Question

GL (n, R), the set of invertible nxn matrices with real entries, is a group under matrix multiplication. Determine if the nxn matrices with determinant -1 or 1 is a subgroup. If not, which property does not hold?

Answer 1

Answer:

Let $S=\{M\in GL(n,r): det(M)=\pm1\}$

The subset S is a subgroup of GL(n,R) if satisfies:

1. The identity matrix $I$ belong to S.

2. If A and B are in S then AB is in S.

3. If A is belong to S then $A^{-1}$ belongs to S.

Let's see if S satisfies these conditions.

1.  We know that $det(I)=1$, then $I\in S$.

2. Let A and B in S. $det(AB)=det(A)det(B)=(\pm1)( \pm 1)=\pm 1$

Then AB is in S.

3. Let $A\in S$, $det(A^{-1})=\frac{1}{det(A)}=\frac{1}{\pm 1}=\pm 1)$, then $A^{-1}\in S$.

Since S satisfies all conditions then S is a subgroup of GL(n,R).