We already know that the set of all invertible \(2 \times 2\) matrices with entries from R is a group under multiplication — notation: GL(2,R). The given set (of all matrices of determinant 1) is a subset of the set of all invertible matrices. Thus, it is sufficient to prove that this is a subgroup of GL(2,R).

Let A, B be two matrices with determinant 1. Then \(\displaystyle{B}^{{-{{1}}}}\) exists, and using Binet-Cauchy Theorem,

\(\displaystyle{B}^{{-{{1}}}}={I}\Rightarrow{\det{{\left({B}^{{-{{1}}}}\right)}}}={\det{{\left({I}\right)}}}\Rightarrow{\det{{B}}}\ {{\det{{B}}}^{{-{{1}}}}=}{1}\Rightarrow{\det{{B}}}=^{{-{{1}}}}\)

Therefore, det \(\displaystyle{B}^{{-{{1}}}}={1}\). This means that \(\displaystyle{\det{{\left({A}{B}^{{-{{1}}}}\right)}}}={\det{{A}}}\ {{\det{{B}}}^{{-{{1}}}}=}{1}\cdot{1}={1}\)

Therefore, \(\displaystyle{A}{B}^{{-{{1}}}}!\) is a \(2 \times 2\) matrix with entries from R and determinant 1, which proves that this is a subgroup of GL(2, B), and it is itself a group.