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+12 votes

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.
asked in Set Theory & Algebra by Veteran (52.1k points) | 1.4k views
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.

3 Answers

+20 votes
Best answer
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 !=0) for the inverse to exist.

C: False. Singular matrices do not form a group under multiplication.

D. False as C is false.
answered by Boss (33.8k points)
selected by
two non singular matrics multiplication is always non-singular??
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
Sir, if the order of matrices is not applicable to Anxm  Bpxq then how it can be a group sir?
0 votes
Option b
answered by Active (3.3k points)
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..
0 votes
  1. If a relation is group then it must be





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

answered by Boss (34.2k points)

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


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