If a matrix is such that its inverse is equal to its transpose, does it belong to the special orthogonal group

Question

I'm trying to solve 3 problems on special orthogonal groups, and I need proof verification of the first 2 and help with the proof of the 3rd.

Consider $SO(n)$ the set of all $n \times n$ matrices with determinant equal to 1:

1. Show that If $A \in SO(n)$ then $A^{-1} \in SO(n)$

Resolution:

$A \in SO(n) \implies det(A) = 1$

$A \times A^{-1} = I \implies det(A \times A^{-1}) = det(I) = 1$

$det(A^{-1}) = 1 \implies A^{-1} \in SO(n)$

2. Show thatIf $A, B \in SO(n)$, then $AB \in SO(n)$

Resolution:

$A, B \in SO(n) \implies det(A) = det(B) = 1$

$det(AB) = det(A) \times det(B) \implies det(AB) = 1 \implies AB \in SO(n)$

2. If $A^{-1} = A^T$, then $A \in SO(3)$?

Resolution:

This is were I'm stuck...

If anybody could point me in the right direction I would be extremely grateful.

Thanks!

Answer 1

The answers to the first two questions are correct. For $3)$ remember that (for a square matrix) $\det A^T=\det A$ and use $A^TA=I \rightarrow (\det A)^2=1$. what you can say from this?