Is the transformation matrix in Goldstein's problem an improper orthogonal one?

[Irid]

In Goldstein there is a problem asking to find a vector representation for a reflection in a plane of a unit normal $$\mathbf{\hat{n}}$$. I find it to be

$$\mathbf{r'} = \mathbf{r} - 2(\mathbf{r\cdot \hat{n}})\mathbf{\hat{n}}$$

and it has a corresponding transformation matrix with elements

$$A_{ij} = \delta_{ij} - 2l_i l_j$$

where $$l_i\, , i=1,2,3$$ are the direction cosines for the orientation of the plane. Goldstein then asks to show that this matrix is an improper orthogonal one. I can show orthogonality by simply noting that $$A^T = A$$, and then I multiply $$A^2 = I$$, which shows that $$A^T = A^{-1}$$, which is the condition for orthogonality.

However, the improper nature of the matrix is unclear to me. If I compute the determinant, by explicitly writing out the form of the matrix, the result I obtain is +1, instead of -1:

$$\text{det}(A) = \begin{vmatrix}<br /> 1-l_1^2 & -l_1 l_2 & -l_1 l_3\\<br /> -2l_1 l_2 & 1-2l_2^2 & -2l_2 l_3\\<br /> -2l_1 l_3 & -2l_2 l_3 & 1-2l_3^2\end{vmatrix} = 1$$

Does it mean that the matrix is a proper one? Or is there an error in the problem statement, or am I missing something?

 

[Avodyne]

If I set l1=l2=0, and l3=1, your matrix is diagonal with entries 1, 1, -1, which obviously has det = -1. So I think you must have made a mistake taking the determinant.

 

[Irid]

Oh yeah, that's right. Conclusion: don't drink behind the table.