= Solution
Since $A^2=-I$, the <matrix inverse> is $A^{-1}=-A$. The <orthogonal matrix> condition also gives $A^{-1}=A^{\mathsf T}$, so $A^{\mathsf T}=-A$. Thus $A$ is <skew-symmetric>, forcing $a=0$ and $c=-b$. Its columns must be <unit vectors>, giving $b^2=1$. Therefore the two possibilities are
$$
\boxed{A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\quad\text{or}\quad A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.}
$$
Both satisfy the original conditions by direct multiplication. They are the two <orthogonal skew-symmetric matrices> in two dimensions, representing opposite quarter-turn <rotations>.
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