Solving ODE $F(t)=A(t)F'(t) $ How to solve 
  $F(t)=A(t)F'(t) ,F(0)= I\tag 1$


*

*All are $3 \times 3$ matrices except variable t

*A(t) is given and has determinant $0$.  $A(t)=(I-tC_1)^{-1}t^3C_2 \tag 2$ 

*I is a constant unit  rotation matrix means  I is unity matrix

*$C_1,C_2$ are constant skew symmetric matrices of  $0$ determinent

*$$C_1=\left( \begin{array}{ccc} 
   0 & -c_0 & b_0 \\
   c_0 & 0 & -a_0 \\
  -b_0 & a_0 & 0 \\
   \end{array} \right).$$
$$C_2=\left( \begin{array}{ccc} 
   0 & -(c_1-c_0) & (b_1-b_0) \\
   (c_1-c_0) & 0 & -(a_1-a_0) \\
   -(b_1-b_0) & (a_1-a_0) & 0 \\
   \end{array} \right).$$ 
NB: All entries of the matrices $C_1$, $C_2$ are constants,can't be altered  

By the product law of determinants we have 
$$\det F(t) = \det A(t) \cdot \det F'(t) = 0$$
if $\det F'(t)$ is finite, but this contradicts $\det F(0) = 1$. This means that $F'(t)$ (if a solution exist at all) is unbounded close to $t=0$ and $\det F(t)$ (and therefore $F(t)$) must be discontinuous at $t=0$.
Lets study the system close to $t=0$. We find
$$\left(\matrix{F_{00}&F_{10}&F_{20}\\F_{01}&F_{11}&F_{21}\\F_{02}&F_{12}&F_{22}}\right) =t^3\left(S+\mathcal{O}(t)Q\right)\left(\matrix{\dot{F_{00}}&\dot{F_{10}}&\dot{F_{20}}\\\dot{F_{01}}&\dot{F_{11}}&\dot{F_{21}}\\\dot{F_{02}}&\dot{F_{12}}&\dot{F_{22}}}\right) $$
where $S$ does not depend on $t$. The $00$-component to lowest order in $t$ gives
$$(c_0 - c_1) \dot{F_{10}} + (-b_0 + b_1) \dot{F_{20}} = \frac{F_{00}}{t^3} \simeq \frac{1}{t^3}$$
Thus one or more of the $F_{ij}$ terms must have a singularity at $t=0$.
