solving matrix equation by matrix rearranging Given ${3}\times{4}$ matrix $K$, ${4}\times{1}$ column vector $m$  and ${3}\times{1}$ column vector $p$, I have the following systems of equations in ${4}\times{4}$ homogeneous transformation matrix $T$
$$K T m = p$$
Apparently, matrix $T$ has $12$ unknown elements but an actual transformation needs only $6$ unknowns. Given several equations ($\geq 3$), can this equation be solved?
If so, I wonder which course or material provides with skills to rearrange and solve matrix equations like this one. Thank you!
 A: Since $T$ is a homogeneous transformation, it can be written as
$$
T=\begin{bmatrix}\tilde{T}\\e_4^T\end{bmatrix},
$$
where $e_4=[0,0,0,1]^T$ and $\tilde{T}$ is $3\times 4$. Hence $p=KTm$ can be written as
$$
p=[\tilde{K},k]\begin{bmatrix}\tilde{T}m\\e_4^Tm\end{bmatrix}=\tilde{K}\tilde{T}m+ke_4^Tm,
$$
where $K=[\tilde{K},k]$ with $\tilde{K}$ being $3\times 3$ and $k$ the trailing column of $K$ ($3\times 1$). Hence we look for $\tilde{T}$ such that
$$
  \tilde{K} \tilde{T} m = p-ke_4^Tm
$$
(note that $e_4^Tm$ is simply the last component of the vector $m$).
Next, we use the Kronecker product and the matrix vectorisation operator to transform (1) to system of linear equations of the form "matrix $\times$ vector = vector". Using this, we get that
$$
\mathrm{vec}(\tilde{K}\tilde{T}m)=(m^T\otimes\tilde{K})\mathrm{vec}(\tilde{T})
$$
and hence (the entries of) the matrix $\tilde{T}$ can be found by solving the system
$$
(m^T\otimes\tilde{K})\mathrm{vec}(\tilde{T})=p-ke_4^Tm,
$$
where $m^T\otimes\tilde{K}$ is a $3\times 12$ matrix.
