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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
asked in Set Theory & Algebra by Boss (18.3k points) | 1.4k views

2 Answers

+22 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.

answered by Boss (42.9k points)
edited by
+3
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 ?
+1
Yes you should do that and fortunately in this case it does.
0
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
answered by (33 points)
0
what is identity element of the matrix and how?
+2
given set is set of matrices so we need identity matrix not identity element..and identity matrix always exist..
0
In a null matrix identity element not exists. So how can u say for a upper triangular matrix identity element exists
+1
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..
0
But can we prove for this matrix identity matrix is  1  0  0

                                                                          0  1  0

                                                                          0   0  1
+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


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