# Solution to Homogenous Matrix Equation Ax=0 with know parameters in x

Given is a Equation in the Form

$A_{i}\underline{r}=0$ like:

$\begin{bmatrix}\underline{0}^T & -\underline{X}_{i}^T & y_{i}\underline{X}_{i}^T\\\underline{X}_{i}^T & \underline{0}^T &-x_{i}\underline{X}_{i}^T \end{bmatrix} \begin{bmatrix} \underline{r}_{1} \\ \underline{r}_{2}\\ \underline{r}_{3}\end{bmatrix} = \begin{bmatrix}\underline{0}\end{bmatrix}$

$A$ can be extended to be a 12 x 12 matrix and $\underline{r}$ should be a 4x1 matrix

I assume that $\underline{r}^T$ is like [a, b, c, 1]. How can I solve this equation with numpy? It is recommended to solve the Equation using SVD, but I doubt that this is the right way.

• $A$ is $3 \times 12$ and $r$ is $4 \times 1$ you say. Are you sure ? Sep 5, 2018 at 13:11
• You're right. I've corrected it. But r should be 4x1 Sep 5, 2018 at 13:27
• $A$ is $12 \times 12$ and $r$ is $4 \times 1$. Are you sure ? Sep 5, 2018 at 13:47
• This is the formula I read in the paper. A is actually 2x 12 but it is mentioned, that A should be extended to A 12 by 12: See p.52 [link](appliedmaths.sun.ac.za/TW792/chapter5.pdf) Sep 5, 2018 at 13:54
• The $r$s here are the transposed rows of a $4\times3$ camera matrix, not homogeneous coordinates of points. They do not all end in $1$. In theory, the solution set is just the null space of the coefficient matrix. In practice, your system will be overdetermined and, as the text book says, computing the SVD is a fairly efficient way of estimating a solution.
– amd
Sep 5, 2018 at 22:31