" Let $A$ be a symmetric $2 \times 2$ matrix with the property $A^{-1} = A$. Find all possible trace values ​of $\operatorname{tr}A$" I need some help solving this.
I have tried:
$$
    \begin{bmatrix}
    a & b  \\
    c & d  \\
    \end{bmatrix}
=\frac{1}{\operatorname{det}A}\cdot \begin{bmatrix}
    d & -b  \\
    -c & a  \\
    \end{bmatrix}$$
I ended up with $$a=\frac{d}{\operatorname{det}A},$$
and
$$d=\frac{a}{\operatorname{det}A}.$$
Then
$$\operatorname{tr}(A)=a+d=\frac{a+d}{\operatorname{det}A},$$
but I don't really think it works.
 A: This problem is easier if you think about it in terms of the eigenvalues of $A$.  Note that if $x$ is an eigenvector of $A$, we have $A^{-1}x = Ax$. What does this tell you about that eigenvalue? 
A: By the Cayley–Hamilton theorem or direct verification, we have $A^{2}-\operatorname {tr}(A)A+\det(A)I=0$.
From $A^2=I$, we get $\operatorname {tr}(A)A=(\det(A)+1)I$.
Taking traces on both sides, we get $\operatorname {tr}(A)^2=2(\det(A)+1)$.
From $A^2=I$, we also get $\det(A)^2=1$ and so $\det(A)=\pm1$.
If $\det(A)=1$, then $\operatorname {tr}(A)^2=4$ and so $\operatorname {tr}(A)=\pm 2$.
If $\det(A)=-1$, then $\operatorname {tr}(A)^2=0$ and so $\operatorname {tr}(A)=0$.
These three possibilities occur for the matrices below:
$$
\operatorname {tr}\begin{pmatrix}1&0\\0&1\end{pmatrix} = 2
\qquad
\operatorname {tr}\begin{pmatrix}-1&\hphantom-0\\\hphantom-0&-1\end{pmatrix} = -2
\qquad
\operatorname {tr}\begin{pmatrix}1&\hphantom-0\\0&-1\end{pmatrix} = 0
$$
A: You have $A^{-1} = A \implies A^2 = I$. So we just calculate that for the matrix you have $$\begin{pmatrix} 1 & 0 \\ 0& 1 \end{pmatrix} = \begin{pmatrix} a & b \\ c& d \end{pmatrix}\cdot\begin{pmatrix} a & b \\ c& d \end{pmatrix} = \begin{pmatrix} a^2 + bc & (a+d)\cdot b \\ (a+d)\cdot c & bc + d^2 \end{pmatrix}$$
Therefore, you get $(a+d)\cdot b = (a+d)\cdot c = 0$. 
Case 1: $a+d = 0$, we're done.
Case 2: Let's assume $b$ or $c$ are $0$ then $a^2 = d^2 = 1 \implies a^2 - d^2 = 0 \implies a = \pm d$. 
If $a = -d \implies a+d = 0$
If $a = d$, then $A = aI$, and $A^2 = I \implies a = \pm 1 \implies A = \pm I$
Therefore, for $A^2 = I$, we have $\text{tr} A = 0, 2 , -2$

Continuing from your chain of thought
In your calculations, you obtained the following, 
$$a+d = \frac{a+d}{\det A}$$
Now, as we know if $A^2 = I$, then $\det A = \pm1$. 
Therefore, for $\det A = -1$. You get $$a+d = -(a+d) \implies \text{tr} A = 0$$
For $\det A = 1$, you get $$\begin{pmatrix} a & b\\ c & d \end{pmatrix} = \begin{pmatrix} d & -b\\ -c & a \end{pmatrix}$$
$\implies b = c = 0, a = d$. Therefore $A = aI$, and $\text{tr} A = 2a$
$$A^2 = I \implies (aI)^2 = I \implies a^2 = 1 \implies a = \pm 1$$
Hence, $\text{tr} A = \pm 2$
Comment: For a more 'clean' answer, please refer to Omnomnomnom's answer/response. I provided an elementary answer to demonstrate one exists, but the better and faster way, in my opinion, is still using eigenvalues. 
