Ambiguous matrix representation of imaginary unit? This question is triggered by another one:

What does it mean to represent a number in term of a 2x2 matrix?

Suppose we seek a $2\times2$ matrix equivalent of
the imaginary unit, that is a matrix $\,i\,$ such that $\,i^2 = -1\,$ , or,
with $\;a,b,c,d \in \mathbb{R}$ :
$$
i^2 = \begin{bmatrix} a&b\\c&d \end{bmatrix} 
\begin{bmatrix} a&b\\c&d \end{bmatrix} =
\begin{bmatrix} a^2+bc&ab+bd\\ac+cd&bc+d^2 \end{bmatrix} =
- \begin{bmatrix} 1&0\\0&1 \end{bmatrix}
$$
Leading to the following equations:
$$
a^2+bc=-1 \quad ; \quad b(a+d)=0 \quad ; \quad c(a+d)=0 \quad ; \quad bc+d^2=-1
$$
Subtracting the first from the last one gives two possible solutions:
$$
a^2-d^2=0 \quad \Longrightarrow \quad a = \pm\, d
$$
The solution $\,a=d\ne0\,$ leads to $\,b=0\,$ and $\,c=0\,$ , giving $\,a^2 = -1$ ,
which is impossible in the reals. The other possibility is $\,d = -a$ :
$$
i = \begin{bmatrix} a&b\\c&-a \end{bmatrix}
\quad \mbox{with} \quad a^2+bc = -1 \quad \mbox{or} \quad
\begin{vmatrix} a&b\\c&-a \end{vmatrix} = 1
$$
But here I'm stuck. IMO it does not follow that $\,i\,$ is
a special case of the above matrix, namely the one with
$\,a=0\,$ and $\,b=-1$ , $c=1$ :
$$
i = \begin{bmatrix} 0&-1\\1&0 \end{bmatrix}
$$
Am I missing something obvious?

Note. In a comment
by Meelo it is remarked that <quote>One can notice that this representation is not unique, though one must always treat $1$ as the matrix
$$ \pmatrix{1&&0\\0&&1} $$ any matrix with characteristic polynomial $x^2+1$ suffices to represent $i$ . For instance, you could use
$$\pmatrix{1&&-2\\1&&-1}$$ for $i$.</quote> That's right, because
$$
\begin{vmatrix}a-\lambda&b\\c&-a-\lambda\end{vmatrix}=(a-\lambda)(-a-\lambda)-bc=\lambda^2+1
$$
I don't see, however, how this can be matched with my answer at the same place:
$$
e^{\begin{bmatrix} 1 & -2 \\ 1 & -1 \end{bmatrix}\theta} = \mbox{?} =
\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}
$$

I would suggest looking at the problem from another angle. Suppose you wanted to find the matrix representation of the linear map $$ L: \mathbb{C} \rightarrow \mathbb{C},\; z \mapsto i\cdot z $$.
If we write $z$ in the form $z = a + bi$, we see
$$L(z) = i\cdot (a+bi) = -b + ai$$
So L maps the number $(1 + 0i)$ to $(0 + 1i)$ and the number $(0+1i)$ to $(-1+0i)$. Identifying $\mathbb{C}$ with $\mathbb{R}^2$ we see that the matrix representation of this map is 
$$ L = \begin{bmatrix} 0&-1\\1&0 \end{bmatrix}
$$
