Answer
See below
Work Step by Step
According to The Adjoint Method for Computing $A^{-1}$, we have the formula:
$$A^{-1}=\frac{1}{\det(A)}adj(A)$$
Finding $\det(A)=-e^{-2t}(-t.e^{3t}.e^{3t}-e^{3t}.(-te^{3t}))-e^{-2t}(e^{3t}.e^{3t}-9te^{3t}.(-te^{3t}))\\=te^{4t}-te^{4t}-e^{4t}-9t^2e^{4t}\\
=-e^{4t}(1+9t^2)$
Since $adj(A)=M^T_C$, we find:
$C_{11}=(-1)^{1+1}.M_{11}=-e^{t}\\
C_{21}=(-1)^{2+1}.M_{21}=e^{t}\\
C_{31}=(-1)^{3+1}.M_{31}=9te^{t}+e^t\\
C_{12}=(-1)^{1+2}.M_{12}=-e^t\\
C_{22}=(-1)^{2+2}.M_{22}=-e^t\\
C_{32}=(-1)^{3+2}.M_{32}=-e^t+te^t\\
C_{13}=(-1)^{1+3}.M_{13}=0\\
C_{23}=(-1)^{2+3}.M_{23}=-e^{6t}-9t^2e^{6t}\\
C_{33}=(-1)^{3+3}.M_{33}=e^{6t}+9t^2e^{6t}$
Hence, $M_C=\begin{bmatrix}
-e^{t}&-te^t & 0\\e^{t} & -te^t & -e^{6t}(9t^2+1) \\e^t(9t+1)&e^t(t-1)&e^{6t}(1+9t^2)
\end{bmatrix}\\
\rightarrow adj(A)=\begin{bmatrix}
-e^{t}&e^t & e^t(9t+1)\\-te^{t} & -te^t & e^{t}(t-1) \\ 0 &-e^{6t}(1+9t^2)&e^{6t}(1+9t^2)
\end{bmatrix}$
Consequently, $A^{-1}=-\frac{1}{e^{4t}(9t^2+1)} \begin{bmatrix}
-e^{t}&e^t & e^t(9t+1)\\-te^{t} & -te^t & e^{t}(t-1) \\ 0 &-e^{6t}(1+9t^2)&e^{6t}(1+9t^2)
\end{bmatrix}$
