# #set theory #groups

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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

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option b ?
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Answer given is C. You can anyways explain your choice of option.
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Monoid means there is an identity element which in this case should be identity matrix. So I think it should be a Monoid
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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.

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

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.

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213 views
set of all possible diagonal matrix of order n ans given monoid my doubt-why it cannot have inverse??
$\text{Prove that : the union of any two subgroup of 'G' is not subgroup of 'G'}$ $\text{Prove that : the intersection of any two subgroup of 'G' is also a subgroup of 'G'}$