23 votes 23 votes Which of the following is true? The set of all rational negative numbers forms a group under multiplication. The set of all non-singular matrices forms a group under multiplication. The set of all matrices forms a group under multiplication. Both B and C are true. Set Theory & Algebra gatecse-2002 set-theory&algebra group-theory normal + – Kathleen asked Sep 15, 2014 Kathleen 11.3k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Shivam Gairola commented Dec 12, 2015 reply Follow Share 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 votes 3 votes saheb sarkar1997 commented Sep 18, 2021 reply Follow Share 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. 1 votes 1 votes Please log in or register to add a comment.
Best answer 30 votes 30 votes Answer: B False. Multiplication of two negative rational numbers give positive number. So, closure property is not satisfied. True. Matrices have to be non-singular (determinant $\neq0$) for the inverse to exist. False. Singular matrices do not form a group under multiplication. False as C is false. Rajarshi Sarkar answered May 11, 2015 • edited Jan 12, 2023 by shadymademe Rajarshi Sarkar comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Verma Ashish commented Nov 29, 2019 reply Follow Share I have same doubt as- https://gateoverflow.in/2629/gate1995-2-17?show=170821#c170821 0 votes 0 votes Thadymademe commented Aug 9, 2022 reply Follow Share @Deepak Poonia Sir , @Arjun SirAnd Sir do we have to assume in the exam that the size of the matrices are nxn else matrix multiplication is not possible. 0 votes 0 votes Deepak Poonia commented Aug 9, 2022 reply Follow Share @abir_banerjee See this comment: https://gateoverflow.in/2629/gate-cse-1995-question-2-17?show=377082#c377082 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Option b anshu answered Feb 3, 2015 anshu comment Share Follow See all 2 Comments See all 2 2 Comments reply amkrj commented Apr 23, 2015 reply Follow Share C is false because singular matrices doesnot have inverse, But why option A is wrong? 1 votes 1 votes Digvijay Pandey commented Apr 23, 2015 reply Follow Share Multiplication of two negative ll be a positive no.. so given set is not even closed.. thats why A is not a group.. 7 votes 7 votes Please log in or register to add a comment.
0 votes 0 votes 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. abhishekmehta4u answered May 9, 2018 abhishekmehta4u comment Share Follow See 1 comment See all 1 1 comment reply Harshada commented Dec 8, 2018 reply Follow Share If a matrix is Singular(Determinant=0) then Inverse DOES NOT EXIST. 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes (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 imkeshav answered Oct 15, 2021 imkeshav comment Share Follow See 1 comment See all 1 1 comment reply Pranavpurkar commented Oct 21, 2022 reply Follow Share (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. 0 votes 0 votes Please log in or register to add a comment.