Determinant equals 1 whereas Inverse unequal to the transpose?

Question

So briefly: Given the matrix $A$ with det$(A) = 1$:

$\left(\begin{array}{ccc} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{array}\right)$

Its inverse however is set by

$\left(\begin{array}{ccc} 1 & -a & a\,c-b\\ 0 & 1 & -c\\ 0 & 0 & 1 \end{array}\right)$

If I'm not mistaken, that's not $A^T$. Also I hold the opinion that this kind of matrices with the property of $\det= \pm 1$ belong to the orthogonal group, which is why I came for the conclusion.

Answer 1

You are right about $A^{-1}$. Since it is distinct from $A^T$, the conclusion is that $A$ is not an orthogonal matrix. Although the determinant of every orthogonal matrix is indeed $\pm1$, most matrices whose determinant is $\pm1$ are not orthogonal.