GATE CS 2002 Q1.6 — Monoids and Groups

To determine which statement is true, let us analyze each option under the group axioms (closure, associativity, identity, and inverse):

1. The set of all negative rational numbers under multiplication:

   Let $S$ be the set of negative rational numbers. For any $a, b \in S$, we have $a < 0$ and $b < 0$. Their product is $a \cdot b > 0$, which is positive and therefore not in $S$. Since the set is not closed under multiplication, it cannot form a group.

2. The set of all non-singular $n \times n$ matrices under multiplication:

   Let $G$ be the set of all $n \times n$ non-singular matrices (matrices with determinant $\neq 0$).

   • Closure: If $A, B \in G$, then $\det(A) \neq 0$ and $\det(B) \neq 0$. Since $\det(AB) = \det(A)\det(B) \neq 0$, the product $AB \in G$.
   • Associativity: Matrix multiplication is associative.
   • Identity: The identity matrix $I_n$ is non-singular ($\det(I_n) = 1 \neq 0$) and serves as the identity element.
   • Inverse: For any $A \in G$, its multiplicative inverse $A^{-1}$ exists and is also non-singular ($\det(A^{-1}) = \frac{1}{\det(A)} \neq 0$), so $A^{-1} \in G$.

   Thus, this set forms a group (known as the General Linear Group, $GL_n$). This statement is true.

3. The set of all $n \times n$ matrices under multiplication:

   This set includes singular matrices (such as the zero matrix). Since singular matrices do not have multiplicative inverses, the inverse axiom fails. Thus, it does not form a group.

Therefore, only option B is correct.