I have the following equation:
$\Delta \dot{\boldsymbol{x}}_2 = D_2(\boldsymbol{x}_0)\begin{bmatrix} \Phi^T_{G_{21}}(\boldsymbol{x}_0)\Delta \delta_e & \Phi^T_{G_{22}}(\boldsymbol{x}_0) \Delta \delta_a & \Phi^T_{G_{23}}(\boldsymbol{x}_0) \Delta \delta_r \end{bmatrix}\begin{bmatrix} \Theta_{G_{21}} \\ \Theta_{G_{22}}\\ \Theta_{G_{23}} \end{bmatrix}$
where $\Delta \dot{\boldsymbol{x}}_2$ is size $3\times 1$, where $D_2$ is $3\times 3$, where $\Phi^T_{G_{2\star}}$ is $3\times 6$, where $\Delta \delta_\star$ are scalars and where $\Theta_{G_{2\star}}$ is $6\times 1$.
This equation is valid at every time instance, so I want to write the following set of equations:
$\begin{bmatrix} \Delta \dot{\boldsymbol{x}}_{2,1}\\ \Delta \dot{\boldsymbol{x}}_{2,2}\\ \vdots\\ \Delta \dot{\boldsymbol{x}}_{2,N} \end{bmatrix} = \begin{bmatrix} D_2(\boldsymbol{x}_0)\\ D_2(\boldsymbol{x}_1)\\ \vdots\\ D_2(\boldsymbol{x}_N) \end{bmatrix}\begin{bmatrix} \Phi^T_{G_{21}}(\boldsymbol{x}_0)\Delta \delta_{e,1} & \Phi^T_{G_{22}}(\boldsymbol{x}_0) \Delta \delta_{a,1} & \Phi^T_{G_{23}}(\boldsymbol{x}_0) \Delta \delta_{r,1} \\ \Phi^T_{G_{21}}(\boldsymbol{x}_1)\Delta \delta_{e,2} & \Phi^T_{G_{22}}(\boldsymbol{x}_1) \Delta \delta_{a,2} & \Phi^T_{G_{23}}(\boldsymbol{x}_1) \Delta \delta_{r,2} \\ \vdots & \vdots & \vdots \\ \Phi^T_{G_{21}}(\boldsymbol{x}_{N-1})\Delta \delta_{e,N} & \Phi^T_{G_{22}}(\boldsymbol{x}_{N-1}) \Delta \delta_{a,N} & \Phi^T_{G_{23}}(\boldsymbol{x}_{N-1}) \Delta \delta_{r,N} \end{bmatrix}\begin{bmatrix} \Theta_{G_{21}} \\ \Theta_{G_{22}}\\ \Theta_{G_{23}} \end{bmatrix} \,.$
However, I doubt whether this notation is correct (especially the matrix containing $D_2$).
The size of the left side of your equation is $3N \times 1$, so we need the right hand side to be the same.
Right hand side: The first term is $3N\times 3$, the second term is $3N \times 18$ and the third term is $18 \times 1$, which isn't correct.
First guess would be to make the first matrix a diagonal matrix with $D_2(x_i)$ on the diagonal and the rest of the entries $3\times 3$ zero matrices. Haven't worked this out, but seems correct. That would at least give you the correct dimensions: diagonal matrix is $3N\times 3N$.