GATE CS 2005 Q46 — Monoids and Groups

To determine the algebraic structure of the set $H$ under matrix multiplication, we verify the group axioms:

1. Closure: Let $A, B \in H$ be two upper triangular matrices: $$A = \begin{pmatrix} a_1 & f_1 & e_1 \\ 0 & b_1 & d_1 \\ 0 & 0 & c_1 \end{pmatrix}, \quad B = \begin{pmatrix} a_2 & f_2 & e_2 \\ 0 & b_2 & d_2 \\ 0 & 0 & c_2 \end{pmatrix}$$ where $a_1 b_1 c_1 \neq 0$ and $a_2 b_2 c_2 \neq 0$.
   
   Multiplying $A$ and $B$ gives: $$AB = \begin{pmatrix} a_1 a_2 & a_1 f_2 + f_1 b_2 & a_1 e_2 + f_1 d_2 + e_1 c_2 \\ 0 & b_1 b_2 & b_1 d_2 + d_1 c_2 \\ 0 & 0 & c_1 c_2 \end{pmatrix}$$ The product $AB$ is also an upper triangular matrix. The product of its diagonal elements is: $$(a_1 a_2)(b_1 b_2)(c_1 c_2) = (a_1 b_1 c_1)(a_2 b_2 c_2) \neq 0$$ Thus, $AB \in H$, and closure holds.
2. Associativity: Matrix multiplication is associative in general, so it is associative for elements in $H$.
3. Identity Element: The $3 \times 3$ identity matrix is: $$I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ Here, $a=1, b=1, c=1$ (so $abc = 1 \neq 0$) and $d=0, e=0, f=0$. Thus, $I \in H$ and acts as the identity element.
4. Inverse: For any matrix $A \in H$, the determinant is $\det(A) = abc$. Since $abc \neq 0$, the matrix is invertible.
   The inverse of an upper triangular matrix is also upper triangular, with diagonal elements $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$. Since $a, b, c$ are non-zero, their reciprocals exist, and their product $\frac{1}{abc} \neq 0$. Thus, $A^{-1} \in H$.

Since all four group axioms are satisfied, the set $H$ is a group under matrix multiplication.