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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

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.
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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 ?
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Yes you should do that and fortunately in this case it does.
A group
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what is identity element of the matrix and how?
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given set is set of matrices so we need identity matrix not identity element..and identity matrix always exist..
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In a null matrix identity element not exists. So how can u say for a upper triangular matrix identity element exists
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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..
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But can we prove for this matrix identity matrix is  1  0  0

0  1  0

0   0  1
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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