Let \(\{B\}\) be a [Coordinate Frame] in frame \(\{A\}\). The following kinds of transformations are possible between \(\{B\}\) and \(\{A\}\).
Translation
If \(\{B\}\) has the same [Orientation] as \(\{A\}\) but the origin of \(\{B\}\) is at a distance of \(^AP_{Borg}\) then if \(^BP\) is a vector in \(\{B\}\), it can be represented in terms of \(\{A\}\) as \(^AP=^BP + ^AP_{Borg}\).
Rotation
If \(\{B\}\) has a different orientation than \(\{A\}\) but their origin are the same, then we can represent the points in \(\{B\}\) in terms of \(\{A\}\) by multiplying them with the [Rotation Matrix] that describes the orientation of \(\{B\}\) with respect to \(\{A\}\).
Combined transform
If \(\{B\}\) has it’s origin at \(^AP_{Borg}\) and it has a orientation with respect to \(\{A\}\) given by \(^A_BR\), then points in \(\{B\}\) can be represented in terms of \(\{A\}\) by using the homogeneous transformation matrix as
\[\begin{bmatrix}^AP \\ 1 \end{bmatrix} = \begin{bmatrix} ^A_BR && ^AP_{Borg} \\ 0 && 1 \end{bmatrix} \begin{bmatrix}^BP \\ 1 \end{bmatrix}\]Since both the \(^A_BR\) and \(^AP_{Borg}\) are invertible, therefore we can say that the homogeneous transformation matrix is also invertible.
If we have a point \(^AP\) in \(\{A\}\) and we want to represent it in terms of \(\{B\}\) as \(^BP\), then we need to multiply \(^AP\) with \(\begin{bmatrix} ^A_BR && ^AP_{Borg} \\ 0 && 1 \end{bmatrix}^{-1}\).
\[\begin{align*} \begin{bmatrix} ^A_BR && ^AP_{Borg} \\ 0 && 1 \end{bmatrix}^{-1} &= \begin{bmatrix} \begin{bmatrix} I && ^AP_{Borg} \\ 0 && 1 \end{bmatrix} \begin{bmatrix} ^A_BR && 0 \\ 0 && 1 \end{bmatrix} \end{bmatrix}^{-1} \\ &= \begin{bmatrix} ^A_BR && 0 \\ 0 && 1 \end{bmatrix}^{-1} \begin{bmatrix} I && ^AP_{Borg} \\ 0 && 1 \end{bmatrix}^{-1} \\ &= \begin{bmatrix} ^A_BR^T && 0 \\ 0 && 1 \end{bmatrix} \begin{bmatrix} I && -^AP_{Borg} \\ 0 && 1 \end{bmatrix} \\ &= \begin{bmatrix} ^B_AR && 0 \\ 0 && 1 \end{bmatrix} \begin{bmatrix} I && -^AP_{Borg} \\ 0 && 1 \end{bmatrix} \\ &= \begin{bmatrix} ^B_AR && -^B_AR ^AP_{Borg} \\ 0 && 1 \end{bmatrix} \end{align*}\]We know that \(^A_BR^{-1}\) \(=\) \(^A_BR^T\) and \(^AP_{Borg}^{-1} = -^AP_{Borg}\)