Let T and U be the set of all orthogonal matr | vedclass

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

Question

(B) The set $A = \{-1, 0, 1\}$ is a finite set with cardinality $n(A) = 3$. The set $T$ consists of all orthogonal matrices of order $3 	imes 3$. The

Vedclass · Accepted Answer

There does not exist a bijective mapping between $A$ and $T$, or between $A$ and $U$.

Vedclass · Answer

(B) The set $A = \{-1, 0, 1\}$ is a finite set with cardinality $n(A) = 3$. The set $T$ consists of all orthogonal matrices of order $3 	imes 3$. The group of orthogonal matrices $O(3)$ is an infinite set. The set $U$ consists of all non-singular matrices of order $3 	imes 3$, which is the general linear group $GL(3, \mathbb{R})$. This is also an infinite set. $A$ bijective mapping between two sets exists if and only if they have the same cardinality. Since $n(A) = 3$ and both $T$ and $U$ are infinite sets, $n(A) 
eq n(T)$ and $n(A) 
eq n(U)$. Therefore, there does not exist a bijective mapping between $A$ and $T$, nor between $A$ and $U$.

Let $T$ and $U$ be the set of all orthogonal matrices of order $3$ over $\mathbb{R}$ and the set of all non-singular matrices of order $3$ over $\mathbb{R}$ respectively. Let $A = \{-1, 0, 1\}$. Then:

Similar Questions

Let $G(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$. If $x+y=0$,then $G(x) G(y) =$

Which of the following matrices is not a square matrix?

$\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \begin{bmatrix} 2 & 1 & -1 \end{bmatrix} = $

For a matrix $A$,the conditions $AI = A$ and $AA^T = I$ are true for:

Which of the following statements is not correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module