To determine, whether the matrix A is orthogonal we should multiply matrix to its transpose, if the result is equal to the identity matrix, then A is orthogonal:
A = ( 1 3 1 6 − 1 2 − 1 3 2 6 0 1 3 1 6 1 2 ) A=\begin {pmatrix}
\cfrac{1}{\sqrt{3}} & \cfrac{1}{\sqrt{6}} & \cfrac{-1}{\sqrt{2}}\\
\cfrac{-1}{\sqrt{3}}& \cfrac{2}{\sqrt{6}}& 0\\
\cfrac{1}{\sqrt{3}} & \cfrac{1}{\sqrt{6}}& \cfrac{1}{\sqrt{2}}
\end{pmatrix} A = ⎝ ⎛ 3 1 3 − 1 3 1 6 1 6 2 6 1 2 − 1 0 2 1 ⎠ ⎞
A T = ( 1 3 − 1 3 1 3 1 6 2 6 1 6 − 1 2 0 1 2 ) A^T=\begin{pmatrix}
\cfrac{1}{\sqrt{3}} & \cfrac{-1}{\sqrt{3}} & \cfrac{1}{\sqrt{3}}\\
\cfrac{1}{\sqrt{6}}& \cfrac{2}{\sqrt{6}}& \cfrac{1}{\sqrt{6}}\\
\cfrac{-1}{\sqrt{2}} & 0& \cfrac{1}{\sqrt{2}}
\end{pmatrix} A T = ⎝ ⎛ 3 1 6 1 2 − 1 3 − 1 6 2 0 3 1 6 1 2 1 ⎠ ⎞
A A T = AA^T= A A T = ( ( 1 3 + 1 6 + 1 2 ) ( − 1 3 + 2 6 ) ( 1 3 + 1 6 − 1 2 ) ( − 1 3 + 2 6 ) ( 1 3 + 4 6 ) ( − 1 3 + 2 6 ) ( 1 3 + 1 6 − 1 2 ) ( − 1 3 + 2 6 ) ( 1 3 + 1 6 + 1 2 ) ) = ( 1 0 0 0 1 0 0 0 1 ) \begin{pmatrix}
(\cfrac{1}{3}+\cfrac{1}{6}+\cfrac{1}{2}) & (\cfrac{-1}{3} + \cfrac{2}{6})& (\cfrac{1}{3}+\cfrac{1}{6}-\cfrac{1}{2})\\
(\cfrac{-1}{3} + \cfrac{2}{6})& (\cfrac{1}{3}+\cfrac{4}{6})& (\cfrac{-1}{3} + \cfrac{2}{6})\\
(\cfrac{1}{3}+\cfrac{1}{6}-\cfrac{1}{2}) & (\cfrac{-1}{3} + \cfrac{2}{6})& (\cfrac{1}{3}+\cfrac{1}{6}+\cfrac{1}{2})
\end{pmatrix}=
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{pmatrix} ⎝ ⎛ ( 3 1 + 6 1 + 2 1 ) ( 3 − 1 + 6 2 ) ( 3 1 + 6 1 − 2 1 ) ( 3 − 1 + 6 2 ) ( 3 1 + 6 4 ) ( 3 − 1 + 6 2 ) ( 3 1 + 6 1 − 2 1 ) ( 3 − 1 + 6 2 ) ( 3 1 + 6 1 + 2 1 ) ⎠ ⎞ = ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞
So A is an orthogonal matrix.
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