{25(25+1)(2 	imes 25 + 1)}{6}$$

$$	ext{Sum} = 4 	imes \frac{25 	imes 26 	imes 51}{6} = 22100$$

The problem states this sum is equal to $100\lambda$. Therefore:

$$100\lambda = 22100 \Rightarrow \mathbf{\lambda = 221}$$

Animated Solution for Mathematics - Matrices and Determinants: Let $S = {n : 1 \leq n \leq 50 and n is odd}$ . Let $a \in S$ and $A = 1 - 1 - a 010 a 01$ . If $\sum_{a \in S} det (adj A) = 100 λ$ , then $λ$ is equal to

Select Answer:

Visualized Solution

Defining the Set $S$

Calculating the Determinant $∣ A ∣$

Property of $det (adj A)$

Substitution of $a = n$

Setting up the Summation

Simplifying the Series

Applying the Sum of Squares Formula

Final Calculation for $λ$

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Matrix and the Determinant

The Adjoint Property—The JEE Secret Weapon

The Summation

The Final Calculation

Similar Questions

Let A be a $3 \times 3$ matrix such that $A + A^{T} = O$ . If $A 1 - 1 0 = 332, A^{2} 1 - 1 0 = - 3 19 - 24$ and $det (adj (2 adj (A + I))) = (2)^{α} \cdot (3)^{β} \cdot (11)^{γ}, α, β, γ$ are non-negative integers, then $α + β + γ$ is equal to .........

Let $α \in (0, \infty)$ and $A = 110201 α 12$ . If $det (adj (2 A - A^{T}) \cdot adj (A - 2 A^{T})) = 2^{8}$ , then $(det (A))^{2}$ is equal to:

Let $A = {a_{ij}}$ be a $3 \times 3$ matrix, where $a_{ij} = ⎩ ⎨ ⎧ (- 1)^{j - i} 2 (- 1)^{i + j} if i < j if i = j if i > j$ then $det (3 Adj (2 A^{- 1}))$ is equal to

Let $A = 246 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q$ . If $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}, m, n \in N$ , then $m + n$ is equal to

Let A be a $3 \times 3$ matrix such that $X^{T} A X = 0$ for all nonzero $3 \times 1$ matrices $X = x y z$ . If $A 111 = 14 - 5$ and $A 121 = 04 - 8$ , and $det (a d j (2 (A + I))) = 2^{α} 3^{β} 5^{γ}$ , $α, β, γ \in N,$ then $α^{2} + β^{2} + γ^{2}$ is

Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $det (Q) = 2^{10}$ . Then the value of $det (adj (adj P))$ is:

Let $A = 210 12 - 1 0 - 1 2$ . If $∣ a d j (a d j (a d j 2 A)) ∣ = (16)^{n}$ , then $n$ is equal to

Let $X = 000100010, Y = α I + β X + γ X^{2}$ and $Z = α^{2} I - α β X + (β^{2} - α γ) X^{2}, α, β, γ \in R$ . If $Y^{- 1} = 1/5 00 - 2/5 1/5 0 1/5 - 2/5 1/5$ , then $(α - β + γ)^{2}$ is equal to

Let $A$ be a $2 \times 2$ matrix with $det (A) = - 1$ and $det ((A + I) (adj (A) + I)) = 4$ . Then the sum of the diagonal elements of $A$ can be :

Select Answer:

Visualized Solution

Defining the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Calculating the Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣

Property of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(adjA)

Substitution of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=n​