Why inverse of an allignment matrix is the same as its transpose?
To align vectors, you are applying a rotation.

If you consider a simple rotation matrix, e.g. rotation of angle $\theta$ around Z axis, then you will trivially see that its inverse matrix, a rotation of $-\theta$, is the transpose,  since $sin(-\theta)=-sin(\theta)$ and $cos(-\theta)=cos(\theta)$.

An arbitrary rotation matrix, R, can be constructed by multiplication of other rotations, e.g $R = A \cdot B$.  If the inverse of A and B exist, then $R^{-1} = B^{-1} \cdot A^{-1}$.
Similarly we know  $(A \cdot B)^T = (B^T \cdot A^T)$.

Put these together and you have your answer.