I have the following nonlinear and non-autonomous system to solve ($a,b,c$ are specific constants): $$\begin{align} x^\prime &= -a\,\omega\,y+\frac{b}{\cosh t}\,\left(z-\frac{1}{z}\right),\\ y^\prime &= a\,x-c\,\left(z-\frac{1}{z}\right),\\ z^\prime &= -\frac{b}{\cosh t}\,x+c\,\omega\,y, \end{align}$$ which can be written in matrix form as $$\begin{equation} \begin{pmatrix}x^\prime \\ y^\prime \\ z^\prime\end{pmatrix} = \begin{pmatrix} 0 & -a\,\omega & \frac{b}{\cosh t}\\ a & 0 & -c\\ -\frac{b}{\cosh t} & c\,\omega & 0 \end{pmatrix}\,\begin{pmatrix} x \\ y \\ z - \dfrac{1}{z}\end{pmatrix}, \end{equation}$$ or, more succinctly as $$\begin{equation} X^\prime = M(t)\,V, \end{equation}$$ where, due to the nonlinear terms involving $z$, $\,V^\top=\begin{pmatrix}x & y & z - \frac{1}{z}\end{pmatrix} \neq X^\top = \begin{pmatrix}x & y & z\end{pmatrix}\,$. From the literature I have read, it seems that a fixed (i.e, equilibrium, or stationary) point of a non-autonomous system should not exist.
However, it is clear that for any time $t$ there are two choices of $(x, y, z)$ for which $V = 0$, namely, $(0,0,1)$ and $(0,0,-1)$, in which case $X^\prime = 0$. On the other hand, the matrix $M(t)$ is singular for any value of $t$, so there is no unique solution for $V$ if $X^\prime = 0$. I should add that the time-dependent Jacobian matrix $$J(t) = \begin{pmatrix}0 & -a\,\omega & \frac{b}{\cosh t}(1 + 1/z^2)\\a & 0 & -c(1 + 1/z^2)\\-\frac{b}{\cosh t} & c\,\omega & 0\end{pmatrix}$$ is also singular for all $(x,y,z)$.
So I have this situation for my system: if $V = 0$ then $X^\prime = 0$ for $(0,0,1)$ and $(0,0,-1)$, but if $X^\prime = 0$ then I cannot find a unique point $(x,y,z)$ for which $V = 0$.
My questions are: (i) are the points $(0,0,1)$ and $(0,0,-1)$ considered to be fixed points? (ii) regardless, is it still appropriate to linearize the system around one of these points, i.e., would the linearized system obtained still provide meaningful information about the solutions and/or stability of the nonlinear system?