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 $\det(A)=3e^t.2e^{2t}-(e^{2t}.2e^t)\\=6e^{3t}-2e^{3t}\\
=4e^{3t}$
Since $adj(A)=M^T_C$, we find:
$C_{11}=(-1)^{1+1}.M_{11}=e^{2t}\\
C_{21}=(-1)^{2+1}.M_{21}=(-1)e^{2t}=-e^{2t}\\
C_{12}=(-1)^{1+2}.M_{12}=(-1).2e^{t}=-2e^t\\
C_{22}=(-1)^{2+2}.M_{22}=1.3e^{t}=3e^t$
Hence, $M_C=\begin{bmatrix}
2e^{2t}&-2e^t \\-e^{2t} &3e^t
\end{bmatrix}\\
\rightarrow adj(A)=\begin{bmatrix}
2e^{2t}&-e^{2t} \\-2e^{t} &3e^t
\end{bmatrix}$
Consequently, $A^{-1}=\frac{1}{4e^{3t}}\begin{bmatrix}
2e^{2t}&-e^{2t} \\-2e^{t} &3e^t
\end{bmatrix}=\begin{bmatrix}
\frac{1}{e^{2t}}& -\frac{1}{4e^t} \\-\frac{1}{2e^{2t}} & \frac{3}{4e^{2t}}
\end{bmatrix}$
