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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$221$

Vedclass · Answer

(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:

Similar Questions

Let $\omega$ be the complex number $\cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3}$. Then the number of distinct complex numbers $z$ satisfying $\left|\begin{array}{ccc} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{array}\right| = 0$ is equal to

Let $S = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{0, 1, 2, 3, 4 \} \text{ and } A^2 - 4A + 3I = 0 \right\}$ be a set of $2 \times 2$ matrices. Then the number of matrices in $S$, for which the sum of the diagonal elements is equal to $4$, is:

If $C$ and $D$ are two $n \times n$ non-singular matrices over the set of real numbers $\mathbb{R}$ such that $CD = -DC$, then $n$ is:

Which of the following is incorrect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module