<s>[INST] 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.
 [/INST] 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 $\det A = ad-bc = -1$.
</s>