# Identify orthogonal matrices Invert orthogonal matrice begin{bmatrix}frac{1}{3} & frac{2}{3}&frac{2}{3} frac{2}{3} & frac{1}{3}&-frac{2}{3}frac{2}{3}&-frac{2}{3}&frac{1}{3} end{bmatrix}

Identify orthogonal matrices Invert orthogonal matrice
$\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right]$
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Step 1
To identify the orthogonal matrix
$\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right]$
Step 2
Note that a matrix is orthogonal if it satisfies:
$A{A}^{T}=I$
$A=\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right]⇒{A}^{T}=\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right]$
$A{A}^{T}=\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right]$
$=\left[\begin{array}{ccc}\frac{1}{9}+\frac{4}{9}+\frac{4}{9}& \frac{2}{9}+\frac{2}{9}-\frac{4}{9}& \frac{2}{9}-\frac{4}{9}+\frac{2}{9}\\ \frac{2}{9}+\frac{2}{9}-\frac{4}{9}& \frac{4}{9}+\frac{1}{9}+\frac{4}{9}& \frac{4}{9}-\frac{2}{9}-\frac{2}{9}\\ \frac{2}{9}-\frac{4}{9}+\frac{2}{9}& \frac{4}{9}-\frac{2}{9}-\frac{2}{9}& \frac{4}{9}+\frac{1}{9}+\frac{4}{9}\end{array}\right]$
$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$
Step 3
Thus matrix is orthogonal.
Now note that the inverse of orthogonal matrix can be find as:
$A{A}^{T}=I$
$⇒{A}^{-1}={A}^{T}$
$⇒{A}^{-1}=\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right]$
Jeffrey Jordon