Homogeneous transformation matrix

Inverse

Given

$$H=H_{\text{trans}}\cdot H_{\text{rot}}=\left[\begin{array}{cc}R& \vec{d}\\ 0& 1\end{array}\right]$$

what is the inverse, $H^{-1}$? Intuitively, the inverse of the displacement vector is simply translating the object back to its original position. Therefore, the displacement becomes $-\vec{d}$.

$$H_{\text{trans}}^{-1}=\left[\begin{array}{cc}I& -\vec{d}\\ 0& 1\end{array}\right]$$

We know Inverse is just its transpose, so $H_{\text{rot}}^{-1}=H_{\text{rot}}^{\top }$. Multiplying them back together, we get

$$H^{-1}=H_{\text{rot}}^{-1}\cdot H_{\text{trans}}^{-1}=\left[\begin{array}{cc}{R}^{\top }& -R^{\top }\cdot \vec{d}\\ 0& 1\end{array}\right]$$