I have been looking for a while now for a solution to the following problem. I have a (row-)vector field $\boldsymbol{v}\left(\boldsymbol{x}\right)$ and a matrix $M\left(\boldsymbol{x}\right)$. I know that the curl of the following product equals zero:
$$\nabla\times\left(\boldsymbol{v}\left(\boldsymbol{x}\right)\cdot M\left(\boldsymbol{x}\right)\right)=0$$
I am looking to use this fact to find an expression for the curl $\nabla\times\boldsymbol{v}\left(\boldsymbol{x}\right)$. It is okay if this expression contains the vector $\boldsymbol{v}$.
I find the following expression for the first component of $\nabla\times\left(\boldsymbol{v}\left(\boldsymbol{x}\right)\cdot M\left(\boldsymbol{x}\right)\right)$:
$$\begin{align}\left[\nabla\times\left(\boldsymbol{v}\left(\boldsymbol{x}\right)\cdot M\left(\boldsymbol{x}\right)\right)\right]^{1} = & \frac{\partial}{\partial{x^{2}}}\left(\sum_{i}v^{i}M^{i3}\right)-\frac{\partial}{\partial{x^{3}}}\left(\sum_{i}v^{i}M^{i2}\right)\\ = & \sum_{i}\left(\frac{\partial v^{i}}{\partial x^{2}}M^{i3}-\frac{\partial v^{i}}{\partial x^{3}}M^{i2}\right)+\sum_{i}v^{i}\left(\frac{\partial M^{i3}}{\partial x^{2}}-\frac{\partial M^{i2}}{\partial x^{3}}\right) \end{align} $$
Denoting the $i$-th row of the matrix $M$ as $M^{(i)}$:
$$\begin{align}\nabla\times\left(\boldsymbol{v}\left(\boldsymbol{x}\right)\cdot M\left(\boldsymbol{x}\right)\right) = & \sum_{i}\left(\nabla v^{i}\times M^{(i)}+v^{i}\nabla \times M^{(i)}\right)=0 \end{align} $$
I tried manipulating this in many ways, but I see no way to obtain from this an expression for the curl $\nabla \times \boldsymbol{v}\left(\boldsymbol{x}\right)$. What do I miss?
Note: I will use Einstein summation in the following answer. If an index is repeated twice in an expression, for instance $$M_z^{2i}v^i$$ it should be interpreted as $$\sum_{i=1}^3 M_z^{2i}v^i$$
Suppose that $$\nabla \times (v(x)M(x))=0\tag{$\star$}$$
As you know, $$(\nabla \times v(x))^1=v_y^3-v_z^2$$
The first vector component of $(\star)$ tells us that $$M^{3i}v_y^i+M_y^{3i}v^i-M^{2i}v_z^i-M_z^{2i}v^i=0$$ By isolating $M^{33}v_y^3-M^{22}v_z^2$ and adding and subtracting $$M^{33}v_y^3-M^{22}v_z^2\color{blue}{-M^{33}v_z^2+M^{22}v_y^3}=M_z^{2i}v^i-M_y^{3i}v^i+\sum_{i\neq 2}M^{2i}v_z^i -\sum_{i\neq 3}M^{3i}v_y^i \color{blue}{-M^{33}v_z^2+M^{22}v_y^3} $$ one obtains $$(M^{33}+M^{22})(\nabla \times v(x))^1=M_z^{2i}v^i-M_y^{3i}v^i+\sum_{i\neq 2}M^{2i}v_z^i -\sum_{i\neq 3}M^{3i}v_y^i -M^{33}v_z^2+M^{22}v_y^3$$ Supposing that $M^{33}+M^{22}\neq 0$, by completing the sum and dividing you can express $$(\nabla \times v(x))^1= \frac{1}{M^{33}+M^{22}}\left (M_z^{2i}v^i-M_y^{3i}v^i+M^{2i}v_z^i -M^{3i}v_y^i \right )+\left( \frac{M^{22}}{M^{33}+M^{22}} +M^{33}\right )v_y^3 -\left(\frac{M^{33}}{M^{33}+M^{22}}+M^{22}\right )v_z^2 $$
or, indicating with $M^i$ the $i$-th row vector of $M$, $$\frac{1}{M^{22}+M^{33}}\left (\partial_z (M^2 v)-\partial_y (M^3 v))\right ) +\left( \frac{M^{22}}{M^{33}+M^{22}} +M^{33}\right )v_y^3 -\left(\frac{M^{33}}{M^{33}+M^{22}}+M^{22}\right )v_z^2 $$ Since $$\partial_z (M^2 v)-\partial_y (M^3 v)=\partial_z ((Mv)^2)-\partial_y ((Mv)^3)=-(\nabla \times Mv)^1=0$$ we obtain $$(\nabla \times v(x))^1=\frac{M^{22}v_y^3-M^{33}v_z^2}{M^{22}+M^{33}}-(M^{22}v_z^2-M^{33}v_y^3)$$ You can find analogous expressions for $$(\nabla \times v(x))^2=\frac{M^{33}v_z^1-M^{11}v_x^3}{M^{33}+M^{11}}-(M^{33}v_x^3-M^{11}v_z^1)$$ $$(\nabla \times v(x))^3=\frac{M^{11}v_x^2-M^{22}v_y^1}{M^{11}+M^{22}}-(M^{11}v_y^1-M^{22}v_x^2)$$