So there is this problem
A square matrix $R$ is called a reflection matrix if $R^2 = I$. Here are some examples $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} , \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Find a $2 \times 2$ reflection matrix with all entries nonzero. Can you find all $2 \times 2$ reflection matrices? That is, find necessary and suffcient conditions on $a, b, c, d$ so that ${\begin{pmatrix} a & b \\ c & d \end{pmatrix}}^2 = I $
This is what I have done so far
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ = $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
So we have
$\begin{pmatrix} a^2 + bc & ab + bd \\ ca + dc & cb + d^2 \end{pmatrix}$ = $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
So what do I say now?
That's a good start. For convenience, let $M = \pmatrix{a&b\\c&d}$.
If you want the entries of $M$ to be nonzero, notice that
$$ \pmatrix{a^2 + bc & ab + bd \\ ac + cd & b^2 + bc} = \pmatrix{a^2 + bc & b(a + d) \\ c(a + d) & d^2 + bc}. $$
So, if $ab + bd = b(a + d) = 0$ with all of $a, b$, and $d$ nonzero, we must have $d = -a$; that takes care of two out of the four entries in $M^2$.
Now $M = \pmatrix{a&b\\c&-d}$ and $$ M^2 = \pmatrix{a^2 + bc & 0 \\ 0 & a^2 + bc}. $$
In order to have $M^2 = I$, you simply need $a, b, c$ such that $a^2 + bc = 1$. There are lots of solutions here, I'm sure you can find one.
As far as "conditions on $a, b, c, d$ so that $M^2 = I$", we already have four equations putting restrictions on the values of $a, b, c, d$ (that various entries should be $1$ or $0$). I'm not sure it gets much better than that. You can separate into cases (whether $a = -d$, whether $b$ and $c$ are $0$), but when $a = -d$, the only other requirement is that $a^2 + bc = 1$, which has many solutions. You could solve for $b$ or $c$ and use only two variables to write $M$, but that's about it.
As a sidenote, the lone requirement that $M^2 = I$ doesn't make a very good definition for a reflection matrix, in the usual sense (but that's not your fault!). Reflections preserve length, so in addition, we'd want $a^2 + c^2 = 1 = b^2 + d^2$ (since $M(e_1) = \begin{bmatrix}a \\ b\end{bmatrix}$ has length $a^2 + b^2$, likewise for $M(e_2)$). Further, reflections should reverse orientation, so that we should require $\det M = ad - bc = -1$, as well.
If you follow this trail, you'll see that conventional reflection matrices come in exactly one form: $$ M = \pmatrix{ a & \pm \sqrt{1 - a^2}\\ \mp \sqrt{1 - a^2} & -a}, $$ with $a \in [-1, 1]$ (which leads nicely to their parameterization using angles and formatting with trig functions; an angle completely determines a line through the origin, which completely determines the reflection).