GATE CSE 1995 | Question: 2.17

Question

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

• A. $A$ is closed under $*$ but $\langle A, *\rangle$ is not a semigroup.
• B. $\langle A, *\rangle$ is a semigroup but not a monoid.
• C. $\langle A, * \rangle$ is a monoid but not a group.
• D. $\langle A, *\rangle$ is a a group but not an abelian group.

Comments

They messed up in the question @shishir__roy.

One line is missing:

Set of all $n*n$ non-singular matrices over reals. 

From options, it can be inferred. 

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@Deepak Poonia   (DETAIL EXP)

Note : Given Base set must be set of all n x n non-singular matrices over Reals 

To check GROUP the following [1 - 4] property must be satisfied:

1) Closure property 

     Multiplication of two non singular matrix is also a non singular

     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. 

           AB=C ⟹ det(AB) = det(C)

          det(AB)=0  [ ∵ det(C)=0 as C is singular ]

          ⟹  ∣A∣.∣B∣=0

          ⟹  ∣A∣=0 or ∣B∣=0

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

2) Associativity :

          MATRIX Multiplication is always Associative

          (A . B) . C =  A . (B . C) .

3) Identity property :

          For identity element, we need to check

          I.A = A  and A.I = A (Both are true for the given structure)

          e = I , where I = Unit Matirx​​​​​.

4) Inverse property :

          If A is non singluar matrix means, |A|= Non Zero

          Therefore A inverse Exist

5) Commutative property :

           Matirx Multiplication is not Commutative 

            (AB may or may not be equal to BA)   or

            we can say ----->  ∃ A,B  (AB ≠ BA)

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from above 5 property we can conclude that 

<A,*> is a Group but Not Abelian Group

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Note : Above binary operation are not closed as @shishir__roy said..in the same way  there are many group theory question which are actually not closed (given)...Don't worry about closure property if they are not specifically asked ..Just feel the question and try to answer based on option.

Answer 1

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.

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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.

Comments

If A•B (Square Matrix multiplication) is coming Null matrix or Singular Matrix that means either A or B or both is/are Singular Matrix.

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I think the Order of A is not mentioned in the question therefore there can be matrices such that matrix A  is 2 X 3 and B is 4 X 5 so here matrix mutliplication is not possible hence its not Closed under * operation thats why none of options are correct.

Correct me if I am wrong!

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@Umair_Khan but there is no option as none of these, so it means we need to pick a n*n matrix.

Answer 2

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.

Answer 3

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

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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.

Comments

Closed because multiplication of two non-singular matrices is non-singular.

Answer 4

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