Can an orthogonal matrix have a determinent that is not 1 nor -1?

Question

As above. Proofs, examples and counter-examples are appreciated.

Answer 1

An orthogonal matrix $O$ satisfies

$O^TO = I; \tag 1$

thus

$\det O^TO = \det I = 1; \tag 2$

but

$\det O^TO = \det O^T \det O, \tag 3$

and

$\det O^T = \det O, \tag 4$

so (2)-(4) yield

$(\det O)^2 = 1, \tag 5$

which forces

$\det O = \pm 1. \tag 6$

Answer 2

Just to be safe: You have to be careful, whether you mean orthogonal or orthonormal matrices.

The former satisfy that all the columns are orthogonal. With resources to the standard scalar product this is the case for $$\begin{bmatrix} 7&0\\0&1\end{bmatrix},$$ which has determinant $7$.

Orthogonal matrices have pairwise orthogonal and normalized (length 1) columns. Those matrices can only have determinant $\pm 1$, since their inverse is their transpose, which has the same determinant and they have to multiply to 1.