in Set Theory & Algebra edited by
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4 votes
4 votes
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
in Set Theory & Algebra edited by
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4 Comments

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.
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2 Answers

1 vote
1 vote
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
1 vote
1 vote
Answer 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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