The Gateway to Computer Science Excellence
+1 vote
81 views
Consider the set H of all 3 × 3 matrices of the type:

$\begin{bmatrix} a&f&e\\ 0&b&d\\ 0&0&c\\ \end{bmatrix}$

where a, b, c, d, e and f are real numbers and $abc ≠ 0$. Under the matrix multiplication operation, the set H is:

(a) a group

(b) a monoid but not a group

(c) a semigroup but not a monoid

(d) neither a group nor a semigroup
in Set Theory & Algebra by Active (2.8k points)
edited by | 81 views
0
option b ?
0
Answer given is C. You can anyways explain your choice of option.
0
Monoid means there is an identity element which in this case should be identity matrix. So I think it should be a Monoid
+2
It should be a group.

As the inverse of an upper triangular matrix is always upper triangular and exists when all diagonals are non zero.

2 Answers

0 votes
Answer (a). This is a group. This satisfies closure and associativity properties.

Identity element is the 3x3 Identity matrix. The inverse also exists because the determinant, a*b*c, is non-zero.

It satisfies all the properties of a group. Hence, it is a group
by (207 points)
0 votes
Answer is (a). Group.

Matrix operation multiplication and addition satisfies closure, associative, identity ($I_{3}$ ).

For addition inverse will always exist. (Null Matrix).

For Multiplication for singular matrix inverse mayn't exist. But here there is a condition that diagonal element product is non zero. Due to this restriction it will be non singular.
by Boss (11.5k points)
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,313 answers
198,349 comments
105,048 users