The set of all 2 times 2 matrices over the re | vedclass

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

Question

(D) group $(G, \cdot)$ must satisfy four axioms: closure,associativity,identity,and inverse. For the set of all $2 	imes 2$ matrices under multiplica

Vedclass · Accepted Answer

Inverse axiom may not be satisfied

Vedclass · Answer

(D) group $(G, \cdot)$ must satisfy four axioms: closure,associativity,identity,and inverse. For the set of all $2 	imes 2$ matrices under multiplication,the identity element is the identity matrix $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$. However,for an element to have an inverse,its determinant must be non-zero. Since there exist $2 	imes 2$ matrices with a determinant of $0$ (singular matrices),these matrices do not have a multiplicative inverse. Therefore,the inverse axiom is not satisfied for all elements in the set,making it not a group.

The set of all $2 \times 2$ matrices over the real numbers is not a group under matrix multiplication because

Similar Questions

$A$ and $B$ are $n$-rowed square matrices such that $AB = O$ and $B$ is non-singular. Then:

If $A$ and $B$ are invertible matrices of the same order, then which of the following is not correct?

If $A = \begin{bmatrix} 2 & -4 \\ -3 & 6 \end{bmatrix}$,then $A^{-1} =$ . . . . . . .

If $|\operatorname{Adj} A|=x$ and $|\operatorname{Adj} B|=y$, then $\left|(\operatorname{Adj}(AB))^{-1}\right|=$

If $\omega$ is a complex cube root of unity and $A = \begin{bmatrix} \omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$,then $A^{-1} = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module