There is a homeomorphism between the the complex numbers
$$
\color{blue}{a} + \color{red}{b}i
$$
and the rotation matrices
$$
\left(
\begin{array}{rc}
\color{blue}{a} & \color{red}{b} \\
-\color{red}{b} & \color{blue}{a} \\
\end{array}
\right)
$$
where $\color{blue}{a}^{2} + \color{red}{b}^{2} = 1.$ We see the familiar rotation matrix
$$
R(\theta) =
\left(
\begin{array}{rc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta \\
\end{array}
\right)
$$
which in the form
$$
x'=R(\theta)x
$$
rotates the $2-$vector $x$ about the origin by $\theta$, producing the $2-$vector $x'$.
Verify homeomorphism
Start with two complex numbers $z_{1}$ and $z_{1}$. The Cartesian forms are
$$
z_{1} = \color{blue}{a} + \color{red}{b}i, \quad
z_{2} = \color{blue}{c} + \color{red}{d}i
$$
where the numbers $a$, $b$, $c$, and $d$, are all real. Blue numbers signify the real component of $z$, and red the imaginary component.
Equivalent matrix forms are defined as
$$
z_{1} = \left(
\begin{array}{rc}
\color{blue}{a} & \color{red}{b} \\
-\color{red}{b} & \color{blue}{c} \\
\end{array}
\right),
\quad
z_{2} = \left(
\begin{array}{rc}
\color{blue}{c}& \color{red}{d} \\
- \color{red}{d} & \color{blue}{c}\\
\end{array}
\right)
$$
Verify basic properties of of the homeomorphism.
Addition
$$
%
z_{1} +
z_{2} =
(\color{blue}{a} + \color{red}{b}i) +
(\color{blue}{c} + \color{red}{d}i)
=
(\color{blue}{a}+\color{blue}{c}) + (\color{red}{b} + \color{red}{d} )i
$$
$$
%
z_{1} +
z_{2} =
\left(
\begin{array}{rc}
\color{blue}{a} & \color{red}{b} \\
-\color{red}{b} & \color{blue}{c} \\
\end{array}
\right) +
\left(
\begin{array}{rc}
\color{blue}{c} & \color{red}{d} \\
-\color{red}{d} & \color{blue}{c} \\
\end{array}
\right)
= \left(
\begin{array}{rc}
\color{blue}{a}+\color{blue}{c}& \color{red}{b}+ \color{red}{d} \\
-\color{red}{b}- \color{red}{d} & \color{blue}{a}+\color{blue}{c}\\
\end{array}
\right)
$$
Multiplication
$$
z_{1} z_{2} =
(\color{blue}{a} + \color{red}{b}i)
(\color{blue}{c} + \color{red}{d}i)
=
(\color{blue}{ac}- \color{red}{bd}) +
(\color{red}{b}\color{blue}{c}+\color{blue}{a}\color{red}{d})i
%
$$
$$
%
z_{1} z_{2} =
\left(
\begin{array}{rc}
\color{blue}{a} & \color{red}{b} \\
-\color{red}{b} & \color{blue}{c} \\
\end{array}
\right)
\left(
\begin{array}{rc}
\color{blue}{c} & \color{red}{d} \\
-\color{red}{d} & \color{blue}{c} \\
\end{array}
\right)
=
%
\left(
\begin{array}{rc}
\color{blue}{a}\color{blue}{c}- \color{red}{b} \color{red}{d} & \color{red}{b}\color{blue}{c}+\color{blue}{a} \color{red}{d} \\
- \color{red}{b} \color{blue}{c}-\color{blue}{a} \color{red}{d} & \color{blue}{a}\color{blue}{c}- \color{red}{b} \color{red}{d} \\
\end{array}
\right)
$$
Inversion
$$
%
\frac{1}{z}
= \frac{1}{\color{blue}{a} + \color{red}{b} i}
= \left( \frac{\color{blue}{a} - \color{red}{b} i}{\color{blue}{a} - \color{red}{b} i} \right)
\frac{1}{\color{blue}{a} + \color{red}{b} i}
= \left( \color{blue}{a}^{2} + \color{red}{b}^{2} \right)^{-1} \left( \color{blue}{a} - \color{red}{b} i \right) %
$$
$$
z^{-1}
= \left(
\begin{array}{cr}
\color{blue}{a} & - \color{red}{b} \\
\color{red}{b} & \color{blue}{a} \\
\end{array}
\right)^{-1}
= \frac{\text{adj }z}{\det z}
= \left( \color{blue}{a}^{2} + \color{red}{b}^{2} \right)^{-1}
\left(
\begin{array}{cr}
\color{blue}{a} & - \color{red}{b} \\
\color{red}{b} & \color{blue}{a} \\
\end{array}
\right)
$$
where adj $z$ is the adjugate matrix.
The correct answer is$$\begin{bmatrix} \cos (\beta) & \sin (\alpha) \sin (\beta) & \cos (\alpha) \sin (\beta) \\ 0 & \cos (\alpha) & -\sin (\alpha) \\ -\sin (\beta) & \cos (\beta) \sin (\alpha) & \cos (\alpha) \cos (\beta) \end{bmatrix}.$$If this is what you got, then your TA is wrong.
Best Answer
Using the fact that $e^{i\theta}=\cos\theta+i\sin\theta$, what do you get when you multiply $$re^{i\alpha}\cdot se^{i\beta}?$$Given the matrix $$\left[\begin{array}{rr} a & -b \\ b & a \end{array}\right]$$ what complex number would you associate it with?