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:
$$
e^{\begin{bmatrix} 1 & -2 \\ 1 & -1 \end{bmatrix}\theta} = \mbox{?} =
\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}
$$

From:
$$
\left[
\begin{array}{cccc}
a^2+bc&ab+bd\\
ac+cd&bc+d^2
\end {array}
\right]
=
\left[
\begin{array}{cccc}
-1&0\\
0&-1
\end {array}
\right]
$$
we have
$$
\begin{cases}
b(a+d)=0\\
c(a+d)=0\\
a^2+bc=-1\\
d^2+bc=-1
\end{cases}
$$
If $a+d\ne0$ then $b=0$ and $c=0$ and we have $a^2=-1$ that is impossibile. So we must have $$
\begin{cases}
a+d=0\\
a^2+bc=-1
\end{cases}
$$
From the second equation we find $ a^2=-1-bc$ that has real solutions iff  $bc \le -1$ and in such a case $a=\pm\sqrt{-1-bc}$.
The simplest case is when $bc=-1$ so that we have only one value for $a$ i.e. $a=0$ and the matrix we are searching for has the form:
$$
\left[
\begin{array}{cccc}
0&b\\
-1/b&0
\end {array}
\right]
$$
Note that the $i$ matrix cited in OP is of this form.
But, as we see there are many other matrices that represent $i$ and for every such matrix the subring generated by the matrix and the identity is isomorphic to the complex numbers.
