Essentially we are asked to prove these results about ODEs...
(i) If $F : I \rightarrow GL(n, R)$ satisfies the matrix ODE $F' = F A$ , then $ Det(F)$ satisfies the scalar ODE $(Det(F))' = Tr(A) Det(F)$.
(ii) If $F : I \rightarrow GL(n, R)$ satisfies the matrix ODE $F' = F A$ and for some $t_0 \in I$ we have $F(t_0) \in GL(n, R)$, where $GL(n, R)$ denotes the group of invertible matrices, then $F : I \rightarrow GL(n, R)$.
(iii) If $F : I \rightarrow GL(n, R)$ satisfies the matrix ODE $F' = F A$ and $Tr(A) = 0$ then $Det (F(t))$ is constant in $t$. In particular, if $Det(F(t_0)) = 1$ then $Det (F(t)) = 1$ for all $t \in I$.
(iv) If $F : I \rightarrow GL(n, R)$ satisfies the matrix ODE $F' = F A$ and $A: I \rightarrow SO(n, R)$ takes values in skew-symmetric matrices, then $F^T (t)F(t)$ is constant in $t$. In particular, if $F(t_0) \in O(n, R)$ then $F : I → O(n, R)$. The analogous statement holds for $F(t_0) \in SO(n, R$).
So I'm stuck on part i) I defined the matrices $F$ and $A$ as $2 \times 2$ matrices with lower case $f$ and $a$ with $f_{11}$ in top left spot for $F$ and $a_{22}$ in bottom right spot (the normal convention) for $A$. I calculated $Det(FA)$ and after some cancellations I found that to be ${f_{11}*a_{11}*f_{22}*a_{22} + f_{12}*a_{21}*f_{21}*a_{12}} - {f_{11}*a_{12}*f_{22}*a_{21} + f_{12}*a_{22}*f_{21}*a_{11}}$
Then I tried to calculate $Det(F)Tr(A)$ and found...
$|F|*Tr(A) = ({f_{11}*f_{22} - f_{12}*f_{21}})*(α_{11} + α_{22})$ but here I can already see that I'll get terms with only 3 things being multiplied (rather than four per term in the $Det(FA)$) I'm not sure if maybe I did something wrong with my computations (I doubled-checked and feel confident) or perhaps I'm missing some cancellation somewhere but I just don't see how to get these two things to equal one another. Any suggestions?
I will address i)
First, auxiliary lemma:
If $X\in C^1(I;\mathbb R^{n\times n})$ then
$$ \frac{\frac{d}{dt}\det X(t)}{\det X(t)}={\rm tr}\left(X'(t)X(t)^{-1}\right).\tag{$\ast$} $$
to prove it consider $$ X(t_0+h)=X(t_0)+hX'(t_0)+o(h). $$ Take the determinant $$ \det X(t_0+h)=\det X(t_0)\det(I+hX'(t_0)X^{-1}(t_0)+o(h)). $$ Next, use the fact that $$ \det(I+h B+o(h))=1+h\,{\rm tr\,}B+o(h)\quad\mbox{(exercise!)} $$ conclude that $$ \det X(t_0+h)=\det X(t_0)(1+h{\,\rm tr}\left(X'(t_0)X^{-1}(t_0)\right)+o(h)), $$ rearrange, divide by $h$ and take the limit $h\to 0$ to get the required formula.
Now from $F'=AF$ and $(\ast)$ you immediately get $$ \frac{d}{dt}\det F(t)={\rm tr\,}A\det F(t). $$