Symmetric Matrices of $I_{2}$ 
Find $10$ symmetric matrices $ A = \begin{pmatrix}
a &b \\ 
 c&d 
\end{pmatrix}$ such that $A^{2}=I_{2}$

(I'm going to call matrix A the "square root" of $A^{2}$. If this is the incorrect name for it, may someone please tell me what it is actually called?)
My professor posed this question in class and told us there was an infinite amount of square roots. (Assuming I understood him correctly). However I don't see how there would be many of these, as I was under the impression that a matrix only has one inverse, for $A  A^{-1}=I_{n}$. If someone could tell me if I either misunderstood the professor or if I'm thinking about something incorrectly, please correct me.
My other question is other than blatant guess and check, is there a method to think of these symmetric square roots?
Thanks in advance.
 A: Your professor is right, there's an infinite number of square roots, kind of like how there's two square roots of $1$ (namely, $1$ and $-1$).
To see how to get it in general, notice that, for a symmetric matrix, you have
$$
\begin{pmatrix}a&b\\b&c\end{pmatrix}^2 = \begin{pmatrix}a^2+b^2&b(a+c)\\b(a+c)&b^2+c^2\end{pmatrix}
$$
So, for the right side to be equal to the identity, you must have
$$
a^2+b^2=1\\
b^2+c^2=1\\
b(a+c) = 0
$$
What solutions does this system of equations admit?
To demonstrate that multiple solutions exist directly, consider that
$$
\begin{pmatrix}0&1\\1&0\end{pmatrix}^2 = \begin{pmatrix}1&0\\0&1\end{pmatrix}
$$
and so is a square root of the identity matrix.
A: Regarding your question about the inverse, of course $A^2=I$ tells you that $A=A^{-1}$. But for different $A$, no contradiction arises. 
Edit: in the first answer I didn't take care to produce symmetric matrices, and was using arbitrary invertible matrices. What one needs here is unitaries (i.e. orthogonal matrices). 
Here is one to construct all symmetric real matrices with square $I_2$. Let 
$$
A_0=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.
$$
Then $A_0^2=I_2$. Now define
$$
B_t=\begin{bmatrix}t&\sqrt{1-t^2}\\\sqrt{1-t^2}&-t\end{bmatrix}, \ \ \ t\in(-1,1).
$$ and $$
A_t=B_tA_0B_t^{-1}
%=\begin{bmatrix}t&-\sqrt{1-t^2}\\\sqrt{1-t^2}&t\end{bmatrix}\begin{bmatrix}t&\sqrt{1-t^2}\\\sqrt{1-t^2}&-t\end{bmatrix}
=\begin{bmatrix}2t^2-1&2t\sqrt{1-t^2}\\2t\sqrt{1-t^2}&1-2t^2 \end{bmatrix}
$$
Then $A_t\ne A_s$ for any $t,s\in(-1,1)$, and $A_t^2=I_2$ for all $t\in(-1,1)$ (since $A_t^2=B_tA_0B_t^{-1}B_tA_0B_t^{-1}=B_tA_0^2B_t^{-1}=B_tB_t^{-1}=I_2$).
Note that the matrices $B_t$ are all the real rotation matrices. Any symmetric matrix $A$ with $A^2=I_2$ is either the identity or has eigenvalues $1,-1$, and is thus unitarily equivalent to $A_0$.
A: For an easy constuction, consider matrices of the form $A=R_\theta^T\pmatrix{1\\ &-1}R_\theta$, where $R_\theta=\pmatrix{\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta}$ is a rotation matrix for angle $\theta\in[0,\pi)$. These $A$s are all distinct (exercise), real symmetric and their squares are equal to $I_2$.
A: You can do this by writing down equations, writing out the multiplication explicitly: You'll get four quadratic equations in four variables.
For a geometric solution, here's a hint: Reflection across any line through the origin would be such a linear map (matrix). For example, $\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$ and  $\begin{bmatrix} 0&1\\1&0\end{bmatrix}$ are, respectively, reflection across the $x$-axis and the line $y=x$.
A: Consider in the Euclidean plane the orthogonal reflection in any line; its square is always the identity. If you want linear maps, take a Euclidean vector space and choose the line to go through the origin. That still leaves infinitely many possibilities. In fact, you don't need orthogonal reflections, the reflection in any line parallel to any complementary line will have square the identity. In other words there is plenty of freedom, which is why in this setting one never speaks of the square root if the identity. You can call each of them a square root of the identity, though the term (linear) "involution" is more common.
