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Find examples of $2 \times 2$ matrices with $a_{12} = \frac{1}{2}$ for which (a) $A^2 = I$, (b) $A^{-1}=A^T$, and (c) $A^2 = A$.

What I thought was a trivial problem is turning out to be rather irritating. At the moment I am dealing with part (a). To solve the problem, I narrowed my search space of examples by taking $A$ to be invertible with unit determinant. If $A$ is invertible, then $A^2 = I$ implies $A = A^{-1}$, further narrowing the search space.

Taking $A = \begin{bmatrix} a & 1/2 \\ c & d \\ \end{bmatrix}$, $A = A^{-1}$ would imply $a=d$ and $c = - \frac{1}{2}$. From this I got $a = \frac{\sqrt{3}}{2}$. However, the calculations never add up. I have done these calculations just short of a trillion times, and even had wolfram do some of the computations, but $A^2$ never equals $I$. What I am doing wrong? Is there some error in my logic?

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    All this shows is that there is no example in your narrowed search space. Look for a more general kind of thing. – 2017-02-09

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Let's start with what you have:

$$A = \begin{bmatrix} a & \frac{1}{2} \\ -\frac{1}{2} & d \end{bmatrix}$$

So then

$$A^2 = \begin{bmatrix} a & \frac{1}{2} \\ -\frac{1}{2} & d \end{bmatrix}\cdot \begin{bmatrix} a & \frac{1}{2} \\ -\frac{1}{2} & d \end{bmatrix} = \begin{bmatrix} a^2-\frac{1}{4} & \frac{1}{2}(a+d) \\ -\frac{1}{2}(a+d) & -\frac{1}{4}+d^2 \end{bmatrix}=I$$

So $a^2 - \frac{1}{4}=1$, or $a = \pm\frac{\sqrt{5}}{2}$. And as Jason points out in the comments, $d = \mp\frac{\sqrt{5}}{2}$.

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    (And then one must choose $d = \mp \frac{\sqrt{5}}{2}$ in order for the off diagonal entries to work.) – 2017-02-09