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