As I googled, I think I realized what you need. Every row is a 2D point. So "rotation" and "translation" are performed to EVERY row. So given two sets of points written in the matrices $A$ and $B$ you want to find a rotation matrix $R = \left(\begin{matrix} \cos{T} & -\sin{T} \\ \sin{T} & \cos{T} \end{matrix}\right)$ and a translation vector $tr = (t_x, t_y)^t$ such that for every row $B_i$ of $B$ and every row $A_i$ of $A$ you have $B_i^t = R.A_i^t + tr$.
Now, if $R$ and $tr$ exist and if you have two distinct rows from $B$ and $A$ you will have that
from where after subtraction you will obtain $B_2^t-B_1^t = R.(A_2^t-A_1^t)$ which is a $2\times 2$ linear system for the "variables" $\cos{T}$ and $\sin{T}$. If your points are distinct, you will obtain unique solution for $R$ (and you know that an angle $T$ is determined uniquely by its $\sin$ and $\cos$). Once you have $R$ determined you obtain $tr$ from the first equation: $tr = B_1^t-R.A_1^t$.
The SVD decomposes a matrix as a weighted sum of matrices which are themselves outer product of two vectors. Hence you trade $mn$ coefficients for $k(m+n)$, where $k$ is the number of weights retained.
Best Answer
We can transform back and forth without any matrix inversion, since for any unitary matrix A, $A^{-1}=A^*$ where * is the conjugate transpose.
Unitary matrices represent an orthogonal basis, which is useful in image processing.