Solution.
A = ( 1 3 1 6 − 1 2 1 3 − 2 6 0 1 3 1 6 1 2 ) A=\begin{pmatrix}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0 \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\
\end{pmatrix} A = ⎝ ⎛ 3 1 3 1 3 1 6 1 − 6 2 6 1 − 2 1 0 2 1 ⎠ ⎞
An orthogonal matrix is a nondegenerate square matrix A A A (usually with real elements) such that
A A T = A T A = I , AA^{T}=A^{T}A=I, A A T = A T A = I , where A T A^{T} A T is transposed matrix to matrix A A A , and I I I is the identity matrix.
det A = − 1 3 − 1 6 − 1 3 − 1 6 = − 1 ≠ 0. \det A=-\frac{1}{3}-\frac{1}{6}-\frac{1}{3}-\frac{1}{6}=-1\neq0. det A = − 3 1 − 6 1 − 3 1 − 6 1 = − 1 = 0.
A T = ( 1 3 1 3 1 3 1 6 − 2 6 1 6 − 1 2 0 1 2 ) A^{T}=\begin{pmatrix}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\
-\frac{1}{\sqrt{2}} & 0 & \frac{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 = ( 1 3 1 6 − 1 2 1 3 − 2 6 0 1 3 1 6 1 2 ) ⋅ ( 1 3 1 3 1 3 1 6 − 2 6 1 6 − 1 2 0 1 2 ) = = ( 1 0 0 0 1 0 0 0 1 ) = I . AA^{T}=\begin{pmatrix}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0 \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\
\end{pmatrix}
\cdot \begin{pmatrix}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\
-\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
\end{pmatrix}=\newline
=\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}=I. A A T = ⎝ ⎛ 3 1 3 1 3 1 6 1 − 6 2 6 1 − 2 1 0 2 1 ⎠ ⎞ ⋅ ⎝ ⎛ 3 1 6 1 − 2 1 3 1 − 6 2 0 3 1 6 1 2 1 ⎠ ⎞ = = ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ = I . A T A = I . A^{T}A=I. A T A = I . Answer. Yes, A A A is orthogonal matrix.
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