hho that 𝑆𝐿𝑛(𝑅) is subgroup of 𝐺𝐿𝑛(𝑅).

To show that SLn(R) is a subgroup of GLn(R), we need to verify the three properties that define a subgroup.

1. Closure: If A and B are in SLn(R), then their product AB is also in SLn(R). This is because the determinant of the product of two matrices is the product of their determinants, and since det(A) = det(B) = 1, we have det(AB) = det(A)det(B) = 1.

2. Identity: The identity matrix I is in SLn(R) because its determinant is 1.

3. Inverses: If A is in SLn(R), then its inverse A^-1 is also in SLn(R). This is because the determinant of the inverse of a matrix is the reciprocal of the determinant of the matrix, and since det(A) = 1, we have det(A^-1) = 1/det(A) = 1.

Therefore, SLn(R) is a subgroup of GLn(R).