Solution
= Solution
Here $A^2=-I$. Splitting the <matrix exponential> into even and odd powers therefore gives
$$
U=I\cos1+A\sin1=\boxed{\begin{pmatrix}\cos1&\sin1\\-\sin1&\cos1\end{pmatrix}}.
$$
For the second <matrix>, $(I+A)(I-A)=I-A^2=2I$, so $(I+A)^{-1}=(I-A)/2$. Hence
$$
V=\frac12(I-A)^2=\frac12(I-2A+A^2)=-A=\boxed{\begin{pmatrix}0&-1\\1&0\end{pmatrix}}.
$$
Both are real <orthogonal matrices>, and hence also <unitary matrices>.