GATE CSE 2002 | Question: 1.6

Question

Which of the following is true?

• A. The set of all rational negative numbers forms a group under multiplication.
• B. The set of all non-singular matrices forms a group under multiplication.
• C. The set of all matrices forms a group under multiplication.
• D. Both B and C are true.

Comments

How is B correct. Unless we fix the size of a nonsingular matrix to say n*n only then will the multiplication operator be closed. For a set of all non singular matrices, multiplication operator will not be even defined for some pairs.

---

we know under binary operation *

1.Groupoid→ closed

2.Semi group ->Associative

3.Monoide->Identity should present

4.Group->Inverse should present

5.Abelian group→ commutative

For remind this order i use (Ground se mat Gharjao app)

 matrix “A” is non singular then $_{A}-1$ will present

For make a matrix a group we need to give guarantee that inverse is present but for every type of  matrix we can’t give guarantee but only for non singular.

 

 

 

 

Answer 1

Answer: B

• A. False. Multiplication of two negative rational numbers give positive number. So, closure property is not satisfied.
• B. True. Matrices have to be non-singular (determinant $\neq0$) for the inverse to exist.
• C. False. Singular matrices do not form a group under multiplication.
• D. False as C is false.

Comments

I have same doubt as- https://gateoverflow.in/2629/gate1995-2-17?show=170821#c170821

---

@Deepak Poonia  Sir , @Arjun Sir

And Sir do we have to assume in the exam that the size of the matrices are nxn else matrix multiplication is not possible.

---

@abir_banerjee

See this comment:

https://gateoverflow.in/2629/gate-cse-1995-question-2-17?show=377082#c377082

Answer 2

Option b

Comments

C is false because singular matrices doesnot have inverse,

But why option A is wrong?

---

Multiplication of two negative ll be a positive no.. so given  set is not even closed.. thats why A is not a group..

Answer 3

1. If a relation is group then it must be

1)Closed

2)Associative

3)Identity

4)Inverse

if a matrix is non-singular then inverse dose not exist. So option c is wrong.

Comments

If a matrix is Singular(Determinant=0) then Inverse DOES NOT EXIST.

Answer 4

(a) False, because 0 does not have an inverse

(b) True, Non-singular means Determinant!=0 So it is a group

(C) False, Determinant of the matrix can be zero which does not have an identity so not a group

(D) False

Comments

 

(a) False, because 0 does not have an inverse

As we go order wise so we will first check the  closure property for negative rational numbers and as it is not closed so we will not go till inverse.