Positive-pivot Givens QR factorization of an orthogonal matrix ($R=\mathrm{diag}(1,\dots ,1,\det A)$)

Use determinant-one Givens rotations and choose every nonzero eliminated pair to leave a positive pivot. If the input square matrix $A$ is orthogonal, its resulting upper triangular factor $R$ is orthogonal too, and hence diagonal with entries of absolute value one. Positivity fixes the first $n-1$ entries to one, while determinant preservation fixes the last to $\det A$. A negative last entry is unavoidable when the input reverses orientation.