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Consider the set $H$ of all $3 * 3$ matrices of the type $$\left( \begin{array}{ccc} a & f & e \\ 0 & b & d \\ 0 & 0 & c \end{array} \right)$$ where $a,b,c,d,e$ and $f$ are real numbers and $abc ≠ 0.$ Under the matrix multiplication operation, the set $H$ is:

  1. a group
  2. a monoid but not a group
  3. a semi group but not a monoid
  4. neither a group nor a semi group
in Set Theory & Algebra by Boss (17.5k points) | 2.2k views

3 Answers

+28 votes
Best answer

Given Information: Matrix is upper triangular. It's determinant is multiplication of principle diagonal elements. i.e., $abc.$
It is given that $abc \neq 0.$ So, Inverse for every such matrix exists.

Now this set is

  1. Closed - You can see after multiplication matrix is in same format and $|AB| = |A||B| \neq 0$ as $|A|,|B| \neq 0$
  2. Associative - Matrix multiplication is associative
  3. Existence of Identity - Identity Matrix is present
  4. Existence of Inverse - as determinant is non zero there exist inverse for every matrix

So, it is group.

Correct Answer: $A$

by Boss (41.9k points)
edited by
I think You should also check that " A inverse " is of the form of  " Upper triangular matrix ". Because it should belong to the set. Am I right ?
Yes you should do that and fortunately in this case it does.
inverse of the matrix will be as follows:

1/a   -f/ab   (fd-be)/abc

0     1/b        0

0       0         1/c

Since a,b,c,d,e and f are real numbers so, all the numbers in the above matrix is also real and since,      a*b*c != 0, means a!=0 and b!=0 and c!=0 So, 1/a=!0 , 1/b !=0, 1/c!=0  as well , So (1/a)*(1/b)*(1/c)!=0  . Hence , inverse matrix also belongs to set H
0 votes
A group
by (33 points)
what is identity element of the matrix and how?
given set is set of matrices so we need identity matrix not identity element..and identity matrix always exist..
In a null matrix identity element not exists. So how can u say for a upper triangular matrix identity element exists
let A b a null matrix..
A * Identity_Matrix = A = Identity_Matrix * A
I(n*n) satisfy above condition for every square matrix of order n..
for null or singular matrix , inverse not exist .. but it doesn't mean identity matrix not exist..
But can we prove for this matrix identity matrix is  1  0  0

                                                                          0  1  0

                                                                          0   0  1
In the matrix a,b,c,d,e,f all are present. so it is closed

matrix multiplication is associative

H * identity matrix= H ,so identity property satisfied

It is non singular matrix so inverse exist

So it is a group

But matrix multiplication is non commutative ,so it is not abelian group
0 votes


The product of diagonal elements in a triangular matrix is the determinant.

=> determinant of such matrices $\neq0$

=> Matrices are non-singular

=> Matrices are invertible. -----------> #1


Closure holds.

Associativity holds. Matrix Chain Multiplication in Dynamic Programming is an example of this.

Identity holds. (The identity matrix)

Inverse holds. // From #1

Commutativity doesn't hold. As A.B $\neq$ B.A for matrices.


So, this is a group. Option A

by Loyal (7k points)

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