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 $\det(A)=e^{-2t}(e^{t}.te^{t}-2te^{t}.e^{t})-e^{-2t}(e^{t}.te^{t}-te^{t}.e^{t})+2e^{-t}(e^t.2te^t-te^t.e^t)\\=t-2t-t+t+4t-2t\\
=t$
Since $adj(A)=M^T_C$, we find:
$C _{11}=(-1)^{1+1}.M_{11}=3te^{-t}\\
C_{21}=(-1)^{2+1}.M_{21}=-te^{t}\\
C_{31}=(-1)^{3+1}.M_{31}=-te^{-t}\\
C_{12}=(-1)^{1+2}.M_{12}=-e^t(1-2t)\\
C_{22}=(-1)^{2+2}.M_{22}=-e^{-t}(-2t+1)\\
C_{32}=(-1)^{3+2}.M_{32}=0\\
C_{13}=(-1)^{1+3}.M_{13}=0\\
C_{23}=(-1)^{2+3}.M_{23}=te^{2t}\\
C_{33}=(-1)^{3+3}.M_{33}=te^{2t}$
Hence, $M_C=\begin{bmatrix}
3te^{-t}&e^{-t}(1-2t) & 0\\-te^{-t} & -e^t(1-2t) & te^{2t} \\-te^{-t}& 0 &te^{2t}
\end{bmatrix}\\
\rightarrow adj(A)=\begin{bmatrix}
3te^{-t}&-te^{-t} & -te^{-t}\\-e^t(1-2t) & -e^{-t}(1-2t)& 0 \\0 & te^{t} &te^{2t}
\end{bmatrix}$
Consequently, $A^{-1}=\frac{1}{t} \begin{bmatrix}
3te^{-t}&-te^{-t} & -te^{-t}\\-e^t(1-2t) & -e^{-t}(1-2t)& 0 \\0 & te^{t} &te^{2t}
\end{bmatrix}$
