434 views
1 votes
1 votes
Let A be the set of all non-singular matrices over real number and let ∗ be the matrix multiplication operation. Then

answer - A is group but not abelian

DOUBT - in order to be a group it should first first be a monoid ( have identity element)

what will be the identity matrix here ?

Please log in or register to answer this question.

Related questions

3 votes
3 votes
1 answer
1
Avdhesh Singh Rana asked Oct 22, 2015
901 views
$A = \{0,1,2,3,4,5,6, \ldots 23\}$$a*b =(a+b)\mod{24}$How many proper subgroups does the group $G(A,*)$ have?
0 votes
0 votes
0 answers
2
pC asked Jun 25, 2016
241 views
A : R- {2}B: R-{1}f:A->Bg:B->Af(x) = $\frac{x}{x-2}$g(x) = $\frac{2x}{x-1}$Q1) find IA and IBQ2) find fog and gof Q3] what is your comparision
0 votes
0 votes
1 answer
3
Dhananjay15 asked Aug 19, 2018
489 views
$(a+b)^* =a^*(ba^*)^*$As this identity already proved. But $a^*(ba^*)^*$ couldn't generate "bab" . But $(a+b)^*$ could generate all strings over {a,b}. Then the above ide...
0 votes
0 votes
3 answers
4
shivani2010 asked Mar 29, 2017
456 views
For example, if we want to check if a function follow commutativity then we have to check $A * B = B * A $. So what is it for identity?