==Problem 2A==
Find all components of matrix \(e^{i\alpha A}\) where \(A\) is
\begin{equation}
A = \begin{pmatrix}
0&&0&&1 \\
0&&0&&0 \\
1&&0&&0
\end{pmatrix}
\end{equation}
==Solution==
Simply exp will yield,
\begin{equation}
\left(\begin{array}{rrr}
\frac{1}{2} \, {\left(e^{\left(2 i \, y\right)} + 1\right)} e^{\left(-i \, y\right)} & 0 & \frac{1}{2} \, {\left(e^{\left(2 i \, y\right)} - 1\right)} e^{\left(-i \, y\right)} \\
0 & 1 & 0 \\
\frac{1}{2} \, {\left(e^{\left(2 i \, y\right)} - 1\right)} e^{\left(-i \, y\right)} & 0 & \frac{1}{2} \, {\left(e^{\left(2 i \, y\right)} + 1\right)} e^{\left(-i \, y\right)}
\end{array}\right)
\end{equation}
Then applying Demovire's theorem, one gets,
\begin{equation}
\left(\begin{array}{rrr}
\frac{1}{2} \, {\left(\cos\left(2 \, y\right) + i \, \sin\left(2 \, y\right) + 1\right)} {\left(\cos\left(y\right) - i \, \sin\left(y\right)\right)} & 0 & \frac{1}{2} \, {\left(\cos\left(2 \, y\right) + i \, \sin\left(2 \, y\right) - 1\right)} {\left(\cos\left(y\right) - i \, \sin\left(y\right)\right)} \\
0 & 1 & 0 \\
\frac{1}{2} \, {\left(\cos\left(2 \, y\right) + i \, \sin\left(2 \, y\right) - 1\right)} {\left(\cos\left(y\right) - i \, \sin\left(y\right)\right)} & 0 & \frac{1}{2} \, {\left(\cos\left(2 \, y\right) + i \, \sin\left(2 \, y\right) + 1\right)} {\left(\cos\left(y\right) - i \, \sin\left(y\right)\right)}
\end{array}\right)
\end{equation}
==Problem 2B==
If \([A,B]=B\), calculate
\begin{equation}
e^{i\alpha A}B e^{-i\alpha A}
\end{equation}
Using equation $2.44$, and setting $Y=-Z$, we get
\begin{align*}
RHS=X-i[-Z,X]-\frac{1}{2}[-Z,[-Z,X]]+\cdots \\
&&=&& X-i[X,Z]-1/2[[X,Z],Z]+\cdots
\end{align*}
Applying this to equation in our problem,
\begin{align*}
e^{i\alpha A} B e^{-i \alpha A} = B - i\alpha[B,A]-\frac{1}{2}\alpha^2[[B,A],A]+\cdots \\
= B + i\alpha B - \frac{1}{2}\alpha^2[B,A]+\cdots \\
= B + i \alpha B - \frac{1}{2} \alpha^2B + \cdots \\
= B e^{-i \alpha}
\end{align*}
