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27 votes
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Let $A$ be the set of all non-singular matrices over real number and let $*$ be the matrix multiplication operation. Then

  1. $A$ is closed under $*$ but $\langle A, *\rangle$ is not a semigroup.
  2. $\langle A, *\rangle$ is a semigroup but not a monoid.
  3. $\langle A, * \rangle$ is a monoid but not a group.
  4. $\langle A, *\rangle$ is a a group but not an abelian group.
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Best answer
42 votes
42 votes

Answer: D

As the matrices are non singular so their determinant $\neq 0.$ Hence, the inverse matrix always exist.

But for a group to be abelian it should follow commutative property. As, matrix multiplication is not commutative, $\langle A,*\rangle$ is a group but not an abelian group.


Two nonsingular matrix can not give a singular matrix after product. Therefore it satisfies closure property also.

Proof: Let $A$ and $B$ are nonsingular and $C$ is singular.

Claim: For any $A$, $B$ and $C$:- $AB = C$ is NOT possible.

I will prove using contradiction. Let it be possible to have $AB = C.$ 

$\begin{align} AB &= C \\ \implies \det(AB)&=\det(C)\end{align}$

$\qquad\;\; \det(AB)=0 \text{ [}\because \det(C)=0 \;\text{as}\; C\; \text{is singular}]$

$\implies \mid A \mid. \mid B \mid =0$

$\implies \mid A \mid =0 \text{ or } \mid B \mid =0$

Which is contradiction as $A$ and $B$ both are non singular.

edited by
7 votes
7 votes
ans is D.

here we can find identity element and inverse also. so it is a group. but matrix multiplication is not commutative therefore not an abelian group.
3 votes
3 votes

Non-singular matrix means it has a non-zero determinant.

 

Closure? Yes.

Associativity? Yes. The existence of Matrix Chain Multiplication in Dynamic Programming is a proof of associativity.

Identity? Yes. The identity matrix.

Inverse? Yes, because given set is the set of non-singluar matrices.

Commutative? NO.

 

So, Group, but not Abelian. Option D


PS: If just Matrices were given, then it'd fail on inverse, because a singular matrix is not invertible, and an invertible matrix is not singular. A matrix can be either singular or invertible, not none, not both.

So, in that case the answer would be Monoid.

Answer:

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