GATE CS 1995 Q2.17 — Monoids and Groups

To determine the algebraic structure of $\langle A, *\rangle$, where $A$ is the set of all non-singular matrices over real numbers and $*$ is matrix multiplication, we verify the group axioms:

1. Closure: Let $M, N \in A$. Since they are non-singular, their determinants are non-zero: $\det(M) \neq 0$ and $\det(N) \neq 0$.
   The determinant of their product is: $$\det(M * N) = \det(M) \cdot \det(N) \neq 0$$ Thus, $M * N \in A$, so $A$ is closed under $*$.
2. Associativity: Matrix multiplication is inherently associative. For any $M, N, P \in A$: $$(M * N) * P = M * (N * P)$$ Thus, $\langle A, *\rangle$ is a semigroup.
3. Identity: The identity matrix $I$ has $\det(I) = 1 \neq 0$, so $I \in A$. For any $M \in A$: $$M * I = I * M = M$$ Thus, $\langle A, *\rangle$ is a monoid.
4. Inverse: Since every matrix $M \in A$ is non-singular, its inverse $M^{-1}$ exists.
   Since $\det(M^{-1}) = \frac{1}{\det(M)} \neq 0$, we have $M^{-1} \in A$. Also: $$M * M^{-1} = M^{-1} * M = I$$ Thus, $\langle A, *\rangle$ is a group (specifically known as the General Linear Group, $GL_n(\mathbb{R})$).
5. Commutativity (Abelian property): Matrix multiplication is generally non-commutative. For example, let: $$M = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad N = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$$ Both are non-singular (determinant is $1$). Computing their products: $$M * N = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \quad \text{and} \quad N * M = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$$ Since $M * N \neq N * M$, the group is not abelian.

Therefore, $\langle A, *\rangle$ is a group but not an abelian group.