Answer
The given matrice is orthogonal.
Work Step by Step
Find the inverse for matrix $A=\begin{bmatrix}
\cos \alpha& \sin \alpha\\
-\sin \alpha &\cos \alpha
\end{bmatrix}$:
$\begin{bmatrix}
\cos \alpha& \sin \alpha | 1 &0\\
-\sin \alpha &\cos \alpha | 0 & 1
\end{bmatrix} \approx^1\begin{bmatrix}
1& \tan \alpha | \frac{1}{\cos \alpha} &0\\
-\sin \alpha &\cot \alpha | 0 & \frac{1}{\sin \alpha}
\end{bmatrix} \approx^2 \begin{bmatrix}
1& \tan \alpha | \frac{1}{\cos \alpha} &0\\
0 &\frac{1}{\sin \alpha \cos \alpha} | \frac{1}{\cos \alpha} & \frac{1}{\sin \alpha}
\end{bmatrix} \approx^3 \begin{bmatrix}
1& \tan \alpha | \frac{1}{\cos \alpha} &0\\
0 &1 | \sin \alpha& \cos \alpha
\end{bmatrix} \approx^4 \begin{bmatrix}
1& 0 | \cos \alpha&-\sin \alpha\\
0 &1 | \sin \alpha& \cos \alpha
\end{bmatrix} $
Hence, $A^{-1}=\begin{bmatrix}
\cos \alpha& -sin \alpha\\
\sin \alpha & \cos \alpha
\end{bmatrix}$
On the other side, we have $A^T=\begin{bmatrix}
\cos \alpha& -sin \alpha\\
\sin \alpha & \cos \alpha
\end{bmatrix}$
Since $A^T=A^{-1}$, the given matrice is orthogonal.
