Discrete Mathematics Questions and Answers – Groups – Existence of Identity & Inverse

This set of Discrete Mathematics Puzzles focuses on “Groups – Existence of Identity & Inverse”.

1. In a group there must be only __________ element.
a) 1
b) 2
c) 3
d) 5
View Answer

Answer: a
Explanation: There can be only one identity element in a group and each element in a group has exactly one inverse element. Hence, two important consequences of the group axioms are the uniqueness of the identity element and the uniqueness of inverse elements.

2. _____ is the multiplicative identity of natural numbers.
a) 0
b) -1
c) 1
d) 2
View Answer

Answer: c
Explanation: 1 is the multiplicative identity of natural numbers as a⋅1=a=1⋅a ∀a∈N. Thus, 1 is the identity of multiplication for the set of integers(Z), set of rational numbers(Q), and set of real numbers(R).

3. An identity element of a group has ______ element.
a) associative
b) commutative
c) inverse
d) homomorphic
View Answer

Answer: c
Explanation: By the definition of all elements of a group have an inverse. For an element, a in a group G, an inverse of a is an element b such that ab=e, where e is the identity in the group. The inverse of an element is unique and usually denoted as -a.
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4. __________ matrices do not have multiplicative inverses.
a) non-singular
b) singular
c) triangular
d) inverse
View Answer

Answer: b
Explanation: The rational numbers are an extension of the integer numbers in which each non-zero number has an inverse under multiplication. A 3 × 3 matrix may or may not have an inverse under matrix multiplication. The matrices which do not have multiplicative inverses are termed as singular matrices.

5. If X is an idempotent nonsingular matrix, then X must be ___________
a) singular matrix
b) identity matrix
c) idempotent matrix
d) nonsingular matrix
View Answer

Answer: b
Explanation: Since X is idempotent, we have X2=X. As X is nonsingular, it is invertible. Thus, the inverse matrix X-1 exists. Then we have, I=X-1X = X-1X2=IX=X.
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6. If A, B, and C are invertible matrices, the expression (AB-1)-1(CA-1)-1C2 evaluates to ____________
a) BC
b) C-1BC
c) AB-1
d) C-1B
View Answer

Answer: a
Explanation: Using the properties (AB)-1=b-1A-1 and (A-1)-1=A, we may have,
(AB-1)-1(CA-1)-1C2
=(B-1)-1A-1(A-1)-1C-1C2
=BA-1AC-1C2
=BIC=BC [As, A-1A=I].

7. If the sum of elements in each row of an n×n matrix Z is zero, then the matrix is ______________
a) inverse
b) non-singular
c) additive inverse
d) singular
View Answer

Answer: d
Explanation: By the definition, an n×n matrix A is said to be singular if there exists a nonzero vector v such that Av=0. Otherwise, it is known that A is a nonsingular matrix.
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8. ___________ are the symmetry groups used in the Standard model.
a) lie groups
b) subgroups
c) cyclic groups
d) poincare groups
View Answer

Answer: a
Explanation: A symmetry group can encode symmetry features of a geometrical object. The group consists of the set of transformations that leave the object unchanged. Lie groups are such symmetry groups used in the standard model of particle physics.

9. A semigroup S under binary operation * that has an identity is called __________
a) multiplicative identity
b) monoid
c) subgroup
d) homomorphism
View Answer

Answer: b
Explanation: Let P(S) is a commutative semigroup has the identity e, since e*A=A=A*e for any element A belongs to P(S). Hence, P(S) is a monoid.
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10. An element a in a monoid is called an idempotent if ______________
a) a-1=a*a-1
b) a*a2=a
c) a2=a*a=a
d) a3=a*a
View Answer

Answer: c
Explanation: An algebraic structure with a single associative binary operation and an Identity element are termed as a monoid. It is studied in semigroup theory. An element x in a monoid is called idempotent if a2 = a*a = a.

Sanfoundry Global Education & Learning Series – Discrete Mathematics.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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