in Set Theory & Algebra
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Which of the following is true?

  1. The set of all rational negative numbers forms a group under multiplication.
  2. The set of all non-singular matrices forms a group under multiplication.
  3. The set of all matrices forms a group under multiplication.
  4. Both B and C are true.
in Set Theory & Algebra
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2 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.
2

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.

 

 

 

 

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3 Answers

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Best answer

Answer: B

  1. False. Multiplication of two negative rational numbers give positive number. So, closure property is not satisfied.
  2. True. Matrices have to be non-singular (determinant !=0) for the inverse to exist.
  3. False. Singular matrices do not form a group under multiplication.
  4. False as C is false.
edited by

5 Comments

two non singular matrics multiplication is always non-singular??
1
if $A$ and $B$ are non-singular matrices, i.e. $det(A)$ $\neq$$0$ and $det(B)$ $\neq$$0$

then, $det(AB)$ =$det(A)$ $det(B)$ $\neq$$0$

so, $AB$ is also non-singular always
8
Sir, if the order of matrices is not applicable to multiplication...like Anxm  Bpxq then how it can be a group sir?
2
edited by
If matrices are rectangular then how can we check for singular or non sigular..

So i think sigular/non singular matrices  means that they are square mateices.. and there is no rectangular matrics (Anxm).
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Option b
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2 Comments

C is false because singular matrices doesnot have inverse,

But why option A is wrong?
1
Multiplication of two negative ll be a positive no.. so given  set is not even closed.. thats why A is not a group..
7
0 votes
  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.

1 comment

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

1
Answer:

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