Determinant of $2 \times 2$ matrix such that $A = A^{-1}$ 
Let $A$ be a $2 \times 2$ matrix such that $A = A^{-1}$. The value of $\operatorname{det} (A)$ can be:
  
  
*
  
*$\operatorname{det} (A)=-2$
  
*$\operatorname{det} (A)=-1$
  
*$\operatorname{det} (A)=0$
  
*$\operatorname{det} (A)=2$
  

My attempt: 
$$\begin{bmatrix}
a &b \\ 
 c& d
\end{bmatrix} = \frac{1}{ad-bc} \times \begin{bmatrix}
d &-b \\ 
 -c& a
\end{bmatrix}$$
Obviously the determinant cannot be $0$, since an inverse exists, but I'm not sure how to proceed. Any help would be appreciated.

Your approach is workable, but as other have answered it's not the easiest approach.
If you want to continue you only have to calculate the determinants of the matrices:
$$\det\begin{pmatrix}a & b \\ c& d\end{pmatrix} = ad - cb$$
$$\det{1 \over ad-bc}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix} =\det\begin{pmatrix}{d \over ad-bc}& {-b \over ad-bc} \\ {-c \over ad-bc} & {a \over ad-bc}\end{pmatrix} = \left({1\over ad-bc}\right)^2\left(da-\left(-c\right)\left(-b\right)\right) = {1\over ad-bc}$$
Now you see that $\det (A^{-1}) = 1/\det A$, then you have that if $A=A^{-1}$ that $\det(A) = 1/\det(A)$ which is a second degree equation with solutions $\det(A) = \pm 1$

Another overly complicated approach might be to use that since $A=A^{-1}$ we have that $A^2 = AA = AA^{-1} = I$ so calculating $A^2$ we get
$$A^2 = \begin{pmatrix}a & b \\ c& d\end{pmatrix}\begin{pmatrix}a & b \\ c& d\end{pmatrix}=\begin{pmatrix}a^2 + bc & (a+d)b \\ (a+d)c & d^+bc\end{pmatrix}$$
Here we see that $a^2+bc = d^2+bc=1$ which implies that $a^2=d^2$ which means $a=\pm d$. If $a=d$ we have that $b=c=0$ which means that the diagonal elements becomes $a^2=d^2=1$ which means $a=\pm1$ and $d=\pm1$ which means the determinant is $\pm 1$. If $a=-d$ the diagonal elements becomes $a^2+bc = d^2 + bc = -ad+bc = 1$, which means that $\det A = ad-bc = -1$.
