Pardon the simple question, but my memory's failing me on this, and I can't seem to find any answers.
I have a system of differential equations, with a coefficient matrix that looks like this, where a, b, and c are constants:
\begin{bmatrix} 0& 1& 0& 0& 0& 0\\ 0& 0&-a& 0& b& 0\\ 0& a& 0& 0& 0& c\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0&-1& 0&-a\\ 0& 0& 0& 0& a& 0 \end{bmatrix}
Now, it's pretty obvious to see that the coefficient matrix has a zero determinant (expand over the first column). Does this mean that the system is unsolvable? If so, what can I do to make it solvable?
The ODE system $\dot x = Ax$ has always the solution $$ x(t)=\exp(tA)x_0=x_0+\sum_{k=1}^\infty \frac{t^k}{k!}A^kx_0 $$ and the series always converges as it is dominated by $\sum_{k=0}^\infty\frac{|t|^k}{k!}\|A\|^k\|x_0\|=e^{|t|\,\|A\|}\|x_0\|$. There is no admissibility restriction on the matrix $A$.