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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 Veteran (18k points) | 1.1k views

2 Answers

+21 votes
Best answer
Given Information - > Matrix is upper triangular. It's determinant is multiplication of principle diagonal elements. i.e. abc.

It is given thab abc != 0. So Inverse for every such matrix exists.

Now this set is

1. Closed.(You can see after multiplication Matrix is in Same format & |AB| = |A||B| = Non zero as |A|,|B| are non zero)

2. Associative (Matrix multiplication is associative).

3. Identity -> Identity Matrix In

4. Inverse, as determinant is non zero there exist inverse for every matrix.

So it is group.
answered by Veteran (49.5k points)
selected 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.
0 votes
A group
answered by (77 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


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