 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 areBoth satisfy the original conditions by direct multiplication. They are the two orthogonal skew-symmetric matrices in two dimensions, representing opposite quarter-turn rotations.