Question

Prove or disprove : The set M of 2x2 singular marties under matrix multiplication forms a group​

Answer 1

Step-by-step explanation:

It is clear that the set of all invertible real 2×2 matrices form a group under matrix multiplication (if it’s not clear, you can convince yourself by proving it follows the group axioms I’m about to present). This is the group GL(2,R) . Let us check that the group axioms still hold if we are only considering a subset of the set of all real 2×2 matrices which have determinant 1; call this subset S

Closure: Suppose A,B∈S and A∘B=C , then

det(C)=det(AB)=det(A)det(B)=1

since A,B∈S ; thus C∈S .

Associativity: Suppose A,B,C∈S . Matrix multiplication is associative, so one has that (A∘B)∘C=A∘(B∘C) and then

det(AB)det(C)=det(A)det(B)det(C)=det(A)det(BC)=1

since A,B,C∈S and closure holds.

Existence of identity: The identity id will be the 2×2 unit matrix and for all A∈S one has that id∘A=A∘id=A . It follows that

det(A)=det(A∘id)=det(A)det(id)⇒det(id)=1

since A∈S .

Existence of inverse: Suppose A∈S and there exists an A−1 such that A∘A−1=A−1∘A=id . Then one has that

det(A∘A−1)=det(A)det(A−1)=det(A−1)=det(id)=1

and thus A−1∈S , since A∈S and the there exists an identity element in S .

Thus the set S of real 2×2 matrices with determinant 1 form a group under matrix multiplication, which is a subgroup of GL(2,R) , and is denoted by SL(2,R) .