Let's say $A=$ $ \begin{pmatrix} a & c \\ b & d \\ \end{pmatrix}$, then inverse of $A$, $A^{-1}=$ $\frac{1}{ad-bc}\begin{pmatrix} d & -c \\ -b & a \\ \end{pmatrix}$
Regardless of what matrix A is, if it's square matrix, we can show the fact that $$det(A^{-1})=\frac{1}{det(A)}$$ Since, $$det(A^{-1}A)=1$$ $$det(A^{-1}A)=det(A)det(A^{-1}).$$
And If we consider the given form of A $$det(A^{-1})=\frac{1}{ad-bc}$$
However, through the defined notation of determinant, we can also derive the value of $A^{-1}$ $$det(A^{-1})=\frac{1}{ad-bc}\begin{vmatrix} d & -c \\ -b & a \\ \end{vmatrix}$$
$$=1$$ which result seems contradicts result above
What I'd done wrong? I kept check it over and over again, still I end up here.
With considering what matrix transformations could geometrically mean, I believe the result $det(A^{-1})=\frac{1}{ad-bc}$ should be the truth.
Note that$$\det\begin{bmatrix}\lambda a&\lambda b\\\lambda c&\lambda d\end{bmatrix}=\lambda^2\det\begin{bmatrix}a&b\\c&d\end{bmatrix}.$$