Why inverse of an allignment matrix is the same as its transpose? $A_{v}^{-1}$ = $A_{v}^{T}$
$A_{v}$ = Allignment matrix that alligns vector v with z axis
$=$\begin{bmatrix}
    \frac{\lambda}{|v|}       &amp; \frac{-ab}{\lambda|v|} &amp; \frac{-ac}{\lambda|v|} &amp;   0 \\
    0 &amp; \frac{c}{|v|} &amp; \frac{-b}{|v|} &amp;   0 \\
    \frac{a}{|v|}       &amp; \frac{b}{|v|} &amp; \frac{c}{|v|} &amp;   0 \\
0 &amp; 0 &amp; 0 &amp; 1
\end{bmatrix}
here, $\lambda = \sqrt{b^2+c^2} $
$|v| = \sqrt{a^2+b^2+c^2} $
And vector $v = (a, b, c)$
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.