Inverse of $3$ by $3$ matrix with non-constant entries. I'm solving a question in nonhomogenous ordinary differential equation system $x'=Px+q$, and to solve my question I need to compute the inverse of the matrix
$A=\begin{pmatrix}e^{-2t} & e^{-t} & 0 \\ -\frac{5}{4}e^{-2t} & -\frac{4}{3}e^{-t} & e^{2t} \\ -\frac{7}{4}e^{-2t} & -\frac{2}{3}e^{-t} & -e^{2t} \end{pmatrix}$
The reason I need to find the inverse is because the columns of this matrix are the independent solutions to the corresponding homogenous system $x'=Px$ and I'm using the variation of parameters method suggested here http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousSystems.aspx
How would I find the inverse of this matrix? There is no simple formula like in $2$ by $2$ matrices, and since the entries are not constant, I'm unsure how to do this easily.

We can factor out the variable dependence in this particular case:
$$ A=\begin{pmatrix}e^{-2t} & e^{-t} & 0 \\ -\frac{5}{4}e^{-2t} & -\frac{4}{3}e^{-t} & e^{2t} \\ -\frac{7}{4}e^{-2t} & -\frac{2}{3}e^{-t} & -e^{2t} \end{pmatrix} $$
$$ = \begin{pmatrix} 1 & 1 & 0 \\ -\frac{5}{4} & -\frac{4}{3} & 1 \\ -\frac{7}{4} & -\frac{2}{3} & -1 \end{pmatrix} \begin{pmatrix} e^{-2t} & 0 & 0 \\ 0 & e^{-t} & 0 \\ 0 & 0 & e^{2t} \end{pmatrix} = C D $$
The inverse of the diagonal matrix $D$ is found by inspection, namely:
$$ D^{-1} = \begin{pmatrix}  e^{2t} & 0 & 0 \\ 0 & e^{t} & 0 \\ 0 & 0 & e^{-2t} \end{pmatrix} $$
Now $A^{-1} = D^{-1} C^{-1}$, and $C^{-1}$ is simply the inverse of a $3\times 3$ matrix with rational (constant) entries.
