# 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|} & \frac{-ab}{\lambda|v|} & \frac{-ac}{\lambda|v|} & 0 \\ 0 & \frac{c}{|v|} & \frac{-b}{|v|} & 0 \\ \frac{a}{|v|} & \frac{b}{|v|} & \frac{c}{|v|} & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}
here, $\lambda = \sqrt{b^2+c^2}$
$|v| = \sqrt{a^2+b^2+c^2}$
And vector $v = (a, b, c)$

• Are you asking "Why is the inverse of a rotation matrix the same as its transpose"? – Simon F Jan 5 '18 at 17:08
• That is a rotation or reflection matrix. You can check that the four columns are orthogonal to each other (the dot product of the columns is zero). Also they are of unit length, so the determinant is $\pm$1. A well known property of orthonormal square matrices is that transpose is equal to the inverse. – Mauricio Cele Lopez Belon Jan 5 '18 at 21:26
• @MauricioCeleLopezBelon your explanation was truly helpful. Thanks. – Bhattacharjee Jan 6 '18 at 6:51

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 $(A \cdot B)^T = (B^T \cdot A^T)$.