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